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  • Arrange the following in increasing order of spin only magnetic moment.<img alt="\mathrm{Fe}^{2+}(\mathrm{I}), \mathrm{Cu}^{2+}(\mathrm{II}), \mathrm{V}^{2+}(\mathrm{III}), \mathrm{Ti}^{2+}(\mathrm

Arrange the following in increasing order of spin only magnetic moment.\mathrm{Fe}^{2+}(\mathrm{I}), \mathrm{Cu}^{2+}(\mathrm{II}), \mathrm{V}^{2+}(\mathrm{III}), \mathrm{Ti}^{2+}(\mathrm{IV})

Option: 1

II<IV<III<I


Option: 2

II<III<I<IV


Option: 3

I<II<III<IV


Option: 4

I<III<IV<I


Answers (1)

best_answer

\text{Fe}^{2+} has \text{3d}^{6} configuration. Hence, it has 4 unpaired electrons. \text{Ti}^{2+} has \text{3d}^{2} configuration.
So, it has 2 unpaired electrons. \text{Cu}^{2+} has \text{3d}^{9} configuration. So, it has 1 unpaired electron. \text{V}^{2+} has \text{3d}^{3} configuration. So, it has 3 unpaired electrons. More the unpaired electrons more will be the spin only magnetic moment.

Posted by

Rakesh

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