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

How many electrons and neutrons do they have?

For $^{27}$Al, the atomic number is 13 and the mass number is 27. Since it is a neutral atom, the number of electrons is equal to the number of protons, i.e., 13.

$$\begin{aligned} \text{Number of neutrons} &= \text{Mass number} - \text{Atomic number} \\ &= 27 - 13 = 14 \end{aligned}$$

Therefore, $^{27}$Al has 13 electrons and 14 neutrons.

For $^{80}$Br$^{-}$, the atomic number is 35 and the mass number is 80. Since bromide ion carries a negative charge of -1, it has gained one electron.

Number of electrons = 35 + 1 = 36

Number of neutrons = 80 - 35 = 45

Therefore, $^{80}$Br$^{-}$ has 36 electrons and 45 neutrons.

For $^{201}$Hg$^{2+}$, the atomic number is 80 and the mass number is 201. Since mercury ion has a positive charge of +2, it has lost two electrons.

Number of electrons = 80 - 2 = 78

Number of neutrons = 201 - 80 = 121

Therefore, $^{201}$Hg$^{2+}$ has 78 electrons and 121 neutrons.