
Select the correct statements.
(A)- The combination of s-orbit and p-orbital, with the increase of p-character, the bond angle decreases.
(B)- $H-\widehat{C}-H$ bond angle $ >H-\widehat{C}F$ bond angle in $C{{H}_{3}}F$ molecule
(C)- $F-\widehat{C}-F$ bond angle $>H-\widehat{C}F$ bond angle in $CH{{F}_{3}}$ bond angle
(D)- None of the above
Answer
554.4k+ views
Hint: The angle between the two bonds that is the angle between two orbitals that contains a pair of bonding electrons around the central atom in a complex molecule or ion is known as Bond angle.
Complete answer:
-Bond angle mainly depends on the following factors-
(i) Hybridization- Bond angle depends on the hybridization of the central atom. Greater the s-character, greater is the bond angle. For example, a central atom with $s{{p}^{3}}$ has a bond angle as $109{}^\circ 28'$ with $s{{p}^{2}}$ hybridization have a bond angle as $120{}^\circ $ and with sp hybridization have a bond angle as $180{}^\circ $.
(ii) Lone pair repulsion- Presence of lone pairs on the central atom causes repulsion and hence decreases the bond angle. For example, the bond angle of $C{{H}_{4}}$which has no lone pairs is $109{}^\circ $, but the bond angle of $N{{H}_{3}}$ which has a pair of lone pairs have a bond angle $107{}^\circ $ .
(iii) Electronegativity- The bond angle decreases with the decrease in the electronegativity of the atom. For example, the bond angle of $N{{H}_{3}}$ is $107{}^\circ $ and the bond angle of $P{{H}_{3}}$ is $93.5{}^\circ $ .
-Let us now consider the statements given in the options.
-In option A, there is a relation between the bond angle and the s-character. As according to the above concept, s-character is directly proportional to the bond angle. Combination of s and p orbitals gives rise to the hybrid orbitals and more the s-character of the hybrid molecule, greater is the bond angle.
Hence we conclude that based on Bent’s rule equation, which states that,
$-\cos x=\dfrac{s}{s-1}$ where x is a bond angle; s is the % of s-character.
It is clear that with the increase in p-character the bond angle will be minimum and with the increase in s-character, the bond angle will be maximum.
Therefore, the statement is correct.
-In option B, the bond angle is influenced by the electronegativity of the atom. According to Bent’s rules, s-character contributed by three hydrogens towards carbon is more than the fluorine’s contribution. Hence higher bond angle is observed in $H-\widehat{C}-H$ $(110.3{}^\circ )$ than $H-\widehat{C}F$ bond angle in $C{{H}_{3}}F$ molecule $(108{}^\circ )$ .
Therefore, the given statement is correct.
-In option C, is the contradiction of option B. Since option B is correct, therefore option C is incorrect.
Hence, the correct answer is option A and B.
Note: Bond angle is a part of the molecular structure which is very important as it helps in predicting and understanding the molecular structure of a molecule because many of the properties of a substance are determined by its geometry.
Complete answer:
-Bond angle mainly depends on the following factors-
(i) Hybridization- Bond angle depends on the hybridization of the central atom. Greater the s-character, greater is the bond angle. For example, a central atom with $s{{p}^{3}}$ has a bond angle as $109{}^\circ 28'$ with $s{{p}^{2}}$ hybridization have a bond angle as $120{}^\circ $ and with sp hybridization have a bond angle as $180{}^\circ $.
(ii) Lone pair repulsion- Presence of lone pairs on the central atom causes repulsion and hence decreases the bond angle. For example, the bond angle of $C{{H}_{4}}$which has no lone pairs is $109{}^\circ $, but the bond angle of $N{{H}_{3}}$ which has a pair of lone pairs have a bond angle $107{}^\circ $ .
(iii) Electronegativity- The bond angle decreases with the decrease in the electronegativity of the atom. For example, the bond angle of $N{{H}_{3}}$ is $107{}^\circ $ and the bond angle of $P{{H}_{3}}$ is $93.5{}^\circ $ .
-Let us now consider the statements given in the options.
-In option A, there is a relation between the bond angle and the s-character. As according to the above concept, s-character is directly proportional to the bond angle. Combination of s and p orbitals gives rise to the hybrid orbitals and more the s-character of the hybrid molecule, greater is the bond angle.
Hence we conclude that based on Bent’s rule equation, which states that,
$-\cos x=\dfrac{s}{s-1}$ where x is a bond angle; s is the % of s-character.
It is clear that with the increase in p-character the bond angle will be minimum and with the increase in s-character, the bond angle will be maximum.
Therefore, the statement is correct.
-In option B, the bond angle is influenced by the electronegativity of the atom. According to Bent’s rules, s-character contributed by three hydrogens towards carbon is more than the fluorine’s contribution. Hence higher bond angle is observed in $H-\widehat{C}-H$ $(110.3{}^\circ )$ than $H-\widehat{C}F$ bond angle in $C{{H}_{3}}F$ molecule $(108{}^\circ )$ .
Therefore, the given statement is correct.
-In option C, is the contradiction of option B. Since option B is correct, therefore option C is incorrect.
Hence, the correct answer is option A and B.
Note: Bond angle is a part of the molecular structure which is very important as it helps in predicting and understanding the molecular structure of a molecule because many of the properties of a substance are determined by its geometry.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

