
Which is the correct statement?
As the s-character of a hybrid orbital decreases
(I) The bond angle decreases
(II) The bond strength increases
(III) The bond length increases
(IV) Size of orbitals increases
(A) (I), (III) and (IV)
(B) (II), (III) and (IV)
(C) (I) and (II)
(D) All are correct
Answer
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Hint: s-character is the contribution of the sigma type bond in a hybridization. Hybridization affects the bond angle, bond length, bond strength and the size of the orbitals.
Complete step by step answer:
> As the s-character increases, the bond angle also increases and vice versa. This is due to the fact that the electrons are arranged in such a manner so as to experience minimum repulsion. So, the more the s-character, the more will be the electron density across the central atom and hence to avoid that repulsion and to increase the stability, the electrons are arranged at a greater angle.
> As the s-character decreases , the bond strength also decreases and vice versa. This is due to the fact that with a decrease in the s-character, the orbitals move away from the nucleus and hence form weaker bonds.
> As the s-character decreases, the bond length increases and vice versa. This is due to the same fact that with a decrease in the s-character, the orbitals move away from the nucleus and hence form weaker bonds and hence have longer bond length.
> As the s-character decreases, the size of the orbitals increases and vice versa. This is due to the fact that the s-orbitals are closer to the nucleus than the p-orbital, thus lesser the s-character, the greater the orbital will be.
Now keeping in mind these points, let us look at the question.
> As the s-character of a hybrid orbital decreases,
- The bond angle decreases: The bond angles for $sp, s{p}^{2} and s{p}^{3}$ hybrid orbitals are ${180}^{0}, {120}^{0} and {109.5}^{0}$ respectively. Thus, it is true.
- The bond strength increases: Triple bond (sp hybridisation) is stronger than double bond ($s{p}^{2}$ hybridisation) which in turn is stronger than single bond ($s{p}^{3}$ hybridisation). Thus, this statement is false.
- The bond length increases: Triple bond (sp hybridisation) is shorter than double bond ($s{p}^{2}$ hybridisation) which in turn is shorter than single bond ($s{p}^{3}$ hybridisation). Thus, it is true.
- The size of the orbital increases: s orbital is smaller than p orbital. Hence, sp hybrid orbitals are smaller than $s{p}^{2}$ hybrid orbitals which in turn are smaller than $s{p}^{3}$ hybrid orbitals. Thus, it is also true.
Therefore, the statements (I), (III) and (IV) are correct.
Note: The bond angle and the bond length of a molecule are two different things. Bond angle is the angle between the two bonds whereas, the bond length is the average length between the nuclei of the two bonded atoms in a molecule.
Complete step by step answer:
> As the s-character increases, the bond angle also increases and vice versa. This is due to the fact that the electrons are arranged in such a manner so as to experience minimum repulsion. So, the more the s-character, the more will be the electron density across the central atom and hence to avoid that repulsion and to increase the stability, the electrons are arranged at a greater angle.
> As the s-character decreases , the bond strength also decreases and vice versa. This is due to the fact that with a decrease in the s-character, the orbitals move away from the nucleus and hence form weaker bonds.
> As the s-character decreases, the bond length increases and vice versa. This is due to the same fact that with a decrease in the s-character, the orbitals move away from the nucleus and hence form weaker bonds and hence have longer bond length.
> As the s-character decreases, the size of the orbitals increases and vice versa. This is due to the fact that the s-orbitals are closer to the nucleus than the p-orbital, thus lesser the s-character, the greater the orbital will be.
Now keeping in mind these points, let us look at the question.
> As the s-character of a hybrid orbital decreases,
- The bond angle decreases: The bond angles for $sp, s{p}^{2} and s{p}^{3}$ hybrid orbitals are ${180}^{0}, {120}^{0} and {109.5}^{0}$ respectively. Thus, it is true.
- The bond strength increases: Triple bond (sp hybridisation) is stronger than double bond ($s{p}^{2}$ hybridisation) which in turn is stronger than single bond ($s{p}^{3}$ hybridisation). Thus, this statement is false.
- The bond length increases: Triple bond (sp hybridisation) is shorter than double bond ($s{p}^{2}$ hybridisation) which in turn is shorter than single bond ($s{p}^{3}$ hybridisation). Thus, it is true.
- The size of the orbital increases: s orbital is smaller than p orbital. Hence, sp hybrid orbitals are smaller than $s{p}^{2}$ hybrid orbitals which in turn are smaller than $s{p}^{3}$ hybrid orbitals. Thus, it is also true.
Therefore, the statements (I), (III) and (IV) are correct.
Note: The bond angle and the bond length of a molecule are two different things. Bond angle is the angle between the two bonds whereas, the bond length is the average length between the nuclei of the two bonded atoms in a molecule.
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