Question

Consider the following changes:
$A\to \ \ {{A}^{+}}\ +\ {{e}^{-}}$ : ${{E}_{1}}$ and ${{A}^{+}}\ \to \ \ {{A}^{2+}}\ +\ {{e}^{-}}$ : ${{E}_{2}}$
The energy required to pull out the two electrons are ${{E}_{1}}$ and ${{E}_{2}}$ respectively. The correct relationship between two energies would be:
(A) ${{E}_{1}}$ < ${{E}_{2}}$
(B) ${{E}_{1}}$ = ${{E}_{2}}$
(C) ${{E}_{1}}$ > ${{E}_{2}}$
(D) ${{E}_{1}}$ $\ne$ ${{E}_{2}}$

Answer 1

Understand the ionization energy of an element. An attempt to this question can be made by taking any arbitrary element having two ionization energies. Now compare the difference in electronic configuration in the two cases. Based on this, you can arrive at a conclusion that can be applied everywhere and answer the above question as well.

Complete step by step solution:
Ionization energy can be defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule.
It is observed that the reaction is usually endothermic. This means that energy must be supplied for the reaction to occur. The ionization energy for an atom/ion is directly proportional to the effective nuclear charge for the species under consideration.
In the above question, an arbitrary element, A is taken. The energy required for the first ionization is termed as ${{E}_{1}}$. The reaction is given below:
$A\to \ \ {{A}^{+}}\ +\ {{e}^{-}}$
For the neutral atom, let us consider the effective nuclear charge to be a variable, say Z.
The energy required for the second ionization is termed as ${{E}_{2}}$. The reaction is given below:
${{A}^{+}}\ \to \ \ {{A}^{2+}}\ +\ {{e}^{-}}$
For the ion, we can say that the effective nuclear charge is greater than Z. This is because it has already lost one electron.
Since the ionisation energy is directly proportional to effective nuclear charge, the relation between the first and second ionisation energy becomes,
${{E}_{1}}$ < ${{E}_{2}}$

Therefore, the correct answer is option (A).

Note: It is important to know that effective nuclear charge is not the same as atomic number of the element. Effective nuclear charge is always calculated by the Slater's rule.