Practice worksheet for identifying protons, neutrons, and electrons in elements using atomic symbols and the periodic table.

Let's solve this step-by-step.

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Part 1: Fill in the element name and the number of protons and neutrons

We are given isotopic notations in the form:

\[
^A_ZX
\]

Where:
- \( A \) = mass number (protons + neutrons)
- \( Z \) = atomic number (number of protons)
- \( X \) = chemical symbol

Number of protons = atomic number (Z)
Number of neutrons = mass number (A) – atomic number (Z)

---

#### 1. \( ^4_2He \)

- Element name: Helium
- # of protons: 2
- # of neutrons: \( 4 - 2 = 2 \)

#### 2. \( ^{56}_{26}Fe \)

- Element name: Iron
- # of protons: 26
- # of neutrons: \( 56 - 26 = 30 \)

#### 3. \( ^{27}_{13}Al \)

- Element name: Aluminum
- # of protons: 13
- # of neutrons: \( 27 - 13 = 14 \)

#### 4. \( ^{40}_{20}Ca \)

- Element name: Calcium
- # of protons: 20
- # of neutrons: \( 40 - 20 = 20 \)

#### 5. \( ^{56}_{26}Ne \)

Wait! This is a mistake. Neon (Ne) has atomic number 10, not 26.
\( ^{56}_{26}Ne \) is incorrect — it should be Fe (iron), which we already did.

But here, it says \( ^{56}_{26}Ne \). That’s an error. The correct notation for neon with mass 56 would be impossible because neon's atomic number is 10. So likely a typo.

But if we take it as written:
- Atomic number = 26 → that’s Iron (Fe), not Neon.
So probably meant to be \( ^{20}_{10}Ne \)? But it says \( ^{56}_{26}Ne \).

Wait — perhaps it's a typo. Let's check:
If it's \( ^{20}_{10}Ne \), then:
- Element: Neon
- Protons: 10
- Neutrons: 10

But here it says \( ^{56}_{26}Ne \), which is invalid. Since atomic number 26 is Fe, this must be a typo.

But let’s assume the symbol is correct, but the label is wrong. So if it's \( ^{56}_{26}X \), then it's Iron, not Neon.

But the symbol says Ne, so maybe it's a mistake.

Alternatively, perhaps it's supposed to be \( ^{20}_{10}Ne \) or \( ^{22}_{10}Ne \), but not 56.

But since it's written as \( ^{56}_{26}Ne \), we have to go by what's given.

Wait — no, Neon (Ne) has atomic number 10, so \( ^{56}_{26}Ne \) is impossible.

So likely, it's a typo and should be \( ^{20}_{10}Ne \), or perhaps \( ^{22}_{10}Ne \), but not 56.

But let’s suppose it’s actually \( ^{20}_{10}Ne \), and the 56 is a typo.

But wait — there’s another one: \( ^{12}_6Ca \). Calcium has atomic number 20, so \( ^{12}_6Ca \) is also invalid.

Ah! There are multiple typos.

Let’s look again:

- \( ^{56}_{26}Ne \): Atomic number 26 is Fe, not Ne → Invalid
- \( ^{12}_6Ca \): Atomic number 6 is Carbon (C), not Ca → Invalid

So clearly, these are mislabeled.

Let’s fix them based on common isotopes.

But the problem says "for the given symbols", so we must use what's written.

But they can't be right.

Let me re-express each:

1. \( ^4_2He \) → Helium, Z=2, A=4 → valid
2. \( ^{56}_{26}Fe \) → Iron, Z=26, A=56 → valid
3. \( ^{27}_{13}Al \) → Aluminum, Z=13, A=27 → valid
4. \( ^{40}_{20}Ca \) → Calcium, Z=20, A=40 → valid
5. \( ^{56}_{26}Ne \) → This is wrong: Ne has Z=10 → should be \( ^{20}_{10}Ne \) or similar. But here Z=26 → must be Fe, not Ne. So likely typo — perhaps it’s \( ^{20}_{10}Ne \)?
6. \( ^{12}_6Ca \) → Z=6 is Carbon, not Ca → should be \( ^{12}_6C \), not Ca. So typo — probably \( ^{40}_{20}Ca \) already listed?
7. \( ^{19}_9F \) → Fluorine, Z=9, A=19 → valid
8. \( ^{35}_{17}Cl \) → Chlorine, Z=17, A=35 → valid
9. \( ^{23}_{11}Na \) → Sodium, Z=11, A=23 → valid

So the problematic ones are:
- \( ^{56}_{26}Ne \) → Should be \( ^{56}_{26}Fe \), but Fe already listed → maybe typo in symbol
- \( ^{12}_6Ca \) → Should be \( ^{12}_6C \), but C is carbon

Possibility: Maybe it's \( ^{12}_6C \), not Ca?

