Question

Sol. [Key Concept: Spin-only magnetic moment μ = √n(n+2) BM]

$In\left\lbrack Ti\left( H_{2}O \right)_{6} \right\rbrack^{3 +} \Rightarrow {Ti}^{3 +} \Rightarrow \lbrack Ar\rbrack 4{\text{ }s}^{0}3{\text{ }d}^{1}$

Number of unpaired electrons $= 1$

$${ln\left\lbrack Mn(CN)_{6} \right\rbrack^{3 -} \Rightarrow {Mn}^{3 +} \Rightarrow \lbrack Ar\rbrack 4{\text{ }s}^{0}3{\text{ }d}^{4} } {Number\ of\ unpaired\ electrons\ = 2 }$$

$$ln\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -} \Rightarrow {Fe}^{3 +} \Rightarrow \lbrack Ar\rbrack 4{\text{ }s}^{0}3{\text{ }d}^{5} $$

Number of unpaired electrons = 1