# An atom has an atomic number of 79 and an atomic mass of 197. Once this atom enters into an ionic bond, it has how many proton, neutrons, and electrons?

If $Z$, the atomic number, is $79$, then, immediately we know that the element is gold. How do we know? Because we have access to the Periodic Table.
And we don't even have to know that the element is gold. If $Z$ $=$ $79$, there are 79 positively charged nuclear particles, 79 protons, in the nucleus. If 79 nuclear charges, the NEUTRAL atom contains 79 electrons. You specified an ion, and gold forms both $A {u}^{+}$ and $A {u}^{3 +}$ ions so $78$ and $76$ electrons respectively.
Now, it is a fact that the atomic mass of gold is $196.97$ $g \cdot m o {l}^{-} 1$, roughly close to $197$. Because the atomic mass is $197$ $\text{amu}$, there must be $197 - 79 = 118 \text{ neutrons}$, massive, neutrally charged "nucular" particles. Atomic mass is to a first (and second) approximation, the sum of the protons and neutrons (there will be a mixture of isotopes because of different numbers of neutrons), because electrons have negligible mass.
Now if $Z$ $=$ $78$, can you tell me the element, the number of protons, the number of neutrons, and the number of electrons?