Let’s examine the structures of the compounds:

1. Chloral has 3 chlorine atoms.

2. Gammaxene has 6 chlorine atoms.

3. Chloropicrin has 3 chlorine atoms.

4. Freon-12 has 2 chlorine atoms.

Thus, Gammaxene has the maximum number of chlorine atoms in its structure. Quick Tip: When comparing halogenated organic compounds, the number of chlorine atoms is one of the key factors to look at in the structure.

The suitable indicators for different types of acid-base titrations are:

- Methyl orange is generally used for titrations involving strong acid vs weak base because it changes color in the acidic pH range (red at pH < 3.4 and yellow at pH > 4.4).

- Phenolphthalein is suitable for titrations involving weak acids vs strong bases as it changes color in the basic pH range (colorless below pH 4.8 and pink above pH 6.4).

- The statement that methyl orange may be used for weak acid vs weak base titration is incorrect because it is not suitable for that combination. For weak acid vs weak base titrations, neutral red or bromothymol blue would be more appropriate.

Thus, the incorrect statement is option (1). Quick Tip: Always choose the indicator based on the pH range at the equivalence point of the titration.

- Veronal is a neurological medicine, and Prontosil is an antibiotic. Both are not used as disinfectants.

- Chloroxylenol, Bithional, and Terpineol are commonly used as disinfectants.

Thus, the correct answer is option (1). Quick Tip: When selecting compounds used as disinfectants, ensure to focus on their role in microbial control rather than their medical or therapeutic uses.

When the given hydrocarbon ‘X’ undergoes catalytic hydrogenation, two moles of \( H_2 \) are added, suggesting the presence of two double bonds.
Upon ozonolysis, two moles of methane dicarbaldehyde are produced, indicating that the molecule is a conjugated diene.
Thus, the structure of the hydrocarbon is cyclohexa-1, 4-diene. Quick Tip: Ozonolysis of conjugated dienes typically produces aldehydes or ketones depending on the position of the double bonds.

Cyclohexylamine (C6H11NH2) reacts with nitrous acid (HNO2) to form a diazonium salt (P).

On treating this with PCC (Pyridinium chlorochromate), oxidation occurs to form a ketone (Q).

When (Q) is heated with dilute NaOH, it undergoes a condensation reaction forming a cyclohexene derivative (R).

Thus, the final product (R) is the structure shown in option (2). Quick Tip: The reaction with PCC typically involves oxidation to a ketone, and heating with NaOH often leads to a condensation reaction.

In the Borax Bead Test, heating a borax bead dipped in cupric sulphate in a non-luminous flame results in the formation of cupric metaborate, which gives a blue-green colour. The formation of copper(I) metaborate is not the correct explanation for the green colour observed; instead, copper(II) metaborate forms.

Thus, Statement I is true, but Statement II is false.

The correct answer is option (3). Quick Tip: In the Borax Bead Test, the formation of cupric metaborate gives the bead its blue-green colour in the reducing flame, not copper(I) metaborate.

- (A) The elevation in boiling point is dependent on the number of particles in the solution. \( NaCl \) dissociates into 2 ions, while urea doesn’t dissociate. Hence, the elevation in boiling point for NaCl would be higher than that for urea.

- (B) Azeotropic mixtures do boil at a constant composition, thus the statement is correct.

- (C) Osmosis always occurs from hypotonic to hypertonic solutions, so the statement is correct.

- (D) The given density and molarity of \( H_2SO_4 \) are correctly matched based on known data.

- (E) Adding KI to silver nitrate solution results in the formation of a positively charged sol (due to the formation of AgI), so this statement is incorrect.

Thus, the correct answer is option (4). Quick Tip: In colligative properties, the presence of ions and solutes plays a significant role in determining changes in properties like boiling point elevation and freezing point depression.

The compound \( A \) is a sugar derivative, and based on the given reactions:

- The reaction with \( Ac_2O \) suggests the presence of hydroxyl groups.

