Step 1: Understanding the Concept:

Mass percentage is a way of expressing the concentration of a component in a mixture or solution.

It is calculated as the mass of the component (solute) divided by the total mass of the solution, multiplied by 100.


Step 2: Key Formula or Approach:

The formula for mass percentage of a solute is:
\[ Mass percentage = \left( \frac{Mass of solute}{Total mass of solution} \right) \times 100% \]
The total mass of the solution is the sum of the mass of the solute and the mass of the solvent.


Step 3: Detailed Explanation:

Given in the question:

Mass of solute = 4 g

Mass of solvent (ethanol) = 46 g


First, we calculate the total mass of the solution:
\[ Total mass of solution = Mass of solute + Mass of solvent \] \[ Total mass of solution = 4 \, g + 46 \, g = 50 \, g \]

Now, we can calculate the mass percentage of the solute using the formula:
\[ Mass percentage = \left( \frac{4 \, g}{50 \, g} \right) \times 100% \] \[ Mass percentage = 0.08 \times 100% \] \[ Mass percentage = 8% \]

Step 4: Final Answer:

The mass percentage of the solute is 8%.
Quick Tip: Always remember that the denominator in the mass percentage formula is the total mass of the \textbf{solution} (solute + solvent), not just the mass of the solvent. This is a common point of error.

Step 1: Understanding the Concept:

The energy of an electron in a specific orbital of a multi-electron atom is primarily determined by the effective nuclear charge (\(Z_{eff}\)) experienced by that electron.

For the same orbital (in this case, 2s), the energy depends on the attraction from the nucleus. A stronger attraction leads to a lower (more negative) energy, making the orbital more stable.


Step 3: Detailed Explanation:

The elements given are Lithium (Li), Sodium (Na), and Potassium (K).

Their atomic numbers (Z) are:

- Li: Z = 3

- Na: Z = 11

- K: Z = 19


As the atomic number increases, the number of protons in the nucleus increases. This results in a stronger electrostatic attraction between the nucleus and the electrons.

For the same subshell (2s), an electron in Potassium (K) will experience a much stronger pull from its 19 protons compared to an electron in Sodium (Na, 11 protons) or Lithium (Li, 3 protons).

This stronger attraction lowers the energy of the orbital. Therefore, the 2s orbital in K is at the lowest energy level, and the 2s orbital in Li is at the highest energy level among the three.

The order of energy is thus:
\[ E_{2s}(Li) \textgreater E_{2s}(Na) \textgreater E_{2s}(K) \]

Step 4: Final Answer:

The correct order of energy for the 2s orbital is Li \textgreater Na \textgreater K.
Quick Tip: For the same orbital in different atoms, remember that energy decreases as the nuclear charge (atomic number) increases. More protons mean a stronger pull on the electrons, making them more stable and lower in energy.

Step 1: Understanding the Concept:

Half-filled (e.g., p³, d⁵) and completely filled (e.g., p⁶, d¹⁰) subshells exhibit greater stability compared to other configurations. This stability arises from two main quantum mechanical effects: symmetrical distribution of electrons and exchange energy.


Step 3: Detailed Explanation:

Let's analyze the factors contributing to this stability:

1. Symmetrical Distribution: In half-filled and completely filled subshells, electrons are symmetrically distributed around the nucleus. This symmetry leads to two consequences:

- Shielding: The electrons shield each other less effectively from the nucleus. This means each electron experiences a greater effective nuclear charge, leading to increased stability. This corresponds to (i) Relatively small shielding.

- Coulombic Repulsion: The symmetrical arrangement places electrons as far apart as possible, which minimizes the electron-electron repulsion forces. This corresponds to (iv) Smaller coulombic repulsion energy.


2. Exchange Energy: This is a quantum mechanical phenomenon. When two or more electrons with the same spin are present in degenerate orbitals of a subshell, they can exchange their positions. Each such exchange releases energy, which is called exchange energy. The more exchanges possible, the more energy is released, and the greater the stability. Half-filled and completely filled configurations have the maximum number of possible exchanges. This corresponds to (v) Larger exchange energy.


Therefore, the correct factors are relatively small shielding, smaller coulombic repulsion energy, and larger exchange energy.


Step 4: Final Answer:

The combination of correct statements is (i), (iv), and (v).
Quick Tip: For stability of electron configurations, think "Symmetry and Exchange". Symmetrical arrangements lead to low repulsion and better nuclear attraction, while maximum exchange of same-spin electrons leads to the release of energy and thus greater stability.

Step 1: Understanding the Concept:

Ionization enthalpy (IE) is the energy required to remove the most loosely bound electron from an isolated gaseous atom. The general trend across a period (from left to right) is that IE increases due to increasing effective nuclear charge and decreasing atomic size. However, there are exceptions due to electronic configurations.


Step 3: Detailed Explanation:

The elements are Boron (B), Carbon (C), Nitrogen (N), and Oxygen (O), all from Period 2.

Their electronic configurations are:

- B (Z=5): \(1s^2 2s^2 2p^1\)

- C (Z=6): \(1s^2 2s^2 2p^2\)

- N (Z=7): \(1s^2 2s^2 2p^3\) (half-filled p-subshell)

- O (Z=8): \(1s^2 2s^2 2p^4\)


General Trend Analysis:

Based on increasing nuclear charge, the expected order would be B \textless C \textless N \textless O.


Analyzing the Exception (N vs. O):

Nitrogen (N) has a half-filled \(2p^3\) subshell. This is a particularly stable configuration due to maximum exchange energy and symmetry. Removing an electron from this stable configuration requires a large amount of energy.

Oxygen (O) has a \(2p^4\) configuration. Removing one electron results in a stable, half-filled \(2p^3\) configuration for the resulting O\(^+\) ion. This process is relatively easier compared to removing an electron from the already stable Nitrogen atom.

Therefore, the ionization enthalpy of Nitrogen is greater than that of Oxygen (IE\(_{N}\) \textgreater IE\(_{O}\)).


Final Order Construction:

- Boron (B) has the lowest nuclear charge and its \(2p\) electron is shielded by the \(2s\) electrons, so it has the lowest IE.

- Carbon (C) has a higher IE than B, following the general trend.

- Between N and O, N has a higher IE due to its stable half-filled configuration.

- Oxygen (O) has a higher nuclear charge than Carbon (C), so its IE is still higher than C's.


Combining these facts, the correct order is: B \textless C \textless O \textless N.


Step 4: Final Answer:

The correct increasing order of first ionization enthalpy is B \textless C \textless O \textless N.
Quick Tip: When ordering ionization enthalpies across a period, always check for elements with fully-filled (Group 2, 18) or half-filled (Group 15) subshells. These configurations are extra stable and lead to higher-than-expected ionization enthalpies, creating exceptions to the general trend.

Step 1: Understanding the Concept:

Atomic radius is a measure of the size of an atom. The general trend for atomic radius across a period in the periodic table (from left to right) is that it decreases.


Step 3: Detailed Explanation:

The elements given are Carbon (C), Nitrogen (N), Oxygen (O), and Fluorine (F). All of these elements belong to the second period of the periodic table.

Their atomic numbers are:

- C: 6

- N: 7

- O: 8

- F: 9


As we move from left to right across a period:

- The number of protons in the nucleus (nuclear charge) increases.

- The electrons are added to the same principal energy level (the n=2 shell in this case).


The increasing positive charge of the nucleus exerts a stronger pull on the electrons in the same shell, pulling the electron cloud closer to the nucleus. This results in a decrease in the atomic radius.

Therefore, the atomic radius decreases in the order C \textgreater N \textgreater O \textgreater F.


The question asks for the increasing order of atomic radii, so we must reverse this sequence.

The correct increasing order is F \textless O \textless N \textless C.


Step 4: Final Answer:

The correct increasing order of atomic radii is F \textless O \textless N \textless C.
Quick Tip: A simple mnemonic for periodic trends: Atomic Radius \textbf{D}ecreases \textbf{A}cross a period and \textbf{I}ncreases \textbf{D}own a group (DAID). Remember this to quickly solve questions on atomic size.

Step 1: Understanding the Concept:

The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their valence shell. An "expanded octet" occurs when a central atom in a molecule has more than eight valence electrons. This is possible for elements in the third period and below because they have vacant d-orbitals that can accommodate extra electrons.


Step 3: Detailed Explanation:

Let's analyze the valence electrons around the central atom in each molecule:

- BCl\(_3\): The central atom is Boron (B), which is in Group 13. It has 3 valence electrons. It forms three single bonds with three Chlorine atoms. The total number of electrons around Boron is 3 bonds \( \times \) 2 electrons/bond = 6 electrons. This is an incomplete octet.

- NO\(_2\) and NO: These are odd-electron molecules, and do not follow the octet rule strictly, but Nitrogen does not have an expanded octet.

- SF\(_6\): The central atom is Sulfur (S), which is in Group 16 and Period 3. It has 6 valence electrons. It forms six single bonds with six Fluorine atoms. The total number of electrons around Sulfur is 6 bonds \( \times \) 2 electrons/bond = 12 electrons. Since 12 is greater than 8, SF\(_6\) has an expanded octet.

- BeH\(_2\): The central atom is Beryllium (Be), which is in Group 2. It has 2 valence electrons. It forms two single bonds with two Hydrogen atoms. The total number of electrons around Beryllium is 2 bonds \( \times \) 2 electrons/bond = 4 electrons. This is an incomplete octet.


Step 4: Final Answer:

The molecule with an expanded octet is SF\(_6\).
Quick Tip: To spot an expanded octet, look for a central atom from Period 3 or below (like P, S, Cl, Br, I, Xe) bonded to several highly electronegative atoms (like F, O, Cl).

Step 1: Understanding the Concept:

This question requires the application of VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the number of bonding pairs (bond pairs) and non-bonding pairs (lone pairs) of electrons around the central atom of each molecule.


Step 3: Detailed Explanation:

Let's determine the number of bond pairs and lone pairs for the central atom in each molecule. The number of lone pairs can be calculated as: \( \frac{1}{2} \times (Valence electrons of central atom - Number of bonds) \).


- NH\(_3\): Central atom is Nitrogen (N, Group 15, 5 valence electrons). It forms 3 single bonds with H.

- Bond Pairs = 3.

- Lone Pairs = \( \frac{1}{2} \times (5 - 3) = \frac{2}{2} = 1 \).

- Total: 3 bond pairs, 1 lone pair.


- SO\(_2\): Central atom is Sulfur (S, Group 16, 6 valence electrons). It forms two double bonds or a resonance hybrid. We count sigma bonds for VSEPR. There are 2 sigma bonds with O.

- Bond Pairs = 2 (sigma bonds).

- Lone Pairs = \( \frac{1}{2} \times (6 - 2 \times 2) = \frac{2}{2} = 1 \). (Using formal charge method for Lewis structure gives 1 lone pair on S).

- Total: 2 bond pairs, 1 lone pair.


- ClF\(_3\): Central atom is Chlorine (Cl, Group 17, 7 valence electrons). It forms 3 single bonds with F.

- Bond Pairs = 3.

- Lone Pairs = \( \frac{1}{2} \times (7 - 3) = \frac{4}{2} = 2 \).

- Total: 3 bond pairs, 2 lone pairs. This matches the question.


- SF\(_4\): Central atom is Sulfur (S, Group 16, 6 valence electrons). It forms 4 single bonds with F.

- Bond Pairs = 4.

- Lone Pairs = \( \frac{1}{2} \times (6 - 4) = \frac{2}{2} = 1 \).

- Total: 4 bond pairs, 1 lone pair.


- H\(_2\)O: Central atom is Oxygen (O, Group 16, 6 valence electrons). It forms 2 single bonds with H.

- Bond Pairs = 2.

- Lone Pairs = \( \frac{1}{2} \times (6 - 2) = \frac{4}{2} = 2 \).

- Total: 2 bond pairs, 2 lone pairs.


Step 4: Final Answer:

The molecule with 3 bond pairs and 2 lone pairs is ClF\(_3\).
Quick Tip: Quickly determine the electron pair geometry using the formula: Steric Number = (Number of atoms bonded to central atom) + (Number of lone pairs on central atom). For ClF\(_3\), Steric Number = 3 + 2 = 5, which corresponds to a trigonal bipyramidal electron geometry and a T-shaped molecular geometry.

Step 1: Understanding the Concept:

In thermodynamics, physical properties of a system are classified into two types:

- Extensive Properties: These properties depend on the amount of matter or the size of the system. Examples include mass, volume, internal energy, enthalpy, and entropy. If you double the amount of substance, the value of an extensive property also doubles.

- Intensive Properties: These properties do not depend on the amount of matter. Examples include temperature, pressure, density, and concentration. If you divide a system in half, the value of an intensive property remains the same in each half.


Step 3: Detailed Explanation:

Let's analyze the given options:

- (A) Molar volume: This is volume per mole (\(V/n\)). Since both volume and moles are extensive, their ratio is intensive.

- (B) Internal energy (U): This is the total energy (kinetic + potential) of all particles in the system. It is directly proportional to the amount of substance. A larger system has more particles and thus more internal energy. It is an extensive property.

- (C) Temperature (T): This is a measure of the average kinetic energy of the particles. It is an intensive property.

- (D) Density (\(\rho\)): This is mass per unit volume (\(m/V\)). Since both mass and volume are extensive, their ratio is intensive.

- (E) Pressure (P): This is force per unit area. It is an intensive property.


Step 4: Final Answer:

Internal energy is the extensive property among the given options.
Quick Tip: To test if a property is extensive or intensive, imagine dividing the system into two equal halves. If the property's value is also halved (like mass or volume), it's extensive. If the property's value remains unchanged (like temperature or density), it's intensive.

Step 1: Understanding the Concept:

The standard enthalpy change of fusion (\( \Delta_{fus} H^\ominus \)) is the amount of energy required to change one mole of a substance from the solid state to the liquid state at standard pressure. Its value is a direct measure of the strength of the forces holding the particles together in the solid crystal lattice. Stronger forces require more energy to overcome, leading to a higher enthalpy of fusion.


Step 3: Detailed Explanation:

Let's analyze the type of solid and the intermolecular/interionic forces for each substance:

- H\(_2\)O (Water): A molecular solid held together by strong hydrogen bonds.

- CO (Carbon monoxide): A molecular solid held together by weak dipole-dipole interactions and London dispersion forces.

- C\(_6\)H\(_6\) (Benzene): A nonpolar molecular solid held together by only London dispersion forces.

- CCl\(_4\) (Carbon tetrachloride): A nonpolar molecular solid held together by London dispersion forces, which are stronger than in benzene due to more electrons.

- NaCl (Sodium chloride): An ionic solid. Its crystal lattice is held together by very strong electrostatic forces of attraction (ionic bonds) between Na\(^+\) and Cl\(^-\) ions.


Comparison of Forces:

Ionic bonds are significantly stronger than any type of intermolecular force (hydrogen bonds, dipole-dipole, or London dispersion forces). To melt an ionic solid like NaCl, these strong electrostatic forces in the entire crystal lattice must be overcome, which requires a very large amount of energy. Molecular solids are held together by much weaker forces, so they have lower enthalpies of fusion.


Step 4: Final Answer:

NaCl, being an ionic compound with strong ionic bonds, will have the highest enthalpy of fusion among the given choices.
Quick Tip: When comparing properties like melting point or enthalpy of fusion, first classify the substances (ionic, metallic, covalent network, molecular). Ionic compounds almost always have much higher values than molecular compounds due to the strength of ionic bonds versus weaker intermolecular forces.

Step 1: Understanding the Concept:

The equilibrium constant in terms of molar concentrations, \(K_c\), is a ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.


Step 2: Key Formula or Approach:

For the given reaction \( N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \), the expression for \(K_c\) is:
\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \]

Step 3: Detailed Explanation:

We are given the equilibrium concentrations:

- \([N_2] = 5 \times 10^{-3} \, M\)

- \([O_2] = 2.8 \times 10^{-3} \, M\)

- \([NO] = 1.4 \times 10^{-3} \, M\)


Now, we substitute these values into the \(K_c\) expression:
\[ K_c = \frac{(1.4 \times 10^{-3})^2}{(5 \times 10^{-3}) \times (2.8 \times 10^{-3})} \] \[ K_c = \frac{1.96 \times 10^{-6}}{(5 \times 2.8) \times 10^{-6}} \] \[ K_c = \frac{1.96 \times 10^{-6}}{14 \times 10^{-6}} \]
The \(10^{-6}\) terms in the numerator and denominator cancel out.
\[ K_c = \frac{1.96}{14} \] \[ K_c = 0.14 \]

Step 4: Final Answer:

The value of Kc for the reaction is 0.14.
Quick Tip: Always write down the \(K_c\) expression first, paying close attention to the stoichiometric coefficients from the balanced equation. They become the exponents in the expression. A common mistake here would be to forget to square the concentration of NO.

