The species 27Al, 80Br- and 201Hg2+ have 13, 35 and 80 protons, respectively. Class 9

The species ²⁷Al, ⁸⁰Br^- and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. Class 9

Question 1.

The species ²⁷Al, ⁸⁰Br^- and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have? Class 9

Answer:

For ²⁷Al (aluminium atom, neutral): Protons = 13 (given) Electrons = 13 (neutral atom, so electrons = protons) Mass number = 27 Neutrons = Mass number - Protons = 27 - 13 = 14 neutrons Electrons =13

For ⁸⁰Br^- (bromide ion): Protons = 35 (given) Br^- has gained 1 electron → Electrons = 35 + 1 = 36 electrons Mass number = 80 Neutrons = Mass number - Protons = 80 - 35 = 45 neutrons

For ²⁰¹Hg²⁺ (mercury(II) ion): Protons = 80 (given) Hg²⁺ has lost 2 electrons → Electrons = 80 - 2 = 78 electrons Mass number = 201 Neutrons = Mass number - Protons = 201 - 80 = 121 neutrons