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)	$5.92B.M.$
(C)	$[Fe(H_{2}O{)}_6{]}^{3+}$	(iii)	$4.89B.M.$
(D)	$[Co(N{H}_3{)}_6{]}^{3+}$	(iv)	$1.732B.M.$

• 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)

$F{e}^{3+}:3d^5$, since $C{N}^{-}$ is a strong field ligand, pairing will take place.
$[Fe(CN{)}_6{]}^{3-}$ :
$\mu =\sqrt{n\left(n+2\right)}=\sqrt{1\left(1+2\right)}=\sqrt{3}$
$=1.732B.M.$
In $[Co{F}_6{]}^{3-}$ , oxidation state of $Co=+3$
$C{o}^{3+}:3d^6$, since $F^{-}$ is a weak field ligand, pairing will not occur.
$[Co{F}_6{]}^{3-}$ :
$[\because n=4]$
$\mu =\sqrt{4\left(4+2\right)}=\sqrt{24}=4.89B.M.$
In $[Fe(H_{2}O{)}_6{]}^{3+}$, oxidation state of $Fe=+3$
$F{e}^{3+}:3d^5$, since $H_{2}O$ is a weak field ligand, pairing willnot take place.
$[Fe(H_{2}O{)}_6{]}^{3+}$ :
$[\because n=5]$
$\mu =\sqrt{5\left(5+2\right)}=\sqrt{35}=5.92B.M.$
In $[Co(N{H}_3{)}_6{]}^{3+}$, oxidation state of $Co=+3$.
$C{o}^{3+}:3d^6$, since $N{H}_3$ is a strong field ligand, pairing will take place.
$[Co(N{H}_3{)}_6{]}^{3+}$
$[\because n=0]$
$\mu =\sqrt{0\left(0+2\right)}=0$