Step 1: Understanding the Concept:

We must calculate the total number of atoms in a given mass of several substances. This requires converting the mass to moles of the molecule/formula unit, and then multiplying by the atom count per molecule.


Step 2: Key Formula or Approach:

1. Moles \(n = \frac{Given Mass}{Molar Mass}\).

2. Number of molecules = \(n \times N_A\).

3. Total atoms = Number of molecules \(\times\) (atoms per molecule).


Step 3: Detailed Explanation:

A. 1.8 mg water (\(H_2O\)):

Molar mass of \(H_2O = 18 g/mol\).

Moles of \(H_2O = \frac{1.8 \times 10^{-3} g}{18} = 10^{-4} mol\).

One molecule of \(H_2O\) has 3 atoms (2 Hydrogen, 1 Oxygen).

Total atoms = \(3 \times 10^{-4} \times N_A\). (Matches III)


B. 9.8 mg sulphuric acid (\(H_2SO_4\)):

Molar mass of \(H_2SO_4 = 2(1) + 32 + 4(16) = 98 g/mol\).

Moles of \(H_2SO_4 = \frac{9.8 \times 10^{-3} g}{98} = 10^{-4} mol\).

One molecule of \(H_2SO_4\) has 7 atoms (2 Hydrogen, 1 Sulfur, 4 Oxygen).

Total atoms = \(7 \times 10^{-4} \times N_A\). (Matches IV)


C. 1.8 mg carbon (\(C\)):

Molar mass of C = \(12 g/mol\).

Moles of C = \(\frac{1.8 \times 10^{-3} g}{12} = 0.15 \times 10^{-3} mol = 1.5 \times 10^{-4} mol\).

Carbon is an elemental atom.

Total atoms = \(1.5 \times 10^{-4} \times N_A\). (Matches II)


D. 5.85 mg salt (NaCl):

Molar mass of NaCl = \(23 + 35.5 = 58.5 g/mol\).

Moles of NaCl = \(\frac{5.85 \times 10^{-3} g}{58.5} = 10^{-4} mol\).

One formula unit of NaCl has 2 ions/atoms (1 Na, 1 Cl).

Total atoms = \(2 \times 10^{-4} \times N_A\). (Matches I)


Matching summary: A-III, B-IV, C-II, D-I.


Step 4: Final Answer:

The correct matching is A-III, B-IV, C-II, D-I.
Quick Tip: To save time, calculate the easiest options first. Finding A-III and B-IV immediately narrows the choices down, making full calculation of every option unnecessary if options are unique enough.

Step 1: Understanding the Concept:

We must calculate the mole fraction of a solvent in a binary mixture given its percentage by weight. Secondly, we evaluate the intermolecular interactions between polar and non-polar molecules to predict deviation from Raoult's Law.


Step 2: Key Formula or Approach:

1. Moles \(n = \frac{Mass}{Molar Mass}\).

2. Mole fraction \(\chi_A = \frac{n_A}{n_A + n_B}\).

3. Deviation from Raoult's Law: Positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions.


Step 3: Detailed Explanation:

Statement I:

30% (w/w) solution of methanol (\(CH_3OH\)) in \(CCl_4\) means 30 g of methanol is mixed with 70 g of \(CCl_4\).

Molar mass of \(CH_3OH = 12 + 3(1) + 16 + 1 = 32 g/mol\).

Moles of methanol \(n_1 = \frac{30}{32} = 0.9375 mol\).

Molar mass of \(CCl_4 = 12 + 4(35.5) = 12 + 142 = 154 g/mol\).

Moles of \(CCl_4\) \(n_2 = \frac{70}{154} \approx 0.4545 mol\).

Mole fraction of \(CCl_4\) is \(\chi_2 = \frac{n_2}{n_1 + n_2} = \frac{0.4545}{0.9375 + 0.4545} = \frac{0.4545}{1.392} \approx 0.326\).

Rounding to two decimal places gives 0.33. So, Statement I is true.


Statement II:

Methanol molecules exhibit strong intermolecular hydrogen bonding.
\(CCl_4\) is a non-polar molecule. When \(CCl_4\) is added to methanol, it inserts itself between the methanol molecules, effectively breaking their hydrogen bonds.

This results in the A-B intermolecular forces (\(CH_3OH - CCl_4\)) being significantly weaker than the pure A-A forces (\(CH_3OH - CH_3OH\)).

Weaker intermolecular forces make it easier for molecules to escape into the vapor phase, increasing vapor pressure above what Raoult's law predicts.

Thus, the mixture shows a positive deviation from Raoult's law. So, Statement II is true.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: A classic rule of thumb for liquid solutions: Mixing a highly polar liquid (like water or alcohols) with a completely non-polar organic solvent (like benzene, toluene, or \(CCl_4\)) almost always results in a positive deviation from Raoult's law.

Step 1: Understanding the Concept:

To find the shape of a molecule or ion, we use VSEPR (Valence Shell Electron Pair Repulsion) theory. We count the number of valence electrons on the central atom, adjust for the ionic charge, and determine the steric number (number of bond pairs + lone pairs) to deduce hybridization and geometry.


Step 2: Key Formula or Approach:

Steric number = \(\frac{1}{2} (V + M - C + A)\)

where \(V\) is valence electrons of central atom, \(M\) is number of monovalent surrounding atoms, \(C\) is positive charge, \(A\) is negative charge.


Step 3: Detailed Explanation:

For the cation \(BrF_2^+\):

Central atom is Bromine (Br), Group 17 \(\implies V = 7\).

Monovalent atoms (F) \(\implies M = 2\).

Cationic charge \(C = 1\).

Number of electron pairs = \(\frac{1}{2}(7 + 2 - 1) = \frac{8}{2} = 4\).

So, it has \(sp^3\) hybridization.

Out of 4 pairs, 2 are bond pairs (with F) and 2 are lone pairs.

The fundamental geometry for 4 pairs is tetrahedral. The presence of 2 lone pairs causes the two bond pairs to form a "bent" or "V-shape".

So, \(BrF_2^+\) is bent.


For the anion \(BrF_4^-\):

Central atom is Bromine (Br), \(V = 7\).

Monovalent atoms (F) \(\implies M = 4\).

Anionic charge \(A = 1\).

Number of electron pairs = \(\frac{1}{2}(7 + 4 + 1) = \frac{12}{2} = 6\).

So, it has \(sp^3d^2\) hybridization.

Out of 6 pairs, 4 are bond pairs (with F) and 2 are lone pairs.

The fundamental geometry for 6 pairs is octahedral. According to VSEPR, lone pairs occupy trans (opposite) positions to minimize repulsion at \(180^\circ\).

The 4 fluorine atoms will therefore occupy the equatorial plane, creating a "square planar" shape.

So, \(BrF_4^-\) is square planar.


Step 4: Final Answer:

The shapes are bent and square planar respectively.
Quick Tip: To quickly visualize structures with 6 electron pairs (\(sp^3d^2\)), remember the lone pair placement rule: put the first lone pair anywhere, and the second lone pair MUST go exactly \(180^\circ\) opposite to the first to minimize \(90^\circ\) repulsions. This leaves a perfect square plane.

Step 1: Understanding the Concept:

Evaluate colligative properties statements. Knowledge of cryoscopic and ebullioscopic constants for water, the properties defining the measurement choices for macromolecules, and the specific molecular interaction geometry of dimerization is required.


Step 2: Key Formula or Approach:

1. Check standard values: For water, \(K_f = 1.86 K kg/mol\), \(K_b = 0.52 K kg/mol\).

2. Colligative magnitudes: \(\Delta T_b = K_b m\) vs \(\Delta T_f = K_f m\).

3. Dimer structural geometry of carboxylic acids.


Step 3: Detailed Explanation:

Let's analyze each statement:

A. For water, the molal freezing point depression constant is \(K_f = 1.86 K kg/mol\) and the molal boiling point elevation constant is \(K_b = 0.52 K kg/mol\). Thus, \(K_f > K_b\). Statement A is incorrect.

