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The complex [Ni(NH3)6]2+ has Ni2+ = 3d8. According to CFT, -6 Δo 2eg2 therefore, hybridisation is sp3d2 & complex is paramagnetic and outer orbital.

1. [Ni(NH3)6]2+
2. [Ni(CN)4]2-
3. [Ni(CN)6]4-
4. [Ni(H2O)6]2+

Correct answer: [Ni(NH3)6]2+

Solution

The complex [Ni(NH3)6]2+ has Ni2+ in a 3d8 configuration. Ammonia is a weak field ligand, so it does not cause pairing of electrons. This leads to sp3d2 hybridization, making the complex paramagnetic and an outer orbital complex.

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