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Question: Match list-I (complexes) with list-II (Hybridization of central atom) and select the correct answer ...

Match list-I (complexes) with list-II (Hybridization of central atom) and select the correct answer using the codes given below the lists:

List-IList-II
A. Ni(CO)4Ni{{\left( CO \right)}_{4}}1. sp3s{{p}^{3}}
B. Ni(CN)42Ni\left( CN \right)_{4}^{2-}2. dsp2ds{{p}^{2}}
C. Fe(CN)64Fe\left( CN \right)_{6}^{4-}3. sp3d2s{{p}^{3}}{{d}^{2}}
D. MnF64MnF_{6}^{4-}4. d2sp3{{d}^{2}}s{{p}^{3}}
5. sp2ds{{p}^{2}}d...

A. A-1, B-3, C-2, D-4
B. A-5, B-2, C-4, D-3
C. A-5, B-3, C-2, D-4
D. A-1, B-2, C-4, D-3

Explanation

Solution

Hint As we know that hybridization is the process of producing a degenerated type of orbital by mixing two atomic orbitals with the same energy levels. During hybridization, atomic orbitals are mixed and a new hybrid orbital is formed.

Complete Step by step solution:
- In Ni(CO)4Ni{{\left( CO \right)}_{4}}, Ni atom is sp3s{{p}^{3}} hybridised. And this will lead to a tetrahedral geometry. We can see the structure:

- In Ni(CN)42Ni\left( CN \right)_{4}^{2-}, Ni atom is dsp2ds{{p}^{2}} hybridised. And this will lead to a square planar geometry.
Here, Ni(CN)42Ni\left( CN \right)_{4}^{2-} is showing dsp2ds{{p}^{2}} hybridisation, because here CN group is a strong field ligand and it forms inner complexes. We can see the structure:

- In Fe(CN)64Fe\left( CN \right)_{6}^{4-}, Ni atom is d2sp3{{d}^{2}}s{{p}^{3}} hybridised. And this will lead to an octahedral geometry. Here, Fe(CN)64Fe\left( CN \right)_{6}^{4-} is showing d2sp3{{d}^{2}}s{{p}^{3}} hybridisation, because here cyanide group is present. And as we know that cyanide group is a strong field ligand and it forms inner complexes. We can see the structure:

- In MnF64MnF_{6}^{4-}, Ni atom is sp3d2s{{p}^{3}}{{d}^{2}} hybridised. And this will lead to an octahedral geometry. We can see the structure:

Hence, we can say that correct option is (d), that is A-1, B-2, C-4, D-3

Note:
- We must note the key features of hybridization: the atomic orbitals that have equal energies undergo hybridization. Shape of any molecule can be predicted from hybridization.
- The number of hybrid orbitals that are formed during hybridization will be equal to the number of atomic orbitals mixed.