Note: This question was likely cancelled because the provided E\(^\circ\)\(_{cell}\) value (0.295 V) does not match the value calculated from standard electrode potentials (0.46 V). However, we can solve it using the given data.


Step 1: Understanding the Concept:

At equilibrium, the cell potential (E\(_{cell}\)) is zero. There is a direct relationship between the standard cell potential (E\(^\circ\)\(_{cell}\)) and the equilibrium constant (\(K_c\)) derived from the Nernst equation.


Step 2: Key Formula or Approach:

The relationship at standard temperature (298 K) is:
\[ E^\circ_{cell} = \frac{0.059}{n} \log K_c \]
where \(n\) is the number of moles of electrons transferred in the balanced redox reaction.


Step 3: Detailed Explanation:

First, we must determine the value of \(n\). The reaction can be split into two half-reactions:

Oxidation: \( Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^- \)

Reduction: \( 2Ag^{+}_{(aq)} + 2e^- \rightarrow 2Ag_{(s)} \)

The number of electrons transferred, \(n\), is 2.


Now, we use the given values:

- \( E^\circ_{cell} = 0.295 \, V \)

- \( n = 2 \)


Substitute these into the formula:
\[ 0.295 = \frac{0.059}{2} \log K_c \] \[ 0.295 = 0.0295 \log K_c \]
Now, solve for \( \log K_c \):
\[ \log K_c = \frac{0.295}{0.0295} = 10 \]
To find \(K_c\), we take the antilog:
\[ K_c = 10^{10} \]

Step 4: Final Answer:

Based on the data provided in the question, the equilibrium constant is 10\(^{10}\).
Quick Tip: Memorize the key electrochemical formula linking standard potential and the equilibrium constant: \( E^\circ_{cell} = (0.059/n) \log K_c \). Also, remember that at equilibrium, \( E_{cell} = 0 \) and \( \Delta G = 0 \), while \( E^\circ_{cell} \) and \( \Delta G^\circ \) are non-zero constants for the reaction.

Step 1: Understanding the Concept:

Corrosion is the gradual destruction of metals by chemical reaction with their environment. One of the most common methods to prevent corrosion is to apply a protective coating that acts as a barrier between the metal and the corrosive agents like oxygen and moisture.


Step 3: Detailed Explanation:

Let's analyze the options in the context of protective coatings:

- (A) Phenol, (B) Benzene, (C) Acetone, (E) Nitrophenol: These are primarily industrial solvents, chemical intermediates, or simple organic compounds. They do not form durable, adherent films suitable for protecting metal surfaces. In fact, some can be corrosive themselves.

- (D) Bisphenol: Specifically Bisphenol A (BPA), is a key monomer (a building block) used in the synthesis of polymers. Most importantly, it is a precursor to epoxy resins. Epoxy resins are high-performance thermosetting polymers known for their excellent adhesion, chemical resistance, and toughness. They are widely used to create protective coatings (paints and primers) for metals in various applications, from automotive bodies to the lining of food cans, specifically to prevent corrosion.


Therefore, bisphenol is the correct answer as it is the fundamental component of epoxy coatings used for corrosion prevention.


Step 4: Final Answer:

Bisphenol is used in the production of epoxy resins, which serve as anti-corrosion coatings for metallic objects.
Quick Tip: For questions about practical applications, think about the material class. Anti-corrosion coatings are typically durable polymers. Connect "Bisphenol" to "Epoxy Resins," which are famous for their use as strong adhesives and protective coatings.

Step 1: Understanding the Concept:

The solubility of a gas in a liquid is governed by the principle of "like dissolves like." Water (H\(_2\)O) is a highly polar solvent capable of forming strong hydrogen bonds. Therefore, polar gases or gases that can interact strongly with water will be more soluble, while nonpolar gases will have low solubility.


Step 3: Detailed Explanation:

Let's analyze the intermolecular forces and polarity of each gas:

- (C) Formaldehyde (HCHO): This is a polar molecule with a C=O bond. It can form hydrogen bonds with water, making it highly soluble.

- (E) Vinyl chloride (C\(_2\)H\(_3\)Cl): The C-Cl bond makes this molecule polar. It will have moderate solubility due to dipole-dipole interactions with water.

- (B) Carbon dioxide (CO\(_2\)): Although it has polar C=O bonds, the molecule is linear and symmetric, making it nonpolar overall. However, CO\(_2\) reacts with water to a small extent to form carbonic acid (H\(_2\)CO\(_3\)), which increases its solubility compared to inert nonpolar gases.

- (A) Argon (Ar) and (D) Methane (CH\(_4\)): Both are nonpolar. Their interaction with water is limited to weak London dispersion forces.

- Argon (Ar): A noble gas atom.

- Methane (CH\(_4\)): A small, nonpolar hydrocarbon.


Comparing the least soluble candidates, Argon and Methane: Both are nonpolar and have very low solubility. The solubility of nonpolar gases in water is generally very low. Comparing empirical data, at 25\(^\circ\)C (298 K) and 1 atm pressure:

- Solubility of Argon in water is approx. 33.6 mg/L.

- Solubility of Methane in water is approx. 22.7 mg/L.

Methane is less soluble than Argon. The smaller size and non-spherical shape of methane may lead to a larger cavity formation energy in the highly structured water network, thus reducing its solubility compared to the spherical argon atom.


Step 4: Final Answer:

Among the given options, Methane has the lowest solubility in water.
Quick Tip: To determine solubility in water, first check for polarity. Nonpolar molecules (like hydrocarbons and noble gases) will be the least soluble. Polar molecules, especially those that can hydrogen bond, will be the most soluble.

Step 1: Understanding the Concept:

The rate of a reaction can be expressed in terms of the change in concentration of a reactant or a product over time. For a reactant, the rate of disappearance is the decrease in its concentration per unit time. The overall rate of the reaction is related to the rate of disappearance of a reactant by its stoichiometric coefficient in the balanced chemical equation.


Step 2: Key Formula or Approach:

For a general reaction \( aA \rightarrow Products \), the rate of reaction is given by:
\[ Rate = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} \]
where \( \Delta[A] \) is the change in concentration of reactant A, and \( \Delta t \) is the time interval. The negative sign indicates that the concentration of the reactant is decreasing.


Step 3: Detailed Explanation:

Given the reaction: \( 3A \rightarrow Products \). The stoichiometric coefficient for reactant A is \(a = 3\).

The initial concentration of A, \( [A]_{initial} = 0.6 \) mol L\(^{-1}\).

The final concentration of A, \( [A]_{final} = 0.3 \) mol L\(^{-1}\).

The time interval, \( \Delta t = 20 \) minutes.


First, calculate the change in concentration of A:
\[ \Delta[A] = [A]_{final} - [A]_{initial} = 0.3 - 0.6 = -0.3 \, mol L^{-1} \]

Next, calculate the rate of disappearance of A:
\[ Rate of disappearance of A = -\frac{\Delta[A]}{\Delta t} = -\frac{-0.3 \, mol L^{-1}}{20 \, min} = \frac{0.3}{20} = 0.015 \, mol L^{-1}min^{-1} \]

Finally, calculate the rate of the reaction using the stoichiometry:
\[ Rate of reaction = \frac{1}{3} \times (Rate of disappearance of A) \] \[ Rate of reaction = \frac{1}{3} \times 0.015 \, mol L^{-1}min^{-1} \] \[ Rate of reaction = 0.005 \, mol L^{-1}min^{-1} \]

Step 4: Final Answer:

The rate of the reaction during this interval is 0.005 mol L\(^{-1}\)min\(^{-1}\).
Quick Tip: A common mistake is to forget to divide by the stoichiometric coefficient when calculating the overall rate of reaction. Always remember that the rate of reaction is defined per one mole of reaction as written in the balanced equation.

Step 1: Understanding the Concept:

The rate law for a reaction expresses the relationship between the rate of the reaction and the concentrations of the reactants. It must be determined experimentally. The method of initial rates involves comparing the initial rates of a reaction under different starting concentrations.


Step 2: Key Formula or Approach:

Let the general rate law be: \( Rate = k[NO]^x[O_2]^y \), where \(x\) and \(y\) are the orders of the reaction with respect to NO and O\(_2\), respectively. Our goal is to find the values of \(x\) and \(y\) by comparing pairs of experiments.


Step 3: Detailed Explanation:

Finding the order with respect to NO (x):

Compare experiments 1 and 2, where the concentration of O\(_2\) is kept constant ([O\(_2\)] = 0.30 M).

- The concentration of NO is doubled (from 0.30 M to 0.60 M).

- The initial rate increases from 0.096 to 0.384.

Let's find the ratio of the rates:
\[ \frac{Rate_2}{Rate_1} = \frac{0.384}{0.096} = 4 \]
Now, let's find the ratio of the concentrations:
\[ \left( \frac{[NO]_2}{[NO]_1} \right)^x = \left( \frac{0.60}{0.30} \right)^x = (2)^x \]
Since \( \frac{Rate_2}{Rate_1} = \left( \frac{[NO]_2}{[NO]_1} \right)^x \), we have \( 4 = 2^x \). Therefore, \( x=2 \).

The reaction is second order with respect to NO.


Finding the order with respect to O\(_2\) (y):

Compare experiments 1 and 3, where the concentration of NO is kept constant ([NO] = 0.30 M).

- The concentration of O\(_2\) is doubled (from 0.30 M to 0.60 M).

- The initial rate increases from 0.096 to 0.192.

Let's find the ratio of the rates:
\[ \frac{Rate_3}{Rate_1} = \frac{0.192}{0.096} = 2 \]
Now, let's find the ratio of the concentrations:
\[ \left( \frac{[O_2]_3}{[O_2]_1} \right)^y = \left( \frac{0.60}{0.30} \right)^y = (2)^y \]
Since \( \frac{Rate_3}{Rate_1} = \left( \frac{[O_2]_3}{[O_2]_1} \right)^y \), we have \( 2 = 2^y \). Therefore, \( y=1 \).

The reaction is first order with respect to O\(_2\).


Combining the results:

The rate law is \( Rate = k[NO]^2[O_2]^1 \), which is written as \( Rate = k[NO]^2[O_2] \).


Step 4: Final Answer:

The rate law for this reaction is Rate = k[NO]\(^2\)[O\(_2\)].
Quick Tip: To find the order of a specific reactant, pick two experiments where only the concentration of that reactant changes, while all other reactant concentrations are held constant. This isolates the effect of that single reactant on the rate.

Step 1: Understanding the Concept:

Allotropy is the property of some chemical elements to exist in two or more different forms, known as allotropes of that element. For metals, these different forms often correspond to different crystal structures. Common crystal structures for metals include body-centered cubic (bcc), face-centered cubic (fcc, which is equivalent to cubic close-packed or ccp), and hexagonal close-packed (hcp). The question asks which element exhibits both bcc and ccp structures among its allotropes, although the phrase "at normal temperature" is misleading as a single element cannot have two structures simultaneously and these forms exist at different temperatures. The question likely means which element is known to have both structures as part of its allotropic forms.


Step 3: Detailed Explanation:

Let's examine the crystal structures of the given elements:

- (A) Titanium (Ti): Exists as hcp (\(\alpha\)-Ti) at room temperature and transforms to bcc (\(\beta\)-Ti) at high temperatures (above 882 \(^{\circ}\)C). It does not have a ccp structure.

- (B) Vanadium (V): Has a bcc structure.

- (C) Silver (Ag): Has a ccp (or fcc) structure.

- (D) Chromium (Cr): Has a bcc structure.

- (E) Manganese (Mn): Is known for its complex allotropy and exists in four different forms at different temperatures:

- \(\alpha\)-Mn (below 727 \(^{\circ}\)C): A complex cubic structure (not bcc or ccp).

- \(\beta\)-Mn (from 727 to 1095 \(^{\circ}\)C): Another complex cubic structure.

- \(\gamma\)-Mn (from 1095 to 1134 \(^{\circ}\)C): Has a face-centered cubic (fcc or ccp) structure.

- \(\delta\)-Mn (from 1134 \(^{\circ}\)C to melting point): Has a bcc structure.

Therefore, Manganese is the element among the choices that possesses both bcc and ccp structures as its allotropes.


Step 4: Final Answer:

Manganese is the element that has both bcc and ccp structures among its allotropic forms.
Quick Tip: Remember that Iron (Fe) is a more common example of an element with both bcc and ccp (fcc) allotropes. Since it's not an option, consider other elements known for complex allotropy, like Manganese. The question's wording "at normal temperature" is likely an error and should be interpreted as "which element possesses these structures as allotropes".

Step 1: Understanding the Concept:

Transition metals exhibit variable oxidation states due to the participation of both ns and (n-1)d electrons in bonding. The "common" oxidation states are those that are most frequently encountered and are thermodynamically stable under normal conditions.


Step 3: Detailed Explanation:

Chromium (Cr) has an atomic number of 24 and an exceptional electron configuration of [Ar] 3d\(^5\) 4s\(^1\). This configuration, with half-filled d and s subshells, allows it to display a wide range of oxidation states from -2 to +6.

- The +3 oxidation state is the most stable state for chromium. In aqueous solutions, chromium exists as the Cr\(^{3+}\) ion, which forms many stable coordination compounds. Chromium(III) oxide (Cr\(_2\)O\(_3\)) is a very stable green pigment.

- The +6 oxidation state is also very common, although it is strongly oxidizing. Chromium in this state is found in compounds like chromium trioxide (CrO\(_3\)), chromate ion (CrO\(_4\)\(^{2-}\)), and dichromate ion (Cr\(_2\)O\(_7\)\(^{2-}\)), which are powerful oxidizing agents.

- Other oxidation states like +2 (chromous) exist, but it is a strong reducing agent and is readily oxidized to the more stable +3 state. Oxidation states like +4 and +5 are less common.

Therefore, the most common and significant oxidation states of chromium are +3 and +6.


Step 4: Final Answer:

The most common oxidation states of chromium are +3 and +6.
Quick Tip: For first-row transition metals, remember key stable oxidation states. For Chromium, think of the stable green Cr\(^{3+}\) compounds and the strong orange/yellow oxidizing agents containing Cr(VI) like dichromate. These are its two most important states.

Step 1: Understanding the Concept:

The magnetic moment of a transition metal ion is primarily due to the spin of its unpaired electrons. The "spin-only" formula is used to calculate this property. The unit for magnetic moment is the Bohr Magneton (BM).


Step 2: Key Formula or Approach:

The spin-only magnetic moment (\(\mu\)) is calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \, BM \]
where \(n\) is the number of unpaired electrons.


Step 3: Detailed Explanation:

The problem states that the divalent ion has three unpaired electrons.

So, \(n = 3\).

Substitute this value into the formula:
\[ \mu = \sqrt{3(3+2)} \] \[ \mu = \sqrt{3 \times 5} \] \[ \mu = \sqrt{15} \]
To estimate \(\sqrt{15}\), we know that \(\sqrt{9} = 3\) and \(\sqrt{16} = 4\). So the value must be between 3 and 4, closer to 4.

Calculating the value:
\[ \mu \approx 3.87 \, BM \]

Step 4: Final Answer:

The magnetic moment of a divalent ion with three unpaired electrons is 3.87 BM.
Quick Tip: You can quickly estimate the magnetic moment. The value is always slightly greater than the number of unpaired electrons (for n \textgreater 0). For n=1, \(\mu\)=1.73; n=2, \(\mu\)=2.84; n=3, \(\mu\)=3.87; n=4, \(\mu\)=4.90; n=5, \(\mu\)=5.92. The integer part always matches 'n'.

Step 1: Understanding the Concept:

The question asks for the bond angle in the dichromate ion, Cr\(_2\)O\(_7\)\(^{2-}\). This ion consists of two CrO\(_4\) tetrahedra that are joined by sharing a single oxygen atom. The angle in question is the one formed at this bridging oxygen atom.


Step 3: Detailed Explanation:

The structure of the dichromate ion can be visualized as O\(_3\)Cr-O-CrO\(_3\). The bridging oxygen atom is bonded to two chromium atoms. According to VSEPR theory, the bridging oxygen has two bond pairs and two lone pairs, which would suggest an electron geometry that is roughly tetrahedral and a bond angle close to 109.5\(^\circ\).

However, several factors cause the actual bond angle to deviate significantly from the ideal tetrahedral angle:

1. Steric Hindrance: The two CrO\(_3\) groups are bulky and repel each other, which tends to increase the bond angle.

2. p\(\pi\)-d\(\pi\) Bonding: There can be some degree of pi-bonding between the lone pairs on the oxygen atom and the vacant d-orbitals on the chromium atoms. This delocalization of electrons would favor a larger bond angle.

