# The calculated magnetic moments (spin only value) for species

Question:

The calculated magnetic moments (spin only value) for species $\left[\mathrm{FeCl}_{4}\right]^{2-},\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ and $\mathrm{MnO}_{4}^{2-}$ respectively are :

1. $5.92,4.90$ and $0 \mathrm{BM}$

2. $5.82, \mathrm{O}$ and $0 \mathrm{BM}$

3. $4.90,0$ and $1.73 \mathrm{BM}$

4. $4.90,0$ and $2.83 \mathrm{BM}$

Correct Option: , 3

Solution:

$\left[\mathrm{FeCl}_{4}\right]^{2-} \mathrm{Fe}^{2+} 3 \mathrm{~d}^{6} \rightarrow 4$ unpaired electron. as $\mathrm{Cl}^{-}$in a weak field liquid.

$\mu_{\text {spin }}=\sqrt{24} 8 \mathrm{M}$

$=4.9 \mathrm{BM}$

$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-} \mathrm{Co}^{3+} 3 \mathrm{~d}^{6} \rightarrow$ for $\mathrm{Co}^{3+}$ with coodination no. $6 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is strong field ligend $\backslash \&$ causes pairing $\backslash \&$ hence no. unpaired electron

$\mu_{\mathrm{spin}}=0$

$\left[\mathrm{MnO}_{4}\right]^{2-} \mathrm{Mn}^{+6}$ it has one unpaired electron.

$\mu_{\text {spin }}=\sqrt{3} \mathrm{BM}$