MCQ
Inertness of ${N_2}$ gas is due to
- ANo vacant $d$ - orbital
- ✓High dissociation energy
- CHigh electronegativity
- DNone
✓
Answer
Correct option: B.
High dissociation energy
b
As the dissociation energy of $N _2$ is very high, it does not react with any molecule.
As the dissociation energy of $N _2$ is very high, it does not react with any molecule.
So, inertness of $N _2$ is due to high dissociation energy of $N _2$.
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
$C{H_3} - CH = C{H_2}\xrightarrow[{hv\,(low\,conc.)}]{{B{r_2}}}\xrightarrow[{Dry\,ether}]{{Mg}}\xrightarrow[{N{H_4}Cl}]{{\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - C{H_3}}
\end{array}}}\xrightarrow[\Delta ]{{{H^ + }}}\mathop {(X)}\limits_{(major)} $
→O \\
{||} \\
{C{H_3} - C - C{H_3}}
\end{array}}}\xrightarrow[\Delta ]{{{H^ + }}}\mathop {(X)}\limits_{(major)} $
End product $(X)$ of the above reaction is
The compound $ ‘A$’ when treated with ceric ammonium nitrate solution gives yellow ppt. The compound $ ‘A’$ is
→${K_2}C{r_2}{O_7}$ on heating with aqueous $NaOH$ gives
→The compound X is:
The reagent used for converting ethanoic acid to ethanol is
Select the correct statement regarding acid catalysed dehydration of ter-butylalcohol.
Product $(D)$ in above reaction is
→Calculate the $e.m.f.$ of the half cells given below ............... $\mathrm{V}$
→$Fe|FeSO_4$ $E^o_{OP} = 0.44\, V$ $(a = 0.1\, M)$
Which one of the following species is stable in aqueous solution ?
For the reaction
→$2 \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
the observed rate expression is, rate $=\mathrm{k}_{\mathrm{f}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right] .$ The rate expression of the reverse reaction is