Question

$[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}_2[\mathrm{Fe}(\mathrm{CN}){}_6]$
Cationic and anionic species spin only magnetic moment in Bohr magneton is :

• A. 4.89 and 0
• B. 5.9 and 0
• C. 0 and 4.89
• D. 2.82 and 2.82

Answer 1

Answer: (A)

$[\mathrm{Fe}{}^{+2}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{2+}[\mathrm{Fe}{}^{+2}(\mathrm{CN}){}_6]{}^{4-}$
$\mathrm{Fe}{}^{2+}\Rightarrow [\mathrm{Ar}]3\mathrm{d}{}^6\mathrm{Fe}{}^{2+}\Rightarrow [\mathrm{Ar}]3\mathrm{d}{}^6$
$\mathrm{H}{}_2\mathrm{O}\Rightarrow \mathrm{WFL}\mathrm{CN}{}^{-}\Rightarrow \mathrm{SFL}$
Number of unpaired electrons = 4 Number of unpaired electrons = 0
$\mu =\sqrt{n(n+2)}=4.89BM\mu =0$