Search any question & find its solution
Question:
Answered & Verified by Expert
Among the following the compound that is both paramagnetic and coloured is
Options:
Solution:
2611 Upvotes
Verified Answer
The correct answer is:
$\mathrm{VOSO}_{4}$
(d) $\operatorname{In} \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \mathrm{Cu}$ is in $+1$ oxidation state hence has no unpaired electron hence colourless and diamagnetic.
(b) $\mathrm{In}\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{TiCl}_{6}\right] \mathrm{Ti}$ is in $+4$ oxidation state. hence has no unpaired electron hence colourless and diamagnetic.
(c) $\operatorname{In} \operatorname{VOSO}_{4}, V$ is in $+4$ oxidation state hence has one unpaired electron, thus it is coloured and paramagnetic.
(a) $\operatorname{In} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, Cr is in $+6$ oxidation. hence has no unpaired electron and thus it is diamagnetic. Though $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ has no unpaired electron but it is coloured. This is due to charge transfer spectrum.
(b) $\mathrm{In}\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{TiCl}_{6}\right] \mathrm{Ti}$ is in $+4$ oxidation state. hence has no unpaired electron hence colourless and diamagnetic.
(c) $\operatorname{In} \operatorname{VOSO}_{4}, V$ is in $+4$ oxidation state hence has one unpaired electron, thus it is coloured and paramagnetic.
(a) $\operatorname{In} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, Cr is in $+6$ oxidation. hence has no unpaired electron and thus it is diamagnetic. Though $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ has no unpaired electron but it is coloured. This is due to charge transfer spectrum.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.