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

The Correct Option is C

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: \[ \text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}. \]