Experimental data from X-ray crystallography shows that the Cr-O-Cr bond angle in the dichromate ion is approximately 126\(^\circ\). This value is a well-established fact for this structure.


Step 4: Final Answer:

The bond angle of the Cr-O-Cr bond in the dichromate ion is 126\(^\circ\).
Quick Tip: For specific, fact-based questions about molecular geometry like this, it is often necessary to memorize key structural parameters. The Cr-O-Cr angle in dichromate (126\(^\circ\)) and the Si-O-Si angle in silicates (around 140\(^\circ\)) are common examples asked in exams.

Step 1: Understanding the Concept:

A dry cell, such as the Leclanché cell or the zinc-carbon battery, is a common type of primary battery. It consists of an anode, a cathode, and an electrolyte paste. A key component is the depolarizer, a substance used to prevent the buildup of gas bubbles on the cathode, which would otherwise stop the reaction.


Step 3: Detailed Explanation:

In a standard zinc-carbon dry cell:

- Anode (Negative Electrode): Is the zinc casing of the battery. Oxidation occurs here: \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \).

- Cathode (Positive Electrode): Is a central carbon (graphite) rod.

- Electrolyte: Is a moist paste of ammonium chloride (NH\(_4\)Cl) and zinc chloride (ZnCl\(_2\)).

- Depolarizer: The carbon rod is surrounded by a paste of manganese dioxide (MnO\(_2\)) and powdered carbon. At the cathode, ammonium ions are reduced to ammonia and hydrogen gas. The hydrogen gas can insulate the cathode, stopping the battery from working (polarization). Manganese dioxide acts as a depolarizer by oxidizing the hydrogen gas:
\[ 2MnO_2(s) + 2NH_4^+(aq) + 2e^- \rightarrow Mn_2O_3(s) + 2NH_3(aq) + H_2O(l) \]
Therefore, manganese dioxide (MnO\(_2\)) is the essential transition metal oxide used in dry cells.


Step 4: Final Answer:

MnO\(_2\) is the transition metal oxide used in dry battery cells.
Quick Tip: Associate "dry cell battery" with its key chemical components: Zinc anode, Carbon cathode, and Manganese dioxide (MnO\(_2\)) depolarizer. This connection is fundamental to electrochemistry and frequently tested.

Step 1: Understanding the Concept:

Ionization enthalpy (IE) is the energy required to remove an electron from an atom. The first IE (IE\(_1\)) is for the first electron, and the second IE (IE\(_2\)) is for the second. Lanthanoids are elements where the 4f subshell is being filled. Their valence electron configuration is typically [Xe] 4f\(^n\) 6s\(^2\).


Step 3: Detailed Explanation:

- Ionization of Lanthanoids: The first two ionization enthalpies of lanthanoids correspond to the removal of the two outermost 6s electrons. The inner 4f electrons are poor at shielding, so the effective nuclear charge increases across the series. However, the removal of the 6s electrons is relatively easy and does not involve the f-electrons directly. The sum of IE\(_1\) + IE\(_2\) for lanthanoids is fairly constant and relatively low.

- Comparison with other elements: We are looking for an element that also readily loses two electrons from an s-orbital.

- (A) Chromium (Cr): A transition metal. Its IE values are different.

- (B) Calcium (Ca): An alkaline earth metal with electron configuration [Ar] 4s\(^2\). It readily loses its two 4s electrons to form a stable Ca\(^{2+}\) ion with a noble gas configuration. The energy required to remove these two electrons is comparable to the energy required to remove the two 6s electrons from the lanthanoids.

- (C) Germanium (Ge): A p-block element.

- (D) Cesium (Cs): An alkali metal. It has a very low IE\(_1\) but a very high IE\(_2\) because the second electron is removed from a stable, filled inner shell.

- (E) Cadmium (Cd): A post-transition metal. It loses two 5s electrons, but its IE values are higher than those of Calcium due to a higher nuclear charge that is poorly shielded by the 4d electrons.

The chemical behavior and ionization energies of lanthanoids (especially for forming the +2 state) are most similar to the alkaline earth metals, with Calcium being the best comparison among the choices.


Step 4: Final Answer:

The first and second ionization enthalpies of lanthanoids are comparable with Calcium.
Quick Tip: Remember that the chemical properties of Lanthanoids are very similar to each other and are often compared to Group 2 (Alkaline Earth) or Group 3 metals. For the first two ionization steps (forming the M\(^{2+}\) ion), the comparison to Calcium (Group 2) is the most appropriate.

Step 1: Understanding the Concept:

Gemstones often get their color from trace amounts of transition metal ions acting as impurities within a crystal lattice. Ruby is a red variety of the mineral corundum, which is crystalline aluminum oxide (Al\(_2\)O\(_3\)).


Step 3: Detailed Explanation:

The crystal lattice of corundum consists of Al\(^{3+}\) and O\(^{2-}\) ions. The characteristic red color of ruby is produced when a small number of the Al\(^{3+}\) ions are replaced by chromium(III) ions, Cr\(^{3+}\).

The presence of Cr\(^{3+}\) in the Al\(_2\)O\(_3\) lattice alters the way the crystal absorbs light. The Cr\(^{3+}\) ions absorb light in the yellow-green region of the visible spectrum, which results in the transmitted light appearing red. The intensity of the red color depends on the concentration of chromium.

For a stone to be classified as a ruby, the concentration of chromium is typically very low. The accepted range for the percentage of Cr(III) by mass in ruby is generally between 0.5% and 1%. Higher concentrations would make the stone more opaque and less valuable as a gem.


Step 4: Final Answer:

The percentage of Cr(III) in Ruby is 0.5 to 1%.
Quick Tip: Remember that gemstone colors are due to trace impurities. For ruby, the magic ingredient is Cr\(^{3+}\) in an Al\(_2\)O\(_3\) crystal. For sapphire (another corundum variety), the impurities are typically Fe\(^{2+}\) and Ti\(^{4+}\) for the blue color.

Step 1: Understanding the Concept:

According to Valence Bond Theory, coordination complexes can be classified based on the type of d-orbitals used for hybridization by the central metal ion.

- Inner Orbital Complex: Uses inner (n-1)d orbitals for hybridization (e.g., d\(^2\)sp\(^3\)). These are also called low-spin complexes and are typically formed with strong-field ligands that force pairing of d-electrons.

- Outer Orbital Complex: Uses outer nd orbitals for hybridization (e.g., sp\(^3\)d\(^2\)). These are also called high-spin complexes and are typically formed with weak-field ligands that do not cause electron pairing.


Step 3: Detailed Explanation:

Let's analyze each complex:

- (A) [Co(NH\(_3\))\(_6\)]\(^{3+}\): Cobalt is in the +3 state (Co\(^{3+}\)), which is a 3d\(^6\) configuration. NH\(_3\) acts as a strong-field ligand with Co\(^{3+}\). It forces the six d-electrons to pair up in the lower energy orbitals, leaving two (n-1)d orbitals empty. Hybridization is d\(^2\)sp\(^3\). This is an inner orbital complex.

- (B) [Mn(CN)\(_6\)]\(^{3-}\): Manganese is in the +3 state (Mn\(^{3+}\)), which is a 3d\(^4\) configuration. CN\(^-\) is a strong-field ligand. It forces electron pairing. Hybridization is d\(^2\)sp\(^3\). This is an inner orbital complex.

- (C) [Co(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\): Cobalt is in the +3 state (Co\(^{3+}\)), a 3d\(^6\) configuration. Oxalate (C\(_2\)O\(_4\)\(^{2-}\)) acts as a strong-field ligand with Co\(^{3+}\), causing electron pairing. Hybridization is d\(^2\)sp\(^3\). This is an inner orbital complex.

- (D) [MnCl\(_6\)]\(^{3-}\): Manganese is in the +3 state (Mn\(^{3+}\)), a 3d\(^4\) configuration. Cl\(^-\) is a weak-field ligand. It does not cause the four d-electrons to pair up. To form six bonds, the central atom must use the outer 4d orbitals. The hybridization is sp\(^3\)d\(^2\). This is an outer orbital complex.

- (E) [Fe(CN)\(_6\)]\(^{3-}\): Iron is in the +3 state (Fe\(^{3+}\)), a 3d\(^5\) configuration. CN\(^-\) is a strong-field ligand. It forces the electrons to pair up as much as possible, resulting in one unpaired electron. Hybridization is d\(^2\)sp\(^3\). This is an inner orbital complex.


Step 4: Final Answer:

[MnCl\(_6\)]\(^{3-}\) is the outer orbital complex.
Quick Tip: A quick way to solve this is to identify the ligand type. Halide ions (like Cl\(^-\), F\(^-\), Br\(^-\)) are almost always weak-field ligands that lead to high-spin, outer orbital complexes. Ligands like CN\(^-\), CO, and NH\(_3\) are typically strong-field, leading to low-spin, inner orbital complexes.

Step 1: Understanding the Concept:

The color of transition metal complexes is due to the absorption of light, which promotes an electron from a lower energy d-orbital to a higher energy d-orbital (a d-d transition). The color observed is the complementary color of the light absorbed. The energy of the absorbed light depends on the crystal field splitting energy (\(\Delta_o\)), which is determined by the metal ion and the ligands.


Step 3: Detailed Explanation:

The complex is [Ni(en)\(_3\)]\(^{2+}\). The central ion is Ni\(^{2+}\) (a 3d\(^8\) ion). The ligand is ethylenediamine (en), which is a bidentate, strong-field ligand.

Let's compare this to the aquated nickel ion, [Ni(H\(_2\)O)\(_6\)]\(^{2+}\), which is known to be green.

The spectrochemical series ranks ligands according to their ability to cause d-orbital splitting: H\(_2\)O \textless en.

Since 'en' is a stronger field ligand than H\(_2\)O, it causes a larger crystal field splitting (\(\Delta_o\)).

A larger \(\Delta_o\) means that the complex will absorb light of higher energy (and shorter wavelength) to promote an electron.

- [Ni(H\(_2\)O)\(_6\)]\(^{2+}\) is green because it absorbs light in the red part of the spectrum.

- [Ni(en)\(_3\)]\(^{2+}\) has a larger \(\Delta_o\), so it will absorb light of higher energy, shifting the absorption from red towards the yellow-green part of the spectrum.

- If a substance absorbs yellow-green light, the complementary color that is transmitted and observed by our eyes is violet or purple.

Therefore, the color of the [Ni(en)\(_3\)]\(^{2+}\) complex is violet.


Step 4: Final Answer:

The colour of the complex [Ni(en)\(_3\)]\(^{2+}\) in water is Violet.
Quick Tip: Remember the spectrochemical series and the concept of complementary colors. A stronger ligand increases \(\Delta_o\), shifting the absorbed light to shorter wavelengths (higher energy) and thus changing the observed color. Knowing the color of the aqueous ion (e.g., [Ni(H\(_2\)O)\(_6\)]\(^{2+}\) is green) is a useful baseline.

Step 1: Understanding the Concept:

Hardness in water is caused by the presence of dissolved divalent cations, primarily calcium (Ca\(^{2+}\)) and magnesium (Mg\(^{2+}\)). The total hardness is a measure of the total concentration of these ions. This concentration is determined analytically by a complexometric titration.


Step 3: Detailed Explanation:

A complexometric titration involves the formation of a colored complex. The standard method for determining water hardness uses Ethylenediaminetetraacetic acid (EDTA) as the titrant.

- Titrant: EDTA, usually in the form of its disodium salt (Na\(_2\)EDTA), is a hexadentate ligand. It is an excellent chelating agent, meaning it can wrap around a metal ion and form multiple bonds, creating a very stable complex called a metal-EDTA complex. It forms stable 1:1 complexes with Ca\(^{2+}\) and Mg\(^{2+}\).

- Procedure: A sample of hard water is buffered to a pH of about 10, and an indicator (like Eriochrome Black T) is added. The indicator forms a wine-red complex with the Mg\(^{2+}\) and Ca\(^{2+}\) ions. The solution is then titrated with a standard solution of Na\(_2\)EDTA. The EDTA reacts with the free metal ions first, and then it displaces the metal ions from the indicator complex. When all the metal ions are complexed with EDTA, the indicator becomes free and changes color from wine-red to blue, signaling the endpoint.

- Other options: The other compounds listed are also complexing agents but are used for specific metal analysis: DMG for nickel, cupron for copper, and \(\alpha\)-nitroso-\(\beta\)-naphthol for cobalt.


Step 4: Final Answer:

Hardness of water is estimated by titration with Na\(_2\)EDTA.
Quick Tip: Whenever you see a question about the quantitative determination of water hardness, the answer is almost always related to EDTA. Associate "water hardness" with "complexometric titration" and "EDTA".

Step 1: Understanding the Concept:

Ammonium phosphomolybdate is a well-known inorganic compound that is important in the qualitative and quantitative analysis of phosphates. It is formed as a bright canary yellow precipitate when a solution containing phosphate ions is treated with ammonium molybdate solution in the presence of excess nitric acid.


Step 3: Detailed Explanation:

The reaction involves the formation of a complex heteropoly acid, phosphomolybdic acid (H\(_3\)PO\(_4\).12MoO\(_3\)), which then reacts with ammonium ions from the ammonium molybdate and nitric acid to form the insoluble salt, ammonium phosphomolybdate.

The chemical structure is complex, consisting of a central phosphate tetrahedron surrounded by twelve molybdate octahedra. The overall formula of the precipitate is correctly represented as (NH\(_4\))\(_3\)[PMo\(_{12}\)O\(_{40}\)].

This complex formula is often written in a simplified salt-hydrate or salt-oxide form for stoichiometric purposes, which is (NH\(_4\))\(_3\)PO\(_4\).12MoO\(_3\). Let's check the options:

- (A) (NH\(_4\))\(_3\)PO\(_4\).12MoO\(_3\): This formula correctly represents the stoichiometry of ammonium, phosphate, and molybdenum trioxide units in the compound.

- Options (B), (C), and (D) have incorrect stoichiometry for the ammonium and phosphate groups.


Step 4: Final Answer:

The formula of Ammonium phosphomolybdate is (NH\(_4\))\(_3\)PO\(_4\).12MoO\(_3\).
Quick Tip: Remember the "canary yellow precipitate" test for the phosphate ion. This test produces ammonium phosphomolybdate, and its formula, (NH\(_4\))\(_3\)PO\(_4\).12MoO\(_3\), is a key fact to memorize for qualitative inorganic analysis.

Step 1: Understanding the Concept:

Combustion analysis is a method used to determine the elemental composition of an organic compound. When an organic compound containing carbon, hydrogen, and possibly oxygen is burned completely, all the carbon is converted into carbon dioxide (CO\(_2\)) and all the hydrogen is converted into water (H\(_2\)O). By measuring the mass of CO\(_2\) produced, we can calculate the mass and percentage of carbon in the original sample.


Step 2: Key Formula or Approach:

1. Calculate the mass of carbon in the CO\(_2\) produced using the ratio of molar masses.
\[ Mass of C = Mass of CO_2 \times \frac{Molar mass of C}{Molar mass of CO_2} \]
2. Calculate the percentage of carbon in the original organic compound.
\[ % C = \frac{Mass of C}{Mass of organic compound} \times 100% \]

Step 3: Detailed Explanation:

Given data:

- Mass of organic compound = 0.96 g

- Mass of CO\(_2\) produced = 0.88 g


Molar masses:

- Molar mass of Carbon (C) \(\approx\) 12 g/mol

- Molar mass of Carbon Dioxide (CO\(_2\)) = 12 + 2(16) = 44 g/mol


1. Calculate the mass of carbon:

The fraction of carbon by mass in CO\(_2\) is \(\frac{12}{44}\).
\[ Mass of C in 0.88 g of CO_2 = 0.88 \, g \times \frac{12}{44} \] \[ Mass of C = \frac{0.88 \times 12}{44} = \frac{0.02 \times 44 \times 12}{44} = 0.02 \times 12 = 0.24 \, g \]

2. Calculate the percentage of carbon:
\[ % C = \frac{Mass of C}{Mass of organic compound} \times 100% \] \[ % C = \frac{0.24 \, g}{0.96 \, g} \times 100% \]
Since \( 0.96 = 4 \times 0.24 \), the fraction is \(\frac{1}{4}\).
\[ % C = \frac{1}{4} \times 100% = 25% \]

Step 4: Final Answer:

The percentage composition of carbon in the compound is 25%.
Quick Tip: To speed up calculations in combustion analysis, remember the mass fractions: Carbon in CO\(_2\) is 12/44 (or 3/11), and Hydrogen in H\(_2\)O is 2/18 (or 1/9). Multiplying the mass of the product by these fractions gives the mass of the element directly.

