Question

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

• A. 1.73 BM and -2.0 ${\Delta }_0$
• B. 2.84 BM and -1.6 ${\Delta }_0$
• C. 0 BM and - 2.4 ${\Delta }_0$
• D. 5.92 BM and 0

Answer 1

Answer: (A)

${\left[Pb\left(F\right)\left(CI\right)\left(Br\right)\left(I\right)\right]}^{2-}$ have three geometrical isomer so formula for ${\left[Fe{\left(CN\right)}_6\right]}^{n-6}is{\left[Fe{\left(CN\right)}_6\right]}^{3-}$ and CFSE for this complex is
$F{e}^{3\oplus }\Rightarrow 3d^{5}4S^{\circ }$
Magnetic Moment =$\sqrt{3}=1.73B.M$
$CFSE=\left[\left(-0.4\times 5\right)+\left(0.6\times 0\right)\right]{\Delta }_0=-2.0{\Delta }_0$