Question

Which among the following is nonlinear?

• A. $N_3^{+}$
• B. $Cl{F}_2^{-}$
• C. $B{r}_3^{-}$
• D. $BrC{l}_2^{+}$

Answer 1

The correct option is D $BrC{l}_2^{+}$

• Hybridisation is given by $\frac{1}{2}(V+M-C+A)$.
  

where, $V=\text{number of valence electrons of central atom}$

$M=\text{number of monovalent atoms}$

$C=\text{positive charge}$

$A=\text{negative charge}$

• For $N_3^{+}$, $\frac{1}{2}(5+0-1+0)=2$
• This implies $sp$ hybridisation. Thus, it is linear in shape.
  
• For $Cl{F}_2^{-}$, $\frac{1}{2}(7+2-0+1)=5$
• The hybridisation is $s{p}^{3}d$. Three lone pair of electrons occupy the equatorial position and two $F$atoms occupy the axial position. Thus, the shape is linear.
• For $B{r}_3^{-}$, $\frac{1}{2}(7+2-0+1)=5$
• $s{p}^{3}d$ hybridisation is present, where 2 $Br$ atoms occupy the axial position and the lone pair of electrons occupy the 3 equatorial position. Thus, the molecule acquires linear shape.
• For $BrC{l}_2^{+}$, $\frac{1}{2}(7+2-1+0)=4$
• Hybridisation is $s{p}^3$. Two positions are occupied by $Cl$atoms and the other two are occupied by lone pairs. Thus, the shape is bent or V-shaped.