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Answer» `:C Cl_2` `BCl_3` `B_2H_6` `Na_2B_4O_7. 10H_2O` SOLUTION :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.
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