Question

Match List-I with List-II. List-I
(Complex ion)| List-II
(Spin only magnetic moment in B.M.)
---|---
(A) [Cr(NH$_3$)$_6$]$^{3+}$| (I) 4.90
(B) [NiCl$_4$]$^{2-}$| (II) 3.87
(C) [CoF$_6$]$^{3-}$| (III) 0.0
(D) [Ni(CN)$_4$]$^{2-}$| (IV) 2.83

Choose the correct answer from the options given below:

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

Answer 1

Answer: (C)

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

Answer 2

The spin-only magnetic moment ($\mu$) in Bohr Magnetons (B.M.) is calculated using the formula: $\mu = \sqrt{n(n+2)} \, \text{B.M.},$ where $n$ is the number of unpaired electrons.
(A) [Cr(NH$_3$)$_6$]$^{3+}$: $\text{Cr}^{3+}: 3d^3 \, (3 \, \text{unpaired electrons}).$
$\mu = \sqrt{3(3+2)} = 3.87 \, \text{B.M.} \, (\text{II}).$
(B) [NiCl$_4$]$^{2-}$: $\text{Ni}^{2+}: 3d^8 \, (2 \, \text{unpaired electrons}).$
$\mu = \sqrt{2(2+2)} = 2.83 \, \text{B.M.} \, (\text{IV}).$
(C) [CoF$_6$]$^{3-}$: $\text{Co}^{3+}: 3d^6 \, (4 \, \text{unpaired electrons}).$
$\mu = \sqrt{4(4+2)} = 4.90 \, \text{B.M.} \, (\text{I}).$
(D) [Ni(CN)$_4$]$^{2-}$: $\text{Ni}^{2+}: 3d^8 \, (\text{low spin, 0 unpaired electrons}).$
$\mu = \sqrt{0(0+2)} = 0.0 \, \text{B.M.} \, (\text{III}).$
The correct matching is: $\text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}.$