Question

The correct order of magnetic moments (spin only values in (B.M.) among the following is:

[Atomic numbers $Mn$=$25$, $Fe$= $26$, $Co$=$27$]

• A. $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>Fe(CN{)}_6{]}^{4-}$
• B. $[MnC{l}_4{]}^{2-}>[Fe(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}$
• C. $\left[Fe\right(CN{)}_6{]}^{4-}>[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}$
• D. $\left[Fe\right(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}>[MnC{l}_4{]}^{2-}$

Answer 1

Answer: (A)

$[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>Fe(CN{)}_6{]}^{4-}$

Magnetic moment of $[MnC{1}_4{]}^{2-}$ is 5.9 BM.

$Mn$ is in +2 oxidation state having configuration of $\left[Ar\right]3d^5$. It has 5 unpaired electrons.

$\mu =\sqrt{n(n+2)}$

$=\sqrt{\left(5\right(5+2)}$

$=5.9BM$

Magnetic moment of $[CoC{1}_4{]}^{2-}$ is 3.9 BM.

$Co$ is in +2 oxidation state having configuration of $\left[Ar\right]3d^7$. It has 3 unpaired electrons.

$\mu =\sqrt{n(n+2)}$

$=\sqrt{\left(3\right(3+2)}$

$=3.9BM$

Magnetic moment of $[Fe{CN}_4{]}^{2-}$ is 0.0 BM.

$Fe$ is in +2 oxidation state having configuration of $\left[Ar\right]3d^6$. In this complex $C{N}^{-}$ ion is a strong ligand and all electrons are paired and there are no unpaired electrons.

$\mu =\sqrt{n(n+2)}$

$=\sqrt{\left(0\right(0+2)}$

$=0BM$

The decreasing order of the magnetic moments is:

$[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>Fe(CN{)}_6{]}^{4-}$

Hence, option $A$ is correct.