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# Which complex has $ds{{p}^{2}}$ hybridization and square planar shape?a) ${{\left[ Ni{{(CN)}_{4}} \right]}^{2-}}$b) ${{\left[ Cu{{(N{{H}_{3}})}_{4}} \right]}^{2+}}$c) ${{\left[ Pt{{(Cl)}_{4}} \right]}^{2-}}$d) All

Last updated date: 17th Sep 2024
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Hint: When we apply valence bond theory to a coordination compound, the original electrons from the d orbital of the transition metal move into non-hybridized d orbitals. The electrons donated by the ligand move into hybridized orbitals of higher energy, which are then filled by electron pairs donated by the ligand.

Complete step by step solution:

- Now, in ${{\left[ Ni{{(CN)}_{4}} \right]}^{2-}}$the oxidation number of Nickel is +2.
x + 4 × (-1) = -2
x = + 2

- The complex ${{\left[ Ni{{(CN)}_{4}} \right]}^{2-}}$- is diamagnetic, as the $C{{N}^{-}}$ is a strong ligand. So, in an excited state all the electrons of Ni get paired and none of them remains unpaired. Thus, ${{\left[ Ni{{(CN)}_{4}} \right]}^{2-}}$ is diamagnetic. With the oxidation number of +2, Nickel has 8 electrons in its 3d shell and for the formation of square planar shape, two unpaired d-electrons are paired up as the energy will be provided by the approaching ligands, this will make one of the 3d orbitals empty. By this, there is no unpaired electron and the complex would be diamagnetic. Now, $C{{N}^{-}}$ needs four orbitals to form bonds with $N{{i}^{2+}}$ ion, thus, two of 4p orbital along with one of 3d and 4s orbital forms 4 hybrid orbitals, having the hybridization $ds{{p}^{2}}$ As the inner d orbitals are forming the bond here, we will write ‘d’ first and when the inner d orbitals are involved in bond formation the shape will be square planar.

- In ${{\left[ Cu{{(N{{H}_{3}})}_{4}} \right]}^{2+}}$ the oxidation state of Cu is +2 that means there is only one unpaired electron in the 3d orbital. Now, $NH_3$ being a strong field ligand comes to the metal to approach and the one unpaired electron from the last 3d orbital enters to the last 4p orbital. Now, the 4 $NH_3$ ligands form a bond with the one 3d, one 4s and two 4p hybrid orbitals of $Cu^{+2}$ to form the given complex. Hence the hybridization of $Cu^{+2}$ here is $dsp^2$ and as the inner d orbitals are involved, it is an example for the inner d complex and its shape will be square planar.

- In ${{\left[ Pt{{(Cl)}_{4}} \right]}^{2-}}$ The oxidation state of Pt is +2 here, there are eight electrons in the 4d orbital. For 4d and 5d series all ligands act as strong ligands hence pairing occurs. So, the 2 unpaired electrons in 4d of the orbital get paired and then there is one vacant orbital. Thus, one orbital of 4d, 5s and two orbitals of 5p participates in hybridization, resulting in $dsp^2$ type of hybridization and square planar geometry.

So, the correct option is (d).

Note: Valence bond theory in coordination compounds involves overlap of valence atomic orbitals of central metal atom/ion and the ligands. This leads to hybridization and different molecular geometries can be formed.