Question:
Match the Xenon compounds in List-I with their molecular geometries in List-II:
[h]
{|l|l|}
List - I (Compound) & List - II (Geometry)
(A) XeF\(_2\) & (I) Trigonal pyramidal
(B) XeO\(_3\) & (II) Distorted octahedral
(C) XeF\(_6\) & (III) Linear
(D) XeOF\(_4\) & (IV) Square pyramidal
Match the Xenon compounds in List-I with their molecular geometries in List-II:
[h]
{|l|l|}
List - I (Compound) & List - II (Geometry)
(A) XeF\(_2\) & (I) Trigonal pyramidal
(B) XeO\(_3\) & (II) Distorted octahedral
(C) XeF\(_6\) & (III) Linear
(D) XeOF\(_4\) & (IV) Square pyramidal
(A) XeF\(_2\) & (I) Trigonal pyramidal
(B) XeO\(_3\) & (II) Distorted octahedral
(C) XeF\(_6\) & (III) Linear
(D) XeOF\(_4\) & (IV) Square pyramidal
Show Hint
To quickly find the geometry of Xenon compounds, count valence electrons (8), subtract electrons used for bonds (1 per F, 2 per O), and divide the remainder by 2 to get lone pairs.
Ex: XeF\(_2\) \(\rightarrow\) (8 - 2)/2 = 3 lone pairs.
Updated On: Jun 3, 2026
- A-III, B-I, C-II, D-IV
- A-IV, B-III, C-II, D-I
- A-III, B-I, C-IV, D-II
- A-IV, B-III, C-II, D-I
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Molecular geometry is determined using VSEPR (Valence Shell Electron Pair Repulsion) theory. We calculate the number of bonding pairs and lone pairs on the central Xenon atom (which has 8 valence electrons).
• XeF\(_2\): 2 bond pairs + 3 lone pairs. Total 5 electron pairs (sp\(^3\)d). Lone pairs occupy equatorial positions, making the shape Linear.
• XeO\(_3\): 3 bond pairs (double bonds) + 1 lone pair. Total 4 electron domains (sp\(^3\)). The shape is Trigonal pyramidal.
• XeF\(_6\): 6 bond pairs + 1 lone pair. Total 7 electron pairs (sp\(^3\)d\(^3\)). The lone pair causes distortion, resulting in a Distorted octahedral geometry.
• XeOF\(_4\): 5 bond pairs (4 F + 1 O) + 1 lone pair. Total 6 electron domains (sp\(^3\)d\(^2\)). The shape is Square pyramidal.
Step 1: Matching XeF\(_2\) and XeO\(_3\).
XeF\(_2\) has a linear geometry because the three lone pairs cancel each other out in the equatorial plane. Thus,A matches with III. XeO\(_3\) is analogous to ammonia (NH\(_3\)) with one lone pair. Thus,B matches with I.
Step 2: Matching XeF\(_6\) and XeOF\(_4\).
XeF\(_6\) is the classic example of a capped or distorted octahedron. Thus,C matches with II. XeOF\(_4\) has an octahedral electron geometry, but with one oxygen and one lone pair, the atoms form a square pyramid. Thus,D matches with IV.
Step 3: Final Match.
Combining these results gives the sequence A-III, B-I, C-II, D-IV, which corresponds to option (a).
• XeF\(_2\): 2 bond pairs + 3 lone pairs. Total 5 electron pairs (sp\(^3\)d). Lone pairs occupy equatorial positions, making the shape Linear.
• XeO\(_3\): 3 bond pairs (double bonds) + 1 lone pair. Total 4 electron domains (sp\(^3\)). The shape is Trigonal pyramidal.
• XeF\(_6\): 6 bond pairs + 1 lone pair. Total 7 electron pairs (sp\(^3\)d\(^3\)). The lone pair causes distortion, resulting in a Distorted octahedral geometry.
• XeOF\(_4\): 5 bond pairs (4 F + 1 O) + 1 lone pair. Total 6 electron domains (sp\(^3\)d\(^2\)). The shape is Square pyramidal.
Step 1: Matching XeF\(_2\) and XeO\(_3\).
XeF\(_2\) has a linear geometry because the three lone pairs cancel each other out in the equatorial plane. Thus,A matches with III. XeO\(_3\) is analogous to ammonia (NH\(_3\)) with one lone pair. Thus,B matches with I.
Step 2: Matching XeF\(_6\) and XeOF\(_4\).
XeF\(_6\) is the classic example of a capped or distorted octahedron. Thus,C matches with II. XeOF\(_4\) has an octahedral electron geometry, but with one oxygen and one lone pair, the atoms form a square pyramid. Thus,D matches with IV.
Step 3: Final Match.
Combining these results gives the sequence A-III, B-I, C-II, D-IV, which corresponds to option (a).
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