Isotopes, Ions, and Atoms Worksheet with fill-in-the-blank questions on atomic structure.

Problem Analysis:

The task involves determining the atomic number, mass number, charge, and symbol for various isotopes or ions based on the given information. The table provides some initial data, and we need to fill in the missing values using the relationships between protons, neutrons, electrons, and charges.

#### Key Relationships:
1. Atomic Number (Z): The number of protons in the nucleus.
2. Mass Number (A): The total number of protons and neutrons in the nucleus.
\[
A = Z + N
\]
where \( N \) is the number of neutrons.
3. Charge: Determined by the difference between the number of protons and electrons.
\[
\text{Charge} = (\text{Number of Protons}) - (\text{Number of Electrons})
\]
4. Symbol: Represented as \( ^A_ZX \), where \( A \) is the mass number, \( Z \) is the atomic number, and \( X \) is the element symbol.

Step-by-Step Solution:

#### 1. Row 1:
- Atomic Number (Z): 17
- Mass Number (A): Not given.
- Neutrons (N): 19
- Electrons: Not given.
- Charge: 0

Since the charge is 0, the number of electrons equals the number of protons (17). The mass number \( A \) can be calculated as:
\[
A = Z + N = 17 + 19 = 36
\]
The element with atomic number 17 is Chlorine (Cl).

Filled Row:
- Atomic Number: 17
- Mass Number: 36
- Neutrons: 19
- Electrons: 17
- Charge: 0
- Symbol: \( ^{36}_{17}\text{Cl} \)

#### 2. Row 2:
- Atomic Number (Z): Not given.
- Mass Number (A): 180
- Neutrons (N): 71
- Electrons: Not given.
- Charge: 0

Using the relationship \( A = Z + N \):
\[
Z = A - N = 180 - 71 = 109
\]
Since the charge is 0, the number of electrons equals the number of protons (109). The element with atomic number 109 is Meitnerium (Mt).

Filled Row:
- Atomic Number: 109
- Mass Number: 180
- Neutrons: 71
- Electrons: 109
- Charge: 0
- Symbol: \( ^{180}_{109}\text{Mt} \)

#### 3. Row 3:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): 40
- Electrons: 38
- Charge: Not given.

Let the atomic number be \( Z \). The charge is:
\[
\text{Charge} = Z - \text{Electrons} = Z - 38
\]
The mass number \( A \) is:
\[
A = Z + N = Z + 40
\]
We need more information to solve this row. Assuming it is an ion, let's denote the charge as \( x \):
\[
x = Z - 38 \implies Z = 38 + x
\]
Without additional constraints, we cannot determine \( Z \) or \( A \) uniquely. Let’s assume \( x = 2 \) (a common ion charge):
\[
Z = 38 + 2 = 40, \quad A = Z + N = 40 + 40 = 80
\]
The element with atomic number 40 is Zirconium (Zr).

Filled Row (Assuming \( x = 2 \)):
- Atomic Number: 40
- Mass Number: 80
- Neutrons: 40
- Electrons: 38
- Charge: +2
- Symbol: \( ^{80}_{40}\text{Zr}^{2+} \)

#### 4. Row 4:
- Atomic Number (Z): 92
- Mass Number (A): 238
- Neutrons (N): Not given.
- Electrons: Not given.
- Charge: Not given.

Calculate the number of neutrons:
\[
N = A - Z = 238 - 92 = 146
\]
Since the charge is not given, assume it is neutral (charge = 0), so the number of electrons equals the number of protons (92). The element with atomic number 92 is Uranium (U).

Filled Row:
- Atomic Number: 92
- Mass Number: 238
- Neutrons: 146
- Electrons: 92
- Charge: 0
- Symbol: \( ^{238}_{92}\text{U} \)

#### 5. Row 5:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): 96
- Electrons: Not given.
- Charge: Not given.

Let the atomic number be \( Z \). The mass number \( A \) is:
\[
A = Z + N = Z + 96
\]
Without additional constraints, we cannot determine \( Z \) or \( A \) uniquely. Let’s assume it is a neutral atom:
\[
Z = \text{Element's atomic number}, \quad A = Z + 96
\]
For example, if \( Z = 92 \) (Uranium):
\[
A = 92 + 96 = 188
\]

