Question

Which of the following complexes has square planar structure?
1.${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$
2.$Ni{\left( {CO} \right)_4}$
3.${\left[ {Zn{{\left( {N{H_3}} \right)}_4}} \right]^{ + 2}}$
4.${\left[ {NiC{l_4}} \right]^{2 - }}$

Answer 1

Hint: We have to know that the geometry (or) structure of complex compounds is majorly based on the amount of bonded electrons and amount of lone pairs found in the complex. We can predict this with the help of valence bond theory. We have to know in square planar geometry there would be four bonded pairs and the number of lone pairs would be zero.

Complete answer:
We have to know that in square planar molecular geometry, the central atom is enclosed by constituent’s atom that forms the corners of the square which lie in the same plane. We have to know that this geometry is prevalent for transition metal complexes with a ${d^8}$ configuration. We have to know that for square planar, the electronic configuration of the central metal atom has to be $ds{p^2}$ hybridization. This is according to the valence bond theory. We have to know that molecules that exhibit square planar geometry would contain their atoms positioned at the ends. The coordination number of this complex is 4 and the bond angle would be ${90^ \circ }$.
In ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$, the oxidation state of nickel is $ + 2$. We can write the electronic configuration of $N{i^{2 + }}$ as,
$N{i^{2 + }} = \left[ {Ar} \right]4{s^0}3{d^8} = 4{s^0}3{d^2}_{xy}3{d^2}_{yz}3{d^2}_{zx}3{d^2}_{{z^2}}3{d^0}_{{x^2} - {y^2}}$
The ligand is strong, so it would pair up with single electrons in d-orbital. So, its hybridization would be $ds{p^2}$ and the structure is square planar structure. So, Option (1) is correct.
We have to know that the structure of $Ni{\left( {CO} \right)_4}$ is a tetrahedral structure. So, Option (2) is incorrect.
We have to know that the structure of ${\left[ {NiC{l_4}} \right]^{2 - }}$ is a tetrahedral structure. So, Option (4) is incorrect.

So, we can say that a structure which has square planar complex is ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$. Option (1) is correct.

Note:
We have to know that some of the examples that exhibit square planar complexes are Wilkinson’s catalyst, Crabtree’s catalyst, Rh(I), Pd(II), Ir (I), Pt(II), Zeise’s salt, Vaska’s complex. We have to know that square planar complexes generally have low spin. Electrons in square planar complexes prefer to be paired instead of being unpaired because their pairing energy is lower than $\Delta $.