Question

$I.CO$
$II.PC{l}_5$
$III.B{F}_3$
$IV.BC{l}_3$
$V.Si{H}_4$
$VI.Si{F}_4$
$VII.CC{l}_4$
$VIII.BeC{l}_2$
$IX.N(Si{H}_3{)}_3$
$X.H_{2}O$
The number of molecules showing back bonding among the given molecules

Answer 1

Some electron deficient compounds, tend to attain octet by back bonding. Back bonding acceptor must have a vacant orbital and back bonding donor must have atleast one lone pair of electrons.

I. $CO$
It has a complete octet, there is no back bonding.

II. $PC{l}_5$
It has an expanded octet and hence, there is no back bonding.

III. $B{F}_3$
$B\to$ empty 2p orbital and electron deficient.
$F\to$ 3 lone pairs.
$\therefore$ shows back bonding.

IV. $BC{l}_3$
$B\to$ empty 2p orbital.
$Cl\to$ 3 lone pairs.
$\therefore$ shows back bonding.

V. $Si{H}_4$
$Si$ has a complete octet and $H$ has no lone pair. There is no back bonding

VI. $Si{F}_4$
It has a complete octet. However, since $Si$ has vacant d-orbital & flourine has lone pair, hence it undergoes back bonding.

VII. $CC{l}_4$
$C\to$ no empty orbital.
$Cl\to$ 3 lone pairs.
$\therefore$ No back bonding.

VIII. $BeC{l}_2$

Though electron deficient, it does not undergo back bonding. It undergoes polymerization.

IX. $N(Si{H}_3{)}_3$

It has a complete octet. However, since $Si$ has vacant d-orbital & $N$ has lone pair, hence it undergoes back bonding.


X. $H_{2}O$
$O\to$ No empty orbital.
$H\to$ No lone pairs.
$\therefore$ No back bonding.