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Match List - I with List - II.
Column I Column II ( a ) P C l 5 (a) PCl_5 ( a ) PC l 5
(i) Square pyramidal ( b ) S F 6 (b) SF_6 ( b ) S F 6
(ii) Trigonal planar ( c ) B r F 5 (c) BrF_5 ( c ) B r F 5
(iii) Octahedral ( d ) B F 3 (d) BF_3 ( d ) B F 3
(iv) Trigonal bipyramidal
medium
Chemical bonding and Molecular structure
2021
chemistry
(a)-(iv) (b)-(iii) (c)-(i) (d)-(ii)
(a)-(ii) (b)-(iii) (c)-(iv) (d)-(i)
(a)-(iii) (b)-(i) (c)-(iv) (d)-(ii)
(a)-(iv) (b)-(iii) (c)-(ii) (d)-(i) Explanation To solve this problem, we need to determine the molecular geometry of each compound listed in List-I and match it with the correct geometry in List-II. \\
Let's analyze each compound: \\
(a) P C l 5 : PCl_5: \\
PC l 5 : • P C l 5 PCl_5 PC l 5
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has a central phosphorus atom surrounded by five chlorine atoms.
• The electron geometry is based on five bonding pairs, which is
trigonal bipyramidal
• Therefore, the molecular geometry of
is
trigonal bipyramidal
(b)
•
has a central sulfur atom surrounded by six fluorine atoms.
• The electron geometry is based on six bonding pairs, which is
octahedral
• Therefore, the molecular geometry of
is
octahedral
(c)
•
has a central bromine atom surrounded by five fluorine atoms and one lone pair.
• The electron geometry is based on six regions of electron density, which is
octahedral
• With one lone pair, the molecular geometry becomes
square pyramidal
(d)
•
has a central boron atom surrounded by three fluorine atoms.
• The electron geometry is based on three bonding pairs, which is
trigonal planar
• Therefore, the molecular geometry of
is
trigonal planar
Now, let's match each compound with its corresponding geometry:
•
( a ) P C l 5 → (a) PCl_5 \rightarrow ( a ) PC l 5 → (iv) Trigonal bipyramidal
•
( b ) S F 6 → (b) SF_6 \rightarrow ( b ) S F 6 → (iii) Octahedral
•
( c ) B r F 5 → (c) BrF_5 \rightarrow ( c ) B r F 5 → (i) Square pyramidal
•
( d ) B F 3 → (d) BF_3 \rightarrow ( d ) B F 3 → (ii) Trigonal planar
Therefore, the correct option is Option 1:
(a)
(iv), (b)
(iii), (c)
(i), (d)
(ii)