Which of the following options is/ are correct?

P O F_{3} > P O C l_{3} > P O B r_{3} > P O I_{3} (P - O Bond length)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} O < C H_{3} O H < C H_{3} O C H_{3} (Bond angle)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F_{2} > C l_{2} > B r_{2} > I_{2} (Oxidising property)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

X e > K r > A r > N e > H e (Solubility in water)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $X e > K r > A r > N e > H e (Solubility in water)$
(A) For these molecules, we can apply bent rule because all given molecules have same hybridization.
According to this rule, in case of $P O F_{3}$ fluorine is more electronegative hence s - character increases more and more in $P = 0$ bond as compared to the other given molecules. So, $P = 0$ bond in shortest in $P O F_{3}$ than other and also increases with decreasing electronegativity. Hence $P = 0$ bond length order is $A < B < C < D .$

Electronegativity order $F > C l > B r > I$
$P O F_{3} < P O C l_{3} < P O B r_{3} < P O I_{3} (P = 0 bond length)$
(B) Due to steric repulsion bond angle increases.

(C) The Halogens can act as oxidising agents by gaining electrons to form halide ions. The oxidising ability decreases down the group with fluorine being the strongest oxidising agent.
$F_{2} > C l_{2} > B r_{2} > I_{2}$ (Oxidising property)

(D) Solublity of noble gases increases down the group due to increase in van der waal's forces.

Suggest Corrections