Question

Match List-I with List-II:

List-I ( Ions )		List-II ( No. of unpaired electrons )
A	Zn$^{2+}$	(I)	0
B	Cu$^{2+}$	(II)	4
C	Ni$^{2+}$	(III)	1
D	Fe$^{2+}$	(IV)	2

Choose the correct answer from the options given below:

• A. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
• B. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
• C. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
• D. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Answer 1

The Correct Option is C

1. Zn$^{2+}$: Electron configuration of Zn is [Ar] 3d$^{10}$. The ion has 0 unpaired electrons. Therefore, (A) - (I).$\\$2. Cu$^{2+}$: Electron configuration of Cu is [Ar] 3d$^{10}$ 4s$^1$. The ion has 1 unpaired electron after losing 2 electrons. Therefore, (B) - (III).$\\$3. Ni$^{2+}$: Electron configuration of Ni is [Ar] 3d$^8$ 4s$^2$. The ion has 2 unpaired electrons. Therefore, (C) - (IV).$\\$4. Fe$^{2+}$: Electron configuration of Fe is [Ar] 3d$^6$ 4s$^2$. The ion has 4 unpaired electrons. Therefore, (D) - (II).$\\$Thus, the correct matches are: (A) - (I), (B) - (III), (C) - (IV), (D) - (II).

Answer 2

1. Zn²⁺: Electron configuration of Zn is [Ar] 3d¹⁰. The ion has 0 unpaired electrons. Therefore, (A) - (I).

2. Cu²⁺: Electron configuration of Cu is [Ar] 3d¹⁰ 4s¹. The ion has 1 unpaired electron after losing 2 electrons. Therefore, (B) - (III).

3. Ni²⁺: Electron configuration of Ni is [Ar] 3d⁸ 4s². The ion has 2 unpaired electrons. Therefore, (C) - (IV).

4. Fe²⁺: Electron configuration of Fe is [Ar] 3d⁶ 4s². The ion has 4 unpaired electrons. Therefore, (D) - (II).

Thus, the correct matches are: (A) - (I), (B) - (III), (C) - (IV), (D) - (II).