Question

how many protons, neutrons, and electrons are in this ion?
$\ce{_{80}^{204}hg^{2+}}$
$80\\ p^+,\\ 124\\ n^0,\\ 82\\ e^-$
$80\\ p^+,\\ 124\\ n^0,\\ 80\\ e^-$
$102\\ p^+,\\ 102\\ n^0,\\ 104\\ e^-$
$80\\ p^+,\\ 124\\ n^0,\\ 78\\ e^-$

Explanation:

Step1: Determine number of protons

The atomic number (bottom number) is 80, so protons ($p^+$) = 80.

Step2: Calculate number of neutrons

Mass number (top number) is 204. Neutrons ($n^0$) = mass number - atomic number = $204 - 80 = 124$.

Step3: Find number of electrons

The ion is $\text{Hg}^{2+}$, meaning it lost 2 electrons. Electrons ($e^-$) = atomic number - charge magnitude = $80 - 2 = 78$.

Answer:

80 p⁺, 124 n⁰, 78 e⁻ (the last option: 80 p⁺, 124 n⁰, 78 e⁻)