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Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List - II and select the most appropriate option.

List-I List-II

(Shape) (Molecules)

(A) T-Shaped (I) $\mathrm{XeF}_{4}$

(B) Trigonal planar (II) $\mathrm{SF}_{4}$

(C) Square planar (III) $\mathrm{ClF}_{3}$

(D) See-Saw (IV) $\mathrm{BF}_{3}$

Option: 1
$(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})$

Option: 2
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$

Option: 3
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})$

Option: 4
$(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$

Answers (1)

best_answer

Molecules	Structure	Hybridisation	Shape
$\mathrm{(I)\; X{e}F_{4}}$		$\mathrm{sp^{3}d^{2}}$	Square Planar
$\mathrm{(II)\; SF_{4}}$		$\mathrm{sp^{3}d}$	See Saw
$\mathrm{(III)\; ClF_{3}}$		$\mathrm{sp^{3}d}$	T-Shaped
$\mathrm{(IV)\; BF_{3}}$		$\mathrm{sp^{2}}$	Trigonal Planar

Hence, the correct answer is Option (2).

Posted by

jitender.kumar

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