Chapter 6, Problem 42A

Interpretation Introduction

Interpretation:

The large difference between the first and second ionization energy of alkali metal is to be explained.

Concept Introduction :

The first ionization energy is the energy needed for the removal of the outermost electron from the neutral gaseous atom.

  $X\left(g\right)+I{E}_1\to X^{+}\left(g\right)+e^{-}$

While second ionization energy is the energy needed for the removal of the outermost electron from the unipositive gaseous ion.

  $X^{+}\left(g\right)+I{E}_2\to X^{2+}\left(g\right)+e^{-}$

Answer to Problem 42A

After removing one electro, the alkali metals achieve noble gas electronic configuration.

Explanation of Solution

The general electronic configuration of alkali metal is,

  $\begin{array}{l}Li:\left[He\right]1s^1\\ Na:\left[Ne\right]2s^1\\ K:\left[Ar\right]3s^1\end{array}$

So, the general electronic configuration of alkali metals is,

  $\left[\text{noble gas}\right]1s^1$

After removing one electron, alkali metals achieve noble gas electronic configuration and thus, it becomes difficult to remove the next electron.