Step 1: Understanding the Concept:

Kolbe's electrolytic method is an organic reaction used to synthesize alkanes. It involves the electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid. The core of the reaction is the decarboxylation of the carboxylate anion at the anode to form an alkyl radical, followed by the dimerization (coupling) of two such radicals to form an alkane.


Step 2: Key Formula or Approach:

The general reaction at the anode is:
\[ 2 R-COO^- \xrightarrow{electrolysis} R-R + 2CO_2 + 2e^- \]
The product alkane is formed by joining two 'R' groups from the starting carboxylate salt, R-COONa. Therefore, the product alkane always has an even number of carbon atoms and is symmetrical.


Step 3: Detailed Explanation:

The desired product is n-hexane.

The structure of n-hexane is CH\(_3\)CH\(_2\)CH\(_2\)-CH\(_2\)CH\(_2\)CH\(_3\).

This molecule is symmetrical and can be seen as two identical propyl groups joined together: (CH\(_3\)CH\(_2\)CH\(_2\)-).

So, the alkyl group 'R' in the general formula must be the n-propyl group, R = CH\(_3\)CH\(_2\)CH\(_2\)-.

The starting carboxylic acid salt must therefore be the salt of butanoic acid, which is sodium butanoate.

The formula for sodium butanoate is CH\(_3\)CH\(_2\)CH\(_2\)COONa.

Let's check the options:

- (A) Sodium propanoate (CH\(_3\)CH\(_2\)COONa) would give n-butane (C\(_2\)H\(_5\)-C\(_2\)H\(_5\)).

- (B) Sodium acetate (CH\(_3\)COONa) would give ethane (CH\(_3\)-CH\(_3\)).

- (C) Sodium formate (HCOONa) would give hydrogen gas (H-H).

- (D) Sodium pentanoate would give n-octane.

- (E) Sodium butanoate (CH\(_3\)CH\(_2\)CH\(_2\)COONa) would give the n-propyl radical, and two of these would combine to form n-hexane. This is the correct choice.


Step 4: Final Answer:

The sodium salt used for the preparation of n-hexane is sodium butanoate, CH\(_3\)CH\(_2\)CH\(_2\)COONa.
Quick Tip: To quickly solve Kolbe's electrolysis problems, look at the desired alkane product. Cut it in half symmetrically. The resulting alkyl group 'R' is the one that was attached to the -COONa group in the starting material. Remember the starting acid has one more carbon than the radical R.

Step 1: Understanding the Concept:

The direct halogenation of alkanes with iodine is a reversible and very slow reaction. The reaction is:
\[ CH_4 + I_2 \rightleftharpoons CH_3I + HI \]
The hydrogen iodide (HI) formed as a byproduct is a strong reducing agent. It can reduce the product, methyl iodide (CH\(_3\)I), back to the reactant, methane (CH\(_4\)), shifting the equilibrium back to the left.


Step 2: Detailed Explanation:

To make the iodination reaction proceed in the forward direction, the HI produced must be removed from the reaction mixture. This is achieved by adding a strong oxidizing agent that will oxidize HI to I\(_2\) but will not oxidize the alkane.

Let's analyze the options:

- (A) HI: This is the reducing agent that needs to be removed.

- (B) KMnO\(_4\) and (C) K\(_2\)Cr\(_2\)O\(_7\): These are very powerful oxidizing agents and are generally not suitable as they can cause unwanted side reactions, including the oxidation of methane itself.

- (D) HIO\(_3\) (Iodic acid): This is a commonly used oxidizing agent for this specific purpose. It effectively oxidizes the HI produced back into I\(_2\), thus preventing the reverse reaction and shifting the equilibrium to the right, favoring the formation of methyl iodide. The reaction is:
\[ 5HI + HIO_3 \rightarrow 3I_2 + 3H_2O \]
Other suitable oxidizing agents include concentrated nitric acid (HNO\(_3\)) and mercuric oxide (HgO).

- (E) K\(_2\)CrO\(_4\): Similar to permanganate and dichromate, this is a strong oxidizing agent and less suitable than HIO\(_3\).


Step 3: Final Answer:

The oxidizing agent used for the iodination of methane is HIO\(_3\).
Quick Tip: Remember the unique challenges for each type of alkane halogenation. Fluorination is too violent, chlorination and bromination proceed well via free radical mechanism, and iodination is reversible. For iodination, always look for an oxidizing agent like HIO\(_3\) or HNO\(_3\) to remove the HI byproduct.

Step 1: Understanding the Concept:

Ozonolysis is a reaction in which alkenes or alkynes are cleaved by reaction with ozone (O\(_3\)). This reaction breaks the carbon-carbon double (or triple) bond and replaces it with carbon-oxygen double bonds (carbonyl groups). In reductive ozonolysis (the most common type, often involving a workup with zinc or dimethyl sulfide), aldehydes and/or ketones are formed.


Step 2: Key Formula or Approach:

1. Draw the structure of the starting alkene.

2. Identify the carbon-carbon double bond.

3. "Cut" the double bond in half.

4. Add a double-bonded oxygen atom (=O) to each of the two carbons that were part of the original double bond.

5. Name the resulting carbonyl compounds.


Step 3: Detailed Explanation:

1. Structure of 3-Ethylpent-2-ene:

- "pent-2-ene" means a 5-carbon chain with a double bond between C2 and C3.

\( C_1H_3 - C_2H = C_3 - C_4H_2 - C_5H_3 \)

- "3-Ethyl" means an ethyl group (-CH\(_2\)CH\(_3\)) is attached to C3.

- The final structure is:
\[ CH_3 - CH = \underset{\substack{|
CH_2CH_3}}{C} - CH_2 - CH_3 \]

2. Ozonolysis Reaction:

We cleave the double bond between C2 and C3:
\[ CH_3 - CH \quad // \quad \underset{\substack{|
CH_2CH_3}}{C} - CH_2 - CH_3 \]
Now, add "=O" to each fragment:

- The left fragment (\(CH_3 - CH\)) becomes \(CH_3 - CH=O\). This is a two-carbon aldehyde, which is Ethanal.

- The right fragment (\(\underset{\substack{|
CH_2CH_3}}{C} - CH_2 - CH_3\)) becomes \(\underset{\substack{|
CH_2CH_3}}{O=C} - CH_2 - CH_3\). This is a ketone with an ethyl group on both sides of the carbonyl carbon. The longest chain containing the carbonyl is 5 carbons long, with the C=O at position 3. The name is Pentan-3-one (also known as diethyl ketone).


Step 4: Final Answer:

The products obtained on ozonolysis are Ethanal and Pentan-3-one.
Quick Tip: To solve ozonolysis problems quickly, just erase the double bond in the alkene's structure and draw two double-bonded oxygens in its place, pointing away from each other. Then, identify the two new molecules you have created.

Step 1: Understanding the Concept:

The reaction described is the catalytic reforming or aromatization of an alkane. This process involves the cyclization and dehydrogenation of a straight-chain alkane with six or more carbon atoms to form an aromatic compound. n-Hexane is converted to benzene in this process.


Step 2: Detailed Explanation:

The conversion of n-hexane to benzene is a high-temperature industrial process that requires a specific catalyst and conditions.
\[ CH_3(CH_2)_4CH_3 \quad \xrightarrow[\Delta, Pressure]{Catalyst} \quad C_6H_6 + 4H_2 \]
- Catalyst: The catalysts are typically oxides of chromium, vanadium, or molybdenum supported on alumina (Al\(_2\)O\(_3\)). For example, Cr\(_2\)O\(_3\) or V\(_2\)O\(_5\) on Al\(_2\)O\(_3\).

- Temperature: The reaction requires high temperatures to provide the activation energy for bond breaking and formation. The typical temperature range is 773 K to 873 K (500 to 600 \(^{\circ}\)C).

- Pressure: A moderate to high pressure is applied, typically in the range of 10 to 20 atmospheres. This helps to maintain the desired reaction kinetics and phase of the reactants.

Comparing these standard conditions with the given options, the combination of 773 K and 10-20 atm is the correct set of conditions for this reaction.


Step 3: Final Answer:

The temperature and pressure required for reforming benzene from n-hexane are 773K and 10-20 atm.
Quick Tip: Aromatization is a key reaction for converting aliphatic hydrocarbons into valuable aromatic compounds. Memorize the typical conditions: n-alkane (C6-C8), high temperature (around 773 K), moderate pressure (10-20 atm), and a catalyst like Cr\(_2\)O\(_3\)/Al\(_2\)O\(_3\).

Step 1: Understanding the Concept:

This question tests the knowledge of named reactions in organic chemistry, specifically for the preparation of alkyl halides. Each of the options represents a distinct and well-known reaction.


Step 2: Detailed Explanation:

Let's analyze the given reaction and the options:

Given Reaction:
\[ CH_3Br + AgF \xrightarrow{\Delta} CH_3F + AgBr \]
This is a halogen exchange reaction where a bromide is replaced by a fluoride.


Analysis of Options:

- (A) Swarts reaction: This reaction is specifically used for the synthesis of alkyl fluorides. It involves heating an alkyl chloride or alkyl bromide with a metallic fluoride like AgF, Hg\(_2\)F\(_2\), CoF\(_2\), or SbF\(_3\). The given reaction is a classic example of the Swarts reaction.

- (B) Finkelstein reaction: This is also a halogen exchange reaction, but it is used to prepare alkyl iodides by treating alkyl chlorides or bromides with sodium iodide (NaI) in dry acetone.

- (C) Sandmeyer's reaction: This reaction is used to synthesize aryl halides from aryl diazonium salts using a copper(I) halide (CuCl, CuBr).

- (D) Wurtz reaction: This reaction synthesizes alkanes by treating two moles of an alkyl halide with sodium metal in dry ether.

- (E) Kolbe's reaction: This refers to two different reactions, one being the Kolbe electrolysis for alkane synthesis and the other being the Kolbe-Schmitt reaction for synthesizing salicylic acid from phenol. Neither fits the description.


Step 3: Final Answer:

The preparation of methyl fluoride from methyl bromide using AgF is known as the Swarts reaction.
Quick Tip: To remember halogen exchange reactions: Swarts is for Fluorides (S in Swarts, F in Fluoride is not a perfect match, but think Swarts makes Special Fluorides), while Finkelstein is for Iodides (I in Finkelstein, I in Iodide).

Step 1: Understanding the Concept:

This reaction involves the replacement of the diazonium group (-N\(_2\)\(^+\)Cl\(^-\)) on a benzene ring with an iodine atom. This is a standard method for preparing aryl iodides.


Step 2: Detailed Explanation:

The reaction of benzene diazonium chloride to form aryl halides is a key transformation in aromatic chemistry.

- For the preparation of chlorobenzene and bromobenzene, the Sandmeyer reaction (using CuCl/HCl or CuBr/HBr) or the Gattermann reaction (using Cu powder/HCl or Cu powder/HBr) is used.

- However, for the preparation of iodobenzene, the reaction is much simpler. It does not require a copper catalyst. Simply warming an aqueous solution of the benzene diazonium chloride with an aqueous solution of potassium iodide (KI) is sufficient to produce iodobenzene. The diazonium group is replaced by iodide, and nitrogen gas evolves.

The reaction is:
\[ C_6H_5N_2^+Cl^- + KI \xrightarrow{Warm} C_6H_5I + KCl + N_2(g) \]
The iodide ion (I\(^-\)) from KI acts as the nucleophile that replaces the diazonium group. Other reagents listed are not suitable for this direct, high-yield conversion.


Step 3: Final Answer:

The reagent 'X' used to convert benzene diazonium chloride to iodobenzene is KI.
Quick Tip: Remember that the synthesis of aryl iodides from diazonium salts is the "odd one out." Unlike chloro- and bromo- derivatives that need a copper catalyst (Sandmeyer/Gattermann), the iodo- derivative is easily made by just warming with KI solution.

Step 1: Understanding the Concept:

A chiral molecule is a molecule that is non-superimposable on its mirror image. The most common cause of chirality in organic molecules is the presence of a chiral center (or stereocenter), which is a carbon atom bonded to four different groups. An achiral molecule is one that is superimposable on its mirror image, often due to a plane of symmetry or the absence of a chiral center.


Step 2: Detailed Explanation:

Let's analyze the structure of each molecule to find a chiral center.

- (A) 2-Chlorobutane: \( CH_3-C^* H(Cl)-CH_2CH_3 \). The second carbon (C2) is bonded to H, Cl, a methyl group (-CH\(_3\)), and an ethyl group (-CH\(_2\)CH\(_3\)). These four groups are different, so C2 is a chiral center. The molecule is chiral.

- (B) 2,3-Dihydroxy propanal (Glyceraldehyde): \( CHO-C^* H(OH)-CH_2OH \). The second carbon (C2) is bonded to H, OH, an aldehyde group (-CHO), and a hydroxymethyl group (-CH\(_2\)OH). These four groups are different. The molecule is chiral.

- (C) 2-Bromo propionic acid: \( CH_3-C^* H(Br)-COOH \). The second carbon (C2) is bonded to H, Br, a methyl group (-CH\(_3\)), and a carboxyl group (-COOH). These four groups are different. The molecule is chiral.

- (D) Butan-2-ol: \( CH_3-C^* H(OH)-CH_2CH_3 \). The second carbon (C2) is bonded to H, OH, a methyl group (-CH\(_3\)), and an ethyl group (-CH\(_2\)CH\(_3\)). These four groups are different. The molecule is chiral.

- (E) 2-Bromo-2-methoxypropane: \( CH_3-C(Br)(OCH_3)-CH_3 \). The central carbon atom (C2) is bonded to a bromine atom (-Br), a methoxy group (-OCH\(_3\)), and two identical methyl groups (-CH\(_3\)). Since the carbon is not bonded to four different groups, it is not a chiral center. The molecule is achiral.


Step 3: Final Answer:

2-Bromo-2-methoxypropane is not a chiral molecule.
Quick Tip: To check for chirality, find a carbon atom that seems to have many attachments. Systematically list the four groups bonded to it. If any two groups are identical (e.g., two -CH\(_3\) groups, two -H atoms), the carbon is not a chiral center, and the molecule is likely achiral.

Step 1: Understanding the Concept:

This is a Grignard reaction. A Grignard reagent (R-MgX) acts as a strong nucleophile and a strong base. It reacts with the electrophilic carbon of a carbonyl group (in an aldehyde or ketone). The reaction with a ketone, followed by hydrolysis (protonation), yields a tertiary alcohol.


Step 2: Key Formula or Approach:

The reaction proceeds in two steps:

1. Nucleophilic Addition: The Grignard reagent attacks the carbonyl carbon.

\( R_2C=O + R'MgX \rightarrow R_2R'C-O^-MgX^+ \)

2. Hydrolysis: The resulting alkoxide is protonated by adding a weak acid (like H\(_2\)O or dilute H\(^+\)).

\( R_2R'C-O^-MgX^+ + H_2O \rightarrow R_2R'C-OH + Mg(OH)X \)


Step 3: Detailed Explanation:

Reactants:

- Propanone (also known as acetone): \( CH_3-C(=O)-CH_3 \). This is a ketone.

- Methyl magnesium bromide: \( CH_3MgBr \). The nucleophilic part is the CH\(_3\)\(^{\delta-}\) group.


Step 1: Nucleophilic Addition

The nucleophilic methyl group from CH\(_3\)MgBr attacks the electrophilic carbonyl carbon of propanone. The \(\pi\) bond of the C=O group breaks, and the electrons move to the oxygen atom.
\[ CH_3-C(=O)-CH_3 + CH_3MgBr \rightarrow CH_3-C(CH_3)(OMgBr)-CH_3 \]

Step 2: Hydrolysis

The intermediate magnesium alkoxide is treated with water to protonate the oxygen, forming the final alcohol product.
\[ CH_3-C(CH_3)(OMgBr)-CH_3 + H_2O \rightarrow CH_3-C(CH_3)(OH)-CH_3 + Mg(OH)Br \]

Identifying the Product:

The product is \( CH_3-C(CH_3)(OH)-CH_3 \).

To name it, we find the longest carbon chain containing the -OH group. This is a 3-carbon chain (propane).

The -OH group is on carbon 2, and a methyl group is also on carbon 2.

The IUPAC name is 2-Methylpropan-2-ol. This is a tertiary alcohol.


