Question

The $I.E_1$ of an element M(g) is 6000 kJ/mol. The ratio of $\frac{I.E_2}{I.E_3}$ is 3 : 2. If $\Delta H$ for conversion, $M\left(g\right)\to M^{3+}\left(g\right)$ is 10000 kJ/mol, what is the value of $I.E_2$ and $I.E_3$ respectively (in kJ/mol)? ($I.E_1=$ first ionization energy, $I.E_2=$ second ionization energy and $I.E_3=$ third ionization energy)

• A. 1200 and 2800
• B. 2400 and 1600
• C. 1600 and 2400
• D. 2800 and 1200

Answer 1

The correct option is B 2400 and 1600

$\Rightarrow \Delta H=I{E}_1+I{E}_2+I{E}_3$
$10000=6000+I{E}_2+I{E}_3$
$3x+2x=4000$
$x=800$
$I{E}_2=3\times 800=2400$
$I{E}_3=2\times 800=1600$