Question

The first three ionization energies of an element A are \(590,1145,4912 \mathrm{kh} / \mathrm{mol} .\) What is the most likely formula for astable ion of A?

• a) \(A^{+}\)
• b) \(A^{2+}\)
• c) \(A^{3+}\)
• d) \(A^{-}\)

Answer 1

Ionization energy: The amount of energy required to remove an electron from its outer most orbital is known as ionization energy or first ionization energy. The first ionization energy for element \(\mathrm{A}\) is as follows:

\(\mathrm{A} \longrightarrow \mathrm{A}^{+}+\mathrm{e}^{-} \quad\) I. \(\mathrm{E}=590 \mathrm{kh} / \mathrm{mol}\)

Second ionization energy means the amount of energy required to remove an electron from

element with positive charge. The second ionization energy for element \(\mathrm{A}\) is as follows:

\(\mathrm{A}^{+} \longrightarrow \mathrm{A}^{2+}+\mathrm{e}^{-} \quad \text { I.E }=1145 \mathrm{kh} / \mathrm{mol}\)

As per the data given in the problem third ionization energy is very compared to the first and second.

That means dipositive ion will not allow removing another electron. Hence the dipositive ion is

very stable.

The formula for the most stable ion of \(\mathrm{A}\) is \({\mathrm{A}^{2+}}\)