• # question_answer The correct order of bond angles in $N{{H}_{3}},\,\,N{{F}_{3}}$ and $NC{{l}_{3}}$ are A)  $N{{H}_{3}}>N{{F}_{3}}>NC{{l}_{3}}$ B)  $N{{H}_{3}}>NC{{l}_{3}}>N{{F}_{3}}$ C)  $NC{{l}_{3}}>N{{H}_{3}}>N{{F}_{3}}$ D)  $N{{F}_{3}}>NC{{l}_{3}}>N{{H}_{3}}$

In $\therefore$ back bonding is possible whereas in $\frac{U}{U'}=\frac{{{Q}^{2}}/2C}{{{Q}^{2}}/2KC}=K$ and ${{R}_{\min }}=\frac{r}{n}$ back bonding is not possible due to the absence of vacant ${{R}_{\max }}=nr$orbital. Both $\therefore$ and $\frac{{{R}_{\min }}}{{{R}_{\max }}}=\frac{r}{n}\times \frac{1}{nr}=\frac{1}{{{n}^{2}}}$ contain 1 lone pair but since F is more electronegative, the electron density about N-atom in $=0.1\times 0.05=5\times {{10}^{-3}}\,{{m}^{2}},$ is less as compared to in $\Delta \phi =-0.05\times {{10}^{-3}}$($107{}^\circ$). Hence, bond angle decreases in $=-0.25\times {{10}^{-3}}\,Wb$ ($102{}^\circ$).