- Oxidation with \( Br_2 - H_2O \) gives an acid, indicating that a terminal hydroxyl group is present.

- Reduction with HI gives isopentane, suggesting a pentose structure.

The correct structure for \( A \) is option (1). Quick Tip: The oxidation and reduction of sugars can help identify functional groups and their positions in the molecule.

As per Hund’s rule, the energy of an orbital is given by the \( n + l \) value. If the value of \( n + l \) remains the same, energy is given by \( n \) only.

- \( (A) n = 3, l = 0, m = 0 \): \( n + l = 3 \)

- \( (B) n = 4, l = 1, m = 0 \): \( n + l = 5 \)

- \( (C) n = 3, l = 1, m = 0 \): \( n + l = 4 \)

- \( (D) n = 3, l = 2, m = 1 \): \( n + l = 5 \)

Thus, the order is \( D > B > C > A \). Quick Tip: When comparing orbitals, use Hund’s rule and the \( n + l \) value to determine energy levels.

Extent of back bonding reduces down the group leading to more Lewis acidic strength.

For example, \( BF_3 \) has the most extent of back bonding due to the \( 2p - 2p \) interaction, whereas \( BCl_3 \) and \( BBr_3 \) show less back bonding. Therefore, the Lewis acid strength increases from \( BF_3 \) to \( BCl_3 \), \( BBr_3 \), and finally \( B1_3 \).

Thus, the correct order is \( B1_3 > BBr_3 > BCl_3 > BF_3 \). Quick Tip: The strength of Lewis acids can be compared based on the extent of back bonding and the electronegativity of halides.

- Physiosorption involves weak van der Waals forces, and is typically associated with single layer adsorption.

- Chemisorption is associated with stronger bonds and involves an energy range of 20-40 kJ mol\(^{-1}\).

- The reaction \( N_2 + 3H_2 \xrightarrow{Fe} 2NH_3 \) is an example of heterogeneous catalysis in which adsorption plays a key role.

- Chromatography is an analytical technique based on adsorption.

Thus, the correct matching is: A - II, B - I, C - IV, D - III. Quick Tip: In adsorption processes, physisorption involves weak forces while chemisorption involves stronger covalent bonds.

- Assertion (A) is incorrect. The first ionization enthalpy of 3d series elements is actually less than that of group 2 elements due to the effective shielding of the nucleus by the d-electrons.

- Reason (R) is true. In the 3d series, the successive filling of d-orbitals leads to increased shielding, affecting ionization energies.

Thus, (A) is false but (R) is true. Quick Tip: In transition metals, the ionization energies are influenced by the filling of d-orbitals, leading to variations from what is observed in s-block elements.

Calcium (\( Ca \)) plays a vital role in neuromuscular function, interneuronal transmission, and other biological processes like cell signaling and muscle contraction. It is involved in the release of neurotransmitters and the contraction of muscle fibers. Quick Tip: Calcium is crucial for nerve function and muscle contraction due to its ability to influence various signaling pathways.

- Statement I is correct: \( H_2O_2 \) is used in the synthesis of cephalosporin, a class of antibiotics.

- Statement II is also correct: \( H_2O_2 \) is used to restore aerobic conditions to sewage wastes by supplying oxygen and aiding in the breakdown of organic material.

Thus, both statements are true. Quick Tip: Hydrogen peroxide is widely used in organic synthesis and environmental applications, including sewage treatment.

Rainwater is naturally slightly acidic due to the dissolution of carbon dioxide (\( CO_2 \)) in water, forming carbonic acid (\( H_2CO_3 \)):
\[ CO_2 + H_2O \rightarrow H_2CO_3 \]

This weak acid dissociates to give a small concentration of \( H^+ \) ions, lowering the pH of the rainwater to about 5.6.