B. For a given molality \(m\) of a non-volatile solute, the elevation in boiling point is \(\Delta T_b = K_b m\) and the depression in freezing point is \(\Delta T_f = K_f m\). Because \(K_f > K_b\), it follows that \(\Delta T_f > \Delta T_b\). Statement B is incorrect.

C. Macromolecules (like proteins and polymers) have very high molar masses, so their molality is exceptionally small, making \(\Delta T_b\) and \(\Delta T_f\) difficult to measure accurately. However, osmotic pressure (\(\pi\)) relies on molarity and produces a large, easily readable pressure value even at dilute concentrations at room temperature. Thus, it is preferred. Statement C is correct.

D. Benzoic acid dimerizes in non-polar solvents like benzene by forming a stable cyclic 8-membered ring through TWO hydrogen bonds. The text structure shown in option D only depicts a single, incomplete linear hydrogen bond association instead of the classic double H-bonded ring structure. Therefore, Statement D is incorrect.


Since A, B, and D are incorrect, the correct option combining these is A, B, and D only.


Step 4: Final Answer:

Statements A, B, and D are incorrect.
Quick Tip: If you know for a fact that \(K_f\) is typically larger than \(K_b\) for standard solvents, statements A and B fall together instantly. This automatically points to (C) as the most likely answer if options are grouped.

Step 1: Understanding the Concept:

We must algebraically manipulate the given equilibrium equations to yield the target equation. When adding/subtracting equations or multiplying by constants, the equilibrium constants undergo multiplication/division or exponentiation, respectively.


Step 2: Key Formula or Approach:

1. Reversing an equation: \(K_{new} = \frac{1}{K_{old}}\).

2. Multiplying an equation by a factor \(n\): \(K_{new} = (K_{old})^n\).

3. Adding two equations: \(K_{new} = K_{eq1} \times K_{eq2}\).


Step 3: Detailed Explanation:

We are given:

Eq 1: \(2xy \rightleftharpoons x_2 + y_2\) with \(K_1 = 2.5 \times 10^5\).

Eq 2: \(xy + \frac{1}{2}z_2 \rightleftharpoons xyz\) with \(K_2 = 5 \times 10^{-3}\).

Target Eq 3: \(\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz\) with \(K_3\).


Notice that the target equation has \(x_2\) and \(y_2\) on the reactant side, whereas Eq 1 has them on the product side. Also, the coefficients are \(1/2\) instead of \(1\).

First, reverse Eq 1 to get \(x_2\) and \(y_2\) as reactants:
\(x_2 + y_2 \rightleftharpoons 2xy\)

The new equilibrium constant is \(K_1' = \frac{1}{K_1}\).

Next, divide this reversed equation by 2 (multiply by 1/2) to match the coefficients in the target equation:
\(\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy\)

The new equilibrium constant is \(K_1'' = (K_1')^{1/2} = \sqrt{\frac{1}{K_1}}\).

Calculate \(K_1''\):
\(K_1'' = \frac{1}{\sqrt{2.5 \times 10^5}} = \frac{1}{\sqrt{25 \times 10^4}} = \frac{1}{5 \times 10^2} = \frac{1}{500} = 0.002 = 2 \times 10^{-3}\).

Now, add this manipulated Eq 1 to Eq 2:

(Manipulated Eq 1): \(\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy\)

(Eq 2): \(xy + \frac{1}{2}z_2 \rightleftharpoons xyz\)

When we add them, the intermediate \(xy\) cancels out:
\(\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz\). This is exactly the target equation!

When adding equations, multiply their equilibrium constants:
\(K_3 = K_1'' \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3})\).
\(K_3 = 10 \times 10^{-6} = 1.0 \times 10^{-5}\).


Step 4: Final Answer:

The value of \(K_3\) is \(1.0 \times 10^{-5}\).
Quick Tip: To prevent math errors when combining equilibrium equations, always double-check the intermediate species (\(xy\) in this case) to ensure it cancels perfectly on both sides.

Step 1: Understanding the Concept:

Electrode potentials (\(E^\ominus\)) are intensive properties and cannot be added or subtracted directly. We must convert them to standard Gibbs free energy changes (\(\Delta G^\ominus\)), which are extensive thermodynamic properties and can be algebraically manipulated according to Hess's Law.


Step 2: Key Formula or Approach:

1. \(\Delta G^\ominus = -n F E^\ominus\), where \(n\) is the number of electrons transferred.

2. Find the target half-reaction by adding/subtracting the given half-reactions.

3. Add/subtract the corresponding \(\Delta G^\ominus\) values to find \(\Delta G^\ominus\) of the target reaction, then convert back to \(E^\ominus\).


Step 3: Detailed Explanation:

Let's write down the given half-reactions as reductions to elemental iron:

Reaction 1: \(Fe^{2+}(aq) + 2e^- \to Fe(s)\)
\(E^\ominus_1 = X\) Volt, \(n_1 = 2\).
\(\Delta G^\ominus_1 = -2FX\).


Reaction 2: \(Fe^{3+}(aq) + 3e^- \to Fe(s)\)
\(E^\ominus_2 = Y\) Volt, \(n_2 = 3\).
\(\Delta G^\ominus_2 = -3FY\).


We want the standard electrode potential for the couple \(Fe^{3+} / Fe^{2+}\).

The target half-reaction is:

Reaction 3: \(Fe^{3+}(aq) + e^- \to Fe^{2+}(aq)\)

This reaction involves \(n_3 = 1\) electron. Let its potential be \(E^\ominus_3\).
\(\Delta G^\ominus_3 = -1 \cdot F \cdot E^\ominus_3\).


To obtain Reaction 3, we can subtract Reaction 1 from Reaction 2:
\((Fe^{3+} + 3e^- \to Fe) - (Fe^{2+} + 2e^- \to Fe) \implies Fe^{3+} + e^- \to Fe^{2+}\).

Applying Hess's law for free energy:
\(\Delta G^\ominus_3 = \Delta G^\ominus_2 - \Delta G^\ominus_1\).

Substitute the free energy terms:
\(-F E^\ominus_3 = (-3FY) - (-2FX)\).
\(-F E^\ominus_3 = -3FY + 2FX\).

Divide the entire equation by \(-F\):
\(E^\ominus_3 = 3Y - 2X\).


Step 4: Final Answer:

The standard potential is \(3Y - 2X\).
Quick Tip: Never add or subtract voltages directly! Always map them to Gibbs Free Energy (\(n \times E^\ominus\)) before performing algebraic sums, then divide by the target electron count.

Step 1: Understanding the Concept:

Statement I requires checking the Arrhenius equation to see if the specified activation energy precisely causes a doubling of the rate constant over a specific temperature shift. Statement II requires knowledge of how half-life scales with initial concentration for different reaction orders.


Step 2: Key Formula or Approach:

1. Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)\).

2. Half-life for first-order reaction: \(t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}\).


Step 3: Detailed Explanation:

Evaluate Statement I:

We need to verify if \(\frac{k_2}{k_1} = 2\).

Given \(E_a = 12.6 kcal/mol\). Convert this to Joules per mole:
\(E_a = 12.6 \times 10^3 cal/mol \times 4.2 J/cal = 52920 J/mol\).

Temperatures are \(T_1 = 298 K\) and \(T_2 = 308 K\).

Substitute into the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{52920}{8.314} \times \left( \frac{308 - 298}{298 \times 308} \right)\).
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{52920}{8.314} \times \left( \frac{10}{91784} \right)\).

