Given below are two statements:

Statement I : ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$  is square planar and diamagnetic complex, with dsp² hybridizatin for Ni but $\left[Ni{\left(CO\right)}_4\right]$  is tetrahedral, paramagnetic and with sp³ hybridization for Ni.

Statement II : ${\left[NiC{l}_4\right]}^{2-}and\left[Ni{\left(CO\right)}_4\right]$  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

Correct Option - 2


Detailed Solution:

 ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

CN is strong field ligand

$N{i}^{2+}=4s^{0}3d^8$

Here; ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ is square planar and diamagnetic.

$\left[Ni{\left(CO\right)}_4\right]$

Ni = 4s²3d⁸Co is strong field ligand.

Here ; $\left[Ni{\left(CO\right)}_4\right]$ is tetrahedral and diamagnetic

${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ has 3d⁸ configuration while $\left[Ni{\left(CO\right)}_4\right]$ has 3d¹⁰ configuration.

<p>&nbsp;<math><mrow><msup><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>N</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></math></p><p>CN is strong field ligand</p><p><math><mrow><mi>N</mi><msup><mrow><mi>i</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo>=</mo><mn>4</mn><msup><mrow><mi>s</mi></mrow><mrow><mn>0</mn></mrow></msup><mn>3</mn><msup><mrow><mi>d</mi></mrow><mrow><mn>8</mn></mrow></msup></mrow></math></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752466559phpVJikUJ.jpeg" alt="" width="359" height="126" loading="lazy"></picture></div></div><p>Here;&nbsp;<math><mrow><msup><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>N</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></math>&nbsp;is square planar and diamagnetic.</p><p><math><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>O</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow></math></p><p>Ni = 4s²3d⁸Co is strong field ligand.</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752466657phpBZsuUz.jpeg" alt="" width="574" height="169" loading="lazy"></picture></div></div><p>Here ;&nbsp;<math><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>O</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow></math>&nbsp;is tetrahedral and diamagnetic</p><p><math><mrow><msup><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>N</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></math>&nbsp;has 3d⁸ configuration while&nbsp;<math><mrow><mrow><mo> [</mo><mrow><mi>N</mi><mi>i</mi><msub><mrow><mrow><mo> (</mo><mrow><mi>C</mi><mi>O</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>4</mn></mrow></msub></mrow><mo>]</mo></mrow></mrow></math>&nbsp;has 3d¹⁰ configuration.</p>

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