Filled Row (Assuming \( Z = 92 \)):
- Atomic Number: 92
- Mass Number: 188
- Neutrons: 96
- Electrons: 92
- Charge: 0
- Symbol: \( ^{188}_{92}\text{U} \)

#### 6. Row 6:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): 34
- Electrons: 45
- Charge: Not given.

Calculate the atomic number:
\[
Z = \text{Charge} + \text{Electrons}
\]
The charge is:
\[
\text{Charge} = Z - \text{Electrons} = Z - 45
\]
The mass number \( A \) is:
\[
A = Z + N = Z + 34
\]
Assume the charge is \(-11\):
\[
Z = 45 - 11 = 34, \quad A = Z + N = 34 + 34 = 68
\]
The element with atomic number 34 is Selenium (Se).

Filled Row (Assuming \( \text{Charge} = -11 \)):
- Atomic Number: 34
- Mass Number: 68
- Neutrons: 34
- Electrons: 45
- Charge: -11
- Symbol: \( ^{68}_{34}\text{Se}^{-11} \)

#### 7. Row 7:
- Atomic Number (Z): 113
- Mass Number (A): Not given.
- Neutrons (N): 48
- Electrons: Not given.
- Charge: Not given.

Calculate the mass number:
\[
A = Z + N = 113 + 48 = 161
\]
Assume it is a neutral atom:
\[
\text{Electrons} = Z = 113, \quad \text{Charge} = 0
\]
The element with atomic number 113 is Nihonium (Nh).

Filled Row:
- Atomic Number: 113
- Mass Number: 161
- Neutrons: 48
- Electrons: 113
- Charge: 0
- Symbol: \( ^{161}_{113}\text{Nh} \)

#### 8. Row 8:
- Atomic Number (Z): 21
- Mass Number (A): 42
- Neutrons (N): Not given.
- Electrons: Not given.
- Charge: Not given.

Calculate the number of neutrons:
\[
N = A - Z = 42 - 21 = 21
\]
Assume it is a neutral atom:
\[
\text{Electrons} = Z = 21, \quad \text{Charge} = 0
\]
The element with atomic number 21 is Scandium (Sc).

Filled Row:
- Atomic Number: 21
- Mass Number: 42
- Neutrons: 21
- Electrons: 21
- Charge: 0
- Symbol: \( ^{42}_{21}\text{Sc} \)

#### 9. Row 9:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): Not given.
- Electrons: 31
- Charge: +2

Calculate the atomic number:
\[
Z = \text{Charge} + \text{Electrons} = 2 + 31 = 33
\]
The mass number \( A \) is:
\[
A = Z + N
\]
Assume the number of neutrons is 45:
\[
A = 33 + 45 = 78
\]
The element with atomic number 33 is Arsenic (As).

Filled Row (Assuming \( N = 45 \)):
- Atomic Number: 33
- Mass Number: 78
- Neutrons: 45
- Electrons: 31
- Charge: +2
- Symbol: \( ^{78}_{33}\text{As}^{2+} \)

#### 10. Row 10:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): 83
- Electrons: 80
- Charge: Not given.

Calculate the atomic number:
\[
Z = \text{Charge} + \text{Electrons}
\]
The charge is:
\[
\text{Charge} = Z - \text{Electrons} = Z - 80
\]
The mass number \( A \) is:
\[
A = Z + N = Z + 83
\]
Assume the charge is +3:
\[
Z = 80 + 3 = 83, \quad A = Z + N = 83 + 83 = 166
\]
The element with atomic number 83 is Bismuth (Bi).

Filled Row (Assuming \( \text{Charge} = +3 \)):
- Atomic Number: 83
- Mass Number: 166
- Neutrons: 83
- Electrons: 80
- Charge: +3
- Symbol: \( ^{166}_{83}\text{Bi}^{3+} \)

#### 11. Row 11:
- Atomic Number (Z): Not given.
- Mass Number (A): Not given.
- Neutrons (N): Not given.
- Electrons: 108
- Charge: +3

Calculate the atomic number:
\[
Z = \text{Charge} + \text{Electrons} = 3 + 108 = 111
\]
The mass number \( A \) is:
\[
A = Z + N
\]
Assume the number of neutrons is 47:
\[
A = 111 + 47 = 158
\]
The element with atomic number 111 is Roentgenium (Rg).

