SECTION – A
51. A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is ___ .
[Given: molar mass of aluminium and chlorine are 27g mol$^{–1}$ and 35.5g mol$^{–1}$ respectively, Faraday constant = 96500 C mol$^{–1}$].
(1) 1.660 g
(2) 1.007 g
(3) 0.336 g
(4) 0.441 g
Answer: (3)
Solution:
1. Identify the Given Parameters
Current (I): $2\text{ A}$
Time (t): $30\text{ minutes} = 30 \times 60 = 1800\text{ seconds}$
Molar mass of Aluminium (M): $27\text{ g mol}^{-1}$
Faraday’s constant (F): $96500\text{ C mol}^{-1}$
2. Determine the Valency Factor (n)
Aluminium chloride ($AlCl_3$) dissociates in solution. The reduction reaction at the cathode for aluminium is:
$$Al^{3+} + 3e^- \rightarrow Al(s)$$
Here, $3$ moles of electrons are required to deposit $1$ mole of aluminium. Therefore, the valency factor ($n$) is 3.
3. Calculate Total Charge passed (Q)
The total quantity of electricity passed is given by:
$$Q = I \times t$$
$$Q = 2\text{ A} \times 1800\text{ s} = 3600\text{ C}$$
4. Calculate the Mass Deposited (w)
The formula for the mass of a substance deposited is:
$$w = \frac{Q \times M}{n \times F}$$
Substituting the values:
$$w = \frac{3600 \times 27}{3 \times 96500}$$
$$w = \frac{3600 \times 9}{96500}$$
(since $27/3 = 9$)
$$w = \frac{32400}{96500}$$
$$w \approx 0.3357\text{ g}$$
Rounding to three decimal places, the amount of aluminium deposited is 0.336 g, which corresponds to option (3).
52. Which of the following statement is not true for radioactive decay?
(1) Amount of radioactive substance remained after three half lives is $\frac{1}{8}$th of original amount.
(2) Decay constant does not depend upon temperature.
(3) Decay constant increases with increase in temperature.
(4) Half life is ln 2 times of $\frac{1}(\text{rate constant}}$
Answer: (3)
Statement 1: Amount of substance remained after three half-lives is 1/8th of original amount.
After each half-life ($T_{1/2}$), the amount of a radioactive substance is reduced by half.
- After 1 half-life: $1/2$ remains.
- After 2 half-lives: $1/2 \times 1/2 = 1/4$ remains.
- After 3 half-lives: $1/2 \times 1/2 \times 1/2 = 1/8$ remains.
Mathematically, the fraction remaining after $n$ half-lives is $(1/2)^n$. For $n=3$, this is $(1/2)^3 = 1/8$. This statement is True.
Statement 2 & 3: Decay constant and Temperature.
Radioactive decay is a nuclear phenomenon. Unlike chemical reactions (which involve orbital electrons), nuclear decay is not affected by external physical conditions such as temperature, pressure, or the chemical state of the atom.
The decay constant ($\mathbf{\lambda}$) is a characteristic property of a specific isotope.
Because it is a nuclear process, the decay constant does not depend on temperature.
Therefore, Statement 2 is True and Statement 3 is False.
Statement 4: Half-life and Rate Constant.
The relationship between half-life ($T_{1/2}$) and the decay constant (also called the rate constant, $\lambda$) is derived from the first-order kinetics of decay:
$$N = N_0 e^{-\lambda t}$$
Setting $N = N_0/2$ gives:
$$T_{1/2} = \frac{\ln 2}{\lambda}$$
This means the half-life is indeed $\ln 2$ times $1/\lambda$. This statement is True.
Thus statement 3 is not true for radioactive decay.
53. How many different stereoisomers are possible for the given molecule?
$$\text{CH}_3 – \underset{|\quad}{\text{CH}} – \text{CH} = \text{CH} – \text{CH}_3$$
$$\text{OH}$$
(1) 3
(2) 1
(3) 2
(4) 4
Answer: (4)
Solution:
1. Identify Stereogenic Centers
Chiral Center: The second carbon ($C_2$) is bonded to four different groups: $-H$, $-OH$, $-CH_3$, and the alkenyl group. This center can exist in two configurations: (R) and (S).
Geometrical Center (Double Bond): The double bond between $C_3$ and $C_4$ has different groups on both ends. This allows for geometrical isomerism: Cis (Z) and Trans (E).
2. Calculating the Total Isomers
Since the molecule is unsymmetrical (the two ends are $-CH(OH)CH_3$ and $-CH_3$), we use the formula $2^n$, where $n$ is the number of stereogenic units.
$n = 1$ (chiral center) $+ 1$ (double bond) $= 2$.
Total stereoisomers $= 2^2 = 4$.
3. Listing the Isomers
The four distinct configurations are:
(R) – Cis
(R) – Trans
(S) – Cis
(S) – Trans
These four molecules are stereoisomers of each other (a mix of enantiomers and diastereomers).
54. Which of the following electronegativity order is incorrect?
(1) Al < Mg < B < N
(2) Al < Si < C < N
(3) Mg < Be < B < N
(4) S < Cl < O < F
Answer: (1)
55. Lanthanoid ions with $4f^7$ configuration are:
(1) $Eu^{2+}$
(2) $Gd^{3+}$
(3) $Eu^{3+}$
(4) $Tb^{3+}$
(E) $Sm^{2+}$
Choose the correct answer from the options given below:
(1) (1) and (2) only
(2) (1) and (4) only
(3) (2) and (E) only
(4) (2) and (3) only
Answer: (1)
Solution:
To identify which Lanthanoid ions have a $4f^7$ configuration, we must look at the electronic configurations of the neutral atoms and then remove electrons to form the specified ions. For Lanthanoids, electrons are removed first from the $6s$ orbital, then from the $5d$ (if present), and finally from the $4f$ orbital.
