Chemistry - Coordination Chemistry Question with Solution | TestHub

In $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$ (brown-ring complex), NO is present as ${\mathrm{NO}}^{+}$, so iron is Fe(I): $3d^{6}4s^1$.

The 4s electron de-excites into the 3d subshell, giving $3d^7$ arranged as $t_{2g}^5{e}_g^2$ (all seven electrons paired into the inner d orbitals as far as possible). This leaves 4s, 4p and 4d orbitals free for $s{p}^3{d}^2$ hybridisation with the six ligands.

Hence the complex has 5 electrons in $t_{2g}$ orbitals: the assertion is true, the reason is true, and the reason correctly explains the assertion. Answer: (A)