Question

Given below are two statements: Statement I: Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc.
Statement II: The correct order of the following complexes in terms of CFSE is [Co(H₂O)₆]²⁺<[Co(H₂O)₆]³⁺<[Co(en)₃]³⁺.
In light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true.
• B. Both Statement I and Statement II are false.
• C. Statement I is true but Statement II is false.
• D. Statement I is false but Statement II is true.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Statement I relies on the stability of electronic configurations. The Third Ionization Energy (IE\(_3\)) involves removing an electron from a \(+2\) cation. High stability of the \(+2\) ion means a high IE\(_3\).
Statement II relies on Crystal Field Stabilization Energy (CFSE). CFSE depends on the oxidation state of the central metal and the field strength of the ligands.
Step 2: Key Formula or Approach:
Write electronic configurations for the \(+2\) ions to assess IE\(_3\).
For CFSE (\(\Delta_o\)): It increases with higher oxidation state of the central metal ion and with stronger field ligands (Spectrochemical series).
Step 3: Detailed Explanation:
Evaluate Statement I:
Sc: \([\text{Ar}] 3d^1 4s^2 \implies \text{Sc}^{2+}\) is \([\text{Ar}] 3d^1\). Removing the 3rd electron leaves a highly stable noble gas core (\(\text{Ar}\)). Thus, its IE\(_3\) is exceptionally low (lowest in the 3d series).
Zn: \([\text{Ar}] 3d^{10} 4s^2 \implies \text{Zn}^{2+}\) is \([\text{Ar}] 3d^{10}\). This is a completely filled, extremely stable d-subshell. Disrupting this full shell requires enormous energy. Thus, its IE\(_3\) is exceptionally high (highest in the 3d series).
Statement I is TRUE.
Evaluate Statement II:
We compare the Crystal Field Splitting Energy (\(\Delta_o\)) of three complexes.
1) \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\): Cobalt is in +2 oxidation state. \(\text{H}_2\text{O}\) is a weak field ligand.
2) \([\text{Co}(\text{H}_2\text{O})_6]^{3+}\): Cobalt is in +3 oxidation state. A higher oxidation state strongly pulls the ligands closer, significantly increasing the splitting energy (\(\Delta_o\)) compared to the +2 state, even with the same weak ligand.
3) \([\text{Co}(\text{en})_3]^{3+}\): Cobalt is in +3 oxidation state, but Ethylenediamine (\(\text{en}\)) is a much stronger field ligand than \(\text{H}_2\text{O}\) (as per the spectrochemical series). This results in the highest splitting energy.
Therefore, the order of CFSE is: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}<[\text{Co}(\text{H}_2\text{O})_6]^{3+}<[\text{Co}(\text{en})_3]^{3+}\).
Statement II is TRUE.
Step 4: Final Answer:
Both Statement I and Statement II are true.