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
	40	51		36

1. complete the following table for ions.

atomic number	# protons	# electrons	ion charge	ion symbol
				v⁵⁺
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 equals the number of protons. For ions, the number of electrons is adjusted according to the charge.

Step2: Analyze \(Cu^{2+}\)

Copper (\(Cu\)) has an atomic number of 29. In \(Cu^{2+}\), two electrons are lost. So, protons = 29, electrons=29 - 2=27.

Step3: Analyze Silicon

Silicon (\(Si\)) has an atomic number of 14. In a neutral silicon atom, protons = 14, electrons = 14.

Step4: Analyze \(N^{3 -}\)

Nitrogen (\(N\)) has an atomic number of 7. In \(N^{3 -}\), three electrons are gained. So, protons = 7, electrons = 7+3 = 10.

Step5: Analyze oxygen ion

Oxygen (\(O\)) has an atomic number of 8. It typically forms a \(O^{2 -}\) ion by gaining two electrons. So, protons = 8, electrons = 8 + 2=10.

Step6: Complete first table row 1

Given charge = +3 and electrons = 23, the number of protons is 23+3 = 26. Looking at the periodic - table, element with atomic number 57 is Lanthanum (\(La\)). Mass number = protons+neutrons. If mass number = 58+80 = 138.

Step7: Complete first table row 2

Given protons = 40, element is Zirconium (\(Zr\)). Mass number = protons + neutrons=40 + 51=91. Charge = protons - electrons=40 - 36 = +4.

Step8: Complete second table row 1

Given protons = 24, atomic number = 24 (chromium, \(Cr\)). Electrons = 21, charge = 24 - 21=+3, ion symbol \(Cr^{3+}\).

Step9: Complete second table row 2

For \(V^{5+}\), vanadium has atomic number 23. Protons = 23, electrons = 23 - 5=18.

Step10: Complete second table row 3

Given atomic number = 34 (selenium, \(Se\)). Protons = 34. If it forms an ion, assume it gains 2 electrons to form \(Se^{2 -}\), so electrons = 36, charge=-2.

Answer:

1.
a. Protons: 29, Electrons: 27
b. Protons: 14, Electrons: 14
c. Protons: 7, Electrons: 10
d. Oxygen forms \(O^{2 -}\), Protons: 8, Electrons: 10
2.
First row: \(La^{3+}\), Protons: 57, Neutrons: 81, Mass # 138
Second row: \(Zr^{4+}\), Protons: 40, Neutrons: 51, Mass # 91, Charge: +4
3.
First row: Atomic Number 24, Protons 24, Electrons 21, Ion Charge +3, Ion Symbol \(Cr^{3+}\)
Second row: Atomic Number 23, Protons 23, Electrons 18
Third row: Atomic Number 34, Protons 34, Electrons 36, Ion Charge - 2, Ion Symbol \(Se^{2 -}\)