$0.001 \mathrm{~mol}$ of…

$0.001 \mathrm{~mol}$ of $\mathrm{Cr}\left(\mathrm{NH}_{3}ight)_{5}\left(\mathrm{NO}_{3}ight)\left(\mathrm{SO}_{4}ight)$ was passed through a cation exchanger and the acid coming out of it required $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ for neutralisation. Hence, the complex is
  1. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}ight)_{5} \mathrm{SO}_{4}ight] \mathrm{NO}_{3}$
  2. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}ight)_{5} \mathrm{NO}_{3}ight] \mathrm{SO}_{4}$
  3. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}ight)_{5}ight]\left(\mathrm{SO}_{4}ight)\left(\mathrm{NO}_{3}ight)$
  4. None of these

Solution

The complex is $\left[\mathbf{C o N} \mathbf{O}_{3}\left(\mathbf{N} \mathbf{H}_{3}ight)_{5}ight] \mathbf{S O}_{4}$
$
\frac{0.020 \mathrm{~L} \times 0.1 \mathrm{~mol} / \mathrm{LNaOH} \times 1 \mathrm{~mol} \text { acid }}{1 \mathrm{~mol} \mathrm{NaOH}}=
$
$0.0020$ mol acid
$
\frac{0.0020 \mathrm{~mol}}{10^{-3} \mathrm{~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 $\left[\mathbf{C o N} \mathrm{O}_{3}\left(\mathrm{NH}_{3}ight)_{5}ight] \mathrm{SO}_{4}$. /

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more COORDINATION COMPOUNDS questions on Aicharya