Calculate the numerical value:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{529200}{763092.176} \approx 0.6935\).

Since \(\ln 2 \approx 0.693\), we can confidently say that \(\frac{k_2}{k_1} \approx 2\).

Therefore, the rate constant doubles. Statement I is true.


Evaluate Statement II:

For a first-order reaction, the half-life is \(t_{1/2} = \frac{0.693}{k}\).

This formula clearly shows that the half-life is a constant value and depends exclusively on the rate constant \(k\), being entirely independent of the initial concentration \([A]_0\).

A graph of \(t_{1/2}\) versus \([A]_0\) would be a perfectly horizontal line.

The statement describes a graph that is a straight line passing through the origin (\(t_{1/2} \propto [A]_0\)), which is actually characteristic of a zero-order reaction.

Therefore, Statement II is false.


Step 4: Final Answer:

Statement I is true but Statement II is false.
Quick Tip: A useful heuristic in chemical kinetics: for reactions occurring near room temperature, an activation energy of approximately \(50 - 55 kJ/mol\) perfectly yields the "Rule of Thumb" where a \(10^\circC\) rise in temperature doubles the reaction rate.

Step 1: Understanding the Concept:

We need to correlate specific electronic configurations in the second period (\(n=2\)) to their relative 1st Ionization Energy (IE) values. The general trend of IE across a period increases, but there are distinct anomalies due to subshell stability (full \(s\)-orbital vs \(p^1\), and half-filled \(p^3\) vs \(p^4\)).


Step 2: Key Formula or Approach:

Identify the elements based on the configuration:
\(ns^2 \to Beryllium (Be)\)
\(ns^2np^1 \to Boron (B)\)
\(ns^2np^3 \to Nitrogen (N)\)
\(ns^2np^6 \to Neon (Ne)\)

Compare their IE based on trends: Noble gases > half-filled \(p\) > full \(s\) > partial \(p\).


Step 3: Detailed Explanation:

Let's list the expected order of Ionization Energies for these elements:

1. Neon (Ne, \(2s^2 2p^6\)) is a noble gas with a completely filled valence shell. Removing an electron is extremely difficult, making it have the highest IE among the group.

2. Nitrogen (N, \(2s^2 2p^3\)) has a stable exactly half-filled \(2p\) subshell, requiring significant energy to break this symmetry.

3. Beryllium (Be, \(2s^2\)) has a completely filled \(2s\) subshell. Its electrons are highly penetrating and tightly held compared to an unpaired \(2p\) electron.

4. Boron (B, \(2s^2 2p^1\)) has a single electron in a higher-energy \(2p\) orbital. It is easier to remove this lone \(p\) electron to reach the stable \(2s^2\) state.

Thus, the expected order from lowest to highest IE is: Boron < Beryllium < Nitrogen < Neon.

Now map the given numerical values: 800, 899, 1402, 2080.

- Neon (D) = highest = 2080 (I)

- Nitrogen (C) = second highest = 1402 (IV)

- Beryllium (A) = third highest = 899 (II)

- Boron (B) = lowest = 800 (III)

Summary of matches: A \(\to\) II, B \(\to\) III, C \(\to\) IV, D \(\to\) I.


Step 4: Final Answer:

The correct matching is A-II, B-III, C-IV, D-I.
Quick Tip: A classic exception across period 2: The IE of Group 2 (\(s^2\)) is unexpectedly higher than Group 13 (\(s^2 p^1\)) due to shielding effects, and Group 15 (\(p^3\)) is higher than Group 16 (\(p^4\)) due to half-filled orbital stability.

Step 1: Understanding the Concept:

We evaluate the physical and chemical trends of Group 15 hydrides (\(EH_3\)), moving down the group from Nitrogen to Bismuth. Trends in bond length, bond dissociation enthalpy, electron density, and orbital hybridization shape these properties.


Step 2: Key Formula or Approach:

As we go down the group:

- Atomic size of the central atom E increases.

- E-H bond length increases, so E-H bond strength/dissociation enthalpy decreases.

- Electronegativity of central atom decreases.


Step 3: Detailed Explanation:

Let's analyze each statement sequentially:

A. Reducing nature depends on the ease of breaking the E-H bond to liberate hydrogen. Since the atomic size increases down the group (\(N < P < As < Sb < Bi\)), the E-H bond length increases and bond dissociation enthalpy drops rapidly. Thus, it's easier to give off hydrogen, making \(BiH_3\) the strongest reducing agent. Statement A is correct.

B. The Lewis basicity (tendency to donate a lone pair) depends on electron density. While all have one lone pair, the volume of the central atom increases drastically down the group. The lone pair electron density gets highly diffused over a larger volume, lowering the capacity to effectively donate it. So, basicity drops from \(NH_3\) to \(BiH_3\). Statement B is correct.

C. Thermal stability depends entirely on the E-H bond strength. As highlighted earlier, increasing bond length means decreasing bond strength. \(BiH_3\) is highly unstable compared to \(NH_3\). Statement C is correct.

D. Bond angles: For \(NH_3\), \(sp^3\) hybridization with high electronegativity causes bond pairs to repel closer to the central atom, pushing the angle to \(107.8^\circ\). Moving down the group, electronegativity drops, bond pairs move further from the central atom towards H, lowering repulsion. Drago's Rule indicates that for \(PH_3\) and below, hybridization barely occurs, and nearly pure \(p\)-orbitals are used, causing bond angles to drop close to \(90^\circ\) (\(PH_3 \approx 93.6^\circ, SbH_3 \approx 91.3^\circ\)). Statement D is correct.


Since all four statements accurately describe the chemical behaviors, the correct choice includes all of them.


Step 4: Final Answer:

Statements A, B, C, and D are all correct.
Quick Tip: For p-block hydrides moving down any group, practically all key properties (bond angle, thermal stability, basicity) **decrease**, with the sole major exception being **reducing character, which increases**.

Step 1: Understanding the Concept:

To determine if transition metal ions are coloured or possess magnetic properties, we must look at their electronic configurations. Ions with partially filled \(d\) or \(f\) orbitals (unpaired electrons) exhibit colour (via \(d-d\) or \(f-f\) transitions) and are paramagnetic. Ions with empty (\(d^0, f^0\)) or completely full (\(d^{10}, f^{14}\)) orbitals are colourless and diamagnetic.


Step 2: Key Formula or Approach:

Evaluate the electronic configurations for the transition metal ions and the f-block ions to count the number of unpaired electrons (\(n\)).

Coloured/Paramagnetic \(\implies n > 0\).

Colourless/Diamagnetic \(\implies n = 0\).


Step 3: Detailed Explanation:

Evaluate Statement I (d-block ions):

- \(Ti^{4+}\): [Ar] \(3d^0 \implies\) 0 unpaired \(e^- \implies\) Colourless.

- \(V^{2+}\): [Ar] \(3d^3 \implies\) 3 unpaired \(e^- \implies\) Coloured.

- \(Mn^{2+}\): [Ar] \(3d^5 \implies\) 5 unpaired \(e^- \implies\) Coloured.

- \(Fe^{3+}\): [Ar] \(3d^5 \implies\) 5 unpaired \(e^- \implies\) Coloured.

- \(Cr^{2+}\): [Ar] \(3d^4 \implies\) 4 unpaired \(e^- \implies\) Coloured.

Now let's check the given pairs:

1. \([Ti^{4+}, V^{2+}]\): \(Ti^{4+}\) is colourless, so the pair is not "both coloured".

2. \([V^{2+}, Mn^{2+}]\): Both are coloured.

3. \([Mn^{2+}, Fe^{3+}]\): Both are coloured.

