Question

Which of the following pair of ions have equal number of unpaired electrons?

• A. \( \text{V}^{2+} \) and \( \text{Ni}^{2+} \)
• B. \( \text{Cr}^{2+} \) and \( \text{Mn}^{2+} \)
• C. \( \text{Fe}^{2+} \) and \( \text{Sc}^{2+} \)
• D. \( \text{Mn}^{3+} \) and \( \text{Fe}^{2+} \)

Hint:

The number of unpaired electrons in transition metal ions is determined by their electron configuration. For ions, electrons are removed first from the outermost orbitals. To determine the number of unpaired electrons, carefully examine the electron configuration of each ion and consider the d-block elements where unpaired electrons are more common.

Answer 1

The Correct Option is B

To determine the number of unpaired electrons, we must first write the electron configuration of the ions: - V\(^2+\) (Vanadium \( Z = 23 \)): The electron configuration of \( \text{V} \) is \( [\text{Ar}] 3d^3 4s^2 \). For \( \text{V}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^3 \), meaning there are 3 unpaired electrons. - Ni\(^2+\) (Nickel \( Z = 28 \)): The electron configuration of \( \text{Ni} \) is \( [\text{Ar}] 3d^8 4s^2 \). For \( \text{Ni}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^8 \), meaning there are 2 unpaired electrons. - Cr\(^2+\) (Chromium \( Z = 24 \)): The electron configuration of \( \text{Cr} \) is \( [\text{Ar}] 3d^5 4s^1 \). For \( \text{Cr}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^4 \), meaning there are 4 unpaired electrons. - Mn\(^2+\) (Manganese \( Z = 25 \)): The electron configuration of \( \text{Mn} \) is \( [\text{Ar}] 3d^5 4s^2 \). For \( \text{Mn}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^5 \), meaning there are 5 unpaired electrons. - Fe\(^2+\) (Iron \( Z = 26 \)): The electron configuration of \( \text{Fe} \) is \( [\text{Ar}] 3d^6 4s^2 \). For \( \text{Fe}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^6 \), meaning there are 4 unpaired electrons. - Sc\(^2+\) (Scandium \( Z = 21 \)): The electron configuration of \( \text{Sc} \) is \( [\text{Ar}] 3d^1 4s^2 \). For \( \text{Sc}^{2+} \), it loses two electrons, so the configuration is \( [\text{Ar}] 3d^1 \), meaning there is 1 unpaired electron. - Mn\(^3+\) (Manganese \( Z = 25 \)): The electron configuration of \( \text{Mn} \) is \( [\text{Ar}] 3d^5 4s^2 \). For \( \text{Mn}^{3+} \), it loses three electrons, so the configuration is \( [\text{Ar}] 3d^4 \), meaning there are 4 unpaired electrons. - Fe\(^2+\) (Iron \( Z = 26 \)): The electron configuration for \( \text{Fe}^{2+} \) was previously calculated as \( [\text{Ar}] 3d^6 \), and there are 4 unpaired electrons. Comparing the number of unpaired electrons: - Cr\(^2+\) (4 unpaired electrons) and Mn\(^2+\) (5 unpaired electrons) do not match in the number of unpaired electrons. Thus, the correct pair with equal unpaired electrons is option (2) \( \text{Cr}^{2+} \) and \( \text{Mn}^{2+} \).