Animated Solution for Chemistry - Periodicity in Properties: Identify the element for which electronic configuration in +3 oxidation state is $[Ar]3d^5$.

When dealing with transition metals and their ions, the most critical concept to master is the order in which electrons are filled and removed. The question asks us to identify an element that, when stripped of three electrons to form a $+3$ oxidation state, leaves behind a stable $[Ar]3d^5$ configuration.

Let's break down the thought process. We are looking for an element in the 3d series (since the core is Argon and the valence shell is 3d). The final ion has $18$ (from Argon) $+5$ (from 3d) $=23$ electrons. Since it is a $+3$ ion, the neutral atom must have had $23+3=26$ electrons.

An atomic number of $Z=26$ corresponds to Iron (Fe). Let's verify this by writing out the full ground state electronic configuration of Iron.

For neutral Iron ($Z=26$), the electrons fill up to the Argon core, followed by the 4s and 3d orbitals. According to the Aufbau principle, the 4s orbital is filled before the 3d orbital, giving us:

Now, we need to form the $F{e}^{3+}$ ion. Here is the catch where many students make a silly mistake: when transition metals ionize, they lose their outermost 's' electrons before they lose their 'd' electrons. Even though the 4s orbital is filled first, it is physically further from the nucleus (higher principal quantum number $n=4$) than the 3d orbital ($n=3$).

To remove three electrons: 1. We first remove the two electrons from the $4s$ orbital. 2. We then remove one electron from the $3d$ orbital.

This leaves us with:

This perfectly matches the configuration given in the question. The $3d^5$ configuration is exactly half-filled, which grants it extra exchange energy and symmetrical stability, making $F{e}^{3+}$ a very common and stable oxidation state for Iron.