MCQ
$LiAl{H_4}$ converts acetic acid into
- AAcetaldehyde
- BMethane
- ✓Ethyl alcohol
- DMethyl alcohol
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|
List$-I$ (Species) |
List$-II$ (Number of lone pairs of electrons on the central atom) |
| $(a)$ $\mathrm{XeF}_{2}$ | $(i)\, 0$ |
| $(b)$ $\mathrm{XeO}_{2} \mathrm{~F}_{2}$ | $(ii) \,1$ |
| $(c)$ $\mathrm{XeO}_{3} \mathrm{~F}_{2}$ | $(iii) \,2$ |
| $(d)$ $\mathrm{XeF}_{4}$ | $(iv) \,3$ |
Choose the most appropriate answer from the options given below:
$CH_3COOH + PCl_5 \longrightarrow (A)$ $\xrightarrow[(2){{H}_{2}}O]{(1)\,C{{H}_{3}}MgBr}\left( B \right)$ Product $B$ would be