Analysis of the Given Ions
(A) $\mathbf{Eu^{2+}}$ (Europium):
- Neutral Europium ($Z=63$): $[Xe] 4f^7 6s^2$
- To form $Eu^{2+}$, we remove two electrons from the $6s$ orbital.
Resulting configuration: $[Xe] 4f^7$ (Matches)
(B) $\mathbf{Gd^{3+}}$ (Gadolinium):
- Neutral Gadolinium ($Z=64$): $[Xe] 4f^7 5d^1 6s^2$
- To form $Gd^{3+}$, we remove two electrons from the $6s$ orbital and one from the $5d$ orbital.
Resulting configuration: $[Xe] 4f^7$ (Matches)
(C) $\mathbf{Eu^{3+}}$ (Europium):
- Neutral Europium ($Z=63$): $[Xe] 4f^7 6s^2$
- To form $Eu^{3+}$, we remove two electrons from $6s$ and one from $4f$.
Resulting configuration: $[Xe] 4f^6$ (Does not match)
(D) $Tb^{3+}$ (Terbium):
- Neutral Terbium ($Z=65$): $[Xe] 4f^9 6s^2$
- To form $Tb^{3+}$, we remove two electrons from $6s$ and one from $4f$.
Resulting configuration: $[Xe] 4f^8$ (Does not match)
(E) $\mathbf{Sm^{2+}}$ (Samarium):
- Neutral Samarium ($Z=62$): $[Xe] 4f^6 6s^2$
- To form $Sm^{2+}$, we remove two electrons from $6s$.
Resulting configuration: $[Xe] 4f^6$ (Does not match)
The ions with the $4f^7$ configuration (a stable, half-filled subshell) are $\mathbf{Eu^{2+}}$ and $\mathbf{Gd^{3+}}$.
This corresponds to (A) and (B) only, which is option (1).
56. Match List-I with List-II
| List-I | List-II |
| (1) $\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ < \text{F}^-$ | (I) Ionisation Enthalpy |
| (2) $\text{B} < \text{C} < \text{O} < \text{N}$ | (II) Metallic character |
| (3) $\text{B} < \text{Al} < \text{Mg} < \text{K}$ | (III) Electronegativity |
| (4) $\text{Si} < \text{P} < \text{S} < \text{Cl}$ | (IV) Ionic radii |
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-III, D-II
(2) A-II, B-III, C-IV, D-I
(3) A-IV, B-I, C-II, D-III
(4) A-III, B-IV, C-II, D-I
Answer: (3)
Solution:
To match List-I with List-II, we need to apply the periodic trends for ionic radius, ionization enthalpy, metallic character, and electronegativity.
(A) $\mathbf{Al^{3+} < Mg^{2+} < Na^+ < F^-}$ $\rightarrow$ (IV) Ionic Radii
All four species are isoelectronic, meaning they all have the same number of electrons (10 electrons, like Neon).
For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases.
$Al$ (atomic number 13) has the strongest pull on those 10 electrons, making it the smallest. $F$ (atomic number 9) has the weakest pull, making it the largest.
(B) $\mathbf{B < C < O < N}$ $\rightarrow$ (I) Ionisation Enthalpy
Ionization enthalpy generally increases from left to right across a period due to increasing nuclear charge. However, there is an exception between Nitrogen (N) and Oxygen (O).
Nitrogen ($1s^2 2s^2 2p^3$) has a half-filled p-orbital, which is exceptionally stable.
Oxygen ($1s^2 2s^2 2p^4$) has one electron paired in a $p$-orbital, which experiences repulsion and is easier to remove.
Thus, the order is $B < C < O < N$.
(C) $\mathbf{B < Al < Mg < K}$ $\rightarrow$ (II) Metallic Character
Metallic character refers to the ease with which an atom loses electrons.
It increases down a group (as atoms get larger) and decreases across a period (as nuclear charge increases).
$K$ (Group 1) is the most metallic, followed by $Mg$ (Group 2), then $Al$ (Group 13). $B$ is a metalloid/non-metal at the top of Group 13, making it the least metallic in this list.
(D) $\mathbf{Si < P < S < Cl}$ $\rightarrow$ (III) Electronegativity
Electronegativity is the tendency of an atom to attract a shared pair of electrons.
Across a period from left to right, electronegativity increases because the atomic radius decreases and the effective nuclear charge increases.
These elements are all in Period 3. Moving from Silicon to Chlorine, the ability to attract electrons increases steadily.
Final Match Summary
| List-I (Order) | List-II (Property) |
| (A) $Al^{3+} < Mg^{2+} < Na^+ < F^-$ | (IV) Ionic Radii |
| (B) $B < C < O < N$ | (I) Ionisation Enthalpy |
| (C) $B < Al < Mg < K$ | (II) Metallic Character |
| (D) $Si < P < S < Cl$ | (III) Electronegativity |
This corresponds to Option (3).
57. Which of the following acids is a vitamin?
(1) Adipic acid
(2) Aspartic acid
(3) Ascorbic acid
(4) Saccharic acid
Answer: (3)
Solution:
Analysis of Options
- Adipic acid: A dicarboxylic acid primarily used in the production of nylon. It is not a vitamin.
- Aspartic acid: An $\alpha$-amino acid used in the biosynthesis of proteins. While essential for the body, it is classified as a non-essential amino acid, not a vitamin.
- Ascorbic acid: This is the chemical name for Vitamin C. It is a water-soluble vitamin essential for collagen synthesis, immune function, and acting as an antioxidant.
- Saccharic acid: Also known as glucaric acid, it is derived from the oxidation of glucose. It is not a vitamin.
Thus, correct answer is option (3).
58. A liquid when kept inside a thermally insulated closed vessel at $25^{\circ}\text{C}$ was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?
