Question

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

• A. 1.73 BM and $-2{△}_0$
• B. 2.84 BM and $-1.6{△}_0$
  
• C. 0 BM and $-2.4{△}_0$
• D. 5.92 BM and 0

Answer 1

The correct option is A 1.73 BM and $-2{△}_0$

Number of geometrical isomers (n) = 3
$\left[Fe\right(CN{)}_6{]}^{n-6}=\left[Fe\right(CN{)}_6{]}^{3-6}=\left[Fe\right(CN{)}_6{]}^{-3}$
This implies, that iron is in its +3 Oxidation state.
$F{e}^{3+}\left(26\right):\left[Ar\right]3d^5$
$C{N}^{-}$ is a strong ligand in $\left[Fe\right(CN{)}_6{]}^{3-}$ and causes pairing. Hence according to CFT configuration will be $t_2^5{e}_g^o$
Hence, there is only 1 unpaired electron, i.e, n = 1 in $\sqrt{n(n+2)}=\sqrt{3}=1.73B.M$
$CFSE=(-0.4\times n{}_{t2g}+0.6\times n_{eg}){△}_0$
$=(-0.4\times 5+0.6\times 0){△}_0$
$=-2{△}_0$