Question

According to the electron configuration, which atom will be the most reactive?

Answer 1

Answer:

The fourth atom.

Explanation:

Number of valence electrons in each of the four atoms:

• First atom: two ($2\, s^{2}$.)
• Second atom: two ($3\, s^{2}$.)
• Third atom: eight ($3\, s^{2}\; 3\, p^{6}$.)
• Fourth atom: two ($4\, s^{2}$.)

Without referring to a periodic table, the first, second, and fourth atoms would all belong to the alkali-earth metal elements. The third atom would belong to a noble gas element. (The four choices are beryllium, magnesium, argon, and calcium, respectively.)

Noble gas elements are highly stable. Thus, the third atom (argon) would be the least reactive among all four choices.

On the other hand, alkali-earth metals are reactive since they tend to lose their valence electrons. The reactivity of an alkali-earth atom depends on its ionization energy (the energy required to take a valence electron away from the atom.)

Since these atoms are in the same group, ionization energy would be smaller for atoms with a larger radius. In a given group, atoms with more filled electron shells would have a larger radius.

The first atom has two filled main shells, the second has three, whereas the fourth has four.

Thus, among the three alkali-earth metals in the choices:

$\begin{aligned}&\text{Number of filled electron shells:} \\ & a < b < d\end{aligned}$.

$\begin{aligned}&\text{Atomic radius:} \\ & a < b < d\end{aligned}$.

$\begin{aligned}&\text{Ionization energy:} \\ & a > b > d\end{aligned}$.

$\begin{aligned}&\text{Reactivity:} \\ & a < b < d\end{aligned}$.

The third choice (a noble gas) is the least reactive among the four choices:

$\begin{aligned}&\text{Reactivity:} \\ & c < a < b < d\end{aligned}$.

Overall, the fourth choice would be the most reactive among the four choices.