Chemistry - Coordination Chemistry Question with Solution | TestHub

The complex is $\left[\mathrm{Fe}(\mathrm{Phen}{)}_2(\mathrm{NCS}{)}_2\right]$. Phen (1,10-phenanthroline) is a strong field ligand, and NCS⁻ (thiocyanate) is a weak field ligand.

At 200 K, the magnetic moment is 4.9 B.M., which corresponds to 4 unpaired electrons (high spin Fe²⁺, $d^6$).

At 50 K, the magnetic moment is 0 B.M., which corresponds to 0 unpaired electrons (low spin Fe²⁺, $d^6$).

This indicates a spin crossover phenomenon.

Statement (i): At 50 K, with 0 B.M., all electrons are paired. For $d^6$ low spin, the electron configuration is $t_{2g}^6{e}_g^0$. At 200 K, with 4.9 B.M., it's high spin, $t_{2g}^4{e}_g^2$. Thus, as the magnetic moment decreases from 4.9 B.M. to 0 B.M., the number of electrons in $t_{2g}$ orbitals increases from 4 to 6. So, (i) is correct.

Statement (ii): The number of unpaired electrons is calculated from the magnetic moment $\mu =\sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons.

At 50 K, $\mu =0$ B.M., so $n=0$.

At 200 K, $\mu =4.9$ B.M.

$4.9\approx \sqrt{4(4+2)}=\sqrt{24}\approx 4.89$ B.M. So, $n=4$.

Thus, the number of unpaired electrons at 50 K and 200 K are 0 and 4 respectively. So, (ii) is correct.

Statement (iii): Both Phen and NCS⁻ are bidentate ligands. The complex has two Phen and two NCS⁻ ligands, resulting in an octahedral geometry at both temperatures. Spin crossover does not change the coordination geometry. So, (iii) is incorrect.

Therefore, only statements (i) and (ii) are correct.

The final answer is $\boxed{C}$