Question:

Match List-I with List-II:

List-I ( Ions )

List-II ( No. of unpaired electrons )

AZn$^{2+}$(I)

0

BCu$^{2+}$(II)

4

CNi$^{2+}$(III)

1

DFe$^{2+}$(IV)

2

Choose the correct answer from the options given below:

Updated On: Nov 2, 2024
  • (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  • (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  • (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  • (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
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The Correct Option is C

Solution and Explanation

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).
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