Number of complexes from the following with even number of unpaired "d" electrons is [V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+, [Ni(H2O)6]3+, [Cu(H2O)6]2+ [Given atomic numbers: V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29]
- 1
- 5
- 2
- 4
Answer
(3) 2
Solution
H2O is a weak field ligand ⇒ no pairing (high spin) ⇒ unpaired e⁻ = number of singly filled d orbitals
V3+: 3d2 ⇒ 2 unpaired (even)
Cr2+: 3d4 ⇒ 4 unpaired (even)
Fe3+: 3d5 ⇒ 5 unpaired (odd)
Ni3+: 3d7 ⇒ 3 unpaired (odd)
Cu2+: 3d9 ⇒ 1 unpaired (odd)
∴ Even number of unpaired d electrons: [V(H2O)6]3+ and [Cr(H2O)6]2+ = 2
Answer: (3)