Ionic Bonding Worksheet-1 - Name ______ Period - Studocu - Free Printable

Problem Overview:

The task involves analyzing ionic bonding reactions between pairs of elements. For each reaction, you need to:
1. Draw atomic diagrams showing the electron configurations of the atoms.
2. Indicate the transfer of electrons during the reaction.
3. Show the resulting ions formed in the product.
4. Fill in a table with details about the valence electrons, electron transfer, and ions formed.

Solution Explanation:

#### General Approach:
1. Identify Valence Electrons: Determine the number of valence electrons for each atom involved in the reaction.
2. Electron Transfer: Metals tend to lose electrons to achieve a stable electron configuration (usually noble gas configuration), while nonmetals gain electrons.
3. Form Ions: After electron transfer, determine the charge on each ion formed.
4. Balance Charges: Ensure that the total positive charge equals the total negative charge in the compound.

#### Detailed Analysis for Each Reaction:

---

Reaction 1: Li + Cl → LiCl

- Li (Lithium):
- Valence electrons: 1
- Lithium loses 1 electron to achieve a stable configuration (like He).
- Ion formed: \( \text{Li}^+ \)

- Cl (Chlorine):
- Valence electrons: 7
- Chlorine gains 1 electron to achieve a stable configuration (like Ar).
- Ion formed: \( \text{Cl}^- \)

- Resulting Compound: \( \text{Li}^+ \text{Cl}^- \) or \( \text{LiCl} \)

- Table Filling:
- Atoms: Li, Cl
- Valence electrons: 1 (Li), 7 (Cl)
- Electron transfer: Li loses 1, Cl gains 1
- Ions formed: \( \text{Li}^+ \), \( \text{Cl}^- \)

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Reaction 2: Ca + O → CaO

- Ca (Calcium):
- Valence electrons: 2
- Calcium loses 2 electrons to achieve a stable configuration (like Ar).
- Ion formed: \( \text{Ca}^{2+} \)

- O (Oxygen):
- Valence electrons: 6
- Oxygen gains 2 electrons to achieve a stable configuration (like Ne).
- Ion formed: \( \text{O}^{2-} \)

- Resulting Compound: \( \text{Ca}^{2+} \text{O}^{2-} \) or \( \text{CaO} \)

- Table Filling:
- Atoms: Ca, O
- Valence electrons: 2 (Ca), 6 (O)
- Electron transfer: Ca loses 2, O gains 2
- Ions formed: \( \text{Ca}^{2+} \), \( \text{O}^{2-} \)

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Reaction 3: Be + F → BeF₂

- Be (Beryllium):
- Valence electrons: 2
- Beryllium loses 2 electrons to achieve a stable configuration (like He).
- Ion formed: \( \text{Be}^{2+} \)

- F (Fluorine):
- Valence electrons: 7
- Fluorine gains 1 electron to achieve a stable configuration (like Ne).
- Since Be loses 2 electrons, 2 F atoms are needed to accept these electrons.
- Ion formed: \( \text{F}^- \) (for each F atom)

- Resulting Compound: \( \text{Be}^{2+} \text{F}^- \text{F}^- \) or \( \text{BeF}_2 \)

- Table Filling:
- Atoms: Be, F
- Valence electrons: 2 (Be), 7 (F)
- Electron transfer: Be loses 2, F gains 1 (per F atom)
- Ions formed: \( \text{Be}^{2+} \), \( \text{F}^- \) (2 F atoms)

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Reaction 4: Mg + S → MgS

- Mg (Magnesium):
- Valence electrons: 2
- Magnesium loses 2 electrons to achieve a stable configuration (like Ne).
- Ion formed: \( \text{Mg}^{2+} \)

- S (Sulfur):
- Valence electrons: 6
- Sulfur gains 2 electrons to achieve a stable configuration (like Ar).
- Ion formed: \( \text{S}^{2-} \)

- Resulting Compound: \( \text{Mg}^{2+} \text{S}^{2-} \) or \( \text{MgS} \)

- Table Filling:
- Atoms: Mg, S
- Valence electrons: 2 (Mg), 6 (S)
- Electron transfer: Mg loses 2, S gains 2
- Ions formed: \( \text{Mg}^{2+} \), \( \text{S}^{2-} \)

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Reaction 5: K + F → KF

- K (Potassium):
- Valence electrons: 1
- Potassium loses 1 electron to achieve a stable configuration (like Ar).
- Ion formed: \( \text{K}^+ \)

- F (Fluorine):
- Valence electrons: 7
- Fluorine gains 1 electron to achieve a stable configuration (like Ne).
- Ion formed: \( \text{F}^- \)

- Resulting Compound: \( \text{K}^+ \text{F}^- \) or \( \text{KF} \)

- Table Filling:
- Atoms: K, F
- Valence electrons: 1 (K), 7 (F)
- Electron transfer: K loses 1, F gains 1
- Ions formed: \( \text{K}^+ \), \( \text{F}^- \)

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Final Answer:

The completed table and diagrams are as follows:

| Reactions | Atoms | Valence electrons | Electron transfer from/to each atom | Ions formed in the product |
|------------------|-------|-------------------|-------------------------------------|-----------------------------|
| 1) Li + Cl → LiCl | Li | 1 | Li loses 1 | \( \text{Li}^+ \) |
| | Cl | 7 | Cl gains 1 | \( \text{Cl}^- \) |
| 2) Ca + O → CaO | Ca | 2 | Ca loses 2 | \( \text{Ca}^{2+} \) |
| | O | 6 | O gains 2 | \( \text{O}^{2-} \) |
| 3) Be + F → BeF₂ | Be | 2 | Be loses 2 | \( \text{Be}^{2+} \) |
| | F | 7 | F gains 1 (per F atom) | \( \text{F}^- \) (2 F atoms)|
| 4) Mg + S → MgS | Mg | 2 | Mg loses 2 | \( \text{Mg}^{2+} \) |
| | S | 6 | S gains 2 | \( \text{S}^{2-} \) |
| 5) K + F → KF | K | 1 | K loses 1 | \( \text{K}^+ \) |
| | F | 7 | F gains 1 | \( \text{F}^- \) |

Boxed Final Answer:
\[
\boxed{
\begin{array}{|c|c|c|c|c|}
\hline
\text{Reactions} & \text{Atoms} & \text{Valence electrons} & \text{Electron transfer from/to each atom} & \text{Ions formed in the product} \\
\hline
1) \text{Li + Cl → LiCl} & \text{Li} & 1 & \text{Li loses 1} & \text{Li}^+ \\
& \text{Cl} & 7 & \text{Cl gains 1} & \text{Cl}^- \\
\hline
2) \text{Ca + O → CaO} & \text{Ca} & 2 & \text{Ca loses 2} & \text{Ca}^{2+} \\
& \text{O} & 6 & \text{O gains 2} & \text{O}^{2-} \\
\hline
3) \text{Be + F → BeF}_2 & \text{Be} & 2 & \text{Be loses 2} & \text{Be}^{2+} \\
& \text{F} & 7 & \text{F gains 1 (per F atom)} & \text{F}^- \text{ (2 F atoms)} \\
\hline
4) \text{Mg + S → MgS} & \text{Mg} & 2 & \text{Mg loses 2} & \text{Mg}^{2+} \\
& \text{S} & 6 & \text{S gains 2} & \text{S}^{2-} \\
\hline
5) \text{K + F → KF} & \text{K} & 1 & \text{K loses 1} & \text{K}^+ \\
& \text{F} & 7 & \text{F gains 1} & \text{F}^- \\
\hline
\end{array}
}
\]