Question

A hypothetical element X undergoes a reaction as shown:
$X+2e^{-}\to X^{2-}$, energy released = 30.87 eV/atom. If the energy released is used to dissociate 4 g of $H_2$ molecule equally into $H^{+}$ and $H^{\ast }$ , where $H^{\ast }$ is the excited state of H atoms where the electron travels in orbit whose circumference is equal to four times its de-broglie's wavelength. Determine the least moles of X that would be required:
Given: I.E of H = 13.6 eV/atom , bond energy of $H_2$ = 4.526 eV/molecule.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Number of moles of $H_2$ = $\frac{4}{2}$ = 2 mol
Energy released for x mol = -30.87 eV/atom $\times N_A\times x$ eV
Bond energy for 2 mol = $4.526\times N_A\times 2$ eV
Ionisation enthalpy = $2\times 13.6\times N_A$ eV
2 mol of hydrogen is excited from the ground state to the excited state which is given as the fourth orbit.
So, $E_n=13.6\frac{Z^2}{n^2}$
$\therefore$ = $\Delta E_n$ for 2 mol = $2\times N_A\times 13.6(1-\frac{1}{4^2}$)
Balancing the energy released and absorbed,
$4.526\times N_A\times 2+(2\times 13.6\times N_A)+2\times N_A\times 13.6(1-\frac{1}{4^2})+(-30.87\times N_A\times xeV)$
On solving, x = 2 mol