Question

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

• A. $\left(I\right)<\left(II\right)<\left(III\right)<\left(IV\right)$
• B. $\left(I\right)<\left(II\right)<\left(IV\right)<\left(III\right)$
• C. $\left(II\right)<\left(III\right)<\left(I\right)<\left(IV\right)$
• D. $\left(IV\right)<\left(III\right)<\left(II\right)<\left(I\right)$

Answer 1

The correct option is B $\left(I\right)<\left(II\right)<\left(IV\right)<\left(III\right)$

1) [Fe(CN)${}_6{]}^{4-}$

Here, Fe is present as Fe${}^{2+}$

Electronic configuration of Fe${}^{2+}$ is [Ar]3d${}^{6}4s^0$

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) [Fe(CN)${}_6{]}^{3-}$

Here, Fe is present as Fe${}^{3+}$

Electronic configuration of Fe${}^{3+}$ is [Ar]3d${}^{5}4s^0$

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) [Cr(NH${}_3{)}_6{]}^{3+}$

Electronic configuration of Cr${}^{3+}$ is [Ar]3d${}^{3}4s^0$

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)[Ni(H${}_{2}O{)}_4{]}^{2+}$

Electronic configuration of Ni${}^{2+}$ 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)