The other options are related to pollutants like NO\(_2\), SO\(_2\), and N\(_2\)O\(_5\), which are not the primary cause of the natural acidity of rainwater. Quick Tip: The acidity of rainwater is primarily due to carbon dioxide, not industrial pollutants.

Let the molecular formula of the hydrocarbon be \( C_xH_y \). The balanced combustion reaction is: \[ C_xH_y + \left( \frac{x + \frac{y}{4}}{2} \right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]
We are given that 11 equivalents of oxygen are required, and 4 equivalents of water are produced. Using stoichiometry: \[ \frac{y}{2} = 4 \quad \Rightarrow \quad y = 8 \]
Substituting in the oxygen requirement: \[ \frac{x + \frac{8}{4}}{2} = 11 \quad \Rightarrow \quad x + 2 = 22 \quad \Rightarrow \quad x = 20 \]
Thus, the molecular formula is \( C_9H_8 \). Quick Tip: When solving combustion problems, remember that the number of moles of oxygen is related to the number of moles of carbon and hydrogen in the molecule.

The compound \( C_4H_{11}N \) reacts with \( HNO_2 \) to release \( N_2 \), which indicates that it is a primary amine. After reacting with Hinsberg reagent (benzenesulfonyl chloride, \( PhSO_2Cl \)), it forms a compound soluble in KOH, which suggests the amine is primary. The structure that fits all these reactions is option (4), ethylamine. Quick Tip: When identifying amines, consider their reaction with reagents like \( HNO_2 \) and Hinsberg reagent, which help differentiate primary, secondary, and tertiary amines.

- Option (1): The zone refining method is used to purify metals like boron and indium, which have low melting points.

- Option (2): Van Arkel method is used to purify titanium, not tungsten. Tungsten is purified by other methods like hydrogen reduction.

- Option (3): Cast iron is indeed obtained by melting pig iron with scrap iron and coke in a blast furnace.

- Option (4): Malleable iron is prepared from cast iron by heating it in a reverberatory furnace with the proper oxidation of impurities.

Thus, the incorrect statement is option (2). Quick Tip: Zone refining is widely used for the purification of semiconductors and metals like indium, boron, and germanium.

The electron configurations of the elements are as follows:

1. \( ^{62}Sm: 4f^6 6s^2 \)

2. \( ^{63}Eu: 4f^7 5d^1 6s^2 \)

3. \( ^{65}Tb: 4f^9 6s^2 \)

4. \( ^{64}Gd: 4f^7 5d^1 6s^2 \)

5. \( ^{61}Pm: 4f^5 6s^2 \)

Thus, the elements with half-filled \( f \)-orbitals in their ground state are Eu (Europium) and Gd (Gadolinium), corresponding to option (1). Quick Tip: Half-filled orbitals, such as in \( 4f^7 \), lead to more stable electronic configurations due to the symmetry and exchange energy.

In Dumas' method, the nitrogen-containing organic compound is heated with copper oxide in an atmosphere of CO\(_2\). The resulting reaction produces nitrogen gas in addition to CO\(_2\) and H\(_2\)O:
\[ C_xH_yN_z + \left( 2x + \frac{y}{2} \right) CuO \rightarrow x CO_2 + \frac{y}{2} H_2O + \frac{z}{2} N_2 + \left( 2x + \frac{y}{2} \right) Cu \]

Afterward, any nitrogen oxides formed are reduced to nitrogen gas by passing the gaseous mixture over heated copper gauze. Quick Tip: Dumas' method is commonly used for the estimation of nitrogen in organic compounds by releasing nitrogen gas through heating with copper oxide.

For the complex \( [ Ti^{3+} (H_2O)_6 ]^{3+} \), the electronic configuration is \( Ti^{3+}: 3d^1 \). The CFSE (Crystal Field Stabilization Energy) is given as -96.0 kJ/mol.

