Question

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

Short answer

${\mathrm{P}}_{\left(9\right)}\to {\mathrm{P}}_{\left(9\right)}^{+}+{\mathrm{e}}^{-}---\left(1\right)$

${\mathrm{P}}_{\left(9\right)}+{\mathrm{e}}^{-}\to {\mathrm{P}}_{\left(9\right)}^{+}---\left(2\right)$

Step-by-step solution

Step 1:

To determine the equation associated with ionization energy and electron affinity, let us define these terms first.

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.

Step 2:

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

${\mathrm{P}}_{\left(9\right)}\to {\mathrm{P}}_{\left(9\right)}^{+}+{\mathrm{e}}^{-}$

Step 3:

On the other hand, for electron affinity, the reaction starts with a neutral and gaseous phosphorus atom as well. In addition to this, an electron is also a part of the reactants since its added to the atom. The result of the addition of an electron is a phosphorus anion with a $-1$ charge.

${\mathrm{P}}_{\left(9\right)}+{\mathrm{e}}^{-}\to {\mathrm{P}}_{\left(9\right)}^{+}$