Question

${\left[\mathrm{Pd}\right(\mathrm{F}\left)\right(\mathrm{Cl}\left)\right(\mathrm{Br}\left)\right(\mathrm{I}\left)\right]}^{2-}$,has n number of geometrical Isomers. Then, the spin-only magnetic moment and crystal field stabilization energy $\left[\mathrm{CFSE}\right]$of $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{\mathrm{n}-6}$ , respectively,

• A. $1.73\mathrm{BM}\mathrm{and}–2{\mathrm{\Delta }}_0$
• B. $2.84\mathrm{BM}\mathrm{and}–1.6{\mathrm{\Delta }}_0$
• C. $0\mathrm{BM}\mathrm{and}–2.4{\mathrm{\Delta }}_0$
• D. $5.92\mathrm{BM}\mathrm{and}0$

Answer 1

The correct option is A

$1.73\mathrm{BM}\mathrm{and}–2{\mathrm{\Delta }}_0$

Explanation for the correct option:

A: $1.73\mathrm{BM}\mathrm{and}–2{\mathrm{\Delta }}_0$

• Possible Isomers of $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{\mathrm{n}-6}$

• Number of geometrical isomers, $\left(\mathrm{n}\right)=3$

$[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{\mathrm{n}-6}=[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{3-6}=[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{-3}$

• This implies, that Iron is in its $+3$ oxidation state.

${\mathrm{Fe}}^{3+}\left(26\right)=\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$

• ${\mathrm{CN}}^{-}$is a strong ligand in$[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{-3}$ and causes pairing.
• Hence, according to $\mathrm{CFT}$, the configuration will be${{\mathrm{t}}_{2\mathrm{g}}}^5{{\mathrm{e}}^0}_{\mathrm{g}}$
• Hence, there is only 1 unpaired electron, i.e,

$\begin{array}{rcl}\mathrm{n}& =& 1\surd \mathrm{n}(\mathrm{n}+2)\\ & =& \surd 3\\ & =& 1.73\mathrm{B}.\mathrm{M}\\ \mathrm{CFSE}& =& (-0.4\times {\mathrm{nt}}_{2\mathrm{g}}+0.6\times \mathrm{n}{\mathrm{e}}_{\mathrm{g}}){\mathrm{\Delta }}_0\\ & =& (-0.4\times 5+0.6\times 0){\mathrm{\Delta }}_0\\ & =& -2{\mathrm{\Delta }}_0\end{array}$

Therefore, Option (A) $1.73\mathrm{BM}\mathrm{and}–2{\mathrm{\Delta }}_0$ is the correct option.