The formula for the CFSE is:
\[ CFSE = -0.4 \Delta_0 \]

Where \( \Delta_0 \) is the crystal field splitting energy. Using the given data:
\[ CFSE = -96 \times 10^3 \, J/mol \]

Now, solving for \( \Delta_0 \):
\[ \Delta_0 = \frac{96 \times 10^3}{6 \times 10^{23}} \quad \Rightarrow \quad \Delta_0 = 1.6 \times 10^{-19} \, J \]

Now, using the formula for the wavelength of absorption:
\[ \frac{hc}{\lambda} = \Delta_0 \]

Substitute the known values:
\[ \frac{6.4 \times 10^{-34} \times 3.0 \times 10^8}{\lambda} = 1.6 \times 10^{-19} \]

Solving for \( \lambda \):
\[ \lambda = \frac{6.4 \times 10^{-34} \times 3.0 \times 10^8}{1.6 \times 10^{-19}} = 480 \times 10^{-9} \, m \]

Thus, the wavelength is \( 480 \, nm \). Quick Tip: The relationship between the energy of absorbed light and the wavelength is given by \( \frac{hc}{\lambda} = \Delta_0 \), where \( \Delta_0 \) is the crystal field splitting energy.

Let's examine the shape of each species:





\(BCI_2^-\) : Linear shape (sp hybridization)
\(BeCl_2\) : Linear shape (sp hybridization)
\(I_3^-\) : Linear shape (sp hybridization)
\(XeF_2\) : Linear shape (sp hybridization)
\(I_5^-\) : V-shape (not linear)
\(CO_2\) : Linear shape (sp hybridization)
\(SO_2\) : V-shape (not linear)
\(C_3O_2\) (O=C=C=C=O) : Linear shape (sp hybridization)


The species with a linear shape are:
\[ BCI_2^-, BeCl_2, I_3^-, XeF_2, CO_2, C_3O_2 \]

Thus, the number of species with linear shape is 5. Quick Tip: In determining the shape of a molecule, consider the number of bonding pairs and lone pairs on the central atom, and use VSEPR theory.

The molar conductivity \( \Lambda_m \) is given by the formula: \[ \Lambda_m = \frac{k \times 1000}{M} \]
where \( k \) is the resistivity and \( M \) is the molarity of the solution.

Also, we have the formula: \[ \Lambda_m = \frac{1000}{\rho} \quad where \quad \rho = resistivity \]

Now, using the given values: \[ \Lambda_m = \frac{1}{\rho} \times 1000 = \frac{1}{5 \times 10^{-3}} \times 1000 \]

Substitute the value of molarity \( M = 0.8 \): \[ \Lambda_m = \frac{1000}{5 \times 10^{-3}} \times 0.8 = 25 \times 10^4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

Thus, the molar conductivity is \( 25 \times 10^4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \). Quick Tip: The molar conductivity is calculated using the relationship between resistivity and molarity. Keep in mind the dimensions of the quantities involved.

The dissociation of silver chloride in water is: \[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \]

The solubility \( S \) of AgCl is given by: \[ S = 1.434 \times 10^{-3} \, g L^{-1} \]

The molar solubility of AgCl, \( S \), can be calculated as: \[ S = \frac{1.434 \times 10^{-3}}{143.4 \times 10^{-3}} \, mol L^{-1} = 1 \times 10^{-5} \, mol L^{-1} \]

The solubility product \( K_{sp} \) is: \[ K_{sp} = S^2 = (1 \times 10^{-5})^2 = 10^{-10} \]

Thus, \( -\log K_{sp} = 10 \). Quick Tip: The solubility product constant \( K_{sp} \) can be calculated from the square of the molar solubility for simple salts like AgCl.

Let the amount of metal in the \( +2 \) oxidation state be \( x \), and the amount in the \( +3 \) oxidation state be \( 0.83 - x \).

