Correct Answer - A,B,D
In : `C Cl_2` , C is `sp^2` hybridized and has 6 electrons around itself.So, Cl donate its lone pair in empty p orbital of C.Similarily, Cl donote electrons in empty p orbital of B of `BCl_3`.In `B_2H_6` , B is `sp^3` hybridized and hence no back bonding . In `Na_2B_4O_7. 10H_2O` , two B are `sp^2` hybridized and show `ppi-ppi` backbonding with lone pair from oxygen.
