1. We need the element whose electronic configuration in the $+3$ oxidation state is $[Ar]3d^5$

2. In transition metals, electrons are removed first from the $4s$ orbital and then from the $3d$ orbital.

3. Check each option:

Option A: Ru

Ruthenium ($Z=44$) has configuration approximately $[Kr]4d^75s^1$ After removing 3 electrons, it cannot give $[Ar]3d^5$. So, A is incorrect.

Option B: Mn

Manganese ($Z=25$) has ground-state configuration $[Ar]3d^54s^2$ For $Mn^{3+}$, remove two $4s$ electrons and one $3d$ electron: $Mn^{3+}=[Ar]3d^4$ So, B is incorrect.

Option C: Co

Cobalt ($Z=27$) has ground-state configuration $[Ar]3d^74s^2$ For $Co^{3+}$: $Co^{3+}=[Ar]3d^6$ So, C is incorrect.

Option D: Fe

Iron ($Z=26$) has ground-state configuration $[Ar]3d^64s^2$ For $Fe^{3+}$, remove two $4s$ electrons and one $3d$ electron: $Fe^{3+}=[Ar]3d^5$ So, D is correct.

4. Therefore, the element is $\boxed{Fe}$

5. Comparison with stored correct answer:

• Stored correct answer: D
• Derived answer: D
• They agree.