Question

Find shape, hybridisation, magnetic character of
a) [Fe(CN). B- b) Mn(C204)13-2H20 c) [Ni(co).] d) [Ni(CN)P- e) Fe(H20)6**
What is the product and colour when both phenol and aniline are exposed to air
58)​

Answer 1

Answer:

hope it helps

Explanation:

To determine hybridization of Co ordination compound first of all we will have to write electronic configuration of central metal.

Thereafter orbital diagram is sketched.

If strong ligand is present then pairing of electron occurs, otherwise no pairing occurs.(ligand containing donating atom C, N, considered as strong field ligand, rest is weak field ligand)

If co-ordination number is 4 then there is two possibilities : either sp3 or dsp2 hybridization.

(if only one S & three P orbital used :sp3, one s, two p & one d orbital used :dsp2)

If co-ordination number is 6 then again two possibilities arises : either d2sp3 or sp3d2 hybridization.

{one s, three p & two d orbital (inner) :d2sp3

one s, three p & two d orbital (outer) used:sp3d2}

Now coming to the question :

Oxidation number of Fe : x-6=-3 , x=3 (since CN is mono dentate negative ligand)

Fe3+: [Ar] 3d5 4S0

We can see that CN is strong field ligand, so pairing of electrons present in d orbital occurs.

On pairing two(inner) d orbitals are vacant containing one unpaired electron.

4s & 4p orbitals are also present.

So d2sp3 hybridization.