Chemistry Quiz: 8 Key Concepts (JEE/NEET PYQs)

 Chemistry Quiz: 8 Key Concepts (JEE/NEET PYQs)

Master topics from $S_N2$ reactions to bond order calculations.

Question 1: Salt Solution Nature (Ionic Equilibrium)

Given below are two statements:

Statement (I): Aqueous solution of ammonium carbonate is basic.

Statement (II): Acidic/basic nature of salt solution of a weak acid and weak base depends on $K_a$ and $K_b$ value of acid and the base forming it.

(1) Both Statement I and Statement II are correct (2) Statement I is correct but Statement II is incorrect (3) Both Statement I and Statement II are incorrect (4) Statement I is incorrect but Statement II is correct

Solution 1: Salt Hydrolysis

Ammonium carbonate ($\mathbf{(NH_4)_2CO_3}$) is a salt of a **weak base** ($\mathbf{NH_4OH}$) and a **weak acid** ($\mathbf{H_2CO_3}$).

• For this salt, $\mathbf{K_b}$ ($\approx 1.8 \times 10^{-5}$) is slightly greater than $\mathbf{K_a}$ ($\approx 4.3 \times 10^{-7}$). Since $K_b > K_a$, the resulting solution is **basic** ($\mathbf{pH > 7}$). Thus, **Statement (I) is correct**.
• The pH of a salt of weak acid and weak base is determined by the formula: $pH = 7 + \frac{1}{2} (pK_a - pK_b)$. This explicitly shows that the pH (and thus the acidic/basic nature) **depends on the relative strengths ($K_a$ and $K_b$)** of the parent acid and base. Thus, **Statement (II) is correct**.

**Correct Option: (1)** (Both Statement I and Statement II are correct).

Question 2: Stereochemistry in $S_N$ Reactions

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is:

(1) Retention occurs in $S_N1$ reaction and inversion occurs in $S_N2$ reaction. (2) Racemisation occurs in $S_N1$ reaction and retention occurs in $S_N2$ reaction. (3) Racemisation occurs in both $S_N1$ and $S_N2$ reactions. (4) Racemisation occurs in $S_N1$ reaction and inversion occurs in $S_N2$ reaction.

Solution 2: $S_N$ Reactions

• **$\mathbf{S_N1}$ Reaction:** Proceeds via a planar **carbocation intermediate** . The nucleophile can attack from either face, leading to a product mixture of $\sim 50\%$ retention and $\sim 50\%$ inversion. This overall process is called **Racemisation**.
• **$\mathbf{S_N2}$ Reaction:** Proceeds via a **single transition state** where the nucleophile attacks from the side opposite to the leaving group. This results in the complete flip of configuration, known as **Inversion** (or Walden Inversion).

**Correct Option: (4)** (Racemisation occurs in $S_N1$ reaction and inversion occurs in $S_N2$ reaction).

Question 3: Qualitative Analysis (Chromyl Chloride Test)

$\text{NaCl}$ reacts with conc. $\text{H}_2\text{SO}_4$ and $\text{K}_2\text{Cr}_2\text{O}_7$ to give reddish fumes (B), which react with $\text{NaOH}$ to give yellow solution (C). (B) and (C) respectively are:

(1) $\text{CrO}_2\text{Cl}_2$, $\text{Na}_2\text{CrO}_4$ (2) $\text{Na}_2\text{CrO}_4$, $\text{CrO}_2\text{Cl}_2$ (3) $\text{CrO}_2\text{Cl}_2$, $\text{KHSO}_4$ (4) $\text{CrO}_2\text{Cl}_2$, $\text{Na}_2\text{Cr}_2\text{O}_7$

Solution 3: Chromyl Chloride Test

This is the definitive test for **chloride ions** ($\text{Cl}^-$).

1. **Step 1: Formation of Reddish Fumes (B)** $$4\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{H}_2\text{SO}_4 \longrightarrow \mathbf{2\text{CrO}_2\text{Cl}_2} (\mathbf{B}) + 2\text{KHSO}_4 + 4\text{NaHSO}_4 + 3\text{H}_2\text{O}$$ $\mathbf{B}$ is **Chromyl Chloride** ($\text{CrO}_2\text{Cl}_2$), which is a volatile, deep reddish-brown gas.
2. **Step 2: Reaction with Base (C)** $$\text{CrO}_2\text{Cl}_2 (\mathbf{B}) + 4\text{NaOH} \longrightarrow \mathbf{\text{Na}_2\text{CrO}_4} (\mathbf{C}) + 2\text{NaCl} + 2\text{H}_2\text{O}$$ $\mathbf{C}$ is **Sodium Chromate** ($\text{Na}_2\text{CrO}_4$), which forms a characteristic **yellow solution** in basic medium.

(B) and (C) are $\mathbf{\text{CrO}_2\text{Cl}_2}$ and $\mathbf{\text{Na}_2\text{CrO}_4}$. **Correct Option: (1)**.

Question 4: Electronic Configuration of Neodymium (f-block)

The electronic configuration for Neodymium ($\text{Nd}$) is: [Atomic Number for Neodymium 60]

(1) $[\text{Xe}] 4f^4 6s^2$ (2) $[\text{Xe}] 5f^7 6s^2$ (3) $[\text{Xe}] 4f^6 6s^2$ (4) $[\text{Xe}] 4f^3 5d^1 6s^2$

Solution 4: Electronic Configuration

Neodymium ($\text{Nd}$, $Z=60$) belongs to the **Lanthanide series** ($4f$-block).

• The core configuration is that of Xenon ($Z=54$): $[\text{Xe}]$.
• The subsequent electrons fill the $6s$ orbital first (2 electrons), then the $4f$ orbital.
• Total valence electrons to place: $60 - 54 = 6$.
• Standard Lanthanide configuration: $[\text{Xe}] 4f^{n} 5d^{0/1} 6s^2$.
• For $\text{Nd} (Z=60)$, the configuration is $\mathbf{[\text{Xe}] 4f^4 6s^2}$. The $5d$ orbital is not occupied in this case.

**Correct Option: (1)** ($\mathbf{[\text{Xe}] 4f^4 6s^2}$).

Question 5: Electron Count by Quantum Numbers (Integer Type)

The number of electrons present in all the completely filled subshells having $n=4$ and $s = +\frac{1}{2}$ is $\underline{\hspace{2em}}$.

(1) 8 (2) 16 (3) 32 (4) 4

Solution 5: Quantum Numbers

For a principal quantum number $\mathbf{n=4}$, the possible subshells (values of $l$) are $4s$ (l=0), $4p$ (l=1), $4d$ (l=2), and $4f$ (l=3). Since the subshells are **completely filled**, each orbital is occupied by two electrons (one with $s=+\frac{1}{2}$ and one with $s=-\frac{1}{2}$).

Subshell	Number of Orbitals ($2l+1$)	Total Electrons (2(Orbitals))	Electrons with $s = +\frac{1}{2}$
$4s$	1	2	1
$4p$	3	6	3
$4d$	5	10	5
$4f$	7	14	7

Total electrons with $n=4$ and $s=+\frac{1}{2}$ = $1 + 3 + 5 + 7 = \mathbf{16}$.

**Answer: 16**.

Question 6: Sum of Bond Orders (Integer Type)

Sum of bond order of $\text{CO}$ and $\text{NO}^+$ is $\underline{\hspace{2em}}$.

(1) 4 (2) 5 (3) 6 (4) 7