Hybridisation

is hybridised due to and shape is square planar.

is hybridised due to -bond.

Hydrogen atom does not show hybridisation, thus number of hybrid orbitals in :

(a) All six C-atoms show -hybridisation (i.e. 3 -orbitals by each -atom )

Total number of hybrid orbitals

(b) C All five C-atoms show

-hybridisation (i.e. 4-orbitals by each C-atom).

Total number of hybrid orbitals .

(c) Two -atoms belong to

group show -hybridisation (i.e. 4 -orbital by each -atom .

One C-atom, bonded with O-atom and -groups, show -hybridisation (i.e.3-hybrid orbitals).

One - atom also show -hybridisation Thus, total number of hybrid orbitals

show sp-hybridisation (i.e. 2 -hybrid orbitals)

and 3 show -hybridisaton (i.e. 3 -hybrid orbital by each C-atom)

C-4 show -hybridisation (i.e. 4 -hybrid orbitals.

N-atom show sp-hybridisation (i.e. 2 -hybrid orbitals) Thus, total number of hybrid orbitals

The number of molecules having square pyramidal geometry from the given molecules is , i.e. . The geometries of the other given molecules are as follows:

Determine the formula and properties of the complex.

When Ni is complexed with ethane-1,2-diamine (en), it typically forms a coordination complex where ethane-1,2-diamine acts as a bidentate ligand, replacing water molecules. The complete replacement of water by three en ligands forms [Ni(en) ] , which is known for its violet color and paramagnetic properties due to unpaired electrons in the d-orbital of Ni .

Conclusion: The correct formula and properties are given by option (3) [Ni(en) ] , confirming it as paramagnetic.