MCQ
Which molecule has the highest bond order
  • ${N_2}$
  • B
    $L{i_2}$
  • C
    $H{e_2}$
  • D
    ${O_2}$

Answer

Correct option: A.
${N_2}$
a
Bond order is the number of chemical bonds between a pair of atoms.

$N _2$ has the highest bond order which is $3 .$

$Li _2, He _2$ and $O _2$ have bond orders $1,0$ and $2$ respectively.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The species, having bond angles of $120^o $ is
Which of the following can not react with $NaHCO_3$
Ammonia gas dissolves in water to give $N{H_4}OH.$ In this reaction water acts as
The correct decreasing order of energy, for the orbitals having, following set of quantum numbers.

$(A)$ $n=3,1=0, m=0$

$(B)$ $n=4,1=0, m=0$

$(C)$ $n =3,1=1, m =0$

$(D)$ $n=3,1=2, m=1$

Consider the reaction equilibrium, $2S{O_{2(g)}}$+${O_{2(g)}}$ $\rightleftharpoons$ $2S{O_{3(g)}}$;  $\Delta H^\circ =\, -198\,kJ$. On the basis of Le-Chatelier’s principle, the condition favourable for the forward reaction is
$K_{sp}$ for $AgBr$ at $25\,^oC$ is $4.9 × 10^{-13}$. How much $AgBr$ will dissolved in $20\  litres$ of its saturated solution if molecular weight of $AgBr =$  $188$
Combustion of $1$ mole of benzene is expressed at

$\mathrm{C}_6 \mathrm{H}_6(1)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(1) \text {. }$

The standard enthalpy of combustion of $2 \mathrm{~mol}$ of benzene is - ' $x$ ' $k J$.

$\mathrm{x}=$. . . . . . . . . .

$(1)$ standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{C}_6 \mathrm{H}_6(1)$, for the reaction $6 \mathrm{C}$ (graphite) $+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_n(1)$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

$(2)$ Standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{CO}_2(\mathrm{~g})$, for the reaction $\mathrm{C}$ (graphite) $+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_2(\mathrm{~g})$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

$(3)$ Standard and Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}(1)$, for the reaction $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(1)$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Assuming that the degree of hydrolysis is small, the $pH$ of $0.1\, M$ solution of sodium acetate $(K_a\, = 1.0\times10^{- 5})$ will be
(image) $\xrightarrow[{\left( 2 \right)\,{H_2}O,Zn}]{{\left( 1 \right)\,{O_3}}} Product$, Product is
The density (in $g\, mL^{-1}$) of a $3.60\, M$ sulphuric acid solution that is $29\%$ $H_2SO_4$ (molar mass $= 98\, g\, mol^{- 1}$) by mass will be