4. \([V^{2+}, Cr^{2+}]\): Both are coloured.

There are precisely 3 pairs where BOTH ions are coloured. Statement I is correct.


Evaluate Statement II (f-block ions):

- \(La^{3+}\): [Xe] \(4f^0 \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

- \(Yb^{2+}\): [Xe] \(4f^{14} \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

- \(Lu^{3+}\): [Xe] \(4f^{14} \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

- \(Ce^{4+}\): [Xe] \(4f^0 \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

- \(Ac^{3+}\): [Rn] \(5f^0 \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

- \(Lr^{3+}\): [Rn] \(5f^{14} \implies\) 0 unpaired \(e^- \implies\) Diamagnetic.

Now let's check the given pairs:

1. \([La^{3+}, Yb^{2+}]\): Both are diamagnetic.

2. \([Lu^{3+}, Ce^{4+}]\): Both are diamagnetic.

3. \([Ac^{3+}, Lr^{3+}]\): Both are diamagnetic.

There are precisely 3 pairs where BOTH ions are diamagnetic. Statement II is correct.


Step 4: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: For f-block element questions on magnetism and colour, always memorize the boundary state ions. The extremes (\(f^0\) and \(f^{14}\)) are inevitably colourless and diamagnetic. \(La^{3+}, Ce^{4+}, Ac^{3+}\) form stable empty shells, while \(Lu^{3+}, Yb^{2+}, Lr^{3+}\) achieve stable full shells.

Step 1: Understanding the Concept:

We need to evaluate the validity of two statements regarding the catalytic action of transition metals, focusing on the electrons used in bonding and the effect of adsorption on the reacting molecules.


Step 2: Key Formula or Approach:

1. Transition metals use both their \((n-1)d\) and \(ns\) electrons for bond formation during catalysis because the energy difference between these orbitals is very small.

2. Chemisorption on a catalyst surface increases the local concentration of reactants, but the interaction with the surface weakens the internal bonds of the reactant molecules, lowering the activation energy.


Step 3: Detailed Explanation:

Let's evaluate Statement I:

Transition metals have variable oxidation states and can form complexes. When acting as catalysts, they do not exclusively use their \(3d\) electrons. They utilize both their \(3d\) and \(4s\) electrons to form temporary bonds with the reactant molecules on the surface of the catalyst. Therefore, Statement I is incorrect.


Let's evaluate Statement II:

Catalysis involves the adsorption of reactant molecules onto the surface of the catalyst. While it is true that this process increases the local concentration of reactants on the surface, the interaction between the reactant molecules and the catalyst atoms actively weakens the chemical bonds within the reactant molecules. This bond weakening is what lowers the activation energy and facilitates the reaction. Strengthening the bonds would make the reaction harder to proceed. Therefore, Statement II is incorrect.


Step 4: Final Answer:

Both Statement I and Statement II are incorrect.
Quick Tip: Remember that the whole purpose of a catalyst is to break bonds easier (lower activation energy). If adsorption strengthened the bonds, the catalyst would be an inhibitor instead.

Step 1: Understanding the Concept:

Fractional distillation relies on the difference in boiling points (and thus volatilities) of the components in a liquid mixture. The temperature gradient in a fractionating column governs which component vaporizes and which condenses at different heights.


Step 2: Key Formula or Approach:

1. Higher boiling point \(\implies\) Lower volatility \(\implies\) Condenses easier at higher temperatures (lower up the column).

2. Lower boiling point \(\implies\) Higher volatility \(\implies\) Remains as vapor at lower temperatures (rises higher up the column).


Step 3: Detailed Explanation:

Let's evaluate Statement I:

In a fractionating column, the temperature is highest at the bottom and gradually decreases towards the top. Vapours of a liquid with a higher boiling point will reach their condensation temperature sooner as they travel up the cooler column. Therefore, they condense first and return to the flask, whereas the lower boiling point vapours continue to rise. Statement I is true.


Let's evaluate Statement II:

As the mixed vapours rise through the fractionating column, they undergo multiple condensation and vaporization cycles. Because the higher boiling point component condenses and falls back down, the remaining vapours that continue rising become increasingly enriched in the more volatile component (the one with the lower boiling point). Therefore, the rising vapours are richer in the low boiling component, not the high boiling component. Statement II is false.


Step 4: Final Answer:

Statement I is true but Statement II is false.
Quick Tip: Always associate "high boiling point" with "wants to be a liquid" and "low boiling point" with "wants to be a gas". The gas (vapour) rising to the top is naturally the component that wants to be a gas (low boiling point).

Step 1: Understanding the Concept:

We must determine the reaction mechanism for each given chemical reaction. Specifically, we need to identify which process does not strictly fall under the classical definition of an intermediate carbocation rearrangement leading to a substituted/eliminated product, but rather belongs to a distinct reaction class like Isomerization.


Step 2: Key Formula or Approach:

Analyze the intermediate for each reaction:

- Reactions with \(AlCl_3\) and alkyl halides form carbocations that undergo 1,2-shifts if possible.

- Acid-catalyzed dehydration and hydration form carbocations that rearrange to achieve maximum stability.

- Heating \(n\)-alkanes with anhydrous \(AlCl_3/HCl\) causes skeletal structural changes, officially classified as 'Isomerization' in standard curricula.


Step 3: Detailed Explanation:

Let's analyze the reactions one by one:


Option A: Friedel-Crafts Alkylation.

Reactants: Benzene + 1-chloro-2-methylpropane (isobutyl chloride).

The isobutyl cation is a primary carbocation which immediately undergoes a 1,2-hydride shift to form a highly stable tertiary carbocation (tert-butyl cation). The major product is tert-butylbenzene. This is a classic rearrangement reaction.


Option B: Acid-catalyzed Dehydration.

Reactants: 3,3-dimethylbutan-2-ol + \(H_3O^+\).

Protonation and loss of water forms a secondary carbocation. A 1,2-methyl shift occurs to form a more stable tertiary carbocation. The subsequent elimination of a proton yields 2,3-dimethylbut-2-ene as the major product. This is a classic rearrangement reaction.


Option C: Alkane Isomerization.

Reactants: n-hexane + Anhydrous \(AlCl_3\) / HCl.

The reaction produces branched alkanes like 2-methylpentane and 3-methylpentane. While the mechanism fundamentally involves transient carbocation shifts, standard textbook classification (such as NCERT) strictly categorizes this reaction specifically as Isomerisation, separating it from addition/substitution "rearrangement" reactions. The product itself is just a structural isomer of the reactant, not a functionalized product.


Option D: Acid-catalyzed Hydration.

Reactants: 3,3-dimethylbut-1-ene + \(H_3O^+\).

Protonation of the double bond yields a secondary carbocation. A 1,2-methyl shift occurs to form a stable tertiary carbocation, which then captures water to form 2,3-dimethylbutan-2-ol. This involves a carbocation rearrangement.


Following formal chemical classification in competitive exam syllabi, Option C is treated as Isomerisation.


Step 4: Final Answer:

The reaction in option (C) is officially an Isomerization.
Quick Tip: In NCERT-based questions, precise nomenclature matters. Skeletal changes of plain alkanes using \(AlCl_3/HCl\) are termed "Isomerisation", while 1,2-shifts during functional group transformations are termed "Rearrangements".

Step 1: Understanding the Concept:

To identify aromatic species, we apply Hückel's Rule. A compound is aromatic if it is cyclic, completely conjugated (planar, sp2 hybridized atoms in the ring), and has \((4n + 2)\) \(\pi\) electrons delocalized in the ring, where \(n = 0, 1, 2, \dots\).


