Assertion `Ni(CO)_(4)` is tetrahedral in shape
Reason `Ni` atom is in zero oxidation state and undergoes `sp^(3)` -hybridisation in `Ni(CO)_(4)` .

The hybridisation of Ni in [Ni(CO)_(4)] is

The hybridisation of Ni in [Ni(CO)_(4)] is

Assertion: Ni(CO)_4 complex is tetrahedral in shape Reason: Ni atom undergoes sp^3 hybridization in Ni(CO)_4 .

Name the hybridisation of Ni in [Ni(CO)_4]

Assertion : [Ni(CO)_4] is diamagnetic and tetrahedral in shape. Reason : [Ni(CO)_4] contains no unpaired electron and involves dsp^2 hybridisation.

[Ni(CO)_4] is tetrahedral but diamgagnetic why?

In Ni(CO)_(4) the oxidation state of Ni is:

Assertion The number of unpaired electrons in [Ni(CO)_(4)] is zero Reason In this compounds 4s-electrons of Ni atom enter the inner d-orbitals to facilitate the sp^(3) hybridisation in Ni atom .

Assertion The number of unpaired electrons in [Ni(CO)_(4)] is zero Reason In this compounds 4s-electrons of Ni atom enter the inner d-orbitals to facilitate the sp^(3) hybridisation in Ni atom .