For the reaction $\mathrm{N}_{2}+3 \mathrm{X}_{2} \rightarrow 2 \mathrm{NX}_{3}$ where $\mathrm{X}=\mathrm{F}, \mathrm{Cl}$ (the average bond energies are $\mathrm{F}-\mathrm{F}=$ $155 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~N}-\mathrm{F}=272 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{Cl}-\mathrm{Cl} \mathrm{kJ} \mathrm{mol}^{-1}, \mathrm{~N}-\mathrm{Cl} \mathrm{kJ} \mathrm{mol}^{-1}$ and $\mathrm{N} \equiv \mathrm{N}=941 \mathrm{kJmol}^{-1}$ ). The heats of formation of $\mathrm{NF}_{3}$ and $\mathrm{NCl}_{3}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$, respectively, are closest to

• Correct answer
  A

  $-226$ and $+467$
• B

  $+226$ and $-467$
• C

  $-151$ and $+311$
• D

  $+151$ and $-311$

Hence, the correct answer is option: 'A'.