
Assertion: Bond angle of $P{F_3} > PC{l_3}$ but bond angle of $PC{l_3} < PB{r_3}$
Reason: The bond angles show an increase in decreasing electronegativity of attached other atoms on central but in $P{F_3}$ $p\pi - d\pi $ bonding results in an increase in bond angle.
A.Both assertion and reason are correct and the reason is the correct explanation for Assertion.
B.Both assertion and reason are correct but reason is not the correct explanation for assertion.
C.Assertion is correct but reason is incorrect.
D.Assertion is incorrect but reason is correct.
Answer
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Hint:We have to know there will be an effect of electronegativity of the surrounding atom on the bond angle of similar molecules. In similar molecules when electronegativity of the surrounding atom decreases, the bond angle increases. The electronegativity of the surrounding atoms increases, the electron pair would tend to come closer to the central atom and there would be repulsion between the electron pair and the central atom because of which bond angle increases.
Complete step by step answer:
We know that there are two types of electron repulsions, which shape the molecules:
Lone pair-bond pair that decreases the angle
Bond pair-bond pair that increases the angle
The lone pair-bond pair repulsion increases slowly with the decrease in electronegativity of the bonded element, hence fluorine would have a large bond angle due to lone pair-bond pair repulsion.
The bond pair-bond pair repulsion increases with the rise in size of the bounded element. We know that a bigger atom means there is more electron density around it. The bond pair-bond pair repulsion increases more than the lone pair-bond pair repulsion on moving down a group. So because of the bond pair-bond pair the bond angle of bromine will be more than bond angle of chlorine.
Therefore, both assertion and reason are correct and the reason is the correct explanation for assertion.
Therefore, the correct option is (A).
Note:
Bond angles depend on the three factors,
Hybridization: Bond angle is dependent on the state of hybridization of the central atom. The bond angle would be greater, if the s character is greater.
Lone pair: The presence of a lone pair of electrons on the central atom affects the bond angle. The presence of lone pair repels the bond pair leading to decrease in bond angle.
Electronegativity: The bond angle decreases with decrease in electronegativity of the central atom.
Complete step by step answer:
We know that there are two types of electron repulsions, which shape the molecules:
Lone pair-bond pair that decreases the angle
Bond pair-bond pair that increases the angle
The lone pair-bond pair repulsion increases slowly with the decrease in electronegativity of the bonded element, hence fluorine would have a large bond angle due to lone pair-bond pair repulsion.
The bond pair-bond pair repulsion increases with the rise in size of the bounded element. We know that a bigger atom means there is more electron density around it. The bond pair-bond pair repulsion increases more than the lone pair-bond pair repulsion on moving down a group. So because of the bond pair-bond pair the bond angle of bromine will be more than bond angle of chlorine.
Therefore, both assertion and reason are correct and the reason is the correct explanation for assertion.
Therefore, the correct option is (A).
Note:
Bond angles depend on the three factors,
Hybridization: Bond angle is dependent on the state of hybridization of the central atom. The bond angle would be greater, if the s character is greater.
Lone pair: The presence of a lone pair of electrons on the central atom affects the bond angle. The presence of lone pair repels the bond pair leading to decrease in bond angle.
Electronegativity: The bond angle decreases with decrease in electronegativity of the central atom.
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