Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Match the molecules in List-I with the corresponding shape in List-II:

List-I (Molecules): (a) $NH_3$ (b) $ClF_3$ (c) $PCl_5$ (d) $BrF_5$

List-II (Shape): (i) Square pyramidal (ii) Trigonal bipyramidal (iii) Trigonal pyramidal (iv) T-shape

Choose the correct answer from the options given below:

(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

According to the VSEPR theory and hybridisation principles detailed in the sources:

$NH_3$ (a): The central nitrogen atom is $sp^3$ hybridised and has three bond pairs and one lone pair. The repulsion between the lone pair and bond pairs reduces the bond angle and results in a trigonal pyramidal shape .
$ClF_3$ (b): The chlorine atom undergoes $sp^3d$ hybridisation. It has three bond pairs and two lone pairs. To minimise repulsions, the lone pairs occupy equatorial positions, resulting in a T-shape .
$PCl_5$ (c): The phosphorus atom is $sp^3d$ hybridised and forms five bond pairs with no lone pairs. This results in a regular trigonal bipyramidal geometry .
$BrF_5$ (d): Bromine is $sp^3d^2$ hybridised with five bond pairs and one lone pair, which gives the molecule a square pyramidal shape .

Matching these correctly gives: (a)-(iii), (b)-(iv), (c)-(ii), and (d)-(i).

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