Question

topic #3: atomic structure, ions, and isotopes
to learn atomic structure, ions, and isotopes, try the following exercises. as best you can, you should try these problems without using your notes or the textbook.
tier i
these problems are embedded with examples to help you learn techniques. solve these problems first:
check your learning exercises: 2.3 (p.78), 2.4 (p.83), 2.5 (p.83), 2.6 (p. 88), 2.8 (p. 97), 2.9 (p. 97), 2.10 (p. 100), 2.11 (p. 101)
tier ii
these problems tend to be straightforward and will help you practice the techniques you learned in tier i. solutions to odd - numbered exercises are in the back of the textbook. solve these problems second:
end - of - chapter exercises: chapter 2, ex. 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, 43, 47, 49
tier iii
these problems tend to be more advanced. you should attempt them only after mastering the tier i and tier ii problems. the answer key for these problems is on blackboard. solve these problems third:

1. determine the number of protons and electrons in each of the following:

a.) cu²⁺
b.) silicon
c.) n³⁻
d.) the ion formed by oxygen

1. complete the following table.

symbol	protons	neutrons	mass #	electrons	charge
fe³⁺	26	32	58	23	+3
zr⁴⁺	40	51	91	36	+4

1. complete the following table for ions.

atomic number	# protons	# electrons	ion charge	ion symbol
24	24	21	+3	cr³⁺
23	23	18	+5	v⁵⁺
34	34

Explanation:

Step1: Recall atomic - number concept

The atomic number of an element is equal to the number of protons. For a neutral atom, the number of electrons is equal to the number of protons.

Step2: Analyze $Cu^{2+}$

Copper (Cu) has an atomic number of 29. So, the number of protons is 29. Since it has a + 2 charge, it has lost 2 electrons. So, the number of electrons is $29 - 2=27$.

Step3: Analyze Silicon

Silicon (Si) has an atomic number of 14. In a neutral silicon atom, the number of protons and electrons is 14.

Step4: Analyze $N^{3 - }$

Nitrogen (N) has an atomic number of 7. So, the number of protons is 7. Since it has a - 3 charge, it has gained 3 electrons. So, the number of electrons is $7 + 3 = 10$.

Step5: Analyze oxygen ion

Oxygen (O) has an atomic number of 8. The common ion of oxygen is $O^{2 - }$. It has gained 2 electrons. So, the number of protons is 8 and the number of electrons is $8+2 = 10$.

Step6: Complete the third - table row for atomic number 34

The element with atomic number 34 is Selenium (Se). For a negatively - charged ion, if the number of protons is 34, and to have a negative charge, it gains electrons. A common ion of Se is $Se^{2 - }$, where the number of electrons is $34 + 2=36$ and the ion charge is - 2.

Answer:

1.

a. Protons: 29, Electrons: 27
b. Protons: 14, Electrons: 14
c. Protons: 7, Electrons: 10
d. The ion formed by oxygen is usually $O^{2 - }$. Protons: 8, Electrons: 10

3.

For atomic number 34: Protons: 34, Electrons: 36, Ion Charge: - 2, Ion Symbol: $Se^{2 - }$