In B₃N₃H₆ compound, the correct option is (State 'T' for True and 'F' for False):

(I) Three $\mathrm{p\pi }-\mathrm{p\pi }$ bonds present in B₃N₃H₆.
(II) It shows back bonding.
(III) It is non-polar compound with non polar bonds.
(IV) $\mathrm{p\pi }-\mathrm{d\pi }$ bonding is also present.

Borazine has a molecular formula B₃N₃H_6. It has three B and three N atoms present at alternate positions in a hexagonal ring. Each B and N atom has one hydrogen atom. 

The boron atom has one vacant 2p_z orbital and N has one lone pair of electrons present in the 2p_z orbital. There is back bonding that takes place between B and N atoms, this back bonding is also referred to as $\mathrm{p\pi }-\mathrm{p\pi }$ bonding. As there are three B and three N atoms, hence there are three back bonding$(\mathrm{p\pi }-\mathrm{p\pi } \mathrm{bonding})$. 

There is an electronegativity difference between B and N, hence bond between B and N is polar while the compound is nonpolar because of has zero dipole moment. 

B and N have no d-orbitals, therefore $\mathrm{p\pi }-\mathrm{d\pi }$ bonding is not possible.