MCQ
A hydride of nitrogen which is acidic is
- A$N{H_3}$
- B${N_2}{H_2}$
- ✓${N_3}H$
- D${N_2}{H_4}$
${N_3}H \to {H^ + } + N_3^ - $
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take place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{{slow}} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{{fast}} H_2O + Br_2$
The order of the reaction is
|
List $I$ (Molecule Species) |
List $I$ ( Property Shape ) |
| $A$ $\mathrm{SO}_2 \mathrm{Cl}_2$ | $I$ Paramagnetic |
| $B$ $NO$ | $II$ Diamagnetic |
| $C$ $\mathrm{NO}_2^{-}$ | $III$ Tetrahedral |
| $D$ $\mathrm{I}_3^{-}$ | $IV$ Linear |
Choose the correct answer from the options given below :
Statement $I$: Aniline reacts with con. $\mathrm{H}_2 \mathrm{SO}_4$ followed by heating at $453-473 \mathrm{~K}$ gives $\mathrm{p}$ aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement $II$: In Friedel - Craft's alkylation and acylation reactions, aniline forms salt with the $\mathrm{AlCl}_3$ catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below :
| Column $-I$ Catalyst | Column $-II$ >Product |
| $(a)$ $V_2O_5$ | $(i)$ Polyethylene |
| $(b)$ $TiCl_4/Al(Me)_3$ | $(ii)$ Ethanal |
| $(c)$ $PdCl_2$ | $(iii)$ $H_2SO_4$ |
| $(d)$ Iron oxide | $(iv)$ $NH_3$ |