Step 4: Final Answer:

The product of the reaction is 2-Methylpropan-2-ol.
Quick Tip: A useful shortcut for Grignard reactions with carbonyls: - Formaldehyde + Grignard reagent \( \rightarrow \) Primary alcohol - Other Aldehydes + Grignard reagent \( \rightarrow \) Secondary alcohol - Ketones + Grignard reagent \( \rightarrow \) Tertiary alcohol Since propanone is a ketone, the product must be a tertiary alcohol.

Step 1: Understanding the Concept:

This question is about the reduction of functional groups, specifically the conversion of a carboxylic acid (-COOH) to a primary alcohol (-CH\(_2\)OH). Carboxylic acids are at a high oxidation state and require a strong reducing agent for this transformation.


Step 2: Detailed Explanation:

Let's evaluate the given reagents:

- (A) PCC (Pyridinium chlorochromate): This is a mild oxidizing agent. It is used to oxidize primary alcohols to aldehydes and secondary alcohols to ketones. It cannot reduce a carboxylic acid.

- (B) LiAlH\(_4\) / H\(_2\)O (Lithium aluminum hydride, followed by aqueous workup): LiAlH\(_4\) is a very strong, unselective reducing agent. It is one of the few reagents powerful enough to reduce carboxylic acids and their derivatives (like esters and amides) all the way down to the corresponding primary alcohols. The reaction is:

\( R-COOH \xrightarrow{1. LiAlH_4, ether} \xrightarrow{2. H_2O} R-CH_2OH \)

This is the correct reagent for the desired conversion.

- (C) NaNO\(_2\) / HCl: This reagent mixture is used to generate nitrous acid (HONO), which is primarily used to convert primary amines into diazonium salts (diazotization).

- (D) Pd / H\(_2\) and (E) Pt / H\(_2\): These are reagents for catalytic hydrogenation. While catalytic hydrogenation is excellent for reducing alkenes, alkynes, and some other functional groups, it is generally not effective for reducing carboxylic acids directly. Very harsh conditions (high pressure and temperature) would be required, and it's not the standard method.


Step 3: Final Answer:

The reagent used for the conversion of carboxylic acids to primary alcohols is LiAlH\(_4\) followed by an aqueous workup.
Quick Tip: For reductions in organic chemistry, categorize your reagents. NaBH\(_4\) is milder and reduces aldehydes and ketones. LiAlH\(_4\) is a "sledgehammer" that reduces almost all carbonyl-containing functional groups, including carboxylic acids and esters, to alcohols. Catalytic hydrogenation (H\(_2\)/metal) is for \(\pi\) bonds (C=C, C\(\equiv\)C) and nitro groups, but not typically for acids.

Step 1: Understanding the Concept:

Acidity is the ability of a compound to donate a proton (H\(^+\)). The strength of an acid is determined by the stability of its conjugate base (the anion formed after losing the proton). Factors that stabilize the conjugate base, such as resonance or electron-withdrawing groups, increase acidity. Factors that destabilize it, such as electron-donating groups, decrease acidity.


Step 2: Detailed Explanation:

Let's analyze the stability of the conjugate base for each compound:

- (iv) Ethanol (CH\(_3\)CH\(_2\)OH): Its conjugate base is the ethoxide ion (CH\(_3\)CH\(_2\)O\(^-\)). The negative charge is localized on the oxygen atom. The ethyl group is an electron-donating group (+I effect), which intensifies the negative charge and destabilizes the anion. Alcohols are generally much less acidic than phenols. So, ethanol is the least acidic.

- (ii) Phenol (C\(_6\)H\(_5\)OH): Its conjugate base is the phenoxide ion (C\(_6\)H\(_5\)O\(^-\)). The negative charge on the oxygen can be delocalized into the benzene ring through resonance. This resonance stabilization makes the phenoxide ion much more stable than the ethoxide ion, so phenol is more acidic than ethanol.

- (iii) o-Cresol (o-methylphenol): Its conjugate base is the o-cresolate ion. It has a methyl group (-CH\(_3\)) at the ortho position. The methyl group is an electron-donating group due to both the +I effect and hyperconjugation. It donates electron density to the ring, which slightly destabilizes the phenoxide ion compared to the unsubstituted phenoxide ion. Therefore, o-cresol is less acidic than phenol.

- (i) o-Nitrophenol: Its conjugate base is the o-nitrophenoxide ion. It has a nitro group (-NO\(_2\)) at the ortho position. The nitro group is a powerful electron-withdrawing group (-I and -R effects). It withdraws electron density from the ring, strongly stabilizing the negative charge on the phenoxide ion through both induction and resonance. This makes o-nitrophenol the most acidic compound in the list. The ortho-position also allows for intramolecular hydrogen bonding in the undissociated molecule, but the stabilization of the anion is the dominant effect on acidity.


Ordering the compounds by increasing acidity:

Ethanol (least stable anion) \(\textless\) o-Cresol (destabilized anion) \(\textless\) Phenol (resonance-stabilized anion) \(\textless\) o-Nitrophenol (strongly stabilized anion)

(iv) \(\textless\) (iii) \(\textless\) (ii) \(\textless\) (i)


Step 3: Final Answer:

The correct order of acidity is (iv) \(\textless\) (iii) \(\textless\) (ii) \(\textless\) (i).
Quick Tip: To compare acidities of phenols: 1. Any phenol is more acidic than any aliphatic alcohol. 2. Electron-withdrawing groups (EWGs like -NO\(_2\), -CN, -X) increase acidity. 3. Electron-donating groups (EDGs like -CH\(_3\), -OCH\(_3\), -NH\(_2\)) decrease acidity. The order is always: EWG-Phenol \textgreater Phenol \textgreater EDG-Phenol \textgreater Alcohol.

Step 1: Understanding the Concept:

This question asks to identify a specific named reaction in organic chemistry based on the reactants and products described. The reaction is a formylation reaction, which means a formyl group (-CHO) is introduced onto an aromatic ring.


Step 2: Detailed Explanation:

Let's analyze the given reaction and the options:

Given Reaction:
\[ Benzene + CO + HCl \xrightarrow{Anhydrous AlCl_3 / CuCl} Benzaldehyde \]
This describes the direct introduction of a -CHO group onto the benzene ring.


Analysis of Options:

- (A) Etard reaction: This is the oxidation of toluene (or other alkylbenzenes) to benzaldehyde using chromyl chloride (CrO\(_2\)Cl\(_2\)).

- (B) Stephen reaction: This is the reduction of nitriles to aldehydes using tin(II) chloride (SnCl\(_2\)) and HCl.

- (C) Hell-Volhard-Zelinsky reaction: This is the \(\alpha\)-halogenation of carboxylic acids using P/X\(_2\) (where X = Br, Cl).

- (D) Gatterman-Koch reaction: This is a specific formylation reaction where benzene or its derivatives are treated with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of a Lewis acid catalyst (anhydrous AlCl\(_3\)) and a co-catalyst (CuCl). This perfectly matches the description in the question.

- (E) Aldol reaction: This is a reaction between two molecules of an aldehyde or a ketone to form a \(\beta\)-hydroxy aldehyde or ketone.


Step 3: Final Answer:

The reaction described is the Gatterman-Koch reaction.
Quick Tip: To distinguish between similar formylation reactions: - \textbf{Gatterman-Koch:} Uses CO + HCl. (Think "Koch" sounds like "C-O-H"). Works on benzene. - \textbf{Gatterman:} Uses HCN + HCl. Works on more activated rings like phenol. Both are formylation reactions.

Step 1: Understanding the Concept:

This reaction is the nitration of benzaldehyde (C\(_6\)H\(_5\)CHO), which is an electrophilic aromatic substitution. The key is to determine the directing effect of the substituent already on the benzene ring, which in this case is the aldehyde group (-CHO).


Step 2: Detailed Explanation:

- Reactants: Benzaldehyde and a nitrating mixture (concentrated HNO\(_3\) and concentrated H\(_2\)SO\(_4\)). This mixture generates the electrophile, the nitronium ion (NO\(_2\)\(^+\)).

- Directing Effect of the Aldehyde Group (-CHO): The aldehyde group consists of a carbon double-bonded to an oxygen. Oxygen is more electronegative than carbon, so it withdraws electron density from the double bond and, by extension, from the benzene ring through resonance (-R effect) and induction (-I effect).

- Deactivating and Meta-directing: Because the -CHO group withdraws electron density, it deactivates the ring towards electrophilic substitution. The resonance structures of benzaldehyde show that positive charges develop at the ortho and para positions. This makes these positions less favorable for attack by the positive electrophile (NO\(_2\)\(^+\)). The meta position, being relatively less deactivated, is the site where substitution occurs. Therefore, the aldehyde group is a meta-director.

- Product: The nitration of benzaldehyde will yield m-nitrobenzaldehyde as the major product. The reaction is carried out at a low temperature (273-283K or 0-10\(^{\circ}\)C) to prevent over-nitration or oxidation of the aldehyde group.


Step 3: Final Answer:

The reaction gives m-Nitrobenzaldehyde.
Quick Tip: To determine the product of electrophilic aromatic substitution, classify the existing substituent: - \textbf{Activating/Ortho, Para-directing}: -OH, -NH\(_2\), -OR, -R (alkyl) - \textbf{Deactivating/Ortho, Para-directing}: Halogens (-F, -Cl, -Br, -I) - \textbf{Deactivating/Meta-directing}: -NO\(_2\), -CN, -SO\(_3\)H, -CHO, -COR, -COOH, -COOR The -CHO group falls into the third category.

Step 1: Understanding the Concept:

This question requires knowledge of the common industrial and commercial applications of several organic compounds.


Step 2: Detailed Explanation:

Let's match each compound with its primary use from the list.

- (a) Benzaldehyde: Also known as oil of bitter almonds, it has a pleasant almond-like fragrance. Due to this, it is widely used in the food industry as a flavoring agent and in the manufacture of perfumes and dyes. So, Benzaldehyde (a) matches with (iii) perfumery.

- (b) Methanoic acid (Formic acid): It is a simple carboxylic acid. It has applications as a preservative and antibacterial agent in livestock feed. It is also used in the leather tanning industry, in textile dyeing, and in the electroplating industry as a component of plating baths. So, Methanoic acid (b) matches with (iv) Electroplating industry.

- (c) Sodium benzoate: This is the sodium salt of benzoic acid. It is a widely used food preservative, particularly in acidic foods such as salad dressings, carbonated drinks, jams, and fruit juices. It inhibits the growth of bacteria, mold, and yeast. So, Sodium benzoate (c) matches with (i) food preservative.

- (d) Hexanedioic acid (Adipic acid): This is a dicarboxylic acid. Its most significant use by far is as a monomer for the production of nylon 6,6, a common synthetic polymer used in fibers and plastics. It reacts with hexamethylenediamine to form the polymer. So, Hexanedioic acid (d) matches with (ii) nylon 6,6.


Final Matching:

- a \(\rightarrow\) iii

- b \(\rightarrow\) iv

- c \(\rightarrow\) i

- d \(\rightarrow\) ii


Step 3: Final Answer:

The correct matching is a-(iii), b-(iv), c-(i), d-(ii).
Quick Tip: Certain compound-use pairs are very common in chemistry exams. Always remember: Adipic acid \(\rightarrow\) Nylon 6,6; Benzoates \(\rightarrow\) Food preservatives; Benzaldehyde \(\rightarrow\) Almond flavor/perfume.

Step 1: Understanding the Concept:

This process is known as exhaustive alkylation or the Hofmann exhaustive alkylation. It involves the reaction of a primary amine with an excess of an alkyl halide. The reaction proceeds in a stepwise manner, where the amine acts as a nucleophile, and each step adds one alkyl group to the nitrogen atom until a quaternary ammonium salt is formed.


Step 2: Detailed Explanation:

Let's consider a primary amine, R-NH\(_2\), reacting with an alkyl halide, R'-X.

- Step 1: Formation of a secondary amine. The primary amine attacks one mole of the alkyl halide to form a secondary ammonium salt, which is then deprotonated (usually by another amine molecule) to give a secondary amine.

\( R-NH_2 + R'-X \rightarrow R-NH(R') + HX \) (1 mole of R'-X used)

- Step 2: Formation of a tertiary amine. The secondary amine formed is also nucleophilic and reacts with a second mole of the alkyl halide.

\( R-NH(R') + R'-X \rightarrow R-N(R')_2 + HX \) (2nd mole of R'-X used)

- Step 3: Formation of a quaternary ammonium salt. The tertiary amine reacts with a third mole of the alkyl halide. In this final step, the product is a stable salt because the nitrogen atom has four alkyl groups and no hydrogen to lose. It carries a permanent positive charge.

\( R-N(R')_2 + R'-X \rightarrow [R-N(R')_3]^+X^- \) (3rd mole of R'-X used)

In total, three moles of the alkyl halide are required to convert the two N-H bonds of the primary amine into N-R' bonds and then to add the final R' group to form the quaternary salt.


Step 3: Final Answer:

A total of 3 moles of alkyl halide are required.
Quick Tip: To find the number of moles of alkyl halide needed for exhaustive alkylation, count the number of hydrogens on the nitrogen of the starting amine and add one. For a primary amine (R-NH\(_2\)), it's 2 hydrogens + 1 = 3 moles. For a secondary amine (R\(_2\)NH), it's 1 hydrogen + 1 = 2 moles.

Step 1: Understanding the Concept:

The boiling point of a substance depends on the strength of its intermolecular forces. For amines, the primary forces are London dispersion forces, dipole-dipole interactions, and hydrogen bonding. Hydrogen bonding is the strongest of these and has the most significant impact on the boiling point. The extent of hydrogen bonding depends on the number of N-H bonds available.


Step 2: Detailed Explanation:

The three given compounds are isomers, meaning they have the same molecular formula (C\(_4\)H\(_{11}\)N) and similar molar masses. Therefore, the differences in their boiling points will be primarily due to differences in hydrogen bonding.

- (i) Butan-1-amine (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)NH\(_2\)): This is a primary (1\(^\circ\)) amine. It has two hydrogen atoms directly bonded to the nitrogen atom. This allows it to form extensive intermolecular hydrogen bonds, acting as both a hydrogen bond donor and acceptor. The linear structure also allows for strong London dispersion forces.

- (ii) N-Ethylethanamine ((CH\(_3\)CH\(_2\))\(_2\)NH): This is a secondary (2\(^\circ\)) amine. It has one hydrogen atom directly bonded to the nitrogen. It can still form hydrogen bonds, but to a lesser extent than a primary amine because it has fewer N-H bonds.

- (iii) N,N-Dimethylethanamine (CH\(_3\)CH\(_2\)N(CH\(_3\))\(_2\)): This is a tertiary (3\(^\circ\)) amine. It has no hydrogen atoms directly bonded to the nitrogen. Therefore, molecules of a tertiary amine cannot form hydrogen bonds with each other. They only interact via weaker dipole-dipole forces and London dispersion forces.


Comparison:

Since the strength of intermolecular forces is: Hydrogen Bonding (1\(^\circ\)) \textgreater Hydrogen Bonding (2\(^\circ\)) \textgreater Dipole-Dipole (3\(^\circ\)), the order of boiling points will follow the same trend.

Boiling Point: Butan-1-amine \textgreater N-Ethylethanamine \textgreater N,N-Dimethylethanamine

(i) \textgreater (ii) \textgreater (iii)


Step 3: Final Answer:

The correct order of boiling points is (i) \(\textgreater\) (ii) \(\textgreater\) (iii).
Quick Tip: For isomeric amines, the boiling point order is always: Primary \textgreater Secondary \textgreater Tertiary. This is because the number of N-H bonds available for hydrogen bonding decreases from two to one to zero.

Step 1: Understanding the Concept:

This is a structure elucidation problem that can be solved by working backward from the final product and analyzing the described reactions (ammonolysis and alkylation).


Step 2: Detailed Explanation:

Analyze the final product and the last reaction:

- The final product is N,N-Dimethylphenylmethanamine. Let's draw its structure: \( C_6H_5-CH_2-N(CH_3)_2 \).
- This product is formed when compound 'Y' reacts with two moles of methyl chloride (CH\(_3\)Cl). This is an alkylation reaction. The two methyl groups in the final product must have come from the two moles of CH\(_3\)Cl.
- This means compound 'Y' must have been the parent amine before methylation, which is \( C_6H_5-CH_2-NH_2 \). This amine has two hydrogens on the nitrogen, which are replaced by the two methyl groups.
- The name of \( C_6H_5-CH_2-NH_2 \) is Benzylamine (or Phenylmethanamine).
- Let's check the molecular formula of Benzylamine: 7 carbons (6 in ring, 1 in CH\(_2\)), 9 hydrogens (5 in ring, 2 in CH\(_2\), 2 in NH\(_2\)), and 1 nitrogen. The formula is C\(_7\)H\(_9\)N, which matches the given formula for Y. So, Y is Benzylamine.


