Question

Match the species given in Column I with the shape given in Column II and mark the correct option.

	Column I		Column II
a	$S{F}_4$	(i)	Tetrahedral
b	$Br{F}_3$	(ii)	Pyramidal
c	$Br{O}_3^{-}$	(iii)	Sea-saw shaped
d	$N{H}_4^{+}$	(iv)	Bent T-shaped

• A. $a\to \left(iii\right)b\to \left(iv\right)c\to \left(ii\right)d\to \left(i\right)$
• B. $a\to \left(iii\right)b\to \left(ii\right)c\to \left(i\right)d\to \left(iv\right)$
• C. $a\to \left(i\right)b\to \left(ii\right)c\to \left(iii\right)d\to \left(iv\right)$
• D. $a\to \left(i\right)b\to \left(iv\right)c\to \left(iii\right)d\to \left(ii\right)$

Answer 1

The correct option is A $a\to \left(iii\right)b\to \left(iv\right)c\to \left(ii\right)d\to \left(i\right)$

Steric number

Hybridization can be calculated by finding the steric number for a given molecule.
Steric number
= number of sigma bonds + number of lone pairs
Based on steric number, hybridization of molecules can be:

Steric Number	Hybridization
2	sp
3	$s{p}^2$
4	$s{p}^3$
5	$s{p}^{3}d$
6	$s{p}^3{d}^2$
7	$s{p}^3{d}^3$

Part (a)
$S{F}_4$
Structure:


Number of sigma bonds = 4
Number of lone pairs = 1
Steric Number = 4+1 = 5
Thus,
Hybridization = $s{p}^{3}d$
Geometry=trigonal bipyramidal,
Shape: see-saw
$a\to \left(iii\right)$

Part (b)
$Br{F}_3$
Structure:


Number of sigma bonds = 3
Number of lone pairs = 2
Steric Number = 3+2 = 5
Thus,
Hybridization = $s{p}^{3}d$
Geometry=trigonal bipyramidal, Shape: Bent T shape
Note: the structure is bent due to stronger lone pair- lone pair repulsions than bond pair lone pair repulsions.

$b\to \left(iv\right)$

Part (c)
$Br{O}_3^{-}$
Structure:


Number of sigma bonds = 3
Number of lone pairs = 1
Steric Number = 3+1 = 4
Thus,
Hybridization = $s{p}^3$
Geometry=Tetrahedral,
Shape: pyramidal

$c\to \left(ii\right)$

Part (d)
$N{H}_4^{+}$
Structure:


Number of sigma bonds = 4
Number of lone pairs = 0
Steric Number = 4+0 = 4
Thus,
Hybridization = $s{p}^3$
Geometry=tetrahedral,
Shape: Tetrahedral

$d\to \left(i\right)$

Hence, option (B) is correct.