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Bond Angle

The correct o...

Question

The correct order of decreasing $X - O - X$ bond angle is: ( X= H, F or Cl)

H_{2} O > C l_{2} O > F_{2} O

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C l_{2} O > H_{2} O > F_{2} O

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F_{2} O > C l_{2} O > H_{2} O

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F_{2} O > H_{2} O > C l_{2} O

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Solution

The correct option is C $C l_{2} O > H_{2} O > F_{2} O$
Bond angle is affected by the presence of lone pairs of electrons at the central atom.
But in case of $F_{2} O, {C l}_{2} O$ and $H_{2} O$ , they are all similar from the point of view that a single oxygen atom forms two single bonds with other atoms.
In all these cases the molecule is tetrahedral with oxygen in the centre, 2 of its electron l.p. on the 2 corners & the other atoms on the other 2 corners.
The angle is determined by the repulsion between the different electron pairs(electron pairs repel each other and try to be located as far away as possible from each other.)
The strength of repulsion decreases in the manner
$l . p - l . p > l p - b p > b p - b p$
lp-lone pair
bp-bond pair
The bond angle of $X - O - X$ where $X = F, C l, H$ will be determined by the repulsion of the lone pairs and the bonding pairs. The more the lone pair-bond pair repulsion increases, the more the bonding angle increases.
The degree of this type of repulsion depends on the EN of $X$ . Hydrogen is the best EN. Thus bonding pair of $O - H$ bond will be closer to $O$ , resulting in higher electron density near $O$ , at the site of the bond. This increases repulsion with the lone pairs in the outer shell of oxygen, thus increasing bond angle.
In case of $F$ ; it is just the opposite. $F$ is highly EN, thus electron density of the $O - F$ bond near $O$ is less than that of $O - H$ .
Therefore $H_{2} O > {C l}_{2} O > F_{2} O$

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