Question: The transition element which shows both $+1$ and $+2$ oxidation states is:

Step 1 (Governing Principle / Electronic Configuration): [0.5 Mark]
The oxidation states of transition elements are determined by the availability of electrons in the $(n-1)d$ and $ns$ orbitals. For Copper ($\text{Cu}$, atomic number $29$), the ground state electronic configuration is:

The stability of the $+1$ oxidation state arises from the completely filled $3d^{10}$ configuration, while the $+2$ state is common due to the hydration/lattice energy compensation for the removal of the second electron.

Step 2 (Step-by-Step Analysis of Options): [0.25 Mark]

• (A) Sc ($3d^1 4s^2$): Primarily shows $+3$ oxidation state.
• (B) Mn ($3d^5 4s^2$): Shows a wide range of oxidation states from $+2$ to $+7$.
• (C) Cu ($3d^{10} 4s^1$): Can lose the $4s^1$ electron to form $\text{Cu}^+$ ($3d^{10}$) and a further $3d$ electron to form $\text{Cu}^{2+}$ ($3d^9$).
• (D) Zn ($3d^{10} 4s^2$): Shows only $+2$ oxidation state due to the stability of the fully filled $d^{10}$ configuration.

Step 3 (Final Concluded Answer): [0.25 Mark]
Based on the electronic configuration and experimental evidence, Copper ($\text{Cu}$) is the only element among the choices that exhibits both $+1$ and $+2$ oxidation states.

Correct Option: (C) $\text{Cu}$

💡 EL10 Concept Insight: Think of Copper like a person with two pockets of coins; it can easily give away the one coin in its outer pocket to become stable ($\text{Cu}^+$), but it's also willing to give up one more coin from its inner pocket if the environment is right, making it $\text{Cu}^{2+}$.