Question

Arrange the following in increasing value of magnetic moments.
(i) $[Fe(CN)_6]^{4-}$
(ii) $[Fe(CN)_6]^{3-}$
(iii) $[Cr(NH_3)_6]^{3+}$
(iv) $[Ni(H_2O)_4]^{2+}$

• 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 $Fe ^{2+}$
Electronic configuration of $Fe ^{2+}$ is $[ Ar ] 3 d ^{6} 4 s ^{0}$
Since $CN ^{-}$ is a strong field ligand so when it approaches $Fe ^{2+}$ 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 $Fe ^{3+}$
Electronic configuration of $Fe ^{3+}$ is $[ Ar ] 3 d ^{5} 4 s ^{0}$
Since, $CN ^{-}$is a strong field ligand so when it approaches $Fe ^{2+}$ 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) $\left[ Cr \left( NH _{3}\right)_{6}\right]^{3+}$
Electronic configuration of $Cr ^{3}+$ is $[ Ar ] 3 d ^{3} 4 s ^{0}$
Since, $NH _{3}$ is a strong field ligand so it pairs up the electrons when it approaches when it approaches $Cr ^{3+}$
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 $Ni ^{2+}$ is $[ Ar ] 3 d ^{8} 4 s ^{0}$
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)