Question

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

Explanation:

Step1: Analyze Fluorine (F)

Fluorine is a non - metal with 7 valence electrons. To attain a noble - gas configuration, it gains 1 electron. So, the fluoride ion has a charge of - 1, and the formula of the ion is $\ce{F^-}$.

Step2: Analyze Nitrogen (N)

Nitrogen has 5 valence electrons. To achieve a noble - gas configuration, it gains 3 electrons. Thus, the nitride ion has a charge of - 3, and the formula of the ion is $\ce{N^{3-}}$.

Step3: Analyze Oxygen (O)

Oxygen has 6 valence electrons. It gains 2 electrons to get a noble - gas configuration. So, the oxide ion has a charge of - 2, and the formula of the ion is $\ce{O^{2-}}$.

Step4: Analyze Sodium (Na)

Sodium is a metal with 1 valence electron. It loses 1 electron to attain a noble - gas configuration. So, the sodium ion has a charge of + 1, and the formula of the ion is $\ce{Na^+}$.

Step5: Analyze Calcium (Ca)

Calcium is a metal with 2 valence electrons. It loses 2 electrons to achieve a noble - gas configuration. So, the calcium ion has a charge of + 2, and the formula of the ion is $\ce{Ca^{2+}}$.

Step6: Analyze Aluminum (Al)

Aluminum has 3 valence electrons. It loses 3 electrons to get a noble - gas configuration. So, the aluminum ion has a charge of + 3, and the formula of the ion is $\ce{Al^{3+}}$.

Answer:

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