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Cold ferrous sulphate solution on absorptive of NO develops brown colour due to the formation of
ChemistryCoordination CompoundsWBJEEWBJEE 2015
Options:
  • A paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}$
  • B diamagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\left(\mathrm{N}_{3}\right)\right] \mathrm{SO}_{4}$
  • C paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\left(\mathrm{NO}_{3}\right)\right]\left[\mathrm{SO}_{4} \mathrm{l}_{2}\right.$
  • D diamagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{SO}_{4}\right)\right] \mathrm{NO}_{3}$
Solution:
1109 Upvotes Verified Answer
The correct answer is: paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}$
Cold FeSO, solution on absorption of NO develops brown colour due to the formation of $[Fe(H_{2}O)_{5}(NO)]SO_4$
$\mathrm{FeSO}_{4}+\mathrm{NO}+5 \mathrm{H}_{2} \mathrm{O} \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_4$
This complex has 3.89 BM magnetic moment shows that it has 3 unpaired electrons

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