Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Match List-I (Molecules) with List-II (Shapes):

List-I: (a) $PCl_5$ (b) $SF_6$ (c) $BrF_5$ (d) $BF_3$

List-II: (i) Square pyramidal (ii) Trigonal planar (iii) Octahedral (iv) Trigonal bipyramidal

Choose the correct answer from the options given below:

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

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

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

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

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory and the principles of hybridisation described in the sources:

$PCl_5$ (a): The central phosphorus atom undergoes $sp^3d$ hybridisation. With five bonding pairs and no lone pairs, it adopts a trigonal bipyramidal geometry .
$SF_6$ (b): The central sulphur atom undergoes $sp^3d^2$ hybridisation. With six bonding pairs, it forms a regular octahedral geometry .
$BrF_5$ (c): Bromine has seven valence electrons. It forms five sigma bonds with fluorine and has one lone pair ( $sp^3d^2$ hybridisation). This 5 bond pair + 1 lone pair arrangement results in a square pyramidal shape .
$BF_3$ (d): Boron undergoes $sp^2$ hybridisation, forming three bond pairs with no lone pairs. This results in a trigonal planar geometry .

Matching these correctly results in (a)-(iv), (b)-(iii), (c)-(i), and (d)-(ii), which corresponds to option 3.

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