But it says "Ca" — calcium.

Alternatively, maybe it's \( ^{40}_{20}Ca \) already listed.

Let’s assume the symbols are correct, and the element names are to be filled based on the atomic number.

So:

- For \( ^{56}_{26}X \): Z=26 → Iron (Fe), not Ne → so element name is Iron, not Neon
- For \( ^{12}_6X \): Z=6 → Carbon (C), not Ca

So likely, the symbols are correct, but the labels are wrong.

But the problem says: "Fill in the element name and the number of protons and neutrons for the given symbols."

So we go by the symbol and atomic number.

So:

#### 5. \( ^{56}_{26}Ne \) → But symbol is Ne, atomic number is 26 → contradiction.

This cannot exist. So it must be a typo.

Likely, it should be \( ^{20}_{10}Ne \) or \( ^{22}_{10}Ne \), but written as \( ^{56}_{26}Ne \), which is wrong.

But perhaps it's meant to be \( ^{20}_{10}Ne \), and the 56 is a typo.

Alternatively, maybe it's \( ^{56}_{26}Fe \), and the symbol is wrong.

But since the symbol is written as Ne, and Z=26, it's inconsistent.

Given the confusion, let’s assume that the atomic number (Z) determines the element, regardless of symbol.

So:

- If Z=26, element is Iron (Fe), even if labeled Ne → but that’s not possible.

So best interpretation: The symbol is correct, and the atomic number matches it.

So for \( ^{56}_{26}Ne \): Z=26, but Ne has Z=10 → contradiction.

Therefore, it must be a typo.

Similarly, \( ^{12}_6Ca \): Z=6 → Carbon, not Calcium.

So likely:

- \( ^{56}_{26}Ne \) → should be \( ^{56}_{26}Fe \) → but Fe already listed
- \( ^{12}_6Ca \) → should be \( ^{12}_6C \)

But let's proceed carefully.

Looking at the pattern, perhaps:

- \( ^{56}_{26}Ne \) → typo, should be \( ^{20}_{10}Ne \) → Ne, 10 protons, 10 neutrons
- \( ^{12}_6Ca \) → typo, should be \( ^{12}_6C \) → Carbon, 6 protons, 6 neutrons

But the problem says "given symbols", so we must use what's written.

Alternative: Perhaps the mass number is wrong.

But let’s try to interpret based on atomic number.

So for each, we’ll use the Z (bottom number) to determine element, and ignore symbol if inconsistent.

But that defeats the purpose.

Better: Assume the symbol is correct, so:

- \( ^{56}_{26}Ne \): symbol Ne → atomic number should be 10, but it's 26 → invalid
- So skip and assume typo

But since this is a worksheet, likely:

- \( ^{56}_{26}Ne \) → should be \( ^{20}_{10}Ne \), but written wrong
- Or perhaps it's \( ^{56}_{26}Fe \), and symbol is wrong

But we have \( ^{56}_{26}Fe \) already listed.

Wait — in the first row: \( ^{56}_{26}Fe \) is listed, then second row: \( ^{56}_{26}Ne \) — so duplicate?

No, both are written.

So likely, the second one is a typo.

Let’s assume:

- \( ^{56}_{26}Ne \) → should be \( ^{20}_{10}Ne \) → Neon, 10 protons, 10 neutrons
- \( ^{12}_6Ca \) → should be \( ^{12}_6C \) → Carbon, 6 protons, 6 neutrons

But let’s check the next ones:

- \( ^{19}_9F \) → Fluorine, Z=9, A=19 → 9 protons, 10 neutrons → valid
- \( ^{35}_{17}Cl \) → Chlorine, Z=17, A=35 → 17 protons, 18 neutrons → valid
- \( ^{23}_{11}Na \) → Sodium, Z=11, A=23 → 11 protons, 12 neutrons → valid

So only two are problematic.

Perhaps the intended ones are:

- \( ^{20}_{10}Ne \) instead of \( ^{56}_{26}Ne \)
- \( ^{12}_6C \) instead of \( ^{12}_6Ca \)

But since the problem says “given symbols”, we have to go with the numbers.

So for each, we'll use the atomic number (Z) to find the element name, and compute protons and neutrons.

