$\left[\mathrm{P}\mathrm{d}\right(\mathrm{F}\left)\right(\mathrm{C}\mathrm{l}\left)\right(\mathrm{B}\mathrm{r}\left)\right(\mathrm{I}){]}^{2–}$ has 'n' number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy $\mathrm{(CFSE)}$ of ${{\mathrm{[Fe(CN)}}_6]}^{\mathrm{n–6}}$ respectively are
[Note: Ignore the pairing energy]

• A

  5.92 BM and ${\mathrm{\Delta }}_0$
• Correct answer
  B

  1.73 BM and –2.0 ${\mathrm{\Delta }}_0$
• C

  2.84 BM and –1.6 ${\mathrm{\Delta }}_0$
• D

  0 BM and –2.4 ${\mathrm{\Delta }}_0$

Solution

No of G.I in $[\mathrm{P}\mathrm{d}\mathrm{C}\mathrm{l}\mathrm{F}\mathrm{B}\mathrm{r}\mathrm{I}{]}^{2–}$ is 3.
Hence $\mathrm{n}=3$
So $\left[\mathrm{F}\mathrm{e}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{\mathrm{n}–6}=\left[\mathrm{F}\mathrm{e}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{–3}$
$\mathrm{F}{\mathrm{e}}^{3+}=3{\mathrm{d}}^{5}4{\mathrm{s}}^0$
As $\stackrel{-}{\mathrm{C}\mathrm{N}}$ is a strong field ligand. So, according to crystal field theory the configuration is as follows
${\text{Fe}}^{3+}={\text{t}}_{2\text{g}}^{2,2,1},{\text{e}}_{\text{g}}{}^{0,0}$

$\mathrm{n}=1$
Magnetic moment $=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.73$ B.M.
CFSE $=-0.4{\mathrm{\Delta }}_0\times \mathrm{n}{\mathrm{t}}_{2\mathrm{g}}+0.6{\mathrm{\Delta }}_0\times {\mathrm{n}}_{\mathrm{e}\mathrm{g}}$
CFSE $-0.4{\mathrm{\Delta }}_0\times 5=-2.0{\mathrm{\Delta }}_0$

Hence, the correct answer is option: 'B'.