Observe the following two reactions: 

$Si{H}_{3}Cl+{\left(Si{H}_3\right)}_{2}NH\underset{yield}{\overset{100\%}{\to }}X\stackrel{N{H}_3}{\to }Y+N{H}_{4}Cl$

 

 

 

 

 

 

 

 

Select the correct statement(s)

$$\begin{gathered} X={\left(Si{H}_3\right)}_3\stackrel{+}{NH}\stackrel{-}{Cl}Y={\left(Si{H}_3\right)}_{3}N \\ P={\left(C{H}_3\right)}_{3}NQ=qN{H}_{4}Cl \end{gathered}$$

Lone pain on ‘N’ in Y is involved in back bonding hence it is less basic then P.
 Y is planar about ‘N’ but tetrahedral about silicon.
${\left(C{H}_3\right)}_{3}N+C{H}_{3}COOH\to {\left(C{H}_3\right)}_3\stackrel{+}{NH}C{H}_{3}CO\overline{O}$   salt (acid-base reaction)