So:

#### 5. \( ^{56}_{26}Ne \) → Z=26 → Iron (Fe), not Ne → so element name: Iron
- # of protons: 26
- # of neutrons: 56 - 26 = 30

But symbol is Ne — contradiction.

But if we ignore symbol and go by Z, then:

- Element name: Iron
- Protons: 26
- Neutrons: 30

But that’s same as first Fe.

So likely, it’s a typo.

But for consistency, let’s assume the Z is correct, and the symbol is wrong.

So:

- \( ^{56}_{26}Ne \) → Z=26 → Iron
- \( ^{12}_6Ca \) → Z=6 → Carbon

So we’ll fill accordingly.

So:

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Corrected Part 1:

#### 1. \( ^4_2He \)
- Element name: Helium
- Protons: 2
- Neutrons: 4 - 2 = 2

#### 2. \( ^{56}_{26}Fe \)
- Element name: Iron
- Protons: 26
- Neutrons: 56 - 26 = 30

#### 3. \( ^{27}_{13}Al \)
- Element name: Aluminum
- Protons: 13
- Neutrons: 27 - 13 = 14

#### 4. \( ^{40}_{20}Ca \)
- Element name: Calcium
- Protons: 20
- Neutrons: 40 - 20 = 20

#### 5. \( ^{56}_{26}Ne \) → Z=26 → Iron (not Ne)
- Element name: Iron
- Protons: 26
- Neutrons: 56 - 26 = 30

But this duplicates the first Fe.

Alternatively, maybe it’s \( ^{20}_{10}Ne \), with A=20, Z=10.

But it says 56.

Another possibility: \( ^{20}_{10}Ne \) is common, but written as \( ^{56}_{26}Ne \) — typo in both numbers.

But let’s move on.

#### 6. \( ^{12}_6Ca \) → Z=6 → Carbon (not Ca)
- Element name: Carbon
- Protons: 6
- Neutrons: 12 - 6 = 6

#### 7. \( ^{19}_9F \)
- Element name: Fluorine
- Protons: 9
- Neutrons: 19 - 9 = 10

#### 8. \( ^{35}_{17}Cl \)
- Element name: Chlorine
- Protons: 17
- Neutrons: 35 - 17 = 18

#### 9. \( ^{23}_{11}Na \)
- Element name: Sodium
- Protons: 11
- Neutrons: 23 - 11 = 12

So final answers for Part 1:

| Symbol | Element Name | Protons | Neutrons |
|--------|--------------|---------|----------|
| \( ^4_2He \) | Helium | 2 | 2 |
| \( ^{56}_{26}Fe \) | Iron | 26 | 30 |
| \( ^{27}_{13}Al \) | Aluminum | 13 | 14 |
| \( ^{40}_{20}Ca \) | Calcium | 20 | 20 |
| \( ^{56}_{26}Ne \) | Iron (should be Fe) | 26 | 30 |
| \( ^{12}_6Ca \) | Carbon (should be C) | 6 | 6 |
| \( ^{19}_9F \) | Fluorine | 9 | 10 |
| \( ^{35}_{17}Cl \) | Chlorine | 17 | 18 |
| \( ^{23}_{11}Na \) | Sodium | 11 | 12 |

But since the symbols are wrong, we’ll fill based on atomic number.

So for the worksheet:

1.

- \( ^4_2He \): Element: Helium, Protons: 2, Neutrons: 2
- \( ^{56}_{26}Fe \): Element: Iron, Protons: 26, Neutrons: 30
- \( ^{27}_{13}Al \): Element: Aluminum, Protons: 13, Neutrons: 14
- \( ^{40}_{20}Ca \): Element: Calcium, Protons: 20, Neutrons: 20
- \( ^{56}_{26}Ne \): Element: Iron (since Z=26), Protons: 26, Neutrons: 30
- \( ^{12}_6Ca \): Element: Carbon (Z=6), Protons: 6, Neutrons: 6
- \( ^{19}_9F \): Element: Fluorine, Protons: 9, Neutrons: 10
- \( ^{35}_{17}Cl \): Element: Chlorine, Protons: 17, Neutrons: 18
- \( ^{23}_{11}Na \): Element: Sodium, Protons: 11, Neutrons: 12

But this is messy.

Alternatively, perhaps the symbols are correct, and the atomic numbers are wrong.

But that doesn’t make sense.

Best guess: The worksheet has typos.

But for now, we’ll proceed with the values based on atomic number.

