Question

Hybridization is the measure to express the molecular shapes. Hybridization is highly correlated with molecular shapes or inter orbital angles. If there is any departure in geometry (on the basis of VSEPR theory), the apparent departure in hybridization can be observed and the following characteristic relationship is the easiest way to interpret that departure. Any hybrid orbital made by $s$ and $p-$ orbitals is characterized by the following relationship.

$f_s=\frac{1}{i{+}_{}1}$ or $f_p=\frac{1}{i+1}$ and $i=\frac{f_p}{f_s}$

For $s{p}^3,i=3$ and $s{p}^2$, $i=2$; for an orbital $f_s+f_p=1$ while all the s-p hybrids of given atom must satisfy the condition,

$\sum f_s=1.00$

and the inter-orbital angle between two equivalent (same $f_s$ and same $f_p$) hybrid orbitals is given by

$cos\theta =\frac{-1}{i}$

In $BFClBr$ molecule the minimum % $s-$character provided from the central atom is in the bond :

• A. $B-F$
• B. $B-Cl$
• C. $B-Br$
• D. cannot be predicted

Answer 1

The correct option is A $B-F$

Among, $F,Cl$ and $Br,F$ is the most electronegative atom. Hence, the bond pair of electrons of $B-F$ bond is attracted more towards $F.$

Hence, $B-F$ bond has the minimum percentage of $s$ character.