Unpaired Electrons in Complexes – Rankers Physics

Uncategorized: Practice Problem & Solution

Match List-I with List-II. **List-I (Complex)** a. \([\text{Fe}(\text{CN})_6]^{3-}\) b. \([\text{Co}(\text{H}_2\text{O})_6]^{3+}\) c. \([\text{FeF}_6]^{3-}\) d. \([\text{NiCl}_4]^{2-}\) **List-II (Number of unpaired electrons)** (i) Zero (ii) 5 (iii) 2 (iv) 1
a(i), b(iv), c(ii), d(iii)
a(iv), b(i), c(iii), d(ii)
a(ii), b(iii), c(iv), d(i)
a(iv), b(i), c(ii), d(iii)

Solution Explained:

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

\([\text{Fe}(\text{CN})_6]^{3-}\) has strong field ligands and 1 unpaired electron. \([\text{Co}(\text{H}_2\text{O})_6]^{3+}\) is low spin with zero unpaired electrons. \([\text{FeF}_6]^{3-}\) has weak field ligands and 5 unpaired electrons, while \([\text{NiCl}_4]^{2-}\) has 2 unpaired electrons.

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