No lone pair of electrons is available in $$BC{l_3}.$$

One lone pair of electrons is available on $$N$$ atom, it occupies a corner in the tetrahedral arrangement. Therefore, $$NC{l_3}$$  appears pyramidal in shape.

In $$B{F_3},B$$   is $$s{p^2}$$ hybridized with one empty $${p_z}$$ orbital. The empty $${p_z}$$ orbital of $$B{F_3}$$  can be filled by lone pair of molecules such as $$N{H_3}.$$  When this occurs a tetrahedral molecule or ion is formed which is $$s{p^3}$$ hybridized.

$$\eqalign{ & ({\text{A}})\,\,{N_2}:\,\,{\text{bond}}\,{\text{order}}\,\,3,\,\,{\text{paramagnetic}} \cr & \,\,\,\,\,\,\,\,{\text{ }}N_2^ - :\,{\text{bond}}\,{\text{order}}\,2.5,\,{\text{paramagnetic}} \cr & ({\text{B}})\,{C_2}\,:\,{\text{bond}}\,{\text{order}}\,2,\,{\text{diamagnetic}} \cr & \,\,\,\,\,\,\,\,\,C_2^ + :\,{\text{bond}}\,{\text{order}}\,1.5,\,{\text{paramagnetic}} \cr & ({\text{C}})\,NO:\,{\text{bond}}\,{\text{order}}\,2.5,\,{\text{paramagnetic}} \cr & \,\,\,\,\,\,\,\,N{O^ + }:\,{\text{bond}}\,{\text{order}}\,3,\,{\text{diamagnetic}} \cr & ({\text{D}})\,{O_2}:\,{\text{bond}}\,{\text{order}}\,2,\,{\text{paramagnetic}} \cr & \,\,\,\,\,\,\,\,\,O_2^ + :\,{\text{bond}}\,{\text{order}}\,2.5,\,{\text{paramagnetic}} \cr} $$

In structure (ii), $$lp{\text{ - }}bp$$   repulsions are minimum.

NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom.
$$NO_3^ - ;$$  No. of $${e^ - } = 7 + 8 \times 3 + 1 = 32,$$      hybridisation of $$N$$ in $$NO_3^ - $$  is $$s{p^3}$$
$$CO_3^{2 - };$$  No. of $${e^ - } = 6 + 8 \times 3 + 2 = 32,$$      hybridisation of $$C$$ in $$CO_3^{2 - }$$  is $$s{p^3}$$
$$ClO_3^ - ;$$  No. of $${e^ - } = 17 + 8 \times 3 + 1 = 42,$$      hybridisation of $$Cl$$  in $$ClO_3^ - $$  is $$s{p^3}$$
$$S{O_3};$$  No. of $${e^ - } = 16 + 8 \times 3 = 40,$$     hybridisation of $$S$$  in $$S{O_3}$$  is $$s{p^2}$$
$$\therefore NO_3^ - $$   and $$CO_3^{2 - }$$  are isostructural and isoelectronic.

$$S{F_4}$$  has trigonal bipyramidal geometry. $$S{F_6},I{F_5}$$   and $$SiF_6^{2 - }$$  have octahedral geometry.

By sharing of 1 electron each, both the atoms of $$Q$$  will get a complement octet.

More will be the electronegativity of $$X,$$ lesser will be the bond length of $$X-O$$  bond.

Both $$N{O_2}$$  and $${O_3}$$ have angular shape and hence will have net dipole moment.