Answer
410.1k+ views
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 $Ni{{\left( CO \right)}_{4}}$, Ni atom is $s{{p}^{3}}$ hybridised. And this will lead to a tetrahedral geometry. We can see the structure:
- In $Ni\left( CN \right)_{4}^{2-}$, Ni atom is $ds{{p}^{2}}$ hybridised. And this will lead to a square planar geometry.
Here, $Ni\left( CN \right)_{4}^{2-}$ is showing $ds{{p}^{2}}$ hybridisation, because here CN group is a strong field ligand and it forms inner complexes. We can see the structure:
- In $Fe\left( CN \right)_{6}^{4-}$, Ni atom is ${{d}^{2}}s{{p}^{3}}$ hybridised. And this will lead to an octahedral geometry. Here, $Fe\left( CN \right)_{6}^{4-}$ is showing ${{d}^{2}}s{{p}^{3}}$ 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 $MnF_{6}^{4-}$, Ni atom is $s{{p}^{3}}{{d}^{2}}$ 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.
Complete Step by step solution:
- In $Ni{{\left( CO \right)}_{4}}$, Ni atom is $s{{p}^{3}}$ hybridised. And this will lead to a tetrahedral geometry. We can see the structure:
![seo images](https://www.vedantu.com/question-sets/184006ea-3fd0-437f-ab13-7fb2a9b3e4c03551100931805814619.png)
- In $Ni\left( CN \right)_{4}^{2-}$, Ni atom is $ds{{p}^{2}}$ hybridised. And this will lead to a square planar geometry.
Here, $Ni\left( CN \right)_{4}^{2-}$ is showing $ds{{p}^{2}}$ hybridisation, because here CN group is a strong field ligand and it forms inner complexes. We can see the structure:
![seo images](https://www.vedantu.com/question-sets/f66930f8-dac6-4044-8732-66174d9de52d5436341031822663886.png)
- In $Fe\left( CN \right)_{6}^{4-}$, Ni atom is ${{d}^{2}}s{{p}^{3}}$ hybridised. And this will lead to an octahedral geometry. Here, $Fe\left( CN \right)_{6}^{4-}$ is showing ${{d}^{2}}s{{p}^{3}}$ 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:
![seo images](https://www.vedantu.com/question-sets/420c02e8-976b-430c-be49-f2325f9b5e388943050164039841379.png)
- In $MnF_{6}^{4-}$, Ni atom is $s{{p}^{3}}{{d}^{2}}$ hybridised. And this will lead to an octahedral geometry. We can see the structure:
![seo images](https://www.vedantu.com/question-sets/a863f6b3-dc1f-44a6-8b0a-24acdc9d5f516626966221154721743.png)
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.
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