Chemistry - Coordination Chemistry Question with Solution | TestHub

The reaction of $[Ni(H_{2}O{)}_6{]}^{2+}$ with excess ethylenediamine (en) forms the product complex $[Ni(en{)}_3{]}^{2+}$ through the equation:

$$[Ni(H_{2}O{)}_6{]}^{2+}+3en\to [Ni(en{)}_3{]}^{2+}+6H_{2}O$$

Finding $a$: The reactant complex has 1 stereoisomer (achiral, no geometric isomers), while the product complex exists as a d/l pair containing 2 optical isomers, giving $a=\frac{2}{1}=2$.

Finding $b$: Using the balanced reaction coefficients, $b=(\Sigma products-\Sigma reactants)=(1+6)-(1+3)=3$.

Finding $c$: Since en is a strong field bidentate ligand, both reactant and product complexes remain high-spin d⁸ systems with identical $t_{2g}^6{e}_g^2$ electronic configurations, leading to no change in Crystal Field Stabilization Energy ($\Delta CFSE=0$), so $c=0$.

Substituting these values into the expression gives:

$$(\mathbf{a}\times \mathbf{b})+\mathbf{c}=(2\times 3)+0=6$$