Step 2: Key Formula or Approach:

1. Check for continuous cycle of p-orbitals.

2. Count \(\pi\) electrons (double bonds = 2, lone pairs involved in resonance = 2, negative charge = 2, positive charge = 0).

3. Match with Hückel numbers: 2, 6, 10, 14, etc.


Step 3: Detailed Explanation:

Let's evaluate each of the 6 structures given in the image:


1. p-benzoquinone (first image, six-membered ring with two outward C=O and two internal double bonds):

It is cyclic and has double bonds, but it is a cross-conjugated system rather than a continuous loop of delocalized electrons following Hückel's rule. It is fundamentally non-aromatic.


2. Tropylium cation (second image, seven-membered ring with a positive charge):

The positive charge provides an empty p-orbital, making the ring fully conjugated. It has 3 double bonds = 6 \(\pi\) electrons. Since \(6 = 4(1) + 2\), it satisfies Hückel's rule and is aromatic.


3. Phenanthrene (third image, three fused benzene rings in an angular arrangement):

It is fully conjugated and planar. Total number of \(\pi\) bonds is 7, meaning \(14 \pi\) electrons. Since \(14 = 4(3) + 2\), it satisfies Hückel's rule and is aromatic.


4. Naphthalene (fourth image, two fused benzene rings):

It is fully conjugated and planar. Total number of \(\pi\) bonds is 5, meaning \(10 \pi\) electrons. Since \(10 = 4(2) + 2\), it satisfies Hückel's rule and is aromatic.


5. Cyclobutadiene dication (fifth image, four-membered ring with two positive charges):

The two positive charges mean two carbons have empty p-orbitals, allowing full conjugation. The single double bond provides 2 \(\pi\) electrons. Since \(2 = 4(0) + 2\), it satisfies Hückel's rule and is aromatic.


6. Cyclopentadienyl anion (sixth image, five-membered ring with a negative charge):

The negative charge acts as a lone pair participating in the \(\pi\) system. It has 2 double bonds (4 e-) + 1 lone pair (2 e-) = 6 \(\pi\) electrons. Since \(6 = 4(1) + 2\), it satisfies Hückel's rule and is aromatic.


Total aromatic species = Tropylium cation, Phenanthrene, Naphthalene, Cyclobutadiene dication, and Cyclopentadienyl anion.

Total count = 5.


Step 4: Final Answer:

The total number of aromatic species is 5.
Quick Tip: Always count the electrons provided by ions in the ring: Carbocations (+) contribute 0 electrons but allow conjugation. Carbanions (-) contribute 2 electrons and allow conjugation.

Step 1: Understanding the Concept:

We must trace a two-step chlorination reaction. First, identify the unique optically active monochloride product of n-butane. Then, determine all possible unique structural positions where a second chlorine atom can be attached to this monochloride.


Step 2: Key Formula or Approach:

1. Monochlorination of n-butane (\(CH_3-CH_2-CH_2-CH_3\)) yields 1-chlorobutane and 2-chlorobutane.

2. Identify the chiral molecule.

3. Substitute one more hydrogen on the chiral molecule with a chlorine atom systematically at every distinct carbon position.


Step 3: Detailed Explanation:

Step 1: Monochlorination of n-butane.

The possible monochlorides are:

- 1-chlorobutane: \(CH_2Cl-CH_2-CH_2-CH_3\) (Achiral)

- 2-chlorobutane: \(CH_3-CHCl-CH_2-CH_3\) (Chiral, because C2 is attached to H, Cl, Methyl, and Ethyl groups).

The problem states that "P" is an optically active compound. Therefore, compound "P" is 2-chlorobutane.


Step 2: Further chlorination of 2-chlorobutane ("P").

We systematically replace one hydrogen atom on each of the four carbon atoms in \(C^{(1)}H_3 - C^{(2)}HCl - C^{(3)}H_2 - C^{(4)}H_3\) with a chlorine atom to find the structural isomers of the dichloro product:


1. Chlorination at C1:
\(CH_2Cl - CHCl - CH_2 - CH_3 \implies\) 1,2-dichlorobutane.


2. Chlorination at C2:
\(CH_3 - CCl_2 - CH_2 - CH_3 \implies\) 2,2-dichlorobutane.


3. Chlorination at C3:
\(CH_3 - CHCl - CHCl - CH_3 \implies\) 2,3-dichlorobutane.


4. Chlorination at C4:
\(CH_3 - CHCl - CH_2 - CH_2Cl \implies\) 1,3-dichlorobutane.


All four structures are distinct constitutional isomers. The problem explicitly asks to ignore stereoisomers.

Total number of distinct dichloro structural isomers = 4.


Step 4: Final Answer:

The number of dichloro compounds obtained is 4.
Quick Tip: To ensure you don't miss or duplicate structural isomers during systematic substitution, write out the IUPAC name for every product you draw. If the names are different, the structural isomers are unique.

Step 1: Understanding the Concept:

We evaluate trends in physical properties (boiling point and melting point) of alkyl and aryl halides based on molecular mass, surface area, and molecular symmetry/dipole moment.


Step 2: Key Formula or Approach:

1. Boiling point of homologous series increases with molecular mass due to stronger van der Waals dispersion forces.

2. Melting point of isomeric benzene derivatives is highly dependent on crystal lattice packing efficiency, driven by symmetry (para \(>\) ortho \(\approx\) meta).

3. Boiling point of isomers depends on polarity. The ortho isomer has a higher net dipole moment than the para isomer (which is zero), leading to stronger dipole-dipole attractions.


Step 3: Detailed Explanation:

Let's evaluate Statement I:

The given compounds are iodomethane (\(CH_3I\)), iodoethane (\(CH_3CH_2I\)), and 1-iodopropane (\(CH_3CH_2CH_2I\)).

As the length of the alkyl chain increases, the molecular mass and the surface area of the molecule increase.

An increased surface area leads to stronger van der Waals forces (London dispersion forces) between molecules.

Consequently, more energy is required to separate the molecules, resulting in higher boiling points.

The order is \(CH_3CH_2CH_2I > CH_3CH_2I > CH_3I\). Statement I is true.


Let's evaluate Statement II:

Compare the isomers of dichlorobenzene.

Melting Point: p-dichlorobenzene is highly symmetric. This symmetry allows its molecules to pack very closely and efficiently in a solid crystal lattice compared to the ortho and meta isomers. Stronger lattice interactions require higher temperatures to break, giving the para isomer a significantly higher melting point.

Boiling Point: Boiling point in liquids is heavily influenced by dipole-dipole interactions. In p-dichlorobenzene, the two C-Cl dipoles are exactly opposite and cancel each other out (\(\mu = 0\)). In o-dichlorobenzene, the C-Cl dipoles are at a \(60^\circ\) angle, resulting in a strong net dipole moment. The polar o-dichlorobenzene molecules have stronger intermolecular dipole attractions than the non-polar para molecules, giving the ortho isomer a higher boiling point.

Therefore, p-dichlorobenzene has a higher melting point but a lower boiling point than o-dichlorobenzene. Statement II is completely true.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: Symmetry dictates Melting Point (packing efficiency). Polarity (dipole moment) dictates Boiling Point (dipole-dipole interaction). The para-isomer dominates melting point, while the ortho-isomer dominates boiling point.

Step 1: Understanding the Concept:

The given reactant is a molecule containing two key functional groups: an external ketone group and a cyclic amide (lactam) ring. We need to sequentially apply three distinct chemical reagents and determine which functional group reacts at each step to form the final major product.


Step 2: Key Formula or Approach:

1. \(NaBH_4\): A mild reducing agent that readily reduces ketones and aldehydes to alcohols, but is not strong enough to reduce amides or esters.

2. \(NaOH (aq), \Delta\): Strong base and heat will hydrolyze an amide (breaking the C-N bond) to yield a carboxylate salt and an amine.

