Q40 / 180 · KCET 2016 · Chemistry
\( M n^{2+} \) compounds are more stable than \( F e^{2+} \) compounds towards oxidation to their \( +3 \) state,
because
because
- Correct answerA\( M n^{2+} \) is more stable with high \( 3 r d \) lonisation energy.
- B\( M n^{2+} \) is bigger in size.
- C\( M n^{2+} \) has completely filled d-orbitals.
- D\( M n^{2+} \) does not exist.
Solution
Electronic configuration of \( M n^{2+} \) is \( 4 s^{0} 3 d^{5} \) :
\begin{array}{|l|l|l|l|l|} \hline \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) \\ \hline \end{array}
Electronic configuration of \( F e^{2+} \) is \( 4 s^{0} 3 d^{6}: \)
\begin{array}{|ll|l|l|l|l|} \hline \( 1 \) & \( \downarrow \) & \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) \\ \hline \end{array}
Since, in \( M n^{2+} \) the d-orbital is completely half-filled, therefore, its third ionization energy is high and hence, it stable
towards oxidation to \( +3 \) oxidation states, whereas, in \( \mathrm{Fe}^{2+} \), the d-orbital is partially filled and only one electron is paired,
therefore, its third ionization energy is lower than \( \mathrm{Mn}^{2+} \) ion and hence, it is less stable towards \( +3 \) oxidation states.
\begin{array}{|l|l|l|l|l|} \hline \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) \\ \hline \end{array}
Electronic configuration of \( F e^{2+} \) is \( 4 s^{0} 3 d^{6}: \)
\begin{array}{|ll|l|l|l|l|} \hline \( 1 \) & \( \downarrow \) & \( 1 \) & \( 1 \) & \( 1 \) & \( 1 \) \\ \hline \end{array}
Since, in \( M n^{2+} \) the d-orbital is completely half-filled, therefore, its third ionization energy is high and hence, it stable
towards oxidation to \( +3 \) oxidation states, whereas, in \( \mathrm{Fe}^{2+} \), the d-orbital is partially filled and only one electron is paired,
therefore, its third ionization energy is lower than \( \mathrm{Mn}^{2+} \) ion and hence, it is less stable towards \( +3 \) oxidation states.
Hence, the correct answer is option: 'A'.