Incorrect order about bond angle is `:`

Options

A. `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te`
B. `C_(2)H_(2) gt C_(2)H_(4) gt CH_(4) gt NH_(3)`
C. `SF_(6) lt NH_(3) lt H_(2)O lt OF_(2)`
D. `ClO_(2) gt H_(2)O gt H_(2)S gt SF_(6)`

Answer

3

Explanation

`(1)` According to VSEPR as electronegativity of central atom decreases, bond angle decreases. So bond angle of `H__(2)OgtH_(2)SgtH_(2)SegtH_(2)Te`
`(2)` `{:(C_(2)H_(2),gt,C_(2)H_(4),gt,CH_(4),gt,NH_(3)),(sp,,sp^(2),,sp^(3),,sp^(3)):}` with bond pair- lone pair repulsion
`(3)` `SF_(6) lt NH_(3) lt H_(2)O lt OF_(2)` in this case bond angle of `NH_(3)` is highest because `Ip-Ip` repulsion is absent in it.
`(4)` `CIO_(2) gt H_(2)Ogt H_(2)Sgt SF_(6)`
`CIO_(2)` bond angle is highest due to its `sp^(2)` hybridisation, rest all are `sp^(3)` or `sp^(3)d^(2)` hybridised more repulsion in double bond electrons.