In CO, the number of electrons

$$ \Rightarrow $$ 6 + 8 = 14 [ Z of C = 6 and O = 8]

Electronic configuration of molecular orbital of CO :

$${\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$$

CN^- also get (6 + 7 + 1) 14 electrons and the configuration is similar to that CO.

So, CN and CO is isoelectronic.

$${O^ - }\hbox{---}\mathop N\limits^{\mathop \uparrow \limits^O } = O$$

In NO₃^- ion, nitrogen has 4 bond pair of electrons and no lone pair of electrons.

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.