Question

The table below lists the first five ionization energies of a second period element. Identify the element.

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

• A. Lithium $\left(Li\right)$
• B. Beryllium $\left(Be\right)$
• C. Boron $\left(B\right)$
• D. Carbon $\left(C\right)$

Answer 1

The correct option is C Boron $\left(B\right)$

The value of fourth IE is very large compared to third IE. Hence the element achieves inert gas configuration on removal of $3$ electrons and hence excessive energy is needed to remove electron from stable configuration. Thus element belongs to thirteenth group. Among given options such element is Boron (B). Hence answer is option C-Boron (B)