Question

Which of the following outermost electronic configuration of the element shows the highest oxidation state?

• A. \(3d^3 4s^2\)
• B. \(3d^5 4s^1\)
• C. \(3d^5 4s^2\)
• D. \(3d^6 4s^2\)
• E. \(3d^2 4s^2\)

Hint:

Maximum oxidation state in transition metals is often equal to total valence electrons, especially for elements with half-filled \(d^5\) configuration like Mn.

Answer 1

The Correct Option is C

Step 1: Understand oxidation state in transition elements.
The oxidation state of transition metals depends on the number of electrons that can be lost from both \(ns\) and \((n-1)d\) orbitals. Higher the number of available valence electrons, higher the possible oxidation state.

Step 2: Identify total valence electrons.
For transition metals: \[ \text{Valence electrons} = (n-1)d + ns \] So we count total electrons available for bonding.

Step 3: Evaluate each configuration.
(A) \(3d^3 4s^2 \rightarrow 5\) electrons
(B) \(3d^5 4s^1 \rightarrow 6\) electrons
(C) \(3d^5 4s^2 \rightarrow 7\) electrons
(D) \(3d^6 4s^2 \rightarrow 8\) electrons (but stability issues arise)
(E) \(3d^2 4s^2 \rightarrow 4\) electrons

Step 4: Consider stability of half-filled configuration.
The configuration \(3d^5 4s^2\) corresponds to manganese (Mn), which can exhibit the highest oxidation state of \(+7\). This is due to the stable half-filled \(d^5\) configuration.

Step 5: Compare with others.
Even though some configurations appear to have more electrons, not all \(d\)-electrons participate effectively due to stability factors. Mn uniquely shows \(+7\) oxidation state.

Step 6: Identify the highest oxidation state.
Among all options, \(3d^5 4s^2\) gives the maximum oxidation state.

Step 7: Final answer.
\[ \boxed{3d^5 4s^2} \] which matches option \((3)\).