From the charge balance equation: \[ 2x + 3(0.83 - x) = 0.83 \]

Simplifying the equation: \[ 2x + 2.49 - 3x = 0.83 \quad \Rightarrow \quad -x + 2.49 = 0.83 \quad \Rightarrow \quad -x = 0.83 - 2.49 = -1.66 \] \[ x = 0.49 \]

Thus, the percentage of metal ions in the \( +2 \) oxidation state is: \[ \frac{0.49}{0.83} \times 100 = 59% \] Quick Tip: In stoichiometry problems, carefully balance the charges and mole ratios to determine the proportions of each oxidation state.

From the equation: \[ 2C(s) + O_2(g) \rightarrow 2CO(g) \]

Limiting reagent is carbon, as 12 g of carbon is burnt. 1 mole of carbon produces one mole of CO. Hence, at STP, 1 mole of CO occupies 22.7 L.

We have: \[ Moles of carbon = \frac{12}{12} = 1 \, mol \]
Thus, the volume of CO produced at STP is: \[ Volume of CO = 1 \times 22.7 = 22.7 \, L \]
Therefore, the volume of carbon monoxide produced is \( 2.27 \times 10^1 \) L at STP. Quick Tip: At STP, one mole of an ideal gas occupies 22.7 L. Use this fact for calculations involving gases at STP.

K and Rb form stable super oxides but Cs has ionization enthalpy less than 400 kJ/mol.
Thus, the correct answer is 2, as K and Rb are the metals that meet the criteria. Quick Tip: Ionization enthalpy and the formation of super oxides are important in determining the chemical behavior of alkali metals.

The enthalpy change \( \Delta H \) for the reaction is calculated using the following formula: \[ \Delta H = \sum H_p - \sum H_R \]
Where \( \sum H_p \) is the sum of enthalpies of the products and \( \sum H_R \) is the sum of enthalpies of the reactants.

From the question: \[ \sum H_R = Enthalpy of reactants = (-394 + 4 \times -92) = -1056 \, kJ/mol \] \[ \sum H_p = Enthalpy of products = (-105 + (2 \times -242)) = -589 \, kJ/mol \]
Therefore: \[ \Delta H = (-589) - (-1056) = -173 \, kJ/mol \]

Thus, the magnitude of enthalpy of the reaction is \( 173 \, kJ/mol \). Quick Tip: To calculate enthalpy change for a reaction, use the formula \( \Delta H = \sum H_p - \sum H_R \), where \( H_p \) and \( H_R \) are the enthalpies of products and reactants, respectively.

The halform test is positive for compounds that contain a methyl group directly attached to a carbonyl group (i.e., the CH3-CO- functional group) or other reactive groups like CH3-CO-H.

Looking at the structures:

- The molecules \( C = O - CH_3 \) and \( OH - C - CH_3 \) give positive halform test because they both have the required functional groups for the reaction.

- The other molecules do not meet the criteria.

Thus, the number of molecules giving positive halform test is 3. Quick Tip: The halform test is used to identify compounds containing a methyl ketone group (\( -COCH_3 \)) or a similar structure.

The integrated rate equation for a first order reaction is: \[ \ln \left( \frac{C_0}{C} \right) = kt \]
Where \( C_0 \) is the initial concentration, \( C \) is the concentration at time \( t \), \( k \) is the rate constant, and \( t \) is the time.

For the reaction to reduce to \( \frac{1}{32} \) of its original concentration, we have: \[ \frac{C}{C_0} = \frac{1}{32} \]

Taking the natural logarithm: \[ \ln \left( \frac{C_0}{C} \right) = \ln 32 = 5 \ln 2 = 5 \times 0.693 = 3.465 \]

Thus, the time \( t \) is: \[ t = \frac{3.465}{k} = \frac{3.465}{20} = 0.17325 \, min \]

Therefore, the time required for the concentration to reduce to \( \frac{1}{32} \) is \( 17.325 \times 10^{-2} \) min. Quick Tip: For a first-order reaction, the time required to reach a certain concentration can be calculated using the formula \( t = \frac{\ln \left( \frac{C_0}{C} \right)}{k} \).