Question

Match the column I with column II and mark the appropriate choice.

Column I		Column II
$(A)$	$[Fe(CN)_{6}]^{3-}$	(i)	Zero
(B)	$[Co F_{6}]^{3-}$	(ii)	$5.92\, B.M.$
(C)	$[Fe(H_{2}O )_{6}]^{3+}$	(iii)	$4.89\, B.M.$
(D)	$[Co (NH_{3})_{6}]^{3+}$	(iv)	$1.732\, B.M.$

• A. $(A) \to (ii), (B) \to (iii), (C) \to (iv), (D) \to (i)$
• B. $(A) \to (iii), (B) \to (ii), (C) \to (i), (D) \to (iv)$
• C. $(A) \to (i), (B) \to (iii), (C) \to (iv), (D) \to (ii)$
• D. $(A) \to (iv), (B) \to (iii), (C) \to (ii), (D) \to (i)$

Answer 1

Answer: (D)

$Fe^{3+} : 3d^{5}$, since $CN^{-}$ is a strong field ligand, pairing will take place.
$[Fe(CN)_{6}]^{3-}$ :
$\mu=\sqrt{n\left(n+2\right)}=\sqrt{1\left(1+2\right)}=\sqrt{3}$
$= 1.732 \,B.M.$
In $[CoF_{6}]^{3-}$ , oxidation state of $Co = + 3$
$Co^{3+} : 3d^{6}$, since $F^{-}$ is a weak field ligand, pairing will not occur.
$[CoF_{6}]^{3-}$ :
$[\because n=4]$
$\mu=\sqrt{4\left(4+2\right)}=\sqrt{24}=4.89\, B.M.$
In $[Fe(H_{2}O )_{6}]^{3+}$, oxidation state of $Fe = + 3$
$Fe^{3+} : 3d^{5}$, since $H_{2}O$ is a weak field ligand, pairing willnot take place.
$[Fe(H_{2}O)_{6}]^{3+}$ :
$[\because n=5]$
$\mu=\sqrt{5\left(5+2\right)}=\sqrt{35}=5.92 \,B.M.$
In $[Co (NH_{3})_{6}]^{3+}$, oxidation state of $Co = +3$.
$Co^{3+} : 3d^{6}$, since $NH_{3}$ is a strong field ligand, pairing will take place.
$[Co(NH_{3})_{6}]^{3+}$
$[\because n=0]$
$\mu=\sqrt{0\left(0+2\right)}=0$