Question

The ground state and the maximum number of spin-allowed electronic transitions possible in a $Co^{2+}$ tetrahedral complex, respectively, are

• A. $^4A_2$ and 3
• B. $^4 T _1$ and 2
• C. $^4 A _2$ and 2
• D. $^4 T_2$ and 3

Answer 1

The Correct Option is A

To determine the ground state and the maximum number of spin-allowed electronic transitions possible for a \(Co^{2+}\) tetrahedral complex, we need to consider the electronic configuration and crystal field theory.

1. **Electronic Configuration and Ground State:**
   • \(Co^{2+}\) ion is derived from \(Co\) which has the electronic configuration \([Ar] 3d^7 4s^2\). As \(Co^{2+}\), it loses two electrons, resulting in \([Ar] 3d^7\).
   • In a tetrahedral complex, the crystal field splitting pattern is such that the \(e\) orbitals (higher energy) are above the \(t_2\) orbitals (lower energy).
   • \(3d^7\) gives a high spin tetrahedral configuration because tetrahedral field does not provide enough splitting to pair electrons (weak field ligands).
   • The ground state term for a high-spin \(d^7\) configuration in a tetrahedral field is \(^4A_2\).
2. **Determining Spin-Allowed Transitions:**
   • Spin-allowed transitions for any dⁿ configuration are governed by selection rules which allow transitions between terms of the same spin multiplicity.
   • For a \(d^7\) configuration in a tetrahedral field, the possible terms based on Hund's rules and considering weak field are \(^4T_1\), \(^4T_2\), and \(^4A_2\).
   • Transitions possible from the ground state \(^4A_2\) are to \(^4T_1\) and \(^4T_2\), thus counting as three transitions: \(^4A_2 \rightarrow ^4T_1\), \(^4A_2 \rightarrow ^4T_2\), and a transition back to itself, \(^4A_2 \rightarrow ^4A_2\).

Therefore, the correct answer is \(^4A_2\) and 3 spin-allowed transitions. This matches the option $^4A_2$ and 3.