Given below are two statements :
Statement I: ${[Ni {(CN)}_{4}]}^{2 -}$ is square planar and diamagnetic complex, with ${dsp}^{2}$ hybridization for $Ni$ but $[Ni {(CO)}_{4}]$ is tetrahedral, paramagnetic and with ${sp}^{3}$ hybridication for $Ni$ .
Statement II : ${[{NiCl}_{4}]}^{2 -}$ and $[Ni {(CO)}_{4}]$ 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

Question

Given below are two statements :

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

Statement II : ${[{NiCl}_{4}]}^{2 -}$ and $[Ni {(CO)}_{4}]$ 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

Aicharya · Accepted Answer

Ni CN 4 2 - : &#160; d 8 configuration, SFL, sq. planar splitting dsp 2 , diamagnetic. Ni CO 4 &#160; : &#160; d 10 config (after excitation), SFL, tetrahedral splitting sp 3 , diamagnetic. NiCl 4 2 - : &#160; d 8 config, WFL, tetrahedral splitting sp 3 , paramagnetic( 2 unpaired e - ). Electronic configuration of Ni atom in Ni CO 4

Given below are two statements : Statement I: Ni CN 4 2 - is square planar and diamagnetic complex, with dsp…

Solution