- 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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$Mn{O_2}(s)\, + \,4{H^ + }(aq)\, + \,2{e^ - }\, \to \, M{n^{ + 2}}(aq)\, + \,2{H_2}O(l)\,;$ $E_2^o\, = \,1.21\,\,V$
$MnO_4^{ - 1}(aq)\, + \,4{H^ + }(aq)\, + \,3{e^ - }\, \to \, Mn{O_2}(s)\, + \,3{H_2}O(l)\,;$ $E_3^o\,?$
value of $E_3^o$ will be ............ $\mathrm{V}$
(image) $\xrightarrow{{B{r_3} + C{H_3} - COOH}}A\xrightarrow{{Alc.KOH}}B$ (Product with six membered ring) $\xrightarrow[{(2)\,{H_2}O}]{{(1)\,C{H_3} - MgBr}}C\xrightarrow{{{H^ \oplus } + \Delta }}D\xrightarrow{{HBr(1\,mole)}}E$