Hybridisation of [Ni(NH3) 6]2+ is d2sp3 or not??????

Question

Answer

In [Ni(NH3)6]2+, Ni is in +2 state and has configuration 3d8. In presence of NH3, the 3d electrons do not pair up. The hybridization is sp3d2 forming an outer orbital complex.

Answer

Answer

Hybridisation of [Ni(NH3)6]2+

The hybridization of a complex ion can be determined by considering the coordination number and the geometry of the complex. In the case of [Ni(NH3)6]2+, the coordination number is 6, indicating that the central Ni atom is surrounded by six ammonia ligands.

Sigma Bonding and Electron Pair Repulsion Theory
Sigma (σ) bonding occurs when orbitals overlap head-on, resulting in the formation of a bond. In the case of [Ni(NH3)6]2+, the sigma bonds are formed between the Ni atom and the nitrogen atoms of the ammonia ligands.

According to the electron pair repulsion theory, the ammonia ligands will arrange themselves around the central Ni atom in an octahedral geometry. This geometry allows for maximum separation of the ligands, minimizing repulsion between electron pairs.

D2sp3 Hybridization
Based on the coordination number and the octahedral arrangement of the ligands, the hybridization of the central Ni atom can be determined. The d2sp3 hybridization scheme is commonly used to explain the bonding in octahedral complexes.

In the d2sp3 hybridization, the 4s, 4p, and 4d orbitals of the Ni atom undergo hybridization to form a set of six equivalent hybrid orbitals. Two of these hybrid orbitals contain lone pairs of electrons, while the other four hybrid orbitals form sigma bonds with the ammonia ligands.

Explanation of the Hybridization Scheme
1. The 4s orbital of the Ni atom mixes with three 4p orbitals and one 4d orbital to form four sp3 hybrid orbitals. These hybrid orbitals are oriented in an octahedral arrangement around the Ni atom.
2. The remaining two 4d orbitals remain unhybridized and are higher in energy than the hybrid orbitals.
3. The six ammonia ligands each donate a pair of electrons to form sigma bonds with the Ni atom. These sigma bonds are formed by the overlap of the hybrid orbitals on the Ni atom and the nitrogen orbitals of the ammonia ligands.
4. The resulting complex, [Ni(NH3)6]2+, has a d2sp3 hybridization scheme, with the Ni atom using two of its hybrid orbitals for sigma bonding and the other four hybrid orbitals containing lone pairs of electrons.

Conclusion
The hybridization of [Ni(NH3)6]2+ is d2sp3. The central Ni atom undergoes hybridization of its 4s, 4p, and 4d orbitals to form six equivalent sp3 hybrid orbitals. Four of these hybrid orbitals form sigma bonds with the ammonia ligands, while the remaining two hybrid orbitals contain lone pairs of electrons. This hybridization scheme explains the octahedral geometry and the bonding in the complex ion.

Answer

Not acc to CFT

Answer

Correct answer is Sp3d2...

Answer

yes

Answer

Why??

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