Question

For the octahedral complexes of $F{e}^{3+}inSC{N}^{–}$ (thiocyanato-S) and in $C{N}^{–}$ ligand environments, the difference between the spin-only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is
[Atomic number of $Fe=26$]

Answer 1

$Fe\left(26\right)\to [Ar{]}_{18}3d^{6}4s^2$
$F{e}^{3+}\to [Ar{]}_{18}3d^{5}4s^0$
$SC{N}^{–}$ is weak field ligand hence pairing will not occur.
$\therefore F{e}^{3+}\to$

Unpaired electrons $=5$
Magnetic moment
$=\sqrt{5(5+2)}B.M=\sqrt{35}B.M=5.92B.M$
$C{N}^{–}$ is strong field ligand hence pairing will take place.
$\therefore F{e}^{3+}\to$

Unpaired electrons $=1$
Magnetic moment $=\sqrt{1(1+2)}B.M=\sqrt{3}B.M=1.732$
Difference $=5.92–1.732=4.188$
Hence answer is (4).