Question

which of the following ions would result in an element having 8 valence electrons?

□ a cl-
□ b sb3-
□ c na+
□ d p2+
□ e ca2+
□ f na-

Explanation:

• For \(Cl^-\): Chlorine (\(Cl\)) has 7 valence electrons. When it gains one electron to form \(Cl^-\), it has \(7 + 1=8\) valence electrons.
• For \(Sb^{3 - }\): Antimony (\(Sb\)) has 5 valence electrons. When it gains 3 electrons to form \(Sb^{3 - }\), it has \(5+3 = 8\) valence electrons.
• For \(Na^+\): Sodium (\(Na\)) has 1 valence electron. When it loses one electron to form \(Na^+\), its valence - shell is the previous shell. The new valence - shell (the second shell for \(Na^+\)) has 8 electrons.
• For \(Ca^{2+}\): Calcium (\(Ca\)) has 2 valence electrons. When it loses 2 electrons to form \(Ca^{2+}\), its valence - shell is the previous shell. The new valence - shell (the third shell for \(Ca^{2+}\) after losing 2 electrons, considering its electron - configuration) has 8 electrons.

• For \(P^{2+}\): Phosphorus (\(P\)) has 5 valence electrons. Losing 2 electrons gives it \(5 - 2=3\) valence electrons.
• For \(Na^-\): Sodium (\(Na\)) has 1 valence electron. Gaining one electron gives it \(1 + 1 = 2\) valence electrons.

Answer:

a. \(Cl^-\), b. \(Sb^{3 - }\), c. \(Na^+\), e. \(Ca^{2+}\)