Question

Given below are two statements :
Statement I : $[Ni(CN)4{]}^{2-}$ is square planar and diamagnetic complex. with $ds{p}^2$ hybridization for $Ni$ but $\left[Ni(CO{)}_4\right]$ is tetrahedral. paramagnetic and with $s{p}^3$-hybridication for $Ni$.
Statement II: ${\left[NiC{l}_4\right]}^{2-}$ and $\left[Ni(CO{)}_4\right]$ both have same d-electron configuration have same geometry and are paramagnetic.
In light the above statements. choose the correct answer form the options given below:

• A. Both Statement I and Statement II are true.
• B. Both Statement I and Statement II are false.
• C. Statement I is correct but statement II is false.
• D. Statement I is incorrect but statement II is true

Answer 1

Answer: (B)

${\left[Ni(CN{)}_4\right]}^{2-}:d^8$ configuration, SFL, sq. planar splitting $\left(ds{p}^2\right)$, diamagnetic.
$\left[Ni(CO{)}_4\right]:d^{10}$ config (after excitation), SFL, tetrahedral splitting $\left(s{p}^3\right)$, diamagnetic.
${\left[NiC{l}_4\right]}^{2-}:d^8$ config, WFL, tetrahedral
splitting $\left(s{p}^3\right)$, paramagnetic(2 unpaired $e^{-})$.