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Given below are two statements:

Statement I: $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is square planar and diamagnetic complex, with $\mathrm{dsp}^{2}$ hybridization for $\text { Ni but }\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ is tetrahedral, paramagnetic and with $\mathrm{sp}^{3}-$ hybridication for $\mathrm{Ni}$ .

Statement II: $\left[\mathrm{NiCl}_{4}\right]^{2-} \text { and }\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ both have same d-electron configuration, have same geometry and are paramagnetic.

In light the above statements, choose the correct answer from the options given below:
Option: 1
Both Statement I and Statement II are true.

Option: 2
Both Statement I and Statement II are false.

Option: 3
Statement I is correct but statement II is false.

Option: 4
Statement I is incorrect but statement II is true.

Answers (1)

best_answer

$\mathrm{\left[N i(C N)_{4}\right]^{2-}: d^{8}}$ configuration due to SFL $\mathrm{(CN^-)}$

Zero unpaired $\mathrm{e^{\ominus }}$ : (diamagnetic)

$\mathrm{\left[\mathrm{Ni}(CO)_{4}]: d^{10}\right.}$ configuration due to SFL (CO) and excitation of $\mathrm{e^{\ominus }}$

Zero unpaired $\mathrm{e^{\ominus }}$ : diamagnetic

$\mathrm{[\mathrm{NiCl_{4}}]^{2-}: d^{8}}$ configuration due to WFL $\mathrm{\left(Cl^{\ominus }\right)}$

2 unpaired $\mathrm{e^{\ominus }}$ : Paramagnetic

Both statements I and II are false.

Hence, Option (2) is correct

Posted by

manish painkra

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