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1g of Mg atoms in the vapour phase absorbs 50.0 kJ of energy. Find the composition of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] formed as a result of absorption of energy. $I{E_1}$ and ${\text{I}}{{\text{E}}_{\text{2}}}$ for Mg are 740 and 1450 ${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively.

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Last updated date: 20th Jun 2024
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Answer
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Hint: You know that to remove an electron from an isolated atom, it requires energy. This energy is termed as ionisation energy. The first and second ionization energy is not same here. To form \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] from Mg, the energy required is the sum of this two energies.

Complete step by step answer:
We know that the energy required to convert $M{g^ + }$ and \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions are 740 ${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ and 1450 ${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively.
The energy used for Mg to $M{g^ + }$ and \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] conversion is: 740 + 1450 = 2190 ${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

A mole of Mg is equal to$\dfrac{1}{{24}}$.

We can assume that $x$ grams of $M{g^ + }$ ions and $y$ grams of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions.
$x + y = 1$ (Total mass in 1 grams)

Therefore, moles of $M{g^ + }$ ions = $\dfrac{x}{{24}}$

Mass of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\]ions = $\dfrac{y}{{24}}$

So, the energy absorbed for the formation of $M{g^ + }$ ions = $\dfrac{x}{{24}} \times 740$

The energy absorbed for forming \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions = $\dfrac{y}{{24}} \times 2190$

So, the total energy absorbed (E) = $\left( {\dfrac{x}{{24}} \times 740} \right) + \left( {\dfrac{y}{{24}} \times 2190} \right)$

50 = $\left( {\dfrac{x}{{24}} \times 740} \right) + \left( {\dfrac{y}{{24}} \times 2190} \right)$

$1200 = 740x + 2190y$
$120 = 74x + 219y$
Since $x + y = 1$
 x = 1 - y
120 = 74(1 - y) + 219y
120 = 74 - 74y + 219y
120 = 145y + 74
46 = 145y
y = 0.3172

Therefore, the mass of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions is 0.3172g
Percentage of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions = $0.31721 \times 100 = 31.72$
Also, percentage of $M{g^ + }$ ions = 100 – 31.72 = 68.28.

So, the percentage of \[{\text{M}}{{\text{g}}^{{\text{2 + }}}}\] ions formed is 31.72 %.

Note: You may have noticed that the first ionization enthalpy of magnesium is smaller than the second ionization enthalpy. Reason for this is the difficulty to remove an electron from a positively charged species rather than the neutral atom.