Analyze the first reaction:

- Compound 'Y' (Benzylamine) is formed by the ammonolysis of compound 'X' (C\(_7\)H\(_7\)Cl).
- Ammonolysis is a nucleophilic substitution reaction where ammonia (NH\(_3\)) replaces a halogen.

\( X + NH_3 \rightarrow Y \)

\( R-Cl + NH_3 \rightarrow R-NH_2 + HCl \)
- Since Y is Benzylamine (\( C_6H_5-CH_2-NH_2 \)), X must be the corresponding chloride, which is \( C_6H_5-CH_2-Cl \).
- The name of \( C_6H_5-CH_2-Cl \) is Benzyl chloride.
- Let's check the molecular formula of Benzyl chloride: 7 carbons, 7 hydrogens (5 in ring, 2 in CH\(_2\)), and 1 chlorine. The formula is C\(_7\)H\(_7\)Cl, which matches the given formula for X. So, X is Benzyl chloride.


Conclusion:

Compound X is Benzyl chloride and compound Y is Benzylamine. This matches option (C). Ammonolysis of an aryl halide like chlorobenzene is very difficult and does not occur under these conditions, ruling out options B and D.


Step 3: Final Answer:

The compounds 'X' and 'Y' are Benzylchloride and Benzylamine.
Quick Tip: When solving multi-step synthesis problems, working backward from the final product (retrosynthesis) is often the most effective strategy. Identify the last reaction and deduce the structure of the precursor, then repeat the process for the previous step.

Step 1: Understanding the Concept:

This question deals with the oxidation of carbohydrates. The product depends on the starting material and the strength of the oxidizing agent. Nitric acid (HNO\(_3\)) is a strong oxidizing agent capable of oxidizing both aldehyde and primary alcohol functional groups to carboxylic acids.


Step 2: Detailed Explanation:

- Starting material: Gluconic acid. Gluconic acid is formed by the mild oxidation of glucose (e.g., with bromine water), where the aldehyde group (-CHO) at the C-1 position is oxidized to a carboxylic acid group (-COOH). Its structure is:

COOH-(CHOH)\(_4\)-CH\(_2\)OH.

- Reaction with Nitric Acid: When gluconic acid is treated with a strong oxidizing agent like nitric acid, the primary alcohol group (-CH\(_2\)OH) at the C-6 position is also oxidized to a carboxylic acid group (-COOH).

- Product: The resulting molecule has carboxylic acid groups at both ends of the carbon chain (C-1 and C-6). This dicarboxylic acid is known as saccharic acid (or glucaric acid). Its structure is:

COOH-(CHOH)\(_4\)-COOH.

The overall reaction is:

\ce{HOOC-(CHOH)4-CH2OH ->[HNO3] HOOC-(CHOH)4-COOH


Step 3: Final Answer:

The oxidation of gluconic acid with nitric acid gives saccharic acid.
Quick Tip: Remember the oxidation ladder for glucose: Glucose + Mild Oxidant (Br\(_2\) water) \(\rightarrow\) Gluconic acid (aldehyde oxidized) Glucose + Strong Oxidant (HNO\(_3\)) \(\rightarrow\) Saccharic acid (aldehyde and primary alcohol oxidized) Knowing this helps predict products quickly.

Step 1: Understanding the Concept:

Living organisms store excess carbohydrates in the form of polysaccharides for later use as energy. The specific storage polysaccharide differs between plants and animals.


Step 2: Detailed Explanation:

- In Animals: Animals, including humans, store glucose in the form of glycogen. Glycogen is a multi-branched polysaccharide of glucose, often referred to as "animal starch." It is mainly stored in the cells of the liver and muscles. When the body needs energy, glycogen is broken down to release glucose.

- In Plants: Plants store glucose as starch, which is a mixture of two polysaccharides: amylose (a linear polymer) and amylopectin (a branched polymer).

- Cellulose: This is also a polysaccharide of glucose, but it serves as a structural component of plant cell walls and cannot be digested by most animals for energy.

- Amylase: This is an enzyme that catalyzes the hydrolysis (breakdown) of starch into smaller sugars. It is not a storage carbohydrate.


Step 4: Final Answer:

The carbohydrate storage form in the animal body is glycogen.
Quick Tip: A simple association to remember: Plants \(\rightarrow\) Starch (storage) and Cellulose (structure). Animals \(\rightarrow\) Glycogen (storage).

Step 1: Understanding the Concept:

To find the dimensions of the ratio B/E, we can use the Lorentz force equation, which relates the electric and magnetic forces on a charged particle. The magnitudes of the electric and magnetic forces can be compared.


Step 2: Key Formula or Approach:

The force on a charge \(q\) moving with velocity \(v\) in an electric field \(E\) is \(F_E = qE\).

The maximum force on the same charge in a magnetic field \(B\) is \(F_B = qvB\).

The dimensions of force must be consistent. Another useful relation is that for an electromagnetic wave in vacuum, \(E = cB\), where \(c\) is the speed of light. From this, \(\frac{B}{E} = \frac{1}{c}\).

Let's use the Lorentz force approach as it is more fundamental. In a situation where the electric and magnetic forces balance (e.g., in a velocity selector), \(F_E = F_B\).
\[ qE = qvB \] \[ E = vB \]
Rearranging to find the ratio \(\frac{B}{E}\):
\[ \frac{B}{E} = \frac{1}{v} \]

Step 3: Detailed Explanation:

The dimension of the ratio \(\frac{B}{E}\) is the same as the dimension of the reciprocal of velocity (\(\frac{1}{v}\)).

The dimensions of velocity (\(v\)) are distance per time, which is [L T\(^{-1}\)].

Therefore, the dimensions of \(\frac{1}{v}\) are:
\[ \left[ \frac{1}{v} \right] = \frac{1}{[L T^{-1}]} = [L^{-1} T^{1}] \]
Expressing this in the standard M, L, T format:
\[ [M^0 L^{-1} T^{1}] \]

Step 4: Final Answer:

The dimensions of \(\frac{B}{E}\) are M\(^0\)L\(^{-1}\)T\(^1\).
Quick Tip: For ratios involving E and B fields, remember the simple relationship for electromagnetic waves: \(E = cB\). This instantly tells you that \(B/E = 1/c\), so the dimensions are those of inverse speed, [L\(^{-1}\)T].

Step 1: Understanding the Concept:

Impulse is defined as the change in momentum of an object. It is also equal to the product of the average force acting on the object and the time interval over which the force acts.


Step 2: Key Formula or Approach:

The relationship between impulse (\(J\)), average force (\(F_{avg}\)), and time interval (\(\Delta t\)) is given by:
\[ J = F_{avg} \times \Delta t \]
We can rearrange this formula to solve for the average force.


Step 3: Detailed Explanation:

Given values are:

- Impulse, \(J = 15\) Ns

- Time of hit, \(\Delta t = 0.2\) s

We need to find the average force, \(F_{avg}\).
\[ F_{avg} = \frac{J}{\Delta t} \]
Substituting the given values:
\[ F_{avg} = \frac{15 \, Ns}{0.2 \, s} \] \[ F_{avg} = \frac{15}{1/5} \, N = 15 \times 5 \, N = 75 \, N \]

Step 4: Final Answer:

The average force exerted by the player on the ball is 75 N.
Quick Tip: Remember the impulse-force relationship \(J = F\Delta t\). If you are given impulse and time, finding the force is a simple division. Be careful with decimal calculations; 0.2 is the same as 1/5, which can make the division easier.

Step 1: Understanding the Concept:

In kinematics, velocity is the first time derivative of the position vector, and acceleration is the second time derivative of the position vector (or the first time derivative of the velocity vector).


Step 2: Key Formula or Approach:

Given the position vector \( \vec{r}(t) \):

1. Find the velocity vector: \( \vec{v}(t) = \frac{d\vec{r}}{dt} \)

2. Find the acceleration vector: \( \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2} \)

3. Find the magnitude of the acceleration vector \( |\vec{a}(t)| \).


Step 3: Detailed Explanation:

The given position vector is:
\[ \vec{r}(t) = 8t\hat{i} + 3t^2\hat{j} + 3\hat{k} \]
1. Differentiate to find velocity:
\[ \vec{v}(t) = \frac{d}{dt} (8t\hat{i} + 3t^2\hat{j} + 3\hat{k}) \] \[ \vec{v}(t) = 8\hat{i} + (2 \cdot 3t)\hat{j} + 0\hat{k} = 8\hat{i} + 6t\hat{j} \]
2. Differentiate velocity to find acceleration:
\[ \vec{a}(t) = \frac{d}{dt} (8\hat{i} + 6t\hat{j}) \] \[ \vec{a}(t) = 0\hat{i} + 6\hat{j} = 6\hat{j} \, ms^{-2} \]
The acceleration vector is constant and points in the positive y-direction.

3. Find the magnitude of the acceleration:

The magnitude of the vector \( \vec{a} = 6\hat{j} \) is:
\[ |\vec{a}| = \sqrt{0^2 + 6^2 + 0^2} = \sqrt{36} = 6 \, ms^{-2} \]

Step 4: Final Answer:

The acceleration of the particle is 6 ms\(^{-2}\).
Quick Tip: When differentiating vectors, treat each component (\(\hat{i}, \hat{j}, \hat{k}\)) separately. Remember the power rule of differentiation: \(\frac{d}{dt}(ct^n) = nct^{n-1}\).

Step 1: Understanding the Concept:

According to Newton's Second Law of Motion, the force acting on an object is equal to the product of its mass and its acceleration (\(F=ma\)). To find the force, we first need to determine the acceleration of the particle from its displacement equation.


Step 2: Key Formula or Approach:

1. Given the displacement equation \(x(t)\).

2. Find the velocity by differentiating displacement with respect to time: \( v = \frac{dx}{dt} \).

3. Find the acceleration by differentiating velocity with respect to time: \( a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \).

4. Calculate the force using Newton's Second Law: \( F = ma \).


Step 3: Detailed Explanation:

The given displacement equation is:
\[ x(t) = 3t + 7t^2 \]
1. Differentiate to find velocity:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} (3t + 7t^2) = 3 + 14t \, m/s \]
2. Differentiate velocity to find acceleration:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt} (3 + 14t) = 14 \, m/s^2 \]
The acceleration is constant.

3. Calculate the force:

Given mass, \(m = 0.2\) kg.
\[ F = m \times a \] \[ F = 0.2 \, kg \times 14 \, m/s^2 \] \[ F = 2.8 \, N \]

Step 4: Final Answer:

The force acting on the particle is 2.8 N.
Quick Tip: This problem combines kinematics with dynamics. The key is to remember the differentiation sequence: displacement \(\rightarrow\) velocity \(\rightarrow\) acceleration. Once you have the acceleration, applying F=ma is straightforward.

Step 1: Understanding the Concept:

This problem is an application of the principle of conservation of linear momentum. The total momentum of an isolated system (in this case, the gun and the bullet) remains constant. Since the gun and bullet are initially at rest, the total initial momentum is zero. Therefore, the total final momentum must also be zero.


Step 2: Key Formula or Approach:

Let \(m_b\) and \(v_b\) be the mass and velocity of the bullet, and \(m_g\) and \(v_g\) be the mass and velocity of the gun.

Conservation of momentum:
\[ Initial Momentum = Final Momentum \] \[ 0 = m_b v_b + m_g v_g \]
We can solve for the recoil velocity of the gun, \(v_g\).


Step 3: Detailed Explanation:

First, ensure all units are consistent (SI units).

- Mass of bullet, \(m_b = 10 \, g = 0.01 \, kg\)

- Speed of bullet, \(v_b = 50 \, m/s\)

- Mass of gun, \(m_g = 2 \, kg\)

Using the conservation of momentum equation:
\[ 0 = (0.01 \, kg)(50 \, m/s) + (2 \, kg) v_g \] \[ 0 = 0.5 \, kg \cdot m/s + 2 v_g \] \[ 2 v_g = -0.5 \, kg \cdot m/s \] \[ v_g = -\frac{0.5}{2} \, m/s = -0.25 \, m/s \]
The negative sign indicates that the gun's velocity is in the opposite direction to the bullet's velocity. The question asks for the recoil speed, which is the magnitude of the velocity.

Recoil speed = \(|v_g| = 0.25\) m/s.


Step 4: Final Answer:

The recoil speed of the gun is 0.25 ms\(^{-1}\).
Quick Tip: In recoil problems, the momentum of the bullet in one direction must equal the momentum of the gun in the opposite direction. You can set up the equation as \(p_{gun} = p_{bullet}\) in terms of magnitude: \(m_g v_g = m_b v_b\). Always convert mass to kilograms before calculating.

Step 1: Understanding the Concept:

The work done in lifting an object against the force of gravity is equal to the change in its gravitational potential energy. This work is calculated by multiplying the force required to lift the object (its weight) by the vertical distance it is lifted.


Step 2: Key Formula or Approach:

The formula for gravitational potential energy (\(E_p\)) and the work done (\(W\)) against gravity is:
\[ W = E_p = mgh \]
where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the vertical height.


Step 3: Detailed Explanation:

Given values are:

- Mass, \(m = 60\) kg

- Height, \(h = 5\) m

- Acceleration due to gravity, \(g = 10\) ms\(^{-2}\)

Substitute these values into the formula:
\[ W = (60 \, kg) \times (10 \, m/s^2) \times (5 \, m) \] \[ W = 600 \times 5 \, J \] \[ W = 3000 \, J \]

Step 4: Final Answer:

The work done to lift the mass is 3000 J.
Quick Tip: Work done against gravity only depends on the vertical displacement (height), not the path taken. The formula \(W=mgh\) is fundamental for these types of problems.

Step 1: Understanding the Concept:

This question is based on Albert Einstein's principle of mass-energy equivalence, which states that mass is a form of energy and can be converted into energy, and vice versa.


Step 2: Key Formula or Approach:

The relationship between energy (\(E\)), mass (\(m\)), and the speed of light (\(c\)) is given by the famous equation:
\[ E = mc^2 \]

Step 3: Detailed Explanation:

Given values are:

- Mass, \(m = 0.5\) kg

- The speed of light, \(c \approx 3 \times 10^8\) m/s

Substitute these values into the equation:
\[ E = (0.5 \, kg) \times (3 \times 10^8 \, m/s)^2 \] \[ E = 0.5 \times (9 \times 10^{16} \, m^2/s^2) \] \[ E = 4.5 \times 10^{16} \, kg \cdot m^2/s^2 \]
Since 1 Joule (J) = 1 kg\( \cdot \)m\(^2\)/s\(^2\),
\[ E = 4.5 \times 10^{16} \, J \]

Step 4: Final Answer:

The energy equivalent of a 0.5 kg mass is 4.5 \( \times \) 10\(^{16}\) J.
Quick Tip: When using \(E=mc^2\), be careful with the squaring of the speed of light. \((3 \times 10^8)^2\) is \(9 \times 10^{16}\), not \(3 \times 10^{16}\) or \(9 \times 10^8\). This is a common point of error.

Step 1: Understanding the Concept:

The center of mass of a system of particles is a weighted average of the positions of the particles, where the weighting factor for each particle is its mass. Since the particles lie on a line, we can use the one-dimensional formula for the center of mass.


Step 2: Key Formula or Approach:

The position of the center of mass (\(X_{CM}\)) for a system of particles along a line is given by:
\[ X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + \dots}{m_1 + m_2 + m_3 + \dots} \]
where \(m_i\) and \(x_i\) are the mass and position of the \(i\)-th particle, respectively.


Step 3: Detailed Explanation:

Given information:

- There are three particles.

- They have equal mass, so let \(m_1 = m_2 = m_3 = m\).

- Their positions from the origin are \(x_1 = 1\) cm, \(x_2 = 2\) cm, and \(x_3 = 3\) cm.

Substitute these values into the formula:
\[ X_{CM} = \frac{(m \times 1) + (m \times 2) + (m \times 3)}{m + m + m} \] \[ X_{CM} = \frac{m(1 + 2 + 3)}{3m} \]
The mass \(m\) cancels out from the numerator and denominator:
\[ X_{CM} = \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2 \, cm \]

Step 4: Final Answer:

The distance of their centre of mass from the origin is 2 cm.
Quick Tip: For a system of particles with equal mass, the center of mass is simply the geometric center, or the average of the positions. In this case, it's just the average of 1, 2, and 3.

Step 1: Understanding the Concept:

The angular momentum (\(\vec{L}\)) of a particle about a point (the origin) is a measure of its rotational motion. It is defined as the cross product of the particle's position vector (\(\vec{r}\)) relative to that point and its linear momentum vector (\(\vec{p}\)).


