MCQ
Electrolysis temperature is maximum for
- A$As{H_3}$
- ✓$N{H_3}$
- C$P{H_3}$
- D$Sb{H_3}$
✓
Answer
Correct option: B.
$N{H_3}$
b
(b)$N{H_3}$ is most thermally stable hydride. Hence, electrolysis temperature is maximum.
(b)$N{H_3}$ is most thermally stable hydride. Hence, electrolysis temperature is maximum.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The correct order of bond angle
Most transition metals
→$I :$ form sets of compounds which display different oxidation states of the metal
$II :$ form coloured ions in solution
$III :$ burn vigorously in oxygen
$IV :$ act as catalyst
In the following transition elements, the lowest $M.P.$ and $B.P.$ is exhibited by
→Which best describes the product of the following reaction ?
Glucose when heated with $C{H_3}OH$ in presence of dry $HCl$ gas gives $\alpha $ and $\beta - {\rm{methyl}}$ glucosides because it contains
→Consider the dissociation of the weak acid $\mathrm{HX}$ as given below
→$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$
$\left[\mathrm{K}_{\mathrm{n}}:\right.$ dissociation constant]
The osmotic pressure of $0.03 \mathrm{M}$ aqueous solution of $\mathrm{HX}$ at $300 \mathrm{~K}$ is ............... $\times 10^{-2}$ bar (nearest integer).
$\left[\right.$ Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{Mol}^{-1} \mathrm{~K}^{-1}$ ]
$[Mn(CO)_4 NO]$ is diamagnetic because
→Benzene hexachloride is prepared from benzene and chlorine in sunlight by
Major product is $(A)$ is
→The results given in the below table were obtained during kinetic studies of the following reaction:
→$2 A + B \longrightarrow C + D$
| Experiment | $[ A ] / molL ^{-1}$ | $[ B ] / molL ^{-1}$ | Initial $rate/molL$ $^{-1}$ $\min ^{-1}$ |
| $I$ | $0.1$ | $0.1$ | $6.00 \times 10^{-3}$ |
| $II$ | $0.1$ | $0.2$ | $2.40 \times 10^{-2}$ |
| $III$ | $0.2$ | $0.1$ | $1.20 \times 10^{-2}$ |
| $IV$ | $X$ | $0.2$ | $7.20 \times 10^{-2}$ |
| $V$ | $0.3$ | $Y$ | $2.88 \times 10^{-1}$ |
$X$ and $Y$ in the given table are respectively :