Question

The table above shows the first five ionization energies for a second-period element. Which identifies the element and provides the best explanation?

	Ionization Energy $\left(kJ/mol\right)$
First	$801$
Second	$2,430$
Third	$3,660$
Fourth	$25,000$
Fifth	$32,820$

• A. Boron, because it has five electrons.
• B. Boron, because is has three valence electrons.
• C. Nitrogen, because it has five valence electrons.
• D. Nitrogen, because it has three electrons in a $2p$ sublevel.

Answer 1

Answer: (B)

Boron, because is has three valence electrons.

The fourth IE is very high compared to third IE, thus on removal of $3$ electrons, the element attains inert gas configuration. Thus the element has $3$ valence electrons. Hence among given options, Boron is correct because it has $3$ valence electrons.