3. \(H_3O^+\): Acid workup protonates the resulting carboxylate salt into a carboxylic acid.


Step 3: Detailed Explanation:

Let's analyze the starting material structure carefully from the image:

- It has a 6-membered ring containing a Nitrogen atom. Adjacent to the Nitrogen in the ring is a carbonyl group (\(C=O\)). This is a \(\delta\)-valerolactam structure (a 6-membered cyclic amide). Counting carbons, the lactam portion contributes 5 carbons (including the carbonyl carbon) and 1 nitrogen.

- The Nitrogen is substituted with a chain. Tracing the chain from N: \(-CH_2-CH_2-C(=O)-CH_3\). This is a 3-oxobutyl group.


Reaction Step 1: (i) \(NaBH_4\)/MeOH

Sodium borohydride reduces the external ketone group (\(-C(=O)-CH_3\)) to a secondary alcohol.

Reaction: \(-CH_2-CH_2-C(=O)-CH_3 \to -CH_2-CH_2-CH(OH)-CH_3\).

The lactam ring remains completely unaffected because \(NaBH_4\) cannot reduce amides.


Reaction Step 2: (ii) \(NaOH (aq), \Delta\)

Aqueous sodium hydroxide with heating will hydrolyze the cyclic amide (lactam).

The amide bond (\(O=C-N\)) breaks. The carbonyl carbon becomes a sodium carboxylate (\(-COO^-Na^+\)), and the nitrogen picks up a hydrogen to become a secondary amine (\(-NH-\)).

The ring opens up into a straight chain.

The resulting structure is: \(Na^+ \ ^-OOC - CH_2 - CH_2 - CH_2 - CH_2 - NH - CH_2 - CH_2 - CH(OH) - CH_3\).


Reaction Step 3: (iii) \(H_3O^+\)

Acidification protonates the carboxylate ion to form a stable carboxylic acid.

Final product structure: \(HOOC - CH_2 - CH_2 - CH_2 - CH_2 - NH - CH_2 - CH_2 - CH(OH) - CH_3\).


Comparing this sequence with the given options:

- Options A and B still contain the cyclic ring, meaning the base hydrolysis step was ignored.

- Option D shows a primary alcohol at the left end (\(HO-CH_2-\)), implying the lactam carbonyl was fully reduced. \(NaBH_4\) cannot do this.

- Option C perfectly matches our derived acyclic amino-acid-alcohol structure.


Step 4: Final Answer:

The major product is the acyclic molecule shown in Option 3.
Quick Tip: Always respect the chemoselectivity of reagents! \(NaBH_4\) is specifically chosen to selectively reduce ketones/aldehydes while explicitly leaving amides, esters, and carboxylic acids completely intact.

Step 1: Understanding the Concept:

The Hoffmann bromamide degradation is a classic reaction used to convert primary amides into primary amines with one less carbon atom. Knowing the balanced chemical equation, suitable reactants, and the exact reaction products is essential.


Step 2: Key Formula or Approach:

The overall balanced chemical equation for the Hoffmann bromamide degradation is:
\(R-CO-NH_2 + Br_2 + 4NaOH \to R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O\)


Step 3: Detailed Explanation:

Let's evaluate each statement against the standard reaction:


Statement A: "4 moles of NaOH and 2 moles of \(Br_2\) are consumed per mole of an amide."

According to the balanced equation, 1 mole of primary amide reacts with 1 mole of \(Br_2\) (not 2 moles) and 4 moles of NaOH. Therefore, Statement A is False.


Statement B: "Hoffmann bromamide reaction is not given by alkyl amides."

The reaction works perfectly well for both alkyl (aliphatic) primary amides (e.g., propanamide \(\to\) ethanamine) and aryl primary amides (e.g., benzamide \(\to\) aniline). Therefore, Statement B is False.


Statement C: "Primary amines can be synthesized by Hoffmann bromamide degradation."

The primary purpose of this reaction is precisely the synthesis of pure primary amines (\(R-NH_2\)). Therefore, Statement C is True.


Statement D: "Secondary amide on reaction with \(Br_2\) and NaOH will give secondary amine."

The mechanism requires the nitrogen atom to lose two protons to form the crucial nitrene intermediate. A secondary amide (\(R-CO-NHR'\)) only has one proton on the nitrogen and cannot undergo this specific rearrangement. The reaction strictly requires primary amides. Therefore, Statement D is False.


Statement E: "The by-products of Hoffmann degradation are \(Na_2CO_3\), NaBr and \(H_2O\)."

Looking at the balanced chemical equation, the species generated alongside the primary amine are indeed Sodium carbonate, Sodium bromide, and Water. Therefore, Statement E is True.


The only true statements are C and E.


Step 4: Final Answer:

C and E only.
Quick Tip: To easily remember the stoichiometry of the Hoffmann bromamide reaction, memorize the ratio 1:1:4. 1 Amide : 1 Bromine : 4 Base.

Step 1: Understanding the Concept:

We must evaluate basic definitions and properties of carbohydrates, including classifications (monosaccharides, oligosaccharides, polysaccharides), their reducing nature, and structural equilibria.


Step 2: Key Formula or Approach:

Analyze definitions:

- Reducing sugars have a free aldehyde or ketone group (or hemiacetal/hemiketal).

- Oligosaccharides yield 2 to 10 monosaccharides upon hydrolysis, which can be identical or entirely different.

- Polysaccharides are large polymers \(>10\) units.

- Monosaccharides like glucose exist in a dynamic equilibrium between open and cyclic forms (mutarotation).


Step 3: Detailed Explanation:

Let's evaluate each statement:


Option A: "All monosaccharides are reducing sugars."

Since all monosaccharides (like glucose, fructose, galactose) have either a free aldehyde or a free ketone group (capable of tautomerizing to an aldose in alkaline mediums like Tollen's reagent), they can reduce oxidizing agents. This statement is True.


Option B: "The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same."

Oligosaccharides are polymers made of 2 to 10 monosaccharide units. Hydrolysis can yield the same units (e.g., Maltose hydrolyzes into two Glucose units) OR different units (e.g., Sucrose hydrolyzes into one Glucose and one Fructose unit; Lactose yields Glucose and Galactose). Therefore, they are NOT always the same. This statement is False (and thus the correct answer to the question).


Option C: "Starch and cellulose are typical examples of polysaccharides..."

Polysaccharides are defined as polymeric carbohydrates containing a large number of monosaccharide units joined by glycosidic linkages (far more than 10). Starch and cellulose are the standard textbook examples. This statement is True.


Option D: "Open chain and cyclic structures co-exist at equilibrium..."

In aqueous solution, D-(+)-glucose undergoes mutarotation, existing as an equilibrium mixture of \(\alpha\)-D-glucopyranose (\(\sim 36%\)), the open-chain form (\(<0.1%\)), and \(\beta\)-D-glucopyranose (\(\sim 64%\)). This equilibrium explains its chemical behavior. This statement is True.


Step 4: Final Answer:

The incorrect statement is that monosaccharide units from oligosaccharide hydrolysis are always the same.
Quick Tip: When defining sugar classes, prefix terminology is key: Homopolysaccharides/Homooligosaccharides yield identical units (like Maltose), while Heteropolysaccharides/Heterooligosaccharides yield different units (like Sucrose).

Step 1: Understanding the Concept:

The neutral ferric chloride (\(FeCl_3\)) test is a classic qualitative analytical test specifically used to detect the presence of a phenolic hydroxyl group. The reaction forms a violet (or sometimes green/blue) coordination complex.


Step 2: Key Formula or Approach:

We must examine the side chain (R-group) of each given amino acid to identify which one contains a phenol ring.


