Solution

$_{22}Ti \rightarrow [Ar]\,4s^2\,3d^2$ ; $Ti^{2+} \rightarrow [Ar]\,3d^2$ Number of unpaired electrons, n = 2 Spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.84$ BM

Coordination Compounds in past papers

104 questions from this chapter have appeared across 27 exam years.