Consider the Fe^ 3+ ion in its ground state. Which of the following...

The atomic number of $\mathrm{Fe}$ is $26$.

The electronic configuration of the $\mathrm{Fe}$ atom is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{6}$.

To form the $\mathrm{Fe}^{3+}$ ion, three electrons are removed (two from the $4s$ subshell and one from the $3d$ subshell).

The electronic configuration of $\mathrm{Fe}^{3+}$ is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5}$.

Evaluating the given options:

The occupied subshells are $1s, 2s, 2p, 3s, 3p,$ and $3d$. The total number of occupied subshells is $6$.

The completely filled atomic orbitals are one in $1s$, one in $2s$, three in $2p$, one in $3s$, and three in $3p$. The total number of completely filled orbitals is $1 + 1 + 3 + 1 + 3 = 9$.

The electrons are present in the principal quantum shells $n=1, n=2,$ and $n=3$. The $n=4$ shell is empty. Thus, electrons are distributed across $3$ principal quantum shells, not $4$.

The $3d^{5}$ configuration has $5$ unpaired electrons according to Hund's rule.

Therefore, the statement regarding the $4$ principal quantum shells is incorrect.

Answer: Electrons are distributed across $4$ principal quantum shells.