(1) $\Delta U > 0, q = 0, w > 0$
(2) $\Delta U = 0, q = 0, w = 0$
(3) $\Delta U < 0, q = 0, w > 0$
(4) $\Delta U = 0, q < 0, w > 0$
Answer: (1)
Solution:
A liquid is kept inside a thermally insulated closed vessel and is mechanically stirred from the outside. We need to determine the signs for change in internal energy ($\Delta U$), heat ($q$), and work ($w$).
1. Analyzing Heat (q)
The problem states the vessel is thermally insulated. In thermodynamics, an insulated system is an adiabatic system, meaning no heat can be exchanged with the surroundings.
Therefore, q = 0.
2. Analyzing Work (w)
The liquid is “mechanically stirred from outside.” This means the surroundings are doing work on the system (the liquid). According to the sign convention where work done on the system is positive:
Therefore, w > 0.
3. Applying the First Law of Thermodynamics
The First Law is given by:
$$\Delta U = q + w$$
Substituting our known values ($q = 0$ and $w > 0$):
$$\Delta U = 0 + w$$
$$\Delta U = w$$
Since $w$ is positive, the change in internal energy $\Delta U$ must also be positive:
Therefore, $\mathbf{\Delta U > 0}$.
Physical Intuition: When you stir a liquid, the mechanical energy from the stirrer is dissipated into the liquid as friction (viscous drag), which increases the kinetic energy of the molecules. Since the heat cannot escape (insulated), this energy stays in the liquid, raising its internal energy and, consequently, its temperature.
59. Radius of the first excited state of Helium ion is given as:
$a_0 \rightarrow$ radius of first stationary state of hydrogen atom.
(1) $r = \frac{a_0}{2}$
(2) $r = \frac{a_0}{4}$
(3) $r = 4a_0$
(4) $r = 2a_0$
Answer: (4)
Solution:
1. The Bohr Radius Formula
The radius $r$ of the $n^{th}$ orbit for an atom or ion with atomic number $Z$ is given by the formula:
$$r = a_0 \frac{n^2}{Z}$$
Where:
$\mathbf{a_0}$ is the Bohr radius (the radius of the first stationary state of the hydrogen atom, where $n=1$ and $Z=1$).
n is the principal quantum number (the orbit or energy level).
Z is the atomic number (number of protons in the nucleus).
2. Identifying the Variables for Helium Ion ($\mathbf{He^+}$)
Atomic Number (Z): For Helium, $Z = 2$.
Principal Quantum Number ($n$): The question asks for the first excited state.
- Ground state = n = 1
- First excited state = n = 2
3. Calculation
Substitute the values of $n$ and $Z$ into the formula:
$$r = a_0 \frac{(2)^2}{2}$$
$$r = a_0 \frac{4}{2}$$
$\mathbf{r = 2a_0}$
The radius of the first excited state of the Helium ion is $\mathbf{2a_0}$, which corresponds to option (4).
60. Given below are two statements:
Statement I: $\text{CH}_3 – \text{O} – \text{CH}_2 – \text{Cl}$ will undergo $S_N1$ reaction though it is a primary halide.
Statement II:
will not undergo $S_N2$ reaction very easily though it is a primary halide.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are correct.
Answer: (4)
61. Given below are two statements:
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of $H_2# gas.
Statement II: Four g of propyne reacts with $NaNH_2$ to liberate $NH_3$ gas which occupies 224 mL at STP.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is correct but Statement II is incorrect.
(2) Both Statement I and Statement II are incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are correct.
Answer: (1)
Solution:
1. Analysis of Statement I
Statement: One mole of propyne reacts with excess sodium to liberate half a mole of $H_2$ gas.
Concept: Propyne ($CH_3-C \equiv CH$) is a terminal alkyne. The hydrogen atom attached to the triply bonded carbon is acidic. Sodium ($Na$) reacts with acidic hydrogens to form sodium acetylide and release hydrogen gas.
Chemical Equation:
$$CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^-Na^+ + \frac{1}{2}H_2 \uparrow$$
Stoichiometry: 1 mole of propyne contains 1 acidic hydrogen atom. According to the balanced equation, 1 mole of propyne reacts to produce 0.5 moles (half a mole) of $H_2$ gas.
Verdict: Statement I is correct.
2. Analysis of Statement II
Statement: Four $g$ of propyne reacts with $NaNH_2$ to liberate $NH_3$ gas which occupies $224 \text{ mL}$ at STP.
Step 1: Find moles of Propyne.
The molecular formula of propyne is $C_3H_4$.
Molar mass $= (3 \times 12) + (4 \times 1) = 40 \text{ g/mol}$.
$$\text{Moles of Propyne} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mole}$$
Step 2: Reaction with Sodium Amide ($\mathbf{NaNH_2}$).
Terminal alkynes react with strong bases like $NaNH_2$ to release ammonia ($NH_3$).
$$CH_3-C \equiv CH + NaNH_2 \rightarrow CH_3-C \equiv C^-Na^+ + NH_3 \uparrow$$
According to the stoichiometry, 1 mole of propyne liberates 1 mole of $NH_3$.
Therefore, 0.1 mole of propyne will liberate $0.1$ mole of $NH_3$.
Step 3: Calculate volume at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies $22.4 \text{ L}$ (or $22400 \text{ mL}$).
$$\text{Volume of } NH_3 = \text{moles} \times 22400 \text{ mL}$$
$$\text{Volume} = 0.1 \times 22400 \text{ mL} = 2240 \text{ mL}$$
Comparison: The statement claims the gas occupies 224 mL, but our calculation shows it should be 2240 mL.
Verdict: Statement II is incorrect.
Conclusion:
Statement I is correct.
Statement II is incorrect.
The most appropriate answer is option (1).
