Problem 96
Question
The correct increasing bond angle among \(\mathrm{BF}_{3}, \mathrm{PF}_{3}\) and \(\mathrm{C} 1 \mathrm{~F}_{3}\) follows the order (a) \(\mathrm{BF}_{3}<\mathrm{PF}_{3}<\mathrm{CIF}_{3}\) (b) \(\mathrm{PF}_{3}<\mathrm{BF}_{3}<\mathrm{CIF}_{3}\) (c) \(\mathrm{CIF}_{3}<\mathrm{PF}_{3}<\mathrm{BF}_{3}\) (d) \(\mathrm{BF}_{3}<\mathrm{PF}_{3}<\mathrm{CIF}_{3}\)
Step-by-Step Solution
Verified Answer
The correct order is (c) \(\mathrm{CIF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3\).
1Step 1: Analyze the Geometry
Begin by determining the molecular geometry of each compound. \(\mathrm{BF}_3\) has a trigonal planar structure with a bond angle of 120 degrees. \(\mathrm{PF}_3\) forms a trigonal pyramidal geometry, influenced by the presence of a lone pair, leading to bond angles slightly less than 109.5 degrees. \(\mathrm{CIF}_3\) adopts a T-shaped geometry due to the presence of two lone pairs on chlorine, resulting in bond angles of about 90 degrees.
2Step 2: Compare the Bond Angles
Now compare the bond angles of each compound. The bond angle of \(\mathrm{CIF}_3\) (approximately 90 degrees) is less than that of \(\mathrm{PF}_3\) (slightly less than 109.5 degrees), which is itself less than that of \(\mathrm{BF}_3\) (120 degrees). Therefore, the correct order of increasing bond angles is \(\mathrm{CIF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3\).
3Step 3: Identify the Correct Answer Option
Find the answer choice that matches the increasing order of bond angles. Option (c) \(\mathrm{CIF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3\) matches the correct order.
Key Concepts
Bond Angle ComparisonTrigonal Planar StructureLone Pairs and Bond Angles
Bond Angle Comparison
Understanding bond angles is key in recognizing how the shapes of different molecules influence their geometry. Bonds exist at different angles depending on the arrangement of atoms around a central atom. The angle measures the space between two bonds originating from the same atom.
In molecules like \(\mathrm{BF}_3\), \(\mathrm{PF}_3\), and \(\mathrm{CIF}_3\), each has different molecular geometries, which directly influence the bond angles.
In molecules like \(\mathrm{BF}_3\), \(\mathrm{PF}_3\), and \(\mathrm{CIF}_3\), each has different molecular geometries, which directly influence the bond angles.
- \(\mathrm{BF}_3\) has a bond angle of 120 degrees due to its trigonal planar structure.
- \(\mathrm{PF}_3\) has smaller bond angles, slightly less than 109.5 degrees, due to its trigonal pyramidal geometry.
- \(\mathrm{CIF}_3\) has bond angles close to 90 degrees because of its T-shaped geometry.
Trigonal Planar Structure
A trigonal planar structure is a form of molecular geometry where three bonds are evenly spaced around a central atom, lying on the same plane. It forms a flat shape and is noted for its symmetry.
\(\mathrm{BF}_3\) is a classic example of a molecule with a trigonal planar structure.
\(\mathrm{BF}_3\) is a classic example of a molecule with a trigonal planar structure.
- The boron atom is central, with three fluorine atoms evenly spaced around it.
- This configuration results in bond angles that are exactly 120 degrees.
Lone Pairs and Bond Angles
Lone pairs are non-bonding pairs of electrons that reside on atoms. These pairs can exert greater repulsion than bonding pairs, affecting bond angles in a molecule.
In \(\mathrm{PF}_3\), a lone pair on phosphorus reduces the bond angle from the typical tetrahedral angle of 109.5 degrees to slightly less than this value.
In \(\mathrm{PF}_3\), a lone pair on phosphorus reduces the bond angle from the typical tetrahedral angle of 109.5 degrees to slightly less than this value.
- The presence of a lone pair distorts the geometry by pushing the bonded electron pairs closer together.
- Lone pairs prevent bonded atoms from achieving angles of 120 or 180 degrees because they occupy more space.
Other exercises in this chapter
Problem 94
The element with the highest first ionization potential is (a) nitrogen (b) oxygen (c) boron (d) carbon
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The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is (a) \(\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}\) (b) \(\mathrm{O
View solution Problem 97
Which ionization potential in the following equations involves the greatest amount of energy? (a) \(\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+\mathrm{e}^{-}\)
View solution Problem 98
Among the elements \(\mathrm{W}, \mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) having atomic numbers \(9,10,11\) and 12 respectively, the correct order of ionizat
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