Answer :
For the ion [tex]197Au^{3+[/tex], there are 79 protons, 118 neutrons, and 76 electrons. correct option is C 79 p⁺, 118 n⁰, 76 e⁻
For the ion [tex]197Au^{3+[/tex], there are 79 protons (as indicated by the atomic number of gold, Au) and the charge of 3+ indicates a loss of 3 electrons.
So, there are 76 electrons.
To find the number of neutrons, we subtract the number of protons from the mass number.
The mass number is 197, so there are 197 - 79 = 118 neutrons.
Therefore, correct option is C 79 p⁺, 118 n⁰, 76 e⁻
The probable question may be:
How many protons, neutrons and electrons are in this
197Au^{3+}
79
A 79 p⁺, 118 n⁰, 82 e⁻
B 79 p⁺, 118 n⁰, 79 e⁻
C 79 p⁺, 118 n⁰, 76 e⁻
D 98 p⁺, 99 n⁰, 95 e⁻