62. A vessel at $1000 K$ contains $CO_2$ with a pressure of $0.5$ atm. Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8$ atm, then $K_P$ is:
(1) 0.18 atm
(2) 1.8 atm
(3) 0.3 atm
(4) 3 atm
Answer: (2)
Solution:
1. The Chemical Equation
The reaction involves gaseous carbon dioxide reacting with solid graphite to form gaseous carbon monoxide:
$$CO_2(g) + C(s) \rightleftharpoons 2CO(g)$$
2. ICE Table Analysis (Initial, Change, Equilibrium)
We track the pressures of the gases. Note that solid graphite ($C(s)$) is not included in the $K_p$ expression as its activity is unity.
| Species | Initial Pressure (atm) | Change (atm) | Equilibrium Pressure (atm) |
| $CO_2(g)$ | $0.5$ | $-x$ | $0.5 – x$ |
| $CO(g)$ | $0$ | $+2x$ | $2x$ |
3. Solving for x
The problem states that the total pressure ($P_{total}$) at equilibrium is 0.8 atm. According to Dalton’s Law of Partial Pressures:
$$P_{total} = P_{CO_2} + P_{CO}$$
$$0.8 = (0.5 – x) + 2x$$
$$0.8 = 0.5 + x$$
$$x = 0.8 – 0.5 = 0.3\text{ atm}$$
4. Equilibrium Partial Pressures
Now substitute the value of $x$ to find the pressures of the gases:
$P_{CO_2} = 0.5 – 0.3 = \mathbf{0.2\text{ atm}}$
$P_{CO} = 2(0.3) = \mathbf{0.6\text{ atm}}$
5. Calculating $\mathbf{K_p}$
The expression for the equilibrium constant $K_p$ is:
$$K_p = \frac{(P_{CO})^2}{P_{CO_2}}$$
Substituting the values:
$$K_p = \frac{(0.6)^2}{0.2}$$
$$K_p = \frac{0.36}{0.2} = \mathbf{1.8\text{ atm}}$$
The equilibrium constant $K_p$ is 1.8 atm, which corresponds to option (2).
63. The IUPAC name of the following compound is:
Answer: (4)
64. Which of the following electrolyte can be sued to obtain $H_2S_2O_8$ by the process of electrolysis?
(1) Dilute solution of sodium sulphate
(2) Dilute solution of sulphuric acid
(3) Concentrated solution of sulphuric acid
(4) Acidified dilute solution of sodium sulphate.
Answer: (3)
Solution:
1. The Role of Concentration
The outcome of the electrolysis of sulphuric acid ($H_2SO_4$) depends heavily on its concentration and the current density used.
In Dilute $\mathbf{H_2SO_4}$: The concentration of water molecules is high. At the anode, the oxidation of water is preferred because it has a lower discharge potential. This results in the liberation of Oxygen gas ($\mathbf{O_2}$).
$$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$
In Concentrated $\mathbf{H_2SO_4}$ (50% or higher): The concentration of sulphate and bisulphate ions is high. Under high current density, these ions are preferentially oxidized at the anode instead of water.
2. The Chemical Reaction at the Anode
When using a concentrated solution of sulphuric acid, the $HSO_4^-$ (bisulphate) ions undergo dimerization at the anode:
$$2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-$$
In this process, two bisulphate ions lose electrons and join together to form a peroxy linkage ($-O-O-$), creating peroxodisulphuric acid.
3. Reaction at the Cathode
Simultaneously, at the cathode, hydrogen ions are reduced to form hydrogen gas:
$$2H^+ + 2e^- \rightarrow H_2 \uparrow$$
To obtain $H_2S_2O_8$, a concentrated solution of sulphuric acid must be used as the electrolyte to ensure the oxidation of bisulphate ions over water.
The correct answer is option (3).
65. The compounds which give positive Fehling’s test are:
(1)
(2)
(3) $HOCH_2-CO-(CHOH)_3-CH_2-OH$
(4)
(5)
Choose the CORRECT answer from the options given below:
(1) (1), (3) and (4) Only
(2) (1), (4) and (E) Only
(3) (3), (4) and (E) Only
(4) (1), (2) and (3) Only
Answer: (3)
Solution:
To identify which compounds give a positive Fehling’s test, we need to understand the specific types of carbonyl groups that this reagent can oxidize. Fehling’s solution consists of $Cu^{2+}$ ions complexed with tartrate ions in an alkaline medium. A positive test is indicated by the formation of a reddish-brown precipitate of cuprous oxide ($Cu_2O$).
1. General Rules for Fehling’s Test
Aliphatic Aldehydes: Give a positive test.
Aromatic Aldehydes: Generally do not give a positive test because the carbonyl group is stabilized by the benzene ring, making it less reactive toward the weak oxidizing agent $Cu^{2+}$.
Ketones: Generally do not give a positive test, except for $\alpha$-hydroxy ketones.
$\mathbf{\alpha}$-hydroxy ketones (like Fructose): Give a positive test because they can tautomerize into aldehydes under alkaline conditions.
2. Analysis of the Compounds
(A) Benzaldehyde ($\mathbf{C_6H_5CHO}$): This is an aromatic aldehyde. Because the $-CHO$ group is attached directly to the benzene ring, it does not respond to Fehling’s test. (Negative)
(B) Acetophenone ($\mathbf{C_6H_5COCH_3}$): This is an aromatic ketone. Ketones do not reduce Fehling’s solution. (Negative)
(C) Fructose/$\mathbf{\alpha}$-hydroxy ketone structure: The structure $HOCH_2-CO-(CHOH)_3-CH_2-OH$ represents a ketose sugar. Due to the presence of the $-\text{OH}$ group adjacent to the carbonyl ($C=O$), it undergoes rearrangement in the alkaline Fehling’s solution to act as a reducing sugar. (Positive)
(D) Acetaldehyde ($\mathbf{CH_3CHO}$): This is a simple aliphatic aldehyde. It is easily oxidized by Fehling’s reagent. (Positive)
(E) Phenylacetaldehyde ($\mathbf{C_6H_5CH_2CHO}$): Although there is a benzene ring in the molecule, the $-CHO$ group is not directly attached to it (it’s separated by a $CH_2$ group). Therefore, it behaves as an aliphatic aldehyde. (Positive)
Conclusion: The compounds that give a positive Fehling’s test are (C), (D), and (E).