Step 2: Key Formula or Approach:

The formula for angular momentum is:
\[ \vec{L} = \vec{r} \times \vec{p} \]
The magnitude of the angular momentum is given by:
\[ L = |\vec{r}| |\vec{p}| \sin(\theta) = rp \sin(\theta) \]
where \(\theta\) is the angle between the vectors \(\vec{r}\) and \(\vec{p}\).


Step 3: Detailed Explanation:

For the angular momentum \(L\) to be non-zero, all three factors in the magnitude equation (\(r, p, \sin(\theta)\)) must be non-zero. Let's analyze the given options:

- (A) angle is 0\(^\circ\): If \(\theta = 0^\circ\), then \(\sin(0^\circ) = 0\), which makes \(L = 0\). This happens when the particle is moving directly away from the origin.

- (B) particle is at the origin: If the particle is at the origin, its position vector \(\vec{r}\) is zero, so \(r=0\), which makes \(L = 0\).

- (C) angle is 90\(^\circ\): If \(\theta = 90^\circ\), then \(\sin(90^\circ) = 1\). In this case, \(L = rp\). As long as the particle is not at the origin (\(r \neq 0\)) and is moving (\(p \neq 0\)), the angular momentum will be non-zero. This condition allows for non-zero angular momentum.

- (D) linear momentum vanishes: If the linear momentum is zero (\(p=0\)), then \(L=0\). This means the particle is at rest.

- (E) angle is 180\(^\circ\): If \(\theta = 180^\circ\), then \(\sin(180^\circ) = 0\), which makes \(L = 0\). This happens when the particle is moving directly towards the origin.


Step 4: Final Answer:

The angular momentum will not be zero if the angle between the position vector and linear momentum is 90\(^\circ\) (assuming \(r\) and \(p\) are non-zero).
Quick Tip: The angular momentum is zero if the momentum vector points directly towards or away from the origin (i.e., it is collinear with the position vector). It is maximum when the momentum is perpendicular to the position vector.

Step 1: Understanding the Concept:

Weightlessness is the sensation of having no weight. It is experienced when there are no contact forces acting on a body to support it. It's a common misconception that there is no gravity in space. In low Earth orbit, the force of gravity is still about 90% of what it is on the surface.


Step 2: Detailed Explanation:

- A satellite in orbit is constantly "falling" towards the Earth under the influence of gravity. However, it also has a high tangential velocity, so as it falls, the Earth's surface curves away from it. The combination of its forward motion and its falling motion results in a stable orbit.

- Everything inside the satellite, including the astronaut, is also falling towards the Earth at the same rate of acceleration as the satellite itself.

- Since the astronaut and the satellite are falling together, the astronaut does not press against the floor, walls, or any object inside the satellite. There is no normal contact force acting on the astronaut.

- Our sensation of weight is due to this normal contact force (e.g., the floor pushing up on our feet). In the absence of this force, we feel "weightless." This state is more accurately described as a state of continuous free fall.

- Option (A) is incorrect because gravity is still significant. Options (C) and (D) are incorrect. Option (E) is incorrect as the Earth's gravity is overwhelmingly dominant.


Step 3: Final Answer:

An astronaut experiences weightlessness because both the astronaut and the satellite are in a constant state of free fall towards Earth.
Quick Tip: Think of being in an elevator when the cable snaps. Both you and the elevator would fall at the same rate, and you would float inside it, feeling weightless. A satellite in orbit is like a perpetually falling elevator that never hits the ground.

Step 1: Understanding the Concept:

The statement describes Hooke's Law of Elasticity. This fundamental principle relates the stress (force per unit area) applied to a material to the strain (relative deformation) it experiences, within its elastic limit.


Step 2: Key Formula or Approach:

Hooke's Law is mathematically expressed as:
\[ Stress \propto Strain \]
To turn the proportionality into an equation, a constant is introduced:
\[ Stress = (Constant) \times Strain \]

Step 3: Detailed Explanation:

- This constant of proportionality is a measure of the material's stiffness or resistance to elastic deformation. It is known as the Modulus of Elasticity.

- There are different types of moduli of elasticity depending on the type of stress and strain:

1. Young's Modulus (Y): Ratio of tensile/compressive stress to longitudinal strain.

2. Shear Modulus (G): Ratio of shear stress to shear strain.

3. Bulk Modulus (K): Ratio of volume stress to volume strain.

- The other options are different physical quantities:

- Poisson's ratio is the ratio of transverse strain to axial strain.

- Compressibility is the reciprocal of the bulk modulus.

- Mechanical strength refers to the material's ability to withstand an applied load without failure.


Step 4: Final Answer:

The constant of proportionality between stress and strain is called the modulus of elasticity.
Quick Tip: Remember the relationship: Stress = Modulus \(\times\) Strain. The modulus is an intrinsic property of a material that tells you how stiff it is.

Step 1: Understanding the Concept:

The flow of blood through the cardiovascular system, including arteries, is governed by the principles of fluid dynamics. The question asks which principle is key to explaining this flow.


Step 2: Detailed Explanation:

- Blood Flow: Blood can be modeled as a fluid in motion. Its flow is characterized by properties like velocity and pressure.

- Bernoulli's Principle: This principle states that for an ideal fluid, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. It relates pressure, velocity, and height for a moving fluid. In the context of blood flow, it helps to explain:

- How blood pressure changes as arteries branch or change in diameter. For example, in a region where an artery is narrowed (stenosis), the blood velocity increases, and according to Bernoulli's principle, the pressure in that region decreases. This pressure difference is important in the pathophysiology of vascular diseases.

- The operation of medical devices like a Venturi meter to measure flow rates.

- Other Principles:

- Magnus effect deals with the lift force on a spinning object in a fluid.

- Boyle's law relates pressure and volume for a gas at constant temperature.

- Pascal's law applies to a confined, static fluid, stating that pressure is transmitted undiminished.

- Archimedes' principle deals with the buoyant force on an object submerged in a fluid.

While blood flow is complex, Bernoulli's principle is the most relevant fluid dynamics principle listed for explaining the relationship between pressure and velocity in arteries.


Step 3: Final Answer:

Bernoulli's principle helps to explain the flow of blood in an artery.
Quick Tip: Associate physical principles with their domain: Bernoulli \(\rightarrow\) moving fluids (airplanes, arteries); Pascal \(\rightarrow\) static fluids (hydraulics); Archimedes \(\rightarrow\) buoyancy (ships, balloons).

Step 1: Understanding the Concept:

The efficiency of an ideal Carnot engine, the most efficient possible heat engine, depends only on the absolute temperatures of the hot reservoir (the source) and the cold reservoir (the sink).


Step 2: Key Formula or Approach:

The efficiency (\(\eta\)) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_{sink}}{T_{source}} \]
where \(T_{sink}\) and \(T_{source}\) are the absolute temperatures (in Kelvin) of the sink and source, respectively.


Step 3: Detailed Explanation:

Given information:

- Efficiency, \(\eta = 40% = 0.40\)

We need to find the ratio \(\frac{T_{sink}}{T_{source}}\).

Substitute the given efficiency into the formula:
\[ 0.40 = 1 - \frac{T_{sink}}{T_{source}} \]
Rearrange the equation to solve for the ratio:
\[ \frac{T_{sink}}{T_{source}} = 1 - 0.40 \] \[ \frac{T_{sink}}{T_{source}} = 0.60 \]

Step 4: Final Answer:

The ratio of the temperature of the sink to that of the source is 0.6.
Quick Tip: The efficiency tells you what fraction of the heat is converted to work. The rest is rejected to the sink. If efficiency is 40% (0.4), then 1 - 0.4 = 0.6, or 60% of the heat energy is rejected. The ratio of temperatures \(T_{sink}/T_{source}\) is equal to this fraction of rejected heat \(Q_{sink}/Q_{source}\), so it must be 0.6.

Step 1: Understanding the Concept:

This question describes the energy balance in a heat engine operating in a cycle. The First Law of Thermodynamics governs the relationship between heat, work, and internal energy.


Step 2: Key Formula or Approach:

The First Law of Thermodynamics states:
\[ \Delta U = Q_{net} - W \]
where \(\Delta U\) is the change in internal energy, \(Q_{net}\) is the net heat added to the system, and \(W\) is the work done by the system.

For any process that operates in a cycle (like a heat engine), the system returns to its initial state, so the net change in internal energy is zero (\(\Delta U = 0\)).


Step 3: Detailed Explanation:

For a cyclic process, \(\Delta U = 0\). The First Law becomes:
\[ 0 = Q_{net} - W \] \[ W = Q_{net} \]
The net heat added to the system, \(Q_{net}\), is the heat supplied to the system minus the heat expelled from the system.

- Heat supplied = \(Q_1\) (heat input, positive)

- Heat expelled = \(Q_2\) (heat output, negative contribution to net heat)

So, the net heat is:
\[ Q_{net} = Q_1 - Q_2 \]
Therefore, the work done by the system is:
\[ W = Q_1 - Q_2 \]

Step 4: Final Answer:

The work done by the system is \(Q_1 - Q_2\).
Quick Tip: Think of a heat engine like a budget. \(Q_1\) is your income, \(Q_2\) is your mandatory expenses (waste heat), and \(W\) is the money you have left over to do something useful (work). Your useful output is always income minus expenses.

Step 1: Understanding the Concept:

This question tests the fundamental properties of a spring and its oscillatory motion (Simple Harmonic Motion). We need to identify the statement that is not physically correct.


Step 2: Detailed Explanation:

Let's analyze each statement:

- (A) Stiff springs have high value of k: The spring constant, k, is a measure of stiffness. A stiff spring requires a large force to produce a given displacement (\(F = kx\)). So, a high k means a stiff spring. This statement is TRUE.

- (B) Soft springs have small value of k: A soft spring is easily stretched or compressed, meaning a small force produces a large displacement. This corresponds to a small value of k. This statement is TRUE.

- (C) The spring constant is independent of the elastic properties of the spring: The spring constant k is fundamentally determined by the material and geometry of the spring. It depends on the material's elastic properties (like the Shear Modulus or Young's Modulus) and its physical dimensions (like wire diameter, coil diameter, and number of coils). A spring made of steel will have a different k than an identical one made of copper. Therefore, this statement is FALSE.

- (D) For smaller oscillations the spring executes simple harmonic motion: Simple harmonic motion occurs when the restoring force is directly proportional to the displacement (\(F = -kx\)), which is Hooke's Law. Real springs obey Hooke's Law very well for small deformations but can deviate for large ones. So, for smaller oscillations, the motion is simple harmonic. This statement is TRUE.

- (E) The period of oscillations of the spring depends upon the value of k: The period (\(T\)) of a mass-spring system is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). The period is clearly dependent on the spring constant k. This statement is TRUE.


Step 3: Final Answer:

The false statement is that the spring constant is independent of the elastic properties of the spring.
Quick Tip: Remember that a physical constant like 'k' for a spring is not an abstract number; it's a derived property based on the physical reality of the object—what it's made of (elastic properties) and how it's shaped (geometry).

Step 1: Understanding the Concept:

The given equation represents the superposition of two waves with the same frequency and wavelength but different amplitudes and a phase difference of 90\(^\circ\) (\(\pi/2\)) (since one is a sine function and the other is a cosine function). The resultant wave will also be a sinusoidal wave, and its amplitude can be found using principles of vector addition or trigonometric identities.


Step 2: Key Formula or Approach:

A wave of the form \(y = R_1 \sin(\theta) + R_2 \cos(\theta)\) can be written as a single sinusoidal wave \(y = R_{res} \sin(\theta + \phi)\), where the resultant amplitude \(R_{res}\) is given by:
\[ R_{res} = \sqrt{R_1^2 + R_2^2} \]

Step 3: Detailed Explanation:

In the given wave equation:
\[ y = 3\sin(3x - 5t) + A\cos(3x - 5t) \]
This matches the form \(R_1 \sin(\theta) + R_2 \cos(\theta)\) where:

- The amplitude of the sine component, \(R_1 = 3\) m.

- The amplitude of the cosine component, \(R_2 = A\).

- The phase argument is \(\theta = (3x - 5t)\).

We are given that the resultant amplitude of the wave is \(R_{res} = 5\) m.

Using the formula for the resultant amplitude:
\[ R_{res} = \sqrt{R_1^2 + R_2^2} \] \[ 5 = \sqrt{3^2 + A^2} \]
To solve for A, we square both sides of the equation:
\[ 5^2 = 3^2 + A^2 \] \[ 25 = 9 + A^2 \] \[ A^2 = 25 - 9 \] \[ A^2 = 16 \] \[ A = \sqrt{16} = 4 \, m \] (Amplitude is a positive quantity)

Step 4: Final Answer:

The value of A is 4 m.
Quick Tip: Recognize this problem as finding the hypotenuse of a right-angled triangle. The amplitudes of the sine and cosine components are the two perpendicular sides, and the resultant amplitude is the hypotenuse. This is a classic 3-4-5 Pythagorean triple.

Step 1: Understanding the Concept:

When a dielectric material is placed in an external electric field, its constituent molecules (or atoms) form tiny electric dipoles, or existing permanent dipoles align with the field. This phenomenon is called dielectric polarization. The polarization (\(\vec{P}\)) is a vector quantity that quantifies the extent of this effect.


Step 2: Detailed Explanation:

- An electric dipole moment (\(\vec{p}\)) is a measure of the separation of positive and negative electrical charges within a system.

- In a macroscopic piece of dielectric material, there are many such dipoles. The Polarization vector (\(\vec{P}\)) is defined as the net dipole moment of the material per unit volume.

- Mathematically, if a small volume element \(\Delta V\) has a net dipole moment \(\Delta\vec{p}_{total}\), then the polarization at that point is:

\[ \vec{P} = \frac{\Delta\vec{p}_{total}}{\Delta V} \]
- Therefore, polarization is the dipole moment density, or dipole moment per unit volume. Its SI units are Coulombs per square meter (C/m\(^2\)).


Step 3: Final Answer:

In dielectrics, polarization is the dipole moment per unit volume.
Quick Tip: Think of polarization as a "density" concept, similar to mass density (mass per unit volume) or charge density (charge per unit volume). Polarization is simply the dipole moment density.

Step 1: Understanding the Concept:

The energy density (\(u_E\)) of an electric field is the amount of energy stored in the field per unit volume. It represents how concentrated the energy is in a region of space.


Step 2: Key Formula or Approach:

The formula for the energy density of an electric field in a vacuum or free space is:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
where \(\epsilon_0\) is the permittivity of free space and \(E\) is the magnitude of the electric field.


Step 3: Detailed Explanation:

We are given the following values:

- Electric field strength, \(E = 2\) Vm\(^{-1}\)

We need to calculate the energy density, \(u_E\).

Substitute the value of E into the formula:
\[ u_E = \frac{1}{2} \epsilon_0 (2)^2 \] \[ u_E = \frac{1}{2} \epsilon_0 \times 4 \] \[ u_E = 2 \epsilon_0 \]
The units of energy density are Joules per cubic meter (J/m\(^3\)).


Step 4: Final Answer:

The energy density of the electric field is 2\(\epsilon_0\).
Quick Tip: Remember the parallel formulas for energy density in electric and magnetic fields: \(u_E = \frac{1}{2} \epsilon_0 E^2\) and \(u_B = \frac{1}{2\mu_0} B^2\). These are fundamental in electromagnetism.

Step 1: Understanding the Concept:

The resistor color code is a system to indicate the resistance value and tolerance of a resistor. The last color band on a resistor represents its tolerance, which is the percentage of error in its stated resistance value.


Step 2: Detailed Explanation:

The standard color codes for tolerance are:

- Gold: \(\pm\) 5%

- Silver: \(\pm\) 10%

- No color band (absent): \(\pm\) 20%

- Other colors like brown (\(\pm\)1%) and red (\(\pm\)2%) are also used for higher precision resistors.

The question states that the resistor has a tolerance of 20%. According to the standard color code system, a tolerance of 20% is indicated by the absence of a fourth color band.


Step 3: Final Answer:

For a tolerance of 20%, the last band is absent.
Quick Tip: Remember the three main tolerance bands: Gold (5%), Silver (10%), and None (20%). These are the most common ones you'll encounter in introductory physics problems.

Step 1: Understanding the Concept:

This problem applies Joule's Law of Heating, which describes the heat generated by an electric current passing through a conductor. The heat produced is proportional to the square of the current, the resistance, and the time for which the current flows.


Step 2: Key Formula or Approach:

Joule's Law of Heating is given by the formula:
\[ H = I^2 R t \]
where \(H\) is the heat liberated, \(I\) is the current, \(R\) is the resistance, and \(t\) is the time.


