Animated Solution for Chemistry - d and f-Block Elements: In which of the following pairs, the outer most electronic configuration will be the same?

Welcome to the fascinating world of transition metals! When dealing with d-block elements, writing electronic configurations can feel like a delicate dance between the $3d$ and $4s$ orbitals.

The question asks us to find a pair of ions that share the exact same outermost electronic configuration. To solve this, we need to be systematic and recall a few crucial rules.

The Golden Rule of Ionization: While the $4s$ orbital fills before the $3d$ orbital, it is also the first to lose electrons when a transition metal forms a cation.

This happens because once the $3d$ orbitals begin to fill, they drop lower in energy than the $4s$ orbital, making the $4s$ electrons the true outermost valence electrons.

Let's break down the options one by one, keeping an eye out for those notorious exceptions!

Vanadium ($Z=23$) has a neutral configuration of $[Ar]3d^{3}4s^2$. When it loses two electrons to form $V^{2+}$, they leave the $4s$ orbital, resulting in $[Ar]3d^3$.

Chromium ($Z=24$) is our first exception. To achieve a stable half-filled d-subshell, its neutral configuration is $[Ar]3d^{5}4s^1$. Losing one electron gives $C{r}^{+}$ a configuration of $[Ar]3d^5$.

Clearly, $3d^3$ and $3d^5$ do not match.

We already established that $C{r}^{+}$ has a configuration of $[Ar]3d^5$.

Now, let's look at Manganese ($Z=25$). Its ground state is $[Ar]3d^{5}4s^2$. When it forms the $M{n}^{2+}$ ion, it loses its two $4s$ electrons.

This leaves $M{n}^{2+}$ with a configuration of $[Ar]3d^5$.

Look at that! Both $C{r}^{+}$ and $M{n}^{2+}$ share the exact same stable half-filled $[Ar]3d^5$ configuration. We have found our matching pair!

Even though we found the answer, a good scientist always double-checks their work.

Nickel ($Z=28$) starts as $[Ar]3d^{8}4s^2$. Losing two electrons gives $N{i}^{2+}$ a configuration of $[Ar]3d^8$.

Copper ($Z=29$) is another famous exception, favoring a fully filled d-subshell with $[Ar]3d^{10}4s^1$. Losing its single $4s$ electron leaves $C{u}^{+}$ as $[Ar]3d^{10}$.

These configurations ($3d^8$ and $3d^{10}$) are completely different.

Iron ($Z=26$) is $[Ar]3d^{6}4s^2$, so $F{e}^{2+}$ becomes $[Ar]3d^6$.

Cobalt ($Z=27$) is $[Ar]3d^{7}4s^2$. Losing one electron makes $C{o}^{+}$ $[Ar]3d^{7}4s^1$.

Again, no match here.

By carefully applying the rules of electron removal and remembering the exceptions for Chromium and Copper, the path to the answer is clear.

The only pair with identical outermost electronic configurations is $C{r}^{+}$ and $M{n}^{2+}$.