Electronic structure of S atom in excited state


In 'S' unhybride d- orbital is present, which will involved in bond formation with oxygen atom.

In oxygen two unpaired p- orbital is present in these one is involved in $$\sigma $$ bond formation while other is used in $$\pi $$ bond formation.

So, in SO₃^2$$-$$, p$$\pi $$ $$-$$ d$$\pi $$ bonding present.

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.

Bond angle is maximum in NH₄⁺ tetrahedral molecule with bond angle 109°.

For strong $$\pi $$-bonding, p$$\pi $$ - p$$\pi $$ bonding should be strong. In case of P, due to larger size as compared to N-atom, p$$\pi $$ - p$$\pi $$ bonding is not so strong.