MCQ
$Xe{F_4}$ on partial hydrolysis produces
- A$Xe{F_2}$
- ✓$XeO{F_2}$
- C$XeO{F_4}$
- D$Xe{O_3}$
Complete hydrolysis;
$2Xe{F_4} + 3{H_2}O\, \to $$Xe + Xe{O_3} + {F_2} + 6HF$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$
the correct option is
| $column 1$ | $column 2$ | $column 3$ |
| $(I)$ Toluene | $(i)$ $\mathrm{NaOH} / \mathrm{Br}_2$ | $(P)$ Condensation |
| $(II)$ Acetophenone | $(ii)$ $\mathrm{Br}_2 / \mathrm{h} v$ | $(Q)$ Carboxylation |
| $(III)$ Benzaldehyde | $(iii)$ $\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O} / \mathrm{CH}_3 \mathrm{COOK}$ | $(R)$ Substion |
| $(IV)$ Phenol | $(iv)$ $\mathrm{NaOH} / \mathrm{CO}_2$ | $(S)$ Haloform |
($1$) For the synthesis of benzoic acid, the only $CORRECT$ combination is
$[A] (III) (iv) (R)$ $[B] (IV)(ii) (P)$ $[C] (I) (iv) (Q)$ $[D] (II) (i) (S)$
($2$) The only CORRECT combination in which the reaction proceeds through radical mechanism is
$[A] (I)$ $(ii) (R)$ $[B] (II) (iii) (R)$ $[C] (III) (ii) (P)$ $[D] (IV) (i) (Q)$
($3$) The only CORRECT combination that gives two different carboxylic acids is
$[A] (IV) (iii) (Q)$ $[B] (III) (iii) (P)$ $[C] (II) (iv) (R)$ $[D] (I) (i) (S)$
Given the answer quetion ($1$) ($2$) and ($3$)