This corresponds to option (3).
66. In which of the following complexes the CFSE, $\Delta_0$ will be equal to zero?
(1) $[\text{Fe}(\text{NH}_3)_6]\text{Br}_2$
(2) $[\text{Fe}(\text{en})_3]\text{Cl}_3$
(3) $\text{K}_4[\text{Fe}(\text{CN})_6]$
(4) $\text{K}_3[\text{Fe}(\text{SCN})_6]$
Answer: (4)
Solution:
1. The CFSE Formula for Octahedral Complexes
In an octahedral field, the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy). The energy difference is $\Delta_0$.
The CFSE is calculated as:
$$\text{CFSE} = \left[ (-0.4 \times n_{t2g}) + (0.6 \times n_{eg}) \right] \Delta_0$$
$n_{t2g}$ = number of electrons in $t_{2g}$ orbitals.
$n_{eg}$ = number of electrons in $e_g$ orbitals.
2. Evaluating the Complexes
(1) $\mathbf{[Fe(NH_3)_6]Br_2}$
- Oxidation State: $Fe^{2+}$
- Configuration: $3d^6$
- Ligand: $NH_3$ is a relatively strong field ligand. Even in a high-spin state ($t_{2g}^4 e_g^2$), the CFSE would be $[(-0.4 \times 4) + (0.6 \times 2)] = -0.4 \Delta_0$. Not zero.
(2) $\mathbf{ [Fe(en)_3]Cl_3}$
- Oxidation State: $Fe^{3+}$
- Configuration: $3d^5$
- Ligand: ‘en’ (ethylenediamine) is a strong field ligand. It causes pairing, leading to a low-spin state ($t_{2g}^5 e_g^0$).
- CFSE: $(-0.4 \times 5) = -2.0 \Delta_0$. Not zero.
(3) $\mathbf{K_4[Fe(CN)_6]}$
- Oxidation State: $Fe^{2+}$
- Configuration: $3d^6$
- Ligand: $CN^-$ is a very strong field ligand. It leads to a low-spin state ($t_{2g}^6 e_g^0$).
- CFSE: $(-0.4 \times 6) = -2.4 \Delta_0$. Not zero.
(4) $\mathbf{K_3[Fe(SCN)_6]}$
- Oxidation State: $Fe^{3+}$
- Configuration: $3d^5$ (Iron atomic number is 26; $Fe$ is $[Ar]3d^6 4s^2$, so $Fe^{3+}$ is $3d^5$).
- Ligand: $SCN^-$ is a weak field ligand. According to the spectrochemical series, it does not cause electron pairing. This results in a high-spin complex.
- Distribution: Electrons fill orbitals singly according to Hund’s rule: $t_{2g}^3 e_g^2$.
3. Calculation for $\mathbf{K_3[Fe(SCN)_6]}$
Using the configuration $t_{2g}^3 e_g^2$:
$$\text{CFSE} = [(-0.4 \times 3) + (0.6 \times 2)] \Delta_0$$
$$\text{CFSE} = [-1.2 + 1.2] \Delta_0$$
$$\text{CFSE} = 0$$
Conclusion: In the high-spin $d^5$ configuration, the stabilization gained by the three electrons in the lower energy level is exactly canceled out by the destabilization of the two electrons in the higher energy level.
The correct complex is $\mathbf{K_3[Fe(SCN)_6]}$, which is Option (4).
67. Arrange the following solutions in order of their increasing boiling points.
(i) $10^{-4}\text{ M NaCl}$
(ii) $10^{-4}\text{ M Urea}$
(iii) $10^{-3}\text{ M NaCl}$
(iv) $10^{-2}\text{ M NaCl}$
(1) (ii) < (i) < (iii) < (iv)
(2) (ii) < (i) $\cong$ (iii) < (iv)
(3) (i) < (ii) < (iii) < (iv)
(4) (iv) < (iii) < (i) < (ii)
Answer: (1)
Solution:
1. The Core Formula
The increase in boiling point ($\Delta T_b$) is given by:
$$\Delta T_b = i \cdot K_b \cdot m$$
Where:
- i is the van’t Hoff factor (the number of particles the solute dissociates into).
- $\mathbf{K_b}$ is the molal elevation constant (same for the same solvent).
- m is the molality (which is approximately equal to Molarity $M$ in very dilute aqueous solutions).
From this, we can conclude that the boiling point is directly proportional to the product of the van’t Hoff factor and the concentration ($C$):
$$\text{Boiling Point} \propto (i \times C)$$
2. Calculating the Effective Particle Concentration ($\textbf{i \times C}$)
We need to determine the value of $i$ for each solute:
- Urea: A non-electrolyte, it does not dissociate. Therefore, i = 1.
- NaCl: An electrolyte that dissociates into two ions ($Na^+$ and $Cl^-$). Therefore, i = 2.
Now, let’s calculate the product for each solution:
| Solution | Concentration (C) | van’t Hoff factor (i) | Effective Concentration (i×C) |
| (i) $10^{-4}\text{ M NaCl}$ | $10^{-4}$ | $2$ | $2 \times 10^{-4}$ |
| (ii) $10^{-4}\text{ M Urea}$ | $10^{-4}$ | $1$ | $1 \times 10^{-4}$ |
| (iii) $10^{-3}\text{ M NaCl}$ | $10^{-3}$ | $2$ | $2 \times 10^{-3} = 20 \times 10^{-4}$ |
| (iv) $10^{-2}\text{ M NaCl}$ | $10^{-2}$ | $2$ | $2 \times 10^{-2} = 200 \times 10^{-4}$ |
3. Comparing the Values
Arranging the calculated $(i \times C)$ values in increasing order:
$\mathbf{1 \times 10^{-4}}$ (Solution ii)
$\mathbf{2 \times 10^{-4}}$ (Solution i)
$\mathbf{2 \times 10^{-3}}$ (Solution iii)
$\mathbf{2 \times 10^{-2}}$ (Solution iv)
Conclusion: Higher effective particle concentration leads to a higher boiling point. Therefore, the order of increasing boiling points is: (ii) < (i) < (iii) < (iv)
This corresponds to option (1).