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Part 2: Fill in the missing information

We are given a table:

| Element | Atomic Number | Atomic Mass | # protons | # neutrons | # electrons |
|--------|---------------|-------------|-----------|------------|-------------|
| S | 16 | ? | 16 | ? | ? |
| ? | 17 | 35 | ? | ? | 17 |
| ? | 18 | ? | 18 | 22 | ? |
| ? | 19 | 39 | ? | ? | 19 |
| Ca | 20 | ? | 20 | ? | ? |

We know:

- Atomic number = number of protons = number of electrons (in neutral atom)
- Atomic mass ≈ protons + neutrons
- So neutrons = atomic mass - protons

Let’s fill row by row.

#### Row 1: S (Sulfur)
- Atomic number: 16 → protons = 16, electrons = 16
- Atomic mass: not given, but common isotope is S-32, so likely 32
- Neutrons = 32 - 16 = 16

But not specified, so we need standard value.

Sulfur has atomic mass ~32.07, so most common isotope is 32

So:
- Atomic mass: 32
- Protons: 16
- Neutrons: 32 - 16 = 16
- Electrons: 16

#### Row 2: Atomic number 17, atomic mass 35, electrons 17
- Element: Chlorine (Cl)
- Protons: 17 (same as atomic number)
- Neutrons: 35 - 17 = 18
- Electrons: 17 (neutral atom)

#### Row 3: Atomic number 18, protons 18, neutrons 22
- Element: Argon (Ar)
- Atomic mass = protons + neutrons = 18 + 22 = 40
- Electrons: 18 (neutral)

#### Row 4: Atomic number 19, atomic mass 39, electrons 19
- Element: Potassium (K)
- Protons: 19
- Neutrons: 39 - 19 = 20
- Electrons: 19

#### Row 5: Ca (Calcium), atomic number 20, protons 20
- Element: Calcium
- Atomic mass: common isotope is Ca-40, so 40
- Neutrons: 40 - 20 = 20
- Electrons: 20

Now fill the table:

| Element | Atomic Number | Atomic Mass | # protons | # neutrons | # electrons |
|--------|---------------|-------------|-----------|------------|-------------|
| S | 16 | 32 | 16 | 16 | 16 |
| Cl | 17 | 35 | 17 | 18 | 17 |
| Ar | 18 | 40 | 18 | 22 | 18 |
| K | 19 | 39 | 19 | 20 | 19 |
| Ca | 20 | 40 | 20 | 20 | 20 |

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Final Answers

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Part 1:

1. \( ^4_2He \)
- Element: Helium
- Protons: 2
- Neutrons: 2

2. \( ^{56}_{26}Fe \)
- Element: Iron
- Protons: 26
- Neutrons: 30

3. \( ^{27}_{13}Al \)
- Element: Aluminum
- Protons: 13
- Neutrons: 14

4. \( ^{40}_{20}Ca \)
- Element: Calcium
- Protons: 20
- Neutrons: 20

5. \( ^{56}_{26}Ne \)
- Note: Symbol is Ne, but Z=26 → should be Fe. Assuming typo, element is Iron
- Protons: 26
- Neutrons: 30

6. \( ^{12}_6Ca \)
- Note: Symbol is Ca, but Z=6 → should be C. Element is Carbon
- Protons: 6
- Neutrons: 6

7. \( ^{19}_9F \)
- Element: Fluorine
- Protons: 9
- Neutrons: 10

8. \( ^{35}_{17}Cl \)
- Element: Chlorine
- Protons: 17
- Neutrons: 18

9. \( ^{23}_{11}Na \)
- Element: Sodium
- Protons: 11
- Neutrons: 12

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Part 2:

| Element | Atomic Number | Atomic Mass | # protons | # neutrons | # electrons |
|--------|---------------|-------------|-----------|------------|-------------|
| S | 16 | 32 | 16 | 16 | 16 |
| Cl | 17 | 35 | 17 | 18 | 17 |
| Ar | 18 | 40 | 18 | 22 | 18 |
| K | 19 | 39 | 19 | 20 | 19 |
| Ca | 20 | 40 | 20 | 20 | 20 |

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Summary of Corrections:

- \( ^{56}_{26}Ne \) → likely typo; should be \( ^{20}_{10}Ne \) or \( ^{56}_{26}Fe \)
- \( ^{12}_6Ca \) → likely typo; should be \( ^{12}_6C \)

But based on atomic number, we filled accordingly.

For educational purposes, students should learn:
- Atomic number = protons = electrons (in neutral atom)
- Mass number = protons + neutrons
- Use periodic table to identify element from atomic number

Let me know if you'd like a cleaned-up version without typos.