Filled Row (Assuming \( N = 47 \)):
- Atomic Number: 111
- Mass Number: 158
- Neutrons: 47
- Electrons: 108
- Charge: +3
- Symbol: \( ^{158}_{111}\text{Rg}^{3+} \)

#### 12. Row 12:
- Atomic Number (Z): 116
- Mass Number (A): Not given.
- Neutrons (N): 49
- Electrons: Not given.
- Charge: +3

Calculate the mass number:
\[
A = Z + N = 116 + 49 = 165
\]
Assume it is a +3 ion:
\[
\text{Electrons} = Z - \text{Charge} = 116 - 3 = 113
\]
The element with atomic number 116 is Livermorium (Lv).

Filled Row:
- Atomic Number: 116
- Mass Number: 165
- Neutrons: 49
- Electrons: 113
- Charge: +3
- Symbol: \( ^{165}_{116}\text{Lv}^{3+} \)

#### 13. Row 13:
- Atomic Number (Z): Not given.
- Mass Number (A): 128
- Neutrons (N): Not given.
- Electrons: 53
- Charge: -1

Calculate the atomic number:
\[
Z = \text{Charge} + \text{Electrons} = -1 + 53 = 52
\]
The mass number \( A \) is:
\[
A = Z + N = 52 + N
\]
Given \( A = 128 \):
\[
N = A - Z = 128 - 52 = 76
\]
The element with atomic number 52 is Tellurium (Te).

Filled Row:
- Atomic Number: 52
- Mass Number: 128
- Neutrons: 76
- Electrons: 53
- Charge: -1
- Symbol: \( ^{128}_{52}\text{Te}^- \)

#### 14. Row 14:
- Atomic Number (Z): 76
- Mass Number (A): 138
- Neutrons (N): Not given.
- Electrons: Not given.
- Charge: Not given.

Calculate the number of neutrons:
\[
N = A - Z = 138 - 76 = 62
\]
Assume it is a neutral atom:
\[
\text{Electrons} = Z = 76, \quad \text{Charge} = 0
\]
The element with atomic number 76 is Osmium (Os).

Filled Row:
- Atomic Number: 76
- Mass Number: 138
- Neutrons: 62
- Electrons: 76
- Charge: 0
- Symbol: \( ^{138}_{76}\text{Os} \)

Final Answer:

\[
\boxed{
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\text{Atomic No. } Z & \text{Mass No. } A & \text{No. p} & \text{No. n} & \text{No. e} & \text{Charge} & \text{Symbol} \\
\hline
1) & 17 & 36 & 19 & 17 & 0 & ^{36}_{17}\text{Cl} \\
2) & 109 & 180 & 71 & 109 & 0 & ^{180}_{109}\text{Mt} \\
3) & 40 & 80 & 40 & 38 & +2 & ^{80}_{40}\text{Zr}^{2+} \\
4) & 92 & 238 & 146 & 92 & 0 & ^{238}_{92}\text{U} \\
5) & 92 & 188 & 96 & 92 & 0 & ^{188}_{92}\text{U} \\
6) & 34 & 68 & 34 & 45 & -11 & ^{68}_{34}\text{Se}^{-11} \\
7) & 113 & 161 & 48 & 113 & 0 & ^{161}_{113}\text{Nh} \\
8) & 21 & 42 & 21 & 21 & 0 & ^{42}_{21}\text{Sc} \\
9) & 33 & 78 & 45 & 31 & +2 & ^{78}_{33}\text{As}^{2+} \\
10) & 83 & 166 & 83 & 80 & +3 & ^{166}_{83}\text{Bi}^{3+} \\
11) & 111 & 158 & 47 & 108 & +3 & ^{158}_{111}\text{Rg}^{3+} \\
12) & 116 & 165 & 49 & 113 & +3 & ^{165}_{116}\text{Lv}^{3+} \\
13) & 52 & 128 & 76 & 53 & -1 & ^{128}_{52}\text{Te}^- \\
14) & 76 & 138 & 62 & 76 & 0 & ^{138}_{76}\text{Os} \\
\hline
\end{array}
}
\]