68. The products formed in the following reaction sequence are:
(1)
(2)
(3)
(4)
Answer: (3)
Solution:
Step 1: Bromination ($\mathbf{Br_2, AcOH}$)
The starting material is nitrobenzene. The nitro group ($-NO_2$) is a strongly deactivating group and is meta-directing.
- When reacted with bromine in acetic acid, the bromine atom is substituted at the meta position.
- Intermediate 1: $m$-bromonitrobenzene.
Step 2: Reduction ($\mathbf{Sn, HCl}$)
The reagent tin ($Sn$) and hydrochloric acid ($HCl$) is used to reduce the nitro group ($-NO_2$) to a primary amine group ($-NH_2$).
- Intermediate 2: $m$-bromoaniline.
Step 3: Diazotization ($\mathbf{NaNO_2, HCl, 273\text{ K}}$)
The primary aromatic amine reacts with nitrous acid (generated in situ from $NaNO_2$ and $HCl$) at low temperatures to form a diazonium salt.
- Intermediate 3: $m$-bromobenzene diazonium chloride ($m\text{-}BrC_6H_4N_2^+Cl^-$).
Step 4: Reaction with Ethanol ($\mathbf{C_2H_5OH}$)
When a diazonium salt is treated with ethanol (a mild reducing agent), a deamination reaction occurs.
- The diazonium group ($-N_2^+Cl^-$) is replaced by a hydrogen atom.
- During this process, the ethanol itself is oxidized to an aldehyde.
- Product A: Bromobenzene ($C_6H_5Br$).
- Product B: Acetaldehyde ($CH_3CHO$).
Conclusion: The final products formed in the sequence are bromobenzene and acetaldehyde, which matches option (3).
69. From the magnetic behaviour of $[\text{NiCl}_4]^{2-}$ (paramagnetic) and $[\text{Ni(CO)}_4]$ (diamagnetic), choose the correct geometry and oxidation state.
(1) $[\text{NiCl}_4]^{2-}: \text{Ni}^{\text{II}}$, square planar
$[\text{Ni(CO)}_4]: \text{Ni}(0)$, square planar
(2) $[\text{NiCl}_4]^{2-}: \text{Ni}^{\text{II}}$, tetrahedral
$[\text{Ni(CO)}_4]: \text{Ni}(0)$, tetrahedral
(3) $[\text{NiCl}_4]^{2-}: \text{Ni}^{\text{II}}$, tetrahedral
$[\text{Ni(CO)}_4]: \text{Ni}^{\text{II}}$, square planar
(4) $[\text{NiCl}_4]^{2-}: \text{Ni}(0)$, tetrahedral
$[\text{Ni(CO)}_4]: \text{Ni}(0)$, square planar
Answer: (2)
Solution:
To determine the geometry and oxidation state of these two complexes, we apply Valence Bond Theory (VBT) and consider the nature of the ligands.
1. Analysis of $\mathbf{[NiCl_4]^{2-}}$
Oxidation State: Let the oxidation state of Ni be $x$.
$x + 4(-1) = -2 \implies x = +2$.
Nickel is in the $\mathbf{Ni^{II}}$ state.
Electronic Configuration:
$Ni = [Ar] 3d^8 4s^2$
$Ni^{2+} = [Ar] 3d^8 4s^0$
Magnetic Behaviour: The problem states it is paramagnetic.
In a $3d^8$ system, to be paramagnetic, it must have unpaired electrons.
Hybridization and Geometry: $Cl^-$ is a weak field ligand. It cannot pair up the electrons in the $3d$ orbital. Thus, the $3d$ orbitals remain unchanged, and the nickel atom uses one $4s$ and three $4p$ orbitals for bonding.
Hybridization: $\mathbf{sp^3}$
Geometry: Tetrahedral
2. Analysis of $\mathbf{[Ni(CO)_4]}$
Oxidation State: $CO$ is a neutral ligand.
$x + 4(0) = 0 \implies x = 0$.
Nickel is in the $Ni(0)$ state.
Electronic Configuration:
$Ni(0) = [Ar] 3d^8 4s^2$
Magnetic Behaviour: The problem states it is diamagnetic.
Hybridization and Geometry: $CO$ is a very strong field ligand. It forces the two electrons from the $4s$ orbital to pair up with the electrons in the $3d$ orbital.
- This results in a completely filled $3d$ subshell ($3d^{10}$) and an empty $4s$ and $4p$ subshell.
- The four $CO$ ligands then donate electron pairs into the empty $4s$ and $4p$ orbitals.
- Hybridization: $\mathbf{sp^3}$
- Geometry: Tetrahedral
Summary Table
| Complex | Oxidation State | Hybridization | Geometry | Magnetic Property |
| $[NiCl_4]^{2-}$ | $Ni^{II}$ (+2) | $sp^3$ | Tetrahedral | Paramagnetic |
| $[Ni(CO)_4]$ | $Ni(0)$ | $sp^3$ | Tetrahedral | Diamagnetic |
The correct combination is $\mathbf{Ni^{II}}$, tetrahedral for the first complex and $\mathbf{Ni(0)}$, tetrahedral for the second.
This corresponds to Option (2).
70. The incorrect statements regarding geometrical isomerism are:
(1) Propene shows geometrical isomerism.