Step 3: Detailed Explanation:

Let's analyze the side chains of the given amino acids:


1. Threonine: The side chain is \(-CH(OH)CH_3\). It contains an aliphatic secondary alcohol group. Aliphatic alcohols do not react with neutral \(FeCl_3\).

2. Serine: The side chain is \(-CH_2OH\). It contains an aliphatic primary alcohol group. It does not react with \(FeCl_3\).

3. Tyrosine: The side chain is \(-CH_2-C_6H_4-OH\) (a phenol group). Because the hydroxyl group is directly attached to the aromatic benzene ring, it exhibits phenolic character. It will react with neutral \(FeCl_3\) to form the characteristic violet-coloured complex.

4. Cysteine: The side chain is \(-CH_2SH\). It contains a thiol (sulfhydryl) group, not a phenol.


Therefore, Tyrosine is the only amino acid listed that will give a positive neutral \(FeCl_3\) test.


Step 4: Final Answer:

Tyrosine will give a violet coloured complex.
Quick Tip: Remembering amino acid structures by functional group categories is highly effective. Tyrosine is uniquely the "phenolic" amino acid, making it the answer to any question involving phenol-specific reactions (like coupling reactions or \(FeCl_3\) tests).

Step 1: Understanding the Concept:

A coordination complex is paramagnetic if its central metal ion has at least one unpaired electron in its \(d\)-orbitals after taking into account the oxidation state, coordination geometry, and the strength of the ligand field (Crystal Field Theory).


Step 2: Key Formula or Approach:

1. Determine the oxidation state and \(d\)-electron count of the metal ion.

2. Identify the geometry (tetrahedral or octahedral/square planar).

3. Identify ligand strength (Spectrochemical series): Strong field ligands (like \(CN^-, CO\)) cause pairing (low spin); Weak field ligands (like halides, \(H_2O\)) avoid pairing (high spin).

4. Fill the split \(d\)-orbitals (\(t_{2g}\) and \(e_g\) for octahedral; \(e\) and \(t_2\) for tetrahedral) and count unpaired electrons (\(n\)). If \(n > 0\), it is paramagnetic.


Step 3: Detailed Explanation:

Let's evaluate each complex:


1. \([MnBr_4]^{2-}\): \(Mn^{2+}\) is \(3d^5\). Bromide is a weak field ligand. Tetrahedral geometry. Configuration: \(e^2 t_2^3\). Unpaired electrons = 5. \(\implies\) Paramagnetic.

2. \([NiCl_4]^{2-}\): \(Ni^{2+}\) is \(3d^8\). Chloride is a weak field ligand. Tetrahedral geometry. Configuration: \(e^4 t_2^4\). Unpaired electrons = 2. \(\implies\) Paramagnetic.

3. \([Ni(CN)_4]^{2-}\): \(Ni^{2+}\) is \(3d^8\). Cyanide is a strong field ligand. Square planar geometry (\(dsp^2\)). The 8 electrons pair up in the lower 4 orbitals. Unpaired electrons = 0. \(\implies\) Diamagnetic.

4. \([Ni(CO)_4]\): \(Ni^0\) is \(3d^8 4s^2 \to 3d^{10}\) (due to strong CO ligand pulling \(s\) electrons into \(d\)). Tetrahedral. Completely filled \(d\)-subshell. Unpaired electrons = 0. \(\implies\) Diamagnetic.

5. \([CoF_6]^{3-}\): \(Co^{3+}\) is \(3d^6\). Fluoride is a weak field ligand. Octahedral high-spin. Configuration: \(t_{2g}^4 e_g^2\). Unpaired electrons = 4. \(\implies\) Paramagnetic.

6. \([Fe(CN)_6]^{4-}\): \(Fe^{2+}\) is \(3d^6\). Cyanide is a strong field ligand. Octahedral low-spin. Configuration: \(t_{2g}^6 e_g^0\). Unpaired electrons = 0. \(\implies\) Diamagnetic.

7. \([Mn(CN)_6]^{3-}\): \(Mn^{3+}\) is \(3d^4\). Cyanide is a strong field ligand. Octahedral low-spin. Configuration: \(t_{2g}^4 e_g^0\). Unpaired electrons = 2. \(\implies\) Paramagnetic.

8. \([Ti(CN)_6]^{3-}\): \(Ti^{3+}\) is \(3d^1\). Regardless of ligand strength, 1 electron will always remain unpaired. Configuration: \(t_{2g}^1 e_g^0\). Unpaired electrons = 1. \(\implies\) Paramagnetic.

9. \([Cu(H_2O)_6]^{2+}\): \(Cu^{2+}\) is \(3d^9\). Octahedral. Regardless of ligand strength, a \(d^9\) system will always have 1 unpaired electron. Unpaired electrons = 1. \(\implies\) Paramagnetic.

10. \([Co(C_2O_4)_3]^{3-}\): \(Co^{3+}\) is \(3d^6\). Oxalate (\(C_2O_4^{2-}\)) is an oxygen donor. Generally weak, but specifically with the highly charged \(Co^{3+}\) ion, oxalate acts as a strong field ligand, causing pairing. Octahedral low-spin. Configuration: \(t_{2g}^6 e_g^0\). Unpaired electrons = 0. \(\implies\) Diamagnetic.


Counting the paramagnetic complexes: \([MnBr_4]^{2-}\), \([NiCl_4]^{2-}\), \([CoF_6]^{3-}\), \([Mn(CN)_6]^{3-}\), \([Ti(CN)_6]^{3-}\), and \([Cu(H_2O)_6]^{2+}\).

Total count = 6.


Step 4: Final Answer:

The number of paramagnetic complexes is 6.
Quick Tip: Beware the exception! While oxygen donors (\(H_2O\), oxalate) are generally weak field ligands, they act as strong field ligands (causing pairing) when paired with the \(+3\) oxidation state of Cobalt (\(Co^{3+}\)). Thus, \([Co(C_2O_4)_3]^{3-}\) and \([Co(H_2O)_6]^{3+}\) are famously diamagnetic.

Step 1: Understanding the Concept:

This problem maps a three-step organic synthesis. We need to identify standard name reactions to deduce the structures of intermediates 'x' and 'y', perform a cross-aldol condensation to find 'z', and finally count the delocalized \(\pi\) electrons in the structure of 'z'.


Step 2: Key Formula or Approach:

1. Reaction 1: Gattermann-Koch reaction (formylation of benzene).

2. Reaction 2: Friedel-Crafts acylation of benzene.

3. Reaction 3: Claisen-Schmidt condensation (a type of cross-aldol condensation between an aldehyde lacking \(\alpha\)-hydrogens and a ketone with \(\alpha\)-hydrogens).

4. Each double bond (\(\pi\) bond) contains 2 \(\pi\) electrons. Benzene ring = 3 \(\pi\) bonds = 6 \(\pi\) electrons.


Step 3: Detailed Explanation:

Finding 'x':

Benzene + \(CO\) + \(HCl\) \(\xrightarrow{CuCl / AlCl_3}\) ?

This is the Gattermann-Koch reaction. The electrophile is the formyl cation (\(HC^+=O\)).

The product 'x' is Benzaldehyde (\(C_6H_5-CHO\)).


Finding 'y':

Benzene + Ethanoyl chloride (\(CH_3COCl\)) \(\xrightarrow{anhyd. AlCl_3}\) ?

This is the Friedel-Crafts acylation. The electrophile is the acylium ion (\(CH_3C^+=O\)).

The product 'y' is Acetophenone (\(C_6H_5-CO-CH_3\)).


Finding 'z':

Heating 'x' (Benzaldehyde) and 'y' (Acetophenone) with alkali (e.g., dilute \(NaOH\)):

Benzaldehyde has NO \(\alpha\)-hydrogens. Acetophenone has three \(\alpha\)-hydrogens.

