Question

chemists refer to positively charged ions as cations. chemists refer to negatively charged ions as anions. 7. fill in the following table. symbol: $_{38}^{88}sr^{2 +}$, $_{16}^{32}s^{2 -}$; atomic number: 35; mass number: 70; number of protons: 31; number of electrons: 28, 36; number of neutrons: 45; cation or anion: 8. could a +3 ion of aluminum be made by adding three protons to an aluminum atom? explain. 9. one of your classmates is having trouble understanding ions. he explains the formation of a cation like this: “when you add an electron, you get a positive charge because adding is positive in math.” a. as a group, explain in a grammatically correct sentence why this student is incorrect. b. provide a better description of how math relates to electrons and ion formation.

Explanation:

Step 1: Analyze the ion $\ce{^{88}_{38}Sr^{2+}}$

The atomic number ($Z$) is the sub - script, so $Z = 38$. The mass number ($A$) is the super - script, so $A=88$. The number of protons is equal to the atomic number, so the number of protons $p = 38$. Since it is a $2+$ ion, the number of electrons $e=38 - 2=36$. The number of neutrons $n=A - Z=88 - 38 = 50$. It is a cation because of the positive charge.

Step 2: Analyze the ion $\ce{^{32}_{16}S^{2-}}$

The atomic number $Z = 16$, the mass number $A = 32$. The number of protons $p = 16$. Since it is a $2 -$ ion, the number of electrons $e=16+2 = 18$. The number of neutrons $n=A - Z=32 - 16=16$. It is an anion because of the negative charge.

Step 3: Analyze the ion with $p = 31$ and $A = 70$

The atomic number is equal to the number of protons, so $Z = 31$. The number of electrons $e=28$, so the charge is $31-28 = 3+$. The number of neutrons $n=A - Z=70 - 31 = 39$. The symbol is $\ce{^{70}_{31}Ga^{3+}}$.

Step 4: Analyze the ion with $Z = 35$ and $n = 45$

The mass number $A=Z + n=35 + 45=80$. The number of protons $p = 35$. The number of electrons $e = 36$, so the charge is $35-36=-1$. The symbol is $\ce{^{80}_{35}Br^{-}}$.

Step 5: Answer question 8

No, a + 3 ion of aluminum cannot be made by adding three protons to an aluminum atom. Adding protons would change the element. An aluminum + 3 ion is formed by removing three electrons from an aluminum atom. Aluminum has 13 protons. If we add protons, we will create a different element.

Step 6: Answer question 9a

The student is incorrect because electrons have a negative charge, so adding an electron results in a negative charge, not a positive charge.

Step 7: Answer question 9b

In ion formation, each electron has a charge of - 1 and each proton has a charge of + 1. When an atom loses electrons, the number of protons is greater than the number of electrons, resulting in a positive - charged cation. When an atom gains electrons, the number of electrons is greater than the number of protons, resulting in a negative - charged anion. The net charge of an ion is the algebraic sum of the charges of its protons and electrons.

Answer:

Symbol	$\ce{^{88}_{38}Sr^{2+}}$	$\ce{^{32}_{16}S^{2-}}$	$\ce{^{70}_{31}Ga^{3+}}$	$\ce{^{80}_{35}Br^{-}}$
Mass Number	88	32	70	80
Number of protons	38	16	31	35
Number of electrons	36	18	28	36
Number of neutrons	50	16	39	45
Cation or anion	Cation	Anion	Cation	Anion

1. No, adding protons changes the element; a + 3 aluminum ion is formed by removing 3 electrons.

9a. Electrons have a negative charge, so adding one gives a negative charge.
9b. Each electron has a - 1 charge and each proton has a + 1 charge; net ion charge is the sum of proton and electron charges.