Answer:
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
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
