Question

given the pair of elements below, show
a. the electron dot diagrams for each
b. the flow of electrons
c. state what happens (write the ions produced)
d. write the formula

1. na cl

a.
b.
c.
d.

1. ca cl

a.
b.
c.
d.

1. k f

a.
b.
c.
d.

1. mg i

a.
b.
c.
d.

Explanation:

Step1: Analyze Na and Cl (Part 1)

a. Electron Dot Diagrams

• Sodium (Na) has 1 valence electron. Its electron dot diagram is $\text{Na} \cdot$ (one dot representing the valence electron).
• Chlorine (Cl) has 7 valence electrons. Its electron dot diagram is $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{Cl}}} \cdot$ (seven dots around the symbol, with three pairs and one single dot).

b. Flow of Electrons

Sodium (Na) donates its 1 valence electron to Chlorine (Cl). Na loses the electron to achieve a stable octet (like Ne), and Cl gains the electron to achieve a stable octet (like Ar).

c. Ions Produced

• Na loses 1 electron: $\text{Na}

ightarrow \text{Na}^+ + e^-$

• Cl gains 1 electron: $\text{Cl} + e^-

ightarrow \text{Cl}^-$

d. Formula

The charges are $+1$ (Na⁺) and $-1$ (Cl⁻). To balance, the formula is $\text{NaCl}$.

Step2: Analyze Ca and Cl (Part 2)

a. Electron Dot Diagrams

• Calcium (Ca) has 2 valence electrons. Its electron dot diagram is $\text{Ca} :$ (two dots representing the valence electrons).
• Chlorine (Cl) has 7 valence electrons (as before): $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{Cl}}} \cdot$

b. Flow of Electrons

Calcium (Ca) donates 2 valence electrons (one to each of two Cl atoms, since each Cl needs 1 electron). Ca loses 2 electrons to achieve a stable octet (like Ar), and each Cl gains 1 electron.

c. Ions Produced

• Ca loses 2 electrons: $\text{Ca}

ightarrow \text{Ca}^{2+} + 2e^-$

• Each Cl gains 1 electron: $\text{Cl} + e^-

ightarrow \text{Cl}^-$ (two Cl atoms are involved, so total electrons gained: 2)

d. Formula

The charges are $+2$ (Ca²⁺) and $-1$ (Cl⁻). To balance, we need 2 Cl⁻ for 1 Ca²⁺. The formula is $\text{CaCl}_2$.

Step3: Analyze K and F (Part 3)

a. Electron Dot Diagrams

• Potassium (K) has 1 valence electron: $\text{K} \cdot$
• Fluorine (F) has 7 valence electrons: $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{F}}} \cdot$

b. Flow of Electrons

Potassium (K) donates its 1 valence electron to Fluorine (F). K loses the electron (stable like Ar), F gains the electron (stable like Ne).

c. Ions Produced

• K loses 1 electron: $\text{K}

ightarrow \text{K}^+ + e^-$

• F gains 1 electron: $\text{F} + e^-

ightarrow \text{F}^-$

d. Formula

Charges: $+1$ (K⁺) and $-1$ (F⁻). Formula: $\text{KF}$.

Step4: Analyze Mg and I (Part 4)

a. Electron Dot Diagrams

• Magnesium (Mg) has 2 valence electrons: $\text{Mg} :$
• Iodine (I) has 7 valence electrons: $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{I}}} \cdot$

b. Flow of Electrons

Magnesium (Mg) donates 2 valence electrons (one to each of two I atoms, as each I needs 1 electron). Mg loses 2 electrons (stable like Ne), each I gains 1 electron.

c. Ions Produced

• Mg loses 2 electrons: $\text{Mg}

ightarrow \text{Mg}^{2+} + 2e^-$

• Each I gains 1 electron: $\text{I} + e^-

ightarrow \text{I}^-$ (two I atoms involved)

d. Formula

Charges: $+2$ (Mg²⁺) and $-1$ (I⁻). Formula: $\text{MgI}_2$.

Answer:

1. Na and Cl

a. $\text{Na} \cdot$; $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{Cl}}} \cdot$
b. Na donates 1 electron to Cl.
c. