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Match List I with List II. Column I Column II A. NH₃ I. Trigonal Pyramidal B. BrF₅ II. Square Planar C. XeF₄ III. Octahedral D. SF₆ IV. Square Pyramidal
medium
Chemical bonding and Molecular structure
2024
chemistry
Explanation To solve this problem, we need to determine the molecular geometry of each compound in List I and match it with the correct geometry in List II. Let's analyze each compound: • \newline
• • A. NH 3 : _3: \\
3 ​ : Ammonia (NH 3 _3 3 ​
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) has a central nitrogen atom with three hydrogen atoms and one lone pair.
The electron pair geometry is tetrahedral, but due to the lone pair, the molecular geometry is trigonal pyramidal.
Thus, NH
corresponds to I. Trigonal Pyramidal.
B.
BrF
Bromine pentafluoride (BrF
) has a central bromine atom with five fluorine atoms and one lone pair.
The electron pair geometry is octahedral, but the presence of one lone pair makes the molecular geometry square pyramidal.
Thus, BrF
corresponds to IV. Square Pyramidal.
C.
XeF
Xenon tetrafluoride (XeF
) has a central xenon atom with four fluorine atoms and two lone pairs.
The electron pair geometry is octahedral, but the molecular geometry is square planar due to the two lone pairs.
Thus, XeF
corresponds to II. Square Planar.
D.
SF
Sulfur hexafluoride (SF
) has a central sulfur atom with six fluorine atoms and no lone pairs.
The electron pair geometry and molecular geometry are both octahedral.
Thus, SF
corresponds to III. Octahedral.
Based on the analysis, the correct matches are:
A
I, B
IV, C
II, D
III.
Therefore, the correct option is Option 3.