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Arindam Paul posted an Question
August 26, 2020 • 19:37 pm 30 points
  • IIT JAM
  • Biotechnology (BT)

The correct order of increasing x-0-x bond angle is (x = h, f or cl) : 1 h,o> cl,0 > f,o 2 cl,0> h,o> f,0 3 3 f0> cl,0> h,o 4 f0 > h,o > cl,0

The correct order of increasing X-0-X bond angle is (X = H, F or Cl) : 1 H,O> Cl,0 > F,O 2 Cl,0> H,O> F,0 3 3 F0> Cl,0> H,O 4 F0 > H,O > Cl,0

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    Narayan singh Best Answer

    Bond angle is affected by the presence of lone pairs of electrons at the central atom. But in case of F 2 ​ O,Cl 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>lp−bp>bp−bp lp-lone pair bp-bond pair The bond angle of X−O−X where X=F, Cl, 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.

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    Priyanshu kumar best-answer

    Option A is correct dear student

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    Sir please discuss

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    Dear student please watch this short video https://youtu.be/9Gvly0T2_uI

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    please ask if any doubt after watching this video...we will discuss on this😊

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