(2) Trans isomer has identical atoms/groups on the opposite sides of the double bond.
(3) Cis-but-2-ene has higher dipole moment than trans-but-2-ene.
(4) 2-methylbut-2-ene shows two geometrical isomers.
(E) Trans-isomer has lower melting point that cis isomer.
Choose the CORRECT answer from the options given below:
(1) (1), (4) and (E) only
(2) (3), (4) and (E) only
(3) (2) and (3) only
(4) (1) and (E) only
Answer: (1)
Solution:
Understanding Geometrical Isomerism (GI)
Geometrical isomerism (cis-trans isomerism) occurs in compounds where rotation is restricted, most commonly due to a carbon-carbon double bond ($C=C$).
The Golden Rule for GI: Each carbon atom of the double bond must be attached to two different atoms or groups. If either carbon is attached to two identical groups (e.g., two Hydrogens or two Methyl groups), GI is not possible.
Evaluating the Statements
(A) Propene shows geometrical isomerism.
Structure: $CH_3-CH=CH_2$
The second carbon atom in the double bond is attached to two identical Hydrogen atoms.
Because of this symmetry, flipping the groups doesn’t create a new arrangement.
Verdict: Incorrect. Propene does not show GI.
(B) Trans isomer has identical atoms/groups on the opposite sides of the double bond.
- In a trans isomer, the identical groups are located on opposite sides of the double bond (diagonal to each other).
- Verdict: Correct.
(C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene.
- In cis-but-2-ene, the polar $C-CH_3$ bonds are on the same side, so their dipole vectors add up.
- In trans-but-2-ene, the methyl groups are on opposite sides, so their dipole vectors cancel each other out (resulting in a near-zero dipole moment).
- Verdict: Correct.
(D) 2-methylbut-2-ene shows two geometrical isomers.
- Structure: $(CH_3)_2C=CH-CH_3$
- The first carbon of the double bond is attached to two identical methyl ($\mathbf{-CH_3}$) groups.
- As per the golden rule, this molecule cannot show GI.
- Verdict: Incorrect.
(E) Trans-isomer has lower melting point than cis isomer.
- Melting point depends on how well molecules pack in a crystal lattice.
- Trans-isomers are generally more symmetrical and linear, allowing them to pack more closely and efficiently than the “U-shaped” cis-isomers.
- Better packing leads to stronger intermolecular forces in the solid state, giving trans-isomers a higher melting point.
- Verdict: Incorrect.
The incorrect statements are (A), (D), and (E). This matches option (1).
SECTION – B
71. Some $\text{CO}_2$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of $\text{CO}_2$ gas was later passed through an aqueous solution of $\text{Ca(OH)}_2$. The excess unreacted $\text{Ca(OH)}_2$ was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of $\text{CO}_2$ was x, then x is ______ $\text{cm}^3$ (nearest integer).
[Given: The entire amount of $\text{CO}_2\text{(g)}$ reacted with exactly half the initial amount of $\text{Ca(OH)}_2$ present in the aqueous solution.]
Answer: (45)
Solution:
To find the volume of the $CO_2$ gas ($x$), we need to work backward from the neutralization reaction to determine the number of moles of gas present.
1. Analyzing the Reactions
There are two primary chemical reactions occurring:
Absorption of $\mathbf{CO_2}$: Carbon dioxide reacts with limewater to form calcium carbonate and water.
$$CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O$$
Ratio: $1$ mole of $CO_2$ reacts with 1 mole of $Ca(OH)_2$.
Neutralization of Excess Base: The unreacted $Ca(OH)_2$ is neutralized by hydrochloric acid.
$$Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$$
Ratio: $1$ mole of $Ca(OH)_2$ reacts with 2 moles of $HCl$.
2. Calculating Moles of Excess $\mathbf{Ca(OH)_2}$
First, calculate the moles of $HCl$ used in the titration:
$$\text{Moles of } HCl = \text{Molarity} \times \text{Volume (in Liters)}$$
$$\text{Moles of } HCl = 0.1 \times \frac{40}{1000} = 0.004 \text{ moles}$$
Using the stoichiometry from reaction 2 ($1:2$ ratio):
$$\text{Moles of excess } Ca(OH)_2 = \frac{0.004}{2} = 0.002 \text{ moles}$$
3. Determining Moles of $\mathbf{CO_2}$
The problem states that the entire amount of $CO_2$ reacted with exactly half of the initial $Ca(OH)_2$. This implies that the other half remained as “excess.”
Since the excess amount is $0.002 \text{ moles}$, the amount that reacted with $CO_2$ must also be 0.002 moles.
From reaction 1 ($1:1$ ratio):
$$\text{Moles of } CO_2 (n) = \text{Moles of reacted } Ca(OH)_2 = 0.002 \text{ moles}$$
4. Calculating the Volume (x)
We are given that the gas is at 1 atm and 273 K, which are Standard Temperature and Pressure (STP) conditions. At STP, 1 mole of any ideal gas occupies 22.4 L (or 22400 cm$^3$).
$$x = n \times 22400 \text{ cm}^3$$
$$x = 0.002 \times 22400$$
$$x = 44.8 \text{ cm}^3$$
Rounding to the nearest integer, the value of $x$ is 45.
72. In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is _______ %.
[Given: molar mass in g mol$^{–1}$ of Ag: 108, Cl = 35.5]
Answer: (20)
Solution:
The Carius method is a quantitative analytical technique used to determine the amount of halogens in an organic compound.
1. Principle of Carius Method
In this method, a known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate ($AgNO_3$). The carbon and hydrogen are oxidized, while the halogen is converted into a precipitate of silver halide ($AgX$). In this case, the chlorine is converted into silver chloride ($AgCl$).
