Paper: CSIR-NET June 2013 · Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Molecular Orbital Ligand Field · Marks: 2 · Difficulty: Medium
Which of the pairs will generally result in tetrahedral coordination complexes, when ligands are $\mathrm{Cl}^{-}$ or $\mathrm{OH}^{-}$.(A) Be(II), Ba(II) (B) Ba(II), Co(II) (C) Co(II), Zn(II) (D) Be(II), Zn(II)
(a)A and B
(b)B and C
(c)C and D
(d)A and D
Answer
Answer: D ✓ checked by 4AB · confidence medium
The source book printed C; on checking, D is correct — see the explanation.
Explanation
Small Be(II) and d¹⁰ Zn(II) give tetrahedral [BeCl₄]²⁻/[Be(OH)₄]²⁻ and [ZnCl₄]²⁻/[Zn(OH)₄]²⁻. Large Ba(II) prefers high coordination numbers, and Co(II) hydroxide is octahedral.