• # question_answer Arrange the following in order of decreasing bond angles (i)$C{{H}_{4}},\,N{{H}_{3}},\,{{H}_{2}}O,\,B{{F}_{3}},\,{{C}_{2}}{{H}_{2}}$ (ii) $N{{H}_{3}},\,NH_{2}^{-},\,NH_{4}^{+}$

(i) $N{{H}_{3}}(107{}^\circ )>{{H}_{2}}O>(104.5{}^\circ ).$ (ii)$NH_{4}^{+}>N{{H}_{3}}>NH_{2}^{-}$ This is because all of them involve$s{{p}^{3}}$ hybridization. The number of lone pair of electrons present on N-atom are 0, 1 and 2 respectively. Greater the number of lone pairs, greater are the repulsions on the bond pairs and hence smaller is the angle.