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
636k+ 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
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

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

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