Step 3: Detailed Explanation:

We are given the following values:

- Current, \(I = 2\) A

- Time, \(t = 2.5\) s

- Heat liberated, \(H = 20\) J

We need to find the resistance, \(R\). We can rearrange the formula to solve for \(R\):
\[ R = \frac{H}{I^2 t} \]
Substitute the given values into the rearranged formula:
\[ R = \frac{20}{(2)^2 \times 2.5} \] \[ R = \frac{20}{4 \times 2.5} \] \[ R = \frac{20}{10} \] \[ R = 2 \, \Omega \]

Step 4: Final Answer:

The resistance of the wire is 2 \(\Omega\).
Quick Tip: Be sure to use the correct form of the power/heat formula. If you are given current and resistance, use \(H = I^2Rt\). If you are given voltage and resistance, use \(H = \frac{V^2}{R}t\). Using the right formula for the given variables saves time.

Step 1: Understanding the Concept:

An electron revolving in a circular orbit is equivalent to a tiny current loop. A current loop produces a magnetic field and has an associated magnetic dipole moment. The magnitude of this moment depends on the current and the area of the loop.


Step 2: Key Formula or Approach:

The magnetic dipole moment (\(M\)) of a current loop is given by \(M = IA\), where \(I\) is the current and \(A\) is the area of the loop.

The equivalent current \(I\) due to an electron with charge \(e\) moving with velocity \(v\) in an orbit of radius \(r\) is \(I = \frac{e}{T} = \frac{ev}{2\pi r}\), where \(T\) is the time period.

The area of the orbit is \(A = \pi r^2\).

Combining these, we get the formula for the orbital magnetic moment:
\[ M = I A = \left(\frac{ev}{2\pi r}\right) (\pi r^2) = \frac{evr}{2} \]

Step 3: Detailed Explanation:

We are given the values:

- Charge of an electron, \(e = 1.6 \times 10^{-19}\) C

- Velocity, \(v = 8 \times 10^7\) m/s

- Radius, \(r = 0.5\) m

Substitute these values into the formula:
\[ M = \frac{(1.6 \times 10^{-19}) \times (8 \times 10^7) \times (0.5)}{2} \] \[ M = \frac{1.6 \times 8 \times 0.5}{2} \times 10^{-19+7} \] \[ M = \frac{6.4}{2} \times 10^{-12} \] \[ M = 3.2 \times 10^{-12} \, Am^2 \]
The calculated value is \(3.2 \times 10^{-12}\) Am\(^2\). Option (A) in the exam paper likely contains a typographical error in the exponent (showing -1 instead of -12), but the numerical part (3.2) is correct. Based on the provided answer key, we select option A under the assumption of this typo.


Step 4: Final Answer:

The magnetic moment of the electron is 3.2 \( \times \) 10\(^{-12}\) Am\(^2\).
Quick Tip: The formula \(M = \frac{evr}{2}\) is a direct and useful result for the orbital magnetic moment. It can also be related to the angular momentum \(L = mvr\) as \(M = (\frac{e}{2m_e})L\).

Step 1: Understanding the Concept:

The magnetic force on a moving charged particle is described by the Lorentz force law. The magnitude of this force depends not only on the charge, speed, and magnetic field strength, but also on the angle between the velocity vector of the particle and the magnetic field vector.


Step 2: Key Formula or Approach:

The magnetic force \(\vec{F}\) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by:
\[ \vec{F} = q(\vec{v} \times \vec{B}) \]
The magnitude of this force is:
\[ F = |q| v B \sin(\theta) \]
where \(\theta\) is the angle between the vectors \(\vec{v}\) and \(\vec{B}\).


Step 3: Detailed Explanation:

The magnitude of the force is \(F = |q| v B \sin(\theta)\). To maximize the force \(F\), the term \(\sin(\theta)\) must be maximum, assuming \(q\), \(v\), and \(B\) are non-zero.

The sine function, \(\sin(\theta)\), has a maximum value of 1.

This maximum value occurs when the angle \(\theta = 90^\circ\) or \(\frac{\pi}{2}\) radians.

When \(\theta = 90^\circ\), the velocity vector is perpendicular to the magnetic field vector.

- If \(\theta = 0^\circ\) or \(\theta = 180^\circ\), \(\sin(\theta) = 0\), and the magnetic force is zero.


Step 4: Final Answer:

The magnetic force is maximum when the angle between v and B is 90\(^\circ\).
Quick Tip: Remember that the magnetic force is a result of a cross product (\(\vec{v} \times \vec{B}\)). Cross products are always maximized when the two vectors are perpendicular to each other.

Step 1: Understanding the Concept:

This problem applies Faraday's Law of Electromagnetic Induction, which states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.


Step 2: Key Formula or Approach:

Faraday's Law of Induction is given by the formula:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \(\mathcal{E}\) is the induced emf and \(\Phi\) is the magnetic flux.


Step 3: Detailed Explanation:

We are given the equation for magnetic flux as a function of time:
\[ \Phi(t) = 5t^2 - 25t - 150 \]
1. Differentiate the flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(5t^2 - 25t - 150) \]
Using the power rule for differentiation, we get:
\[ \frac{d\Phi}{dt} = (2 \cdot 5)t - 25 - 0 = 10t - 25 \]
2. Apply Faraday's Law to find the emf:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -(10t - 25) = 25 - 10t \]
3. Calculate the emf at the specific time t = 2 s:
\[ \mathcal{E}(t=2) = 25 - 10(2) \] \[ \mathcal{E}(t=2) = 25 - 20 = 5 \, V \]

Step 4: Final Answer:

The induced emf in the coil at t = 2s is +5 V.
Quick Tip: Don't forget the negative sign in Faraday's Law (\(\mathcal{E} = -d\Phi/dt\)). It represents Lenz's Law, which determines the direction of the induced current. In many calculation problems, it's a crucial part of getting the correct sign for the emf.

Step 1: Understanding the Concept:

Electromagnetic waves are produced by accelerating charges. The frequency of the resulting electromagnetic wave is the same as the frequency of oscillation of the charge. The time period of an oscillation is the reciprocal of its frequency.


Step 2: Key Formula or Approach:

The relationship between time period (\(T\)) and frequency (\(f\)) is:
\[ T = \frac{1}{f} \]

Step 3: Detailed Explanation:

We are given the frequency of the electromagnetic wave:

- \(f = 2\) MHz (MegaHertz)

First, convert the frequency to the base SI unit, Hertz (Hz).
\[ f = 2 \times 10^6 \, Hz \]
Now, calculate the time period using the formula:
\[ T = \frac{1}{2 \times 10^6 \, Hz} \] \[ T = 0.5 \times 10^{-6} \, s \]
To express this in standard scientific notation, we can write it as:
\[ T = 5 \times 10^{-1} \times 10^{-6} \, s = 5 \times 10^{-7} \, s \]

Step 4: Final Answer:

The time period of oscillation is 5 \( \times \) 10\(^{-7}\)s.
Quick Tip: Remember the standard metric prefixes: kilo- (k) is 10\(^3\), Mega- (M) is 10\(^6\), Giga- (G) is 10\(^9\). Being fluent with these conversions is essential for quick calculations.

Step 1: Understanding the Concept:

Astigmatism is a common vision condition that causes blurred vision. It occurs when the cornea (the clear front cover of the eye) or the lens inside the eye has an irregular, non-spherical curvature. This irregular shape prevents light from focusing properly on the retina.


Step 2: Detailed Explanation:

- In an eye with astigmatism, the curvature is different in different meridians (planes). For example, the cornea might be shaped more like a football than a basketball. This causes light rays to focus on two different points instead of one, leading to distorted or blurred vision at all distances.

- To correct this, a special lens is needed that can correct for the different curvatures. A cylindrical lens has curvature in only one direction. By orienting this cylindrical power correctly, it can add or subtract refractive power in a specific meridian to compensate for the eye's irregular shape and bring all light rays to a single focal point on the retina.

- Convex and concave lenses have spherical surfaces and are used to correct farsightedness (hyperopia) and nearsightedness (myopia), respectively.


Step 3: Final Answer:

The eye defect astigmatism can be corrected by using a cylindrical lens.
Quick Tip: Associate the eye defect with its corrective lens: - Myopia (nearsightedness) \(\rightarrow\) Concave lens. - Hyperopia (farsightedness) \(\rightarrow\) Convex lens. - Astigmatism (irregular cornea) \(\rightarrow\) Cylindrical lens. - Presbyopia (age-related) \(\rightarrow\) Bifocal or progressive lenses.

Step 1: Understanding the Concept:

This question deals with controlling the intensity of light using polarizers. It implicitly involves starting with unpolarized light, polarizing it, and then using a second polarizer (an analyzer) to control the final intensity.


Step 2: Detailed Explanation:

The process occurs in two stages:

Stage 1: Creating Polarized Light

- When unpolarized light passes through the first polarizer, it becomes linearly polarized. The intensity of the light is reduced by half in this process.

- If the initial intensity of the unpolarized light is \(I_{unpol}\), the intensity after the first polarizer is \(I_{pol} = \frac{1}{2} I_{unpol}\). This means the maximum possible intensity of the final beam is 50% of the original unpolarized light.


Stage 2: Controlling Intensity with a Second Polarizer (Analyzer)

- The now polarized light with intensity \(I_{pol}\) passes through a second polarizer, called an analyzer. The intensity of the light transmitted by the analyzer (\(I_{final}\)) is given by Malus's Law:

\[ I_{final} = I_{pol} \cos^2(\theta) \]
where \(\theta\) is the angle between the transmission axes of the two polarizers.

- We can control the final intensity by rotating the analyzer, which changes the angle \(\theta\).

- When the axes are aligned (\(\theta = 0^\circ\)), \(\cos^2(0^\circ) = 1\), and the transmitted intensity is maximum: \(I_{final} = I_{pol}\). This corresponds to 50% of the original unpolarized light.

- When the axes are perpendicular (\(\theta = 90^\circ\)), \(\cos^2(90^\circ) = 0\), and the transmitted intensity is minimum: \(I_{final} = 0\).

- Therefore, by rotating the second polarizer, the intensity of the light can be controlled over the range from its maximum possible value (50% of the original) down to zero.


Step 3: Final Answer:

The intensity of a polarized light can be controlled by a second polarizer from 50% to 0% (assuming the process starts with unpolarized light).
Quick Tip: Remember the "rule of halves" for the first polarizer: unpolarized light passing through a polarizer always loses 50% of its intensity. The second polarizer (analyzer) then controls the remaining 50% according to Malus's Law.

Step 1: Understanding the Concept:

This question applies the de Broglie hypothesis, which states that all matter has wave-like properties. The wavelength associated with a particle is inversely proportional to its momentum.


Step 2: Key Formula or Approach:

The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.


Step 3: Detailed Explanation:

We are given the momentum of the particle in terms of Planck's constant:

- Momentum, \(p = (2 \times 10^{10})h\) kgms\(^{-1}\)

Substitute this expression for \(p\) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{(2 \times 10^{10})h} \]
The Planck's constant \(h\) in the numerator and denominator cancels out:
\[ \lambda = \frac{1}{2 \times 10^{10}} \, m \] \[ \lambda = 0.5 \times 10^{-10} \, m \]
The question asks for the wavelength in angstroms (\(\AA\)). We know that 1 \(\AA\) = 10\(^{-10}\) m.

Therefore,
\[ \lambda = 0.5 \, \AA \]

Step 4: Final Answer:

The de Broglie wavelength associated with the particle is 0.5 angstrom.
Quick Tip: When momentum is given as a multiple of 'h', the calculation becomes very simple as 'h' will cancel out. Pay close attention to the final units required (meters, nanometers, angstroms) and perform the conversion correctly.

Step 1: Understanding the Concept:

This question is based on Bohr's model of the atom, specifically his second postulate, which deals with the quantization of angular momentum. It states that an electron can only revolve in orbits for which its angular momentum is an integral multiple of \(\frac{h}{2\pi}\).


Step 2: Key Formula or Approach:

The formula for the quantized angular momentum (\(L\)) of an electron in the \(n\)-th orbit is:
\[ L = \frac{nh}{2\pi} \]
where \(n\) is the principal quantum number (orbit number) and \(h\) is Planck's constant.


Step 3: Detailed Explanation:

We are asked to find the angular momentum for the 2\(^{nd}\) orbit.

So, the principal quantum number is \(n = 2\).

Substitute \(n=2\) into the formula:
\[ L = \frac{2h}{2\pi} \]
Simplify the expression by canceling the factor of 2:
\[ L = \frac{h}{\pi} \]

Step 4: Final Answer:

The angular momentum of the electron in the 2\(^{nd}\) orbit is \(\frac{h}{\pi}\).
Quick Tip: Sometimes the quantity \(\frac{h}{2\pi}\) is written as \(\hbar\) (h-bar). In that notation, the angular momentum is \(L = n\hbar\). For the 2nd orbit, it would be \(2\hbar\), which is equivalent to \(h/\pi\).

Step 1: Understanding the Concept:

This question describes a nuclear decay process. We need to identify the unknown particle 'X' by applying the laws of conservation of mass number, atomic number, and lepton number.


Step 2: Detailed Explanation:

Let's analyze the given nuclear reaction:
\[ \ce{^{22}_{11}Na -> ^{22}_{10}Ne + e+ + X} \]
- Conservation of Mass Number (A): On the left, A = 22. On the right, Ne has A = 22 and the positron (e\(^+\)) has A = 0. So, 22 \(\rightarrow\) 22 + 0. The mass number is conserved. Particle X must have a mass number of 0.

- Conservation of Atomic Number (Z) / Charge: On the left, Z = 11. On the right, Ne has Z = 10 and the positron has a charge of +1. So, 11 \(\rightarrow\) 10 + 1. The charge is conserved. Particle X must have a charge of 0.

- Based on mass number and charge, X could be a neutron or a neutrino/anti-neutrino.

- Conservation of Lepton Number: This decay involves a positron (e\(^+\)), which is an anti-lepton and has a lepton number of L = -1. The reactants (protons, neutrons) have L = 0. To conserve lepton number, the total lepton number on the product side must also be 0. Therefore, a particle with L = +1 must be emitted along with the positron.

- The electron neutrino (\(\nu_e\)) is a lepton with L = +1.

- The electron anti-neutrino (\(\bar{\nu}_e\)) is an anti-lepton with L = -1.

- To balance the lepton number (-1 for the positron), a particle with lepton number +1 is needed. This particle is the neutrino.

- This type of decay, where a proton converts into a neutron, is called positron emission or beta-plus decay, and it is always accompanied by the emission of a neutrino.


Step 3: Final Answer:

The particle X is a neutrino (\(\nu_e\)).
Quick Tip: A simple rule for beta decays: - Beta-minus decay (electron \(e^-\) emitted): An \textbf{anti}-neutrino is also emitted. - Beta-plus decay (positron \(e^+\) emitted): A \textbf{neutrino} is also emitted. This ensures the conservation of lepton number.

Step 1: Understanding the Concept:

This question is about the energy band theory in solids, specifically for a semiconductor like silicon or germanium, which has N atoms. We need to determine the number of available energy states in the valence and conduction bands, not the number of electrons occupying them.


Step 2: Detailed Explanation:

- Consider a crystal with N atoms. Each atom has 4 valence electrons (from its s and p orbitals, e.g., Si is [Ne]3s\(^2\)3p\(^2\)).

- Valence Band: When N atoms come together to form a crystal, their valence atomic orbitals (one s and three p orbitals per atom) combine to form molecular orbitals, which broaden into a continuous band of energy levels. The total number of available valence states is 4 per atom (1 from 's' and 3 from 'p'). For N atoms, this results in a total of 4N available energy states in the valence band.

- Conduction Band: Similarly, the next set of unoccupied atomic orbitals also broadens into an energy band, which is the conduction band. The number of available states in the conduction band is also determined by the number of atomic orbitals from which it is formed. For typical semiconductors, the conduction band also has 4N available energy states.

- At Absolute Zero (0 K): The system is in its lowest energy state. The 4N valence electrons from the N atoms completely fill the 4N available states in the valence band. (Note: Due to spin, 4N electrons can fill 2N states, but the band structure provides 4N states in total, so it is considered full). The conduction band is completely empty of electrons.

- The question asks for the "number of energy states," which refers to the capacity or the total number of available states in each band, regardless of their occupancy.


Step 3: Final Answer:

The number of energy states in the valence band is 4N, and the number of energy states in the conduction band is also 4N.
Quick Tip: Be careful to distinguish between the "number of electrons" in a band and the "number of states" in a band. The number of states is like the number of seats in a theater, while the number of electrons is like the number of people sitting in them. The capacity of both the valence and conduction bands in a typical semiconductor crystal is 4N.