Question

0.001 mol of $Co\left(N{H}_3{)}_5\right(N{O}_3\left)\right(S{O}_4)$ was passed through a cation exchanger and the acid coming out of its required 20 ml of 0.1 M NaOH for neutralisation. Hence, the complex is :

• A. $\left[CoS{O}_4\right(N{H}_3{)}_5]N{O}_3$
• B. $\left[CoN{O}_3\right(N{H}_3{)}_5]S{O}_4$
• C. $\left[Co\right(N{H}_3{)}_5\left]\right(S{O}_4\left)\right(N{O}_3)$
• D. None of these

Answer 1

The correct option is B $\left[CoN{O}_3\right(N{H}_3{)}_5]S{O}_4$

Option (B) is correct.

The complex is $\left[CoN{O}_3\right(N{H}_3{)}_5]S{O}_4$.

$\frac{0.020L\times 0.1mol/LNaOH\times 1molacid}{1molNaOH}=0.0020molacid$

$\frac{0.0020mol}{{10}^{-3}mol}=2$

As $2$ moles of acid in the exchanger replaced each $1$ mole of complex cations, so the complex must have an oxidation state of $+2$, so the complex will be $\left[CoN{O}_3\right(N{H}_3{)}_5]S{O}_4$.