Question

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

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

Hint:

For transition metal ions, always remove electrons from the \( 4s \) orbital before the \( 3d \) orbital. Then apply Hund's rule to count the number of unpaired electrons correctly.

Answer 1

The Correct Option is D

Step 1: Write the electronic configuration of each ion.
To find the number of unpaired electrons, we first write the electronic configuration of each metal ion. While forming cations, electrons are removed first from the \( 4s \) orbital and then from the \( 3d \) orbital.
\[ \mathrm{V} = [Ar]\,3d^3 4s^2 \Rightarrow \mathrm{V^{2+}} = [Ar]\,3d^3 \] \[ \mathrm{Ni} = [Ar]\,3d^8 4s^2 \Rightarrow \mathrm{Ni^{2+}} = [Ar]\,3d^8 \] \[ \mathrm{Cr} = [Ar]\,3d^5 4s^1 \Rightarrow \mathrm{Cr^{2+}} = [Ar]\,3d^4 \] \[ \mathrm{Mn} = [Ar]\,3d^5 4s^2 \Rightarrow \mathrm{Mn^{2+}} = [Ar]\,3d^5 \] \[ \mathrm{Fe} = [Ar]\,3d^6 4s^2 \Rightarrow \mathrm{Fe^{2+}} = [Ar]\,3d^6 \] \[ \mathrm{Sc} = [Ar]\,3d^1 4s^2 \Rightarrow \mathrm{Sc^{2+}} = [Ar]\,3d^1 \] \[ \mathrm{Mn^{3+}} = [Ar]\,3d^4 \] Step 2: Count the unpaired electrons in each ion.
Now we count the number of unpaired electrons in each \( d \)-configuration using Hund's rule:
\begin{itemize} \item \( \mathrm{V^{2+}} = 3d^3 \Rightarrow 3 \) unpaired electrons \item \( \mathrm{Ni^{2+}} = 3d^8 \Rightarrow 2 \) unpaired electrons \item \( \mathrm{Cr^{2+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Mn^{2+}} = 3d^5 \Rightarrow 5 \) unpaired electrons \item \( \mathrm{Fe^{2+}} = 3d^6 \Rightarrow 4 \) unpaired electrons \item \( \mathrm{Sc^{2+}} = 3d^1 \Rightarrow 1 \) unpaired electron \item \( \mathrm{Mn^{3+}} = 3d^4 \Rightarrow 4 \) unpaired electrons \end{itemize} Step 3: Compare the pairs given in the options.
\begin{itemize} \item (A) \( \mathrm{V^{2+}} \) and \( \mathrm{Ni^{2+}} \): \( 3 \) and \( 2 \) unpaired electrons, so not equal. \item (B) \( \mathrm{Cr^{2+}} \) and \( \mathrm{Mn^{2+}} \): \( 4 \) and \( 5 \) unpaired electrons, so not equal. \item (C) \( \mathrm{Fe^{2+}} \) and \( \mathrm{Sc^{2+}} \): \( 4 \) and \( 1 \) unpaired electrons, so not equal. \item (D) \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \): both have \( 4 \) unpaired electrons, so this pair is correct. \end{itemize} Step 4: Conclusion.
Hence, the pair of ions that have equal number of unpaired electrons is \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \). Both possess the \( 3d^4 \) and \( 3d^6 \) configurations respectively, and each has \( 4 \) unpaired electrons.
Final Answer: \( \mathrm{Mn^{3+}} \) and \( \mathrm{Fe^{2+}} \).