The base abstracts an \(\alpha\)-proton from acetophenone to form an enolate. This enolate attacks the highly electrophilic carbonyl carbon of benzaldehyde. Heating causes immediate dehydration of the resulting aldol.
\(C_6H_5-CHO + CH_3-CO-C_6H_5 \xrightarrow{OH^-, \Delta} C_6H_5-CH=CH-CO-C_6H_5 + H_2O\).

The major product 'z' is Chalcone (1,3-diphenylprop-2-en-1-one).


Counting \(\pi\) electrons in 'z':

Structure of chalcone: \(Ph - CH = CH - CO - Ph\).

Let's count the distinct \(\pi\) bonds in this conjugated system:

- First Phenyl ring (\(Ph\)): 3 alternating double bonds = 6 \(\pi\) electrons.

- Second Phenyl ring (\(Ph\)): 3 alternating double bonds = 6 \(\pi\) electrons.

- Alkene double bond (\(C=C\)): 1 double bond = 2 \(\pi\) electrons.

- Carbonyl double bond (\(C=O\)): 1 double bond = 2 \(\pi\) electrons.

Total number of \(\pi\) electrons = \(6 + 6 + 2 + 2 = 16\).


Step 4: Final Answer:

The total number of \(\pi\) electrons in z is 16.
Quick Tip: When counting \(\pi\) electrons for a large molecule, simply count the total number of drawn double bonds and multiply by 2. (8 double bonds \(\times 2 = 16\) electrons). Don't accidentally double-count by counting rings and then re-counting the bonds inside them.

Step 1: Understanding the Concept:

The energy of electromagnetic radiation is inversely proportional to its wavelength, according to Planck's quantum theory. We must establish the ratio of energies using this inverse relationship.


Step 2: Key Formula or Approach:

Planck's equation for the energy of a photon:
\(E = \frac{hc}{\lambda}\)

where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.

From this, \(E \propto \frac{1}{\lambda}\), which implies \(\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}\).


Step 3: Detailed Explanation:

Given wavelengths:
\(\lambda_1 = 2000 \AA\)
\(\lambda_2 = 6000 \AA\)

Using the inverse proportionality relationship derived from Planck's equation:
\(\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}\)

Substitute the given values into the equation:
\(\frac{E_1}{E_2} = \frac{6000 \AA}{2000 \AA}\)
\(\frac{E_1}{E_2} = \frac{6000}{2000} = 3\).

The exact ratio is 3, which is an integer.


Step 4: Final Answer:

The ratio of the energies is 3.
Quick Tip: For ratio questions involving purely proportional or inversely proportional quantities, completely ignore all the constant values (like \(h\), \(c\), or unit conversions from Angstroms to meters) as they will perfectly cancel out anyway.

Step 1: Understanding the Concept:

The standard enthalpy of reaction can be calculated using the standard enthalpies of formation of the reactants and products. It is the difference between the sum of enthalpies of products and the sum of enthalpies of reactants.


Step 2: Key Formula or Approach:

Standard enthalpy of reaction:
\(\Delta_r H^\ominus = \sum \left( n \times \Delta_f H^\ominus \right)_{products} - \sum \left( m \times \Delta_f H^\ominus \right)_{reactants}\)

where \(n\) and \(m\) are the stoichiometric coefficients from the balanced chemical equation.

Remember that the standard enthalpy of formation for any element in its standard reference state (like \(O_2(g)\)) is exactly zero.


Step 3: Detailed Explanation:

The balanced chemical equation is:
\(2H_2S(g) + 3O_2(g) \to 2H_2O(l) + 2SO_2(g)\)

Write the enthalpy change equation using the coefficients:
\(\Delta_r H^\ominus = [2 \times \Delta_f H^\ominus(H_2O) + 2 \times \Delta_f H^\ominus(SO_2)] - [2 \times \Delta_f H^\ominus(H_2S) + 3 \times \Delta_f H^\ominus(O_2)]\)

Substitute the given values into the equation. Note that \(\Delta_f H^\ominus(O_2(g)) = 0\).
\(\Delta_r H^\ominus = [2(-286.0) + 2(-297.0)] - [2(-20.1) + 3(0)]\)

Calculate the products sum:
\(= [-572.0 - 594.0]\)
\(= -1166.0 kJ/mol\)

Calculate the reactants sum:
\(= [-40.2]\)

Now combine them:
\(\Delta_r H^\ominus = -1166.0 - (-40.2)\)
\(\Delta_r H^\ominus = -1166.0 + 40.2\)
\(\Delta_r H^\ominus = -1125.8 kJ/mol\)

The problem specifically asks for the magnitude of the enthalpy change to the nearest integer.

Magnitude = \(|-1125.8| = 1125.8\)

Rounding 1125.8 to the nearest integer gives 1126.


Step 4: Final Answer:

The magnitude of enthalpy change is 1126.
Quick Tip: Always read the fine print in physical chemistry numerical questions. Asking for "magnitude" means dropping the negative sign, and "nearest integer" requires standard rounding rules (\(\geq .5\) rounds up). Neglecting either costs easy marks.

Step 1: Understanding the Concept:

We are dealing with simultaneous chemical equilibria in a single closed vessel. Since all reactants and products are in the same container, any gaseous species present in multiple equilibrium reactions (like \(CO_2\)) must exert the exact same partial pressure in all equilibrium constant expressions simultaneously.


Step 2: Key Formula or Approach:

Write the equilibrium constant expression (\(K_p\)) for each reaction.

The partial pressure of pure solid and liquid phases is taken as unity (1). Thus, they are excluded from the \(K_p\) expression.
\(K_{p1} = P_{CO_2}\)
\(K_{p2} = \frac{(P_{CO})^2}{P_{CO_2}}\)


Step 3: Detailed Explanation:

Let's analyze the first equilibrium reaction:

Reaction 1: \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\)

The equilibrium constant is given by:
\(K_{p1} = P_{CO_2}\)

We are given \(K_{p1} = 0.08 atm\).

Because both \(CaCO_3\) and \(CaO\) are solids, as long as they are present in the vessel, the partial pressure of \(CO_2\) is fixed exclusively by this equilibrium at temperature \(T\).

Therefore, \(P_{CO_2} = 0.08 atm\).


Now let's analyze the second equilibrium reaction:

Reaction 2: \(C(s) + CO_2(g) \rightleftharpoons 2CO(g)\)

The equilibrium constant is given by:
\(K_{p2} = \frac{(P_{CO})^2}{P_{CO_2}}\)

We are given \(K_{p2} = 2 atm\).

Since this is occurring in the same vessel, the \(CO_2\) partial pressure must be the same \(0.08 atm\) we just established. Substitute this into the second equation:
\(2 = \frac{(P_{CO})^2}{0.08}\)

Multiply both sides by 0.08 to solve for \((P_{CO})^2\):
\((P_{CO})^2 = 2 \times 0.08 = 0.16 atm^2\)

Take the square root of both sides to find the partial pressure of CO:
\(P_{CO} = \sqrt{0.16} = 0.4 atm\)


The problem asks for the answer in the specific format of "\(\_\_\_\_\_ \times 10^{-1} atm\)".

Rewrite \(0.4\) in this scientific notation format:
\(0.4 = 4 \times 10^{-1} atm\).

Thus, the value to fill in the blank is 4.


Step 4: Final Answer:

The partial pressure of CO is 4.
Quick Tip: In simultaneous equilibria problems, find the "linker" species that appears in both equations (here, \(CO_2\)). If one equation consists only of solids and a single gas, that equation single-handedly locks the partial pressure of that gas for the entire system!