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Answer the following.
The molecular shape of [Ni(CO)₄] is tetrahedral while that of [Ni(CN)₄]²⁻ is square planar. Justify…
The molecular shape of $[\text{Ni}(\text{CO})_4]$ is tetrahedral while that of $[\text{Ni}(\text{CN})_4]^{2-}$ is square planar.
(a)[1.5]
Justify the shape of the two compounds.
(b)[1.5]
Find the oxidation state of Ni in each complex.
Answer
Answer (a)
Official answer keyOxidation state of Ni is zero in $[\mathrm{Ni(CO)_4}]$. CO is a stronger field ligand, and it can pair the unpaired electrons. The 3d orbital is now totally filled, while the 4s and 4p orbitals remain open. Therefore, exhibits $sp^3$ hybridization and tetrahedral geometry.
The oxidation state of Ni in $[\mathrm{Ni(CN)_4}]^{2-}$ is +2. While in presence of $\mathrm{CN^-}$ being a stronger field ligand all the electrons are paired up. The empty 3d, 4s and two 4p orbitals undergo $dsp^2$ hybridization to make bonds with $\mathrm{CN^-}$ ligands in square planar geometry.
Answer (b)
Official answer keyNi has 0 oxidation state in complex $[\mathrm{Ni(CO)_4}]$ and +2 oxidation state in $[\mathrm{Ni(CN)_4}]^{2-}$