Question

question 7. what is the formula of the ions formed when atoms of the following elements gain or lose valence electrons and attain noble-gas configurations?
a. fluorine (f)
b. nitrogen (n)
c. oxygen (o)
d. sodium (na)
e. calcium (ca)
f. aluminium (al)

Explanation:

Step1: Analyze Fluorine (F)

Fluorine has 9 electrons. To attain noble - gas (Ne) configuration, it gains 1 electron. So the ion is $\ce{F^-}$.

Step2: Analyze Nitrogen (N)

Nitrogen has 7 electrons. To attain noble - gas (Ne) configuration, it gains 3 electrons. So the ion is $\ce{N^{3 - }}$.

Step3: Analyze Oxygen (O)

Oxygen has 8 electrons. To attain noble - gas (Ne) configuration, it gains 2 electrons. So the ion is $\ce{O^{2 - }}$.

Step4: Analyze Sodium (Na)

Sodium has 11 electrons. To attain noble - gas (Ne) configuration, it loses 1 electron. So the ion is $\ce{Na^+}$.

Step5: Analyze Calcium (Ca)

Calcium has 20 electrons. To attain noble - gas (Ar) configuration, it loses 2 electrons. So the ion is $\ce{Ca^{2 + }}$.

Step6: Analyze Aluminium (Al)

Aluminium has 13 electrons. To attain noble - gas (Ne) configuration, it loses 3 electrons. So the ion is $\ce{Al^{3 + }}$.

Answer:

a. $\ce{F^-}$; b. $\ce{N^{3 - }}$; c. $\ce{O^{2 - }}$; d. $\ce{Na^+}$; e. $\ce{Ca^{2 + }}$; f. $\ce{Al^{3 + }}$