Question

Using the element phosphorus as an example, write equations for the process in which the energy change corresponds to the ionisation energy and to the electron affinity.

Short answer

$P_{\left(9\right)}\to P_{\left(9\right)}^{+}+e^{-}\cdots \left(1\right)$

$P_{\left(9\right)}+e^{-}\to P_{\left(9\right)}^{-}\cdots \left(2\right)$

Step-by-step solution

Definition/ Concept

Let us define these terms first to determine the equation associated with ionization energy and electron affinity.

Ionizationenergy is defined as the minimum energy that has to be absorbed to remove an electron from a neutral and gaseous atom.

Electron affinity is defined as the energy released when an electron is added to a neutral atom.

Equations of Ionization energy

Based on these definitions, the reaction starts with a neutral phosphorus atom in the gaseous state for ionization energy. Afterward, this absorbs energy, and the products are an electron and a phosphorus cation with a +1 charge. Therefore, we can write the equation as:-

$P_{\left(9\right)}\to P_{\left(9\right)}^{+}+e^{-}$

Step 3: Equations of Electron Affinity

On the other hand, for electron affinity, the reaction also starts with a neutral and gaseous phosphorus atom. In addition, an electron is also a part of the reactants since its added to the atom. The result of adding an electron is a phosphorus anion with a charge.

$P_{\left(9\right)}+e^{-}\to P_{\left(9\right)}^{-}$