Question

specify the number of protons (#p⁺), neutrons (#n⁰) and electrons (#e⁻) given the following symbol:
₃₀⁶⁴zn²⁺
#p+
#n0
#e-

Explanation:

Step1: Determine number of protons

The atomic number is the number of protons. For $\ce{_{30}^{64}Zn^{2 + }}$, the sub - script 30 is the atomic number. So, $\#p^{+}=30$.

Step2: Determine number of neutrons

The mass number is 64 (the superscript). The number of neutrons $n$ is calculated as $n =$ mass number - atomic number. So, $\#n^{0}=64 - 30=34$.

Step3: Determine number of electrons

A neutral atom has equal number of protons and electrons. This is a $2+$ ion, which means it has lost 2 electrons. So, $\#e^{-}=30 - 2 = 28$.

Answer:

$\#p^{+}:30$
$\#n^{0}:34$
$\#e^{-}:28$