Question

Based on Valence Bond Theory, match the complexes listed in Column I with the number of unpaired electrons on the central metal ion, given in Column II.

• A. A = S B = Q C = R D = P
• B. A = R B = Q C = P D = R
• C. A = R B = P C = Q D = Q
• D. A = S B = C C = Q D = P

Hint:

When determining the number of unpaired electrons, consider the nature of the ligand (strong-field vs weak-field) and the oxidation state of the metal ion. This helps predict the electron pairing behavior.

Answer 1

The Correct Option is C

- Complex ion \[ \text{[Fe}^{3+}\text{]} \] has a 3+ charge, which causes it to have no unpaired electrons (low-spin configuration). Thus, the number of unpaired electrons is 0. This corresponds to (R). - The complex \[ \text{[Fe(CN}_6\text{)}^{3-}] \] is a strong-field ligand (due to cyanide), which leads to pairing of electrons. This results in the central metal ion having 1 unpaired electron. Hence, this matches with (P). - The complex \[ \text{[Fe(H}_2\text{O})_6^{3+}] \] involves water as a weak-field ligand and hence has 5 unpaired electrons on the Fe^3+ ion. Therefore, this corresponds to (Q). - The complex \[ \text{[Fe(CN}_6\text{)}^{4-}] \] also involves cyanide, which is a strong field ligand, and leads to the formation of 4 unpaired electrons on the central ion. This corresponds to (Q). Thus, the correct matching is: A = R, B = P, C = Q, D = Q.