Question

Using condensed electron configurations, write reactions showing the formation of the common ions of the following elements:

(a) Ba (Z = 56)
(b) O (Z = 8)
(c) Pb (Z = 82)

Answer 1

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰

Step-by-step explanation:

The condensed electron configurations of given elements are below

a) Ba (Z = 56), [Xe].6s²

b) O (Z = 8), [He].2s².2p⁴

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p²

Since atoms tend to donate/receive more electrons to achieve the saturated or half-saturated orbital. So in our case it happens as below

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