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) \[N{{F}_{3}}>NC{{l}_{3}}>N{{H}_{3}}\]
Correct Answer: C
Solution :
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 \[=-0.25\times {{10}^{-3}}\,Wb\] (\[102{}^\circ \]).
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