MCQ
The correct statement regarding perxenate ion $(XeO_6^{4-})$ is
- AIt is polar species
- BIt is a planar species
- ✓$'Xe-O'$ bond order is $1.33$
- DMolecular ion has only one type of bond angle
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(Round off to the NearestInteger)
[Use:Atomicmasses:$Na:23.0\,u \mathrm{O}: 16.0 \,\mathrm{u} \quad \mathrm{H}: 1.0 \,\mathrm{u}$, Density of $\mathrm{H}_{2} \mathrm{O}: 1.0 \,\mathrm{~g} \,\mathrm{~cm}^{-3}$ ]
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ on treatment with alc. $\mathrm{KOH}$ gives $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ as a major product. | $(P)$ $E1$ reaction |
| $(B)$ $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3$ reacts faster than $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3$. | $(Q)$ $E2$ reaction |
| $(C)$ $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$ on treatment with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$ gives $\mathrm{Ph}-\mathrm{CD}=\mathrm{CH}_2$ as the major product. | $(R)$ $E1$ cb reaction |
| $(D)$ $\mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{PhCD}_2 \mathrm{CH}_2 \mathrm{Br}$ react with same rate. | $(S)$ First order reaction |
$1-$ Bromopropane is reacted with reagents in List $I$ to give product in List $II$
| List$I-$ Reagent | List$II-$Product |
| $A$ $KOH (\text { alc) }$ | $I$ Nitrile |
| $B$ $KCN \text { (alc) }$ | $II$ Ester |
| $C$ $AgNO _2$ | $III$ Alkene |
| $D$ $H _3 CCOOAg$ | $IV$ Nitroalkane |