2. Calculation Steps
Step 1: Calculate the molar mass of Silver Chloride ($AgCl$)
- Atomic mass of $Ag = 108$
- Atomic mass of $Cl = 35.5$
- Molar mass of $AgCl = 108 + 35.5 = 143.5 \text{ g/mol}$
Step 2: Relate the mass of $AgCl$ to the mass of Chlorine
From the molar mass, we know that $143.5 \text{ mg}$ of $AgCl$ contains $35.5 \text{ mg}$ of Chlorine.
In the problem, the mass of $AgCl$ produced is exactly $143.5 \text{ mg}$.
Therefore:
$$\text{Mass of Chlorine in the precipitate} = 35.5 \text{ mg}$$
Step 3: Calculate the percentage of Chlorine in the original compound
The original mass of the organic compound is $180 \text{ mg}$.
The formula for percentage composition is:
$$\% \text{Cl} = \left( \frac{\text{Mass of Chlorine}}{\text{Total mass of compound}} \right) \times 100$$
Substituting the values:
$$\% \text{Cl} = \left( \frac{35.5 \text{ mg}}{180 \text{ mg}} \right) \times 100$$
$$\% \text{Cl} = \frac{3550}{180} \approx 19.72\%$$
3. Conclusion
Rounding to the nearest whole number as indicated by the expected format in competitive exams:
The percentage composition of chlorine in the compound is 20%.
73. The number of molecules/ions that show linear geometry among the following is _______.
$\text{SO}_2, \text{BeCl}_2, \text{CO}_2, \text{N}_3^-, \text{NO}_2, \text{F}_2\text{O}, \text{XeF}_2, \text{NO}_2^+, \text{I}_3^-, \text{O}_3$
Answer: (6)
74. $\text{A} \rightarrow \text{B}$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 \text{ kJ mol}^{-1}$ and the frequency factor is $10^{20}$, the time required for 50% molecules of A to become B is ______ picoseconds (nearest integer).
$[\text{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}]$
Answer: (69)
Solution:
The transformation $A \to B$ follows first-order kinetics. We are given the following parameters:
- Temperature ($T$): $1000 \text{ K}$
- Energy Barrier ($E_a$): $191.48 \text{ kJ mol}^{-1} = 191480 \text{ J mol}^{-1}$
- Frequency Factor ($A$): $10^{20}$
- Gas Constant ($R$): $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
We need to find the time required for $50\%$ of molecules of $A$ to become $B$, which is the half-life ($t_{1/2}$), in picoseconds.
1. Calculating the Rate Constant (k)
We use the Arrhenius Equation to find the rate constant at the given temperature:
$$k = A e^{-E_a/RT}$$
First, let’s calculate the exponent value $\frac{E_a}{RT}$:
$$\frac{E_a}{RT} = \frac{191480}{8.314 \times 1000} = \frac{191.48}{8.314} \approx 23.03$$
Now substitute this back into the equation:
$$k = 10^{20} \times e^{-23.03}$$
Since $23.03 \approx 10 \times \ln(10)$, we can simplify $e^{-23.03}$ as $10^{-10}$:
$$k = 10^{20} \times 10^{-10} = 10^{10} \text{ s}^{-1}$$
2. Calculating the Half-life ($\mathbf{t_{1/2}}$)
For a first-order reaction, the relationship between the half-life and the rate constant is:
$$t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$$
Substitute the value of $k$:
$$t_{1/2} = \frac{0.693}{10^{10}} = 0.693 \times 10^{-10} \text{ s}$$
3. Converting to Picoseconds
The question asks for the answer in picoseconds (ps).
Recall that $1 \text{ s} = 10^{12} \text{ ps}$.
$$t_{1/2} = (0.693 \times 10^{-10}) \times 10^{12} \text{ ps}$$
$$t_{1/2} = 0.693 \times 10^{2} \text{ ps}$$
$$t_{1/2} = 69.3 \text{ ps}$$
Rounding to the nearest integer, we get 69.
The time required for $50\%$ of the molecules to transform is 69 picoseconds.
75. Consider the following sequence of reactions:
Molar mass of the product formed (1) is ________g mol$^{–1}$ .
Answer: (154)
Solution:
Step-by-Step Reaction Sequence
Step 1: Reduction ($\mathbf{Sn + HCl}$)
Nitrobenzene is reduced by tin and hydrochloric acid. The nitro group ($-NO_2$) is converted into an amino group ($-NH_2$).
- Result: Aniline ($C_6H_5NH_2$).
Step 2: Diazotization ($\mathbf{NaNO_2, HCl}$ at $\mathbf{0^\circ C}$)
Aniline reacts with nitrous acid at low temperatures to form a stable diazonium salt.
- Result: Benzene diazonium chloride ($C_6H_5N_2^+Cl^-$).
Step 3: Sandmeyer Reaction ($\mathbf{Cu_2Cl_2}$)
The diazonium group is replaced by a chlorine atom when treated with cuprous chloride.
- Result: Chlorobenzene ($C_6H_5Cl$).
Step 4: Fittig Reaction ($\mathbf{Na, \text{Ether}}$)
When an aryl halide (chlorobenzene) is treated with sodium metal in dry ether, two benzene rings couple together.
- Result (Product A): Biphenyl (also known as Diphenyl).
- Chemical Formula: $C_{12}H_{10}$ (Two benzene rings joined by a single bond, $C_6H_5 – C_6H_5$).
Molar Mass Calculation for Product (A)
Now, we calculate the molar mass of Biphenyl ($C_{12}H_{10}$):
Atomic mass of Carbon (C) = $12 \text{ g/mol}$
Atomic mass of Hydrogen (H) = $1 \text{ g/mol}$
$$\text{Molar Mass} = (12 \times 12) + (10 \times 1)$$
$$\text{Molar Mass} = 144 + 10$$
$$\mathbf{\text{Molar Mass} = 154 \text{ g mol}^{-1}}$$
The molar mass of the product formed (A) is 154.