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The magnetic moments of the complexes \mathrm{[Ni(CN)_4]^{2-}} and \mathrm{[NiCl_4]^{2-}} respectively were found to be 0 and \mathrm{2.83 \ BM} respectively. The hybridisation of \mathrm{Ni^{2+}} in the complexes is respectively 

Option: 1

\mathrm{sp^3\ and\ sp^3}


Option: 2

\mathrm{sp^3\ and\ dsp^2}


Option: 3

\mathrm{dsp^2\ and\ sp^3}


Option: 4

\mathrm{dsp^2\ and\ dsp^2}


Answers (1)

best_answer

As we have learnt,

VBT helps us to predict the hybridisation of the compound if the magnetic moment is known.

Both the complexes have \mathrm{Ni^{2+}} with a \mathrm{d^8} configuration.

Now, since \mathrm{[Ni(CN)_4]^{2-}}is diamagnetic, hence it must have a \mathrm{dsp^{2}} hybridisation wherein all the d electrons are paired up.

And, since \mathrm{NiCl_4^{2-}} is paramagnetic with 2 unpaired electrons, it must have \mathrm{sp^3} hybridisation

Hence, the correct answer is Option (3)

Posted by

sudhir.kumar

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