Question

complete atomic notation, # of protons, # of neutrons, # of electrons
34, 31, 36
$_{13}^{28}al^{3 + }$
$_{36}^{83}kr$

Explanation:

Step1: Recall atomic notation rules

The general form of atomic notation is $_{Z}^{A}X^{n\pm}$, where $Z$ is the atomic number (number of protons), $A$ is the mass - number (number of protons + neutrons), $X$ is the chemical symbol, and $n$ is the charge.

Step2: First row

The number of protons $Z = 34$, the number of neutrons $N=31$, and the number of electrons $E = 36$. The mass - number $A=Z + N=34 + 31=65$. The element with $Z = 34$ is Selenium (Se). The ion has a charge of $34-36=- 2$. So the complete atomic notation is $_{34}^{65}Se^{2 -}$.

Step3: Second row

For $_{13}^{28}Al^{3+}$, the number of protons $Z = 13$ (from the sub - script in the atomic notation). The number of neutrons $N=A - Z=28 - 13 = 15$. The number of electrons $E=Z - 3=13 - 3 = 10$ (due to the $3+$ charge).

Step4: Third row

For $_{36}^{83}Kr$, the number of protons $Z = 36$ (from the sub - script). The number of neutrons $N=A - Z=83 - 36 = 47$. In a neutral atom, the number of electrons is equal to the number of protons, so $E = 36$.

Complete Atomic Notation	# of Protons	# of Neutrons	# of Electrons
$_{13}^{28}Al^{3+}$	13	15	10
$_{36}^{83}Kr$	36	47	36

Answer:

Complete Atomic Notation	# of Protons	# of Neutrons	# of Electrons
$_{13}^{28}Al^{3+}$	13	15	10
$_{36}^{83}Kr$	36	47	36