MCQ
Bond angle is minimum for
- A${H_2}O$
- B${H_2}S$
- C${H_2}Se$
- ✓${H_2}Te$
All these are the hydrides of 6 th group elements in which the central atom undergo $S P^{3}$ hybridization and should posses the bond angle 109 degree but the bond angle distorts due to the repusions between lone pair and lone pair. But this repulsions are minimum in $H_{2}$ Te as the tellurium has large size which makes the repulsions minimum. Hence option D is correct.
| ${H_2}O$ | ${H_2}S$ | $H_2Se$ | ${H_2}Te$ |
| ${104.5^o}$ | ${92.1^o}$ | ${91^o}$ |
${90^o}$ |
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Above reaction is an example of
$B.E. (H-H) = x_1 ; B.E. (O = O) = x_2$ ;
$B.E. (O -H) = x_3$
Heat of vaporisation of water $= x_4$ then $\Delta {H_f}$ [heat of formation of liquid water] is