Question

What is the geometry of \[{\left[ {NiC{l_4}} \right]^{2 - }},{\left[ {Ni{{(CN)}_4}} \right]^{2 - }}\] and \[\left[ {Ni{{(CO)}_4}} \right]\]
A. \[{\left[ {NiC{l_4}} \right]^{2 - }}\] Is square planar and \[{\left[ {Ni{{(CN)}_4}} \right]^{2 - }},\left[ {Ni{{(CO)}_4}} \right]\] are tetrahedral
B. \[\left[ {Ni{{(CO)}_4}} \right]\] Is square planar and \[{\left[ {NiC{l_4}} \right]^{2 - }},{\left[ {Ni{{(CN)}_4}} \right]^{2 - }}\] are tetrahedral
C. \[{\left[ {Ni{{(CN)}_4}} \right]^{2 - }}\] Is square planar and \[{\left[ {NiC{l_4}} \right]^{2 - }},\left[ {Ni{{(CO)}_4}} \right]\] are tetrahedral
D. None of the above

Answer 1

Hint: Hybridisation is the process of mixing of orbitals of different shapes and energies. When two atomic orbital combines, it forms a hybrid orbital. When one ‘s’ orbital and ‘3’ 3p orbitals are mixed, then the formed hybridization is tetrahedral. When one ‘d’ orbital, one ‘s’ orbital, and two ‘p’ orbitals are mixed, then the formed hybridization is square planar. The hybridization of square planar is \[ds{p^2}\] and for tetrahedral is \[s{p^3}\]

Complete step by step answer:
The outermost electronic configuration of the Nickel atom can be written as,
\[Ni = 3{d^8}4{s^2}\]
In \[{\left[ {NiC{l_4}} \right]^{2 - }}\], the oxidation state of nickel atom is +2 and its valence electronic configuration can be written as, \[N{i^{2 + }} = 3{d^8}4{s^0}\]
In \[{\left[ {Ni{{(CN)}_4}} \right]^{2 - }}\], the oxidation state of nickel atom is +2 and its valence electronic configuration can be written as, \[N{i^{2 + }} = 3{d^8}4{s^0}\]
In \[\left[ {Ni{{(CO)}_4}} \right]\], the outermost electronic configuration of Nickel atom can be written as,
\[Ni = 3{d^8}4{s^2}\]
It is known that \[C{l^ - }\] is a weak ligand and \[C{N^ - }\],\[CO\] are strong ligands. Generally, a strong ligand helps in pairing up the valence shell electron to create the space for ligand whereas a weak field ligand won’t help in the pairing of electrons.
In the given three complexes, each central metal atom is surrounded by four ligands.
In \[{\left[ {NiC{l_4}} \right]^{2 - }}\],\[C{l^ - }\] is a weak ligand and it won’t help in the pairing of electrons. Thus, the electronic configuration of \[N{i^{2 + }} = 3{d^8}4{s^0}\]
Thus, the four ligands occupy an additional one s- subshell and three p-subshells forming the hybridization \[s{p^3}\]. Thus, its geometry is tetrahedral.
In \[{\left[ {Ni{{(CN)}_4}} \right]^{2 - }}\],\[C{N^ - }\] is a strong ligand and thus, it helps in the pairing. Thus, four ligands occupy 1 d-subshell, 1 s-subshell, and 2 p-subshell forming the hybridization \[ds{p^2}\]. Thus, its geometry is square planar.
In \[\left[ {Ni{{(CO)}_4}} \right]\], \[CO\] is a strong ligand and it helps in the pairing. The oxidation state of nickel in \[\left[ {Ni{{(CO)}_4}} \right]\] is 0. Thus, while pairing two electrons present in 4s subshell will shift to 3d subshell, now, it has filled 3d subshell. Thus, the four ligands occupy an additional one s- subshell and three p-subshell forming the hybridization \[s{p^3}\]. Thus, its geometry is tetrahedral.

So, the correct answer is Option C.

Note: The shapes of hybridization are linear, bent, trigonal planar, trigonal bipyramidal, square planar, tetrahedral, and octahedral respectively. In tetrahedral hybridization, four ligands are occupied in four corners of the regular tetrahedron. The angle between \[s{p^3}\] orbitals is $109^0$$28^{'}$ Square planar hybridization can be achieved by flattening a tetrahedron.