Question

Arrange the following in increasing value of magnetic moments.
(i) $[Fe(CN{)}_6{]}^{4-}$
(ii) $[Fe(CN{)}_6{]}^{3-}$
(iii) $[Cr(N{H}_3{)}_6{]}^{3+}$
(iv) $[Ni(H_{2}O{)}_4{]}^{2+}$

• A. $(i)<(ii)<(iii)<(iv)$
• B. $(i)<(ii)<(iv)<(iii)$
• C. $(ii)<(iii)<(i)<(iv)$
• D. $(iii)<(i)<(ii)<(iv)$

Answer 1

Answer: (B)

1) ${\left[Fe(CN{)}_6\right]}^{4-}$
Here, $Fe$ is present as $F{e}^{2+}$
Electronic configuration of $F{e}^{2+}$ is $[Ar]3d^{6}4s^0$
Since $C{N}^{-}$ is a strong field ligand so when it approaches $F{e}^{2+}$ it pair up the $e^{-}s$ and result in the formation of low spin complex.
No. of unpaired $e^{-}s=0$
So, magnetic moment for $n$ unpaired $e^{-s}$
$=\sqrt{n(n+2)}BM$
$=0$
2) ${\left[Fe(CN{)}_6\right]}^{3-}$
Here, $Fe$ is present as $F{e}^{3+}$
Electronic configuration of $F{e}^{3+}$ is $[Ar]3d^{5}4s^0$
Since, $C{N}^{-}$is a strong field ligand so when it approaches $F{e}^{2+}$ it pair up the $e^{-}s$ so,
No. of unpaired $e^{-}s=1$
So, magnetic moment for $n$ unpaired $e^{-s}$
$=\sqrt{1\times (1+2)}$
$=\sqrt{3}BM$.
3) ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$
Electronic configuration of $C{r}^3+$ is $[Ar]3d^{3}4s^0$
Since, $N{H}_3$ is a strong field ligand so it pairs up the electrons when it approaches when it approaches $C{r}^{3+}$
No. of unpaired $e^{-}s=3$
So, magnetic moment $=\sqrt{3(3+2)}$
$=\sqrt{1\times 5}$
$=\sqrt{15}BM$
4) ${\left[Ni{\left(H_{2}O\right)}_4\right]}^{2+}$
Electronic configuration of $N{i}^{2+}$ is $[Ar]3d^{8}4s^0$
Since, $H_{2}O$ is a weak field ligand.
it does not pair up the $e^{-}s$ and forms outer orbital complex.
No. of unpaired $e^{-}s=2$
So, magnetic moment $=\sqrt{2(2+2)}$
$=\sqrt{8}BM$
But because of orbital effect it will be little lesser. So, correct order is iii)>iv)> ii)> i)