JEE Advance - Chemistry (2010 - Paper 1 Offline - No. 19)
The ionisation isomer of $$\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$$ is
$$[Cr{({H_2}O)_4}({O_2}N)]C{l_2}$$
$$[Cr{({H_2}O)_4}C{l_2}](N{O_2})$$
$$[Cr{({H_2}O)_4}Cl(ONO)]Cl$$
$$[Cr{({H_2}O)_4}C{l_2}(N{O_2})]{H_2}O$$
Explanation
The ionisation isomers give different ions in solution. In the complex given in option (B), Cl$$^-$$ is replaced by NO$$_2^ - $$ in ionisation sphere.
$$\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$$ and $$[Cr{({H_2}O)_4}C{l_2}](N{O_2})$$ have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionisation isomers.
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