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. .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-1wigfont-weight:bold;text-align:left;vertical-align:top .tg .tg-0laxtext-align:left;vertical-align:top List - I (Shape) List - II (Molecules) (A) T-shaped (I) XeF_4 (B) Trigonal planar (II) SF_4 (C) Square planar (III) ClF_3 (D) See-saw (IV) BF_3

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

(A) - (III), (B) - (IV), (C) - (I), (D) - (I)

(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

Chemical Bonding and Molecular Structure PYQ — JEE Main 2022 · 27 Jun · Shift 1 · Q2 with Solution

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 (Shape)		List - II (Molecules)
(A)	T-shaped	(I)	$XeF_4$
(B)	Trigonal planar	(II)	$SF_4$
(C)	Square planar	(III)	$ClF_3$
(D)	See-saw	(IV)	$BF_3$

A
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
B
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
C
(A) - (III), (B) - (IV), (C) - (I), (D) - (I)
D
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

View written solutionFree

Correct answer: B

Identify the geometry of each molecule using VSEPR theory

We match each molecule in List-II with its molecular shape.

Analyze each molecule

(I) $XeF_4$

Central atom: Xe
Xe has 8 valence electrons.
It forms 4 bonds with 4 F atoms, leaving 2 lone pairs.
Total electron pairs around Xe $= 6$
Electron pair geometry: octahedral
With 2 lone pairs opposite each other, the molecular shape becomes:

$\text{Square planar}$

So,

$XeF_4 \rightarrow \text{Square planar}$

(II) $SF_4$

Central atom: S
S has 6 valence electrons.
It forms 4 bonds with F and has 1 lone pair.
Total electron pairs around S $= 5$
Electron pair geometry: trigonal bipyramidal
With 1 lone pair, the molecular shape is:

$\text{See-saw}$

So,

$SF_4 \rightarrow \text{See-saw}$

(III) $ClF_3$

Central atom: Cl
Cl has 7 valence electrons.
It forms 3 bonds with F and has 2 lone pairs.
Total electron pairs around Cl $= 5$
Electron pair geometry: trigonal bipyramidal
With 2 lone pairs occupying equatorial positions, the molecular shape is:

$\text{T-shaped}$

So,

$ClF_3 \rightarrow \text{T-shaped}$

(IV) $BF_3$

Central atom: B
B has 3 valence electrons.
It forms 3 bonds and has no lone pair.
Total electron pairs around B $= 3$
Molecular shape:

$\text{Trigonal planar}$

So,

$BF_3 \rightarrow \text{Trigonal planar}$

Now match List-I with List-II

(A) T-shaped $\rightarrow (III)$ $ClF_3$
(B) Trigonal planar $\rightarrow (IV)$ $BF_3$
(C) Square planar $\rightarrow (I)$ $XeF_4$
(D) See-saw $\rightarrow (II)$ $SF_4$

Thus the correct matching is:

(A)-(III),\quad (B)-(IV),\quad (C)-(I),\quad (D)-(II)

Compare with options

This corresponds to Option B.

Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So they agree.

More from Chemical Bonding and Molecular Structure

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 (Shape)		List - II (Molecules)
(A)	T-shaped	(I)	$XeF_4$
(B)	Trigonal planar	(II)	$SF_4$
(C)	Square planar	(III)	$ClF_3$
(D)	See-saw	(IV)	$BF_3$

A
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
B
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
C
(A) - (III), (B) - (IV), (C) - (I), (D) - (I)
D
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

View written solutionFree

Correct answer: B

Identify the geometry of each molecule using VSEPR theory

We match each molecule in List-II with its molecular shape.

Analyze each molecule

(I) $XeF_4$

Central atom: Xe
Xe has 8 valence electrons.
It forms 4 bonds with 4 F atoms, leaving 2 lone pairs.
Total electron pairs around Xe $= 6$
Electron pair geometry: octahedral
With 2 lone pairs opposite each other, the molecular shape becomes:

$\text{Square planar}$

So,

$XeF_4 \rightarrow \text{Square planar}$

(II) $SF_4$

Central atom: S
S has 6 valence electrons.
It forms 4 bonds with F and has 1 lone pair.
Total electron pairs around S $= 5$
Electron pair geometry: trigonal bipyramidal
With 1 lone pair, the molecular shape is:

$\text{See-saw}$

So,

$SF_4 \rightarrow \text{See-saw}$

(III) $ClF_3$

Central atom: Cl
Cl has 7 valence electrons.
It forms 3 bonds with F and has 2 lone pairs.
Total electron pairs around Cl $= 5$
Electron pair geometry: trigonal bipyramidal
With 2 lone pairs occupying equatorial positions, the molecular shape is:

$\text{T-shaped}$

So,

$ClF_3 \rightarrow \text{T-shaped}$

(IV) $BF_3$

Central atom: B
B has 3 valence electrons.
It forms 3 bonds and has no lone pair.
Total electron pairs around B $= 3$
Molecular shape:

$\text{Trigonal planar}$

So,

$BF_3 \rightarrow \text{Trigonal planar}$

Now match List-I with List-II

(A) T-shaped $\rightarrow (III)$ $ClF_3$
(B) Trigonal planar $\rightarrow (IV)$ $BF_3$
(C) Square planar $\rightarrow (I)$ $XeF_4$
(D) See-saw $\rightarrow (II)$ $SF_4$

Thus the correct matching is:

(A)-(III),\quad (B)-(IV),\quad (C)-(I),\quad (D)-(II)

Compare with options

This corresponds to Option B.

Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So they agree.

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

(I) $XeF_4$

(II) $SF_4$

(III) $ClF_3$

(IV) $BF_3$

More from Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

(I) $XeF_4$

(II) $SF_4$

(III) $ClF_3$

(IV) $BF_3$

More from Chemical Bonding and Molecular Structure

(I) XeF4XeF_4XeF4​

(II) SF4SF_4SF4​

(III) ClF3ClF_3ClF3​

(IV) BF3BF_3BF3​

More from Chemical Bonding and Molecular Structure

(I) XeF4XeF_4XeF4​

(II) SF4SF_4SF4​

(III) ClF3ClF_3ClF3​

(IV) BF3BF_3BF3​

More from Chemical Bonding and Molecular Structure

(I) $XeF_4$

(II) $SF_4$

(III) $ClF_3$

(IV) $BF_3$

(I) $XeF_4$

(II) $SF_4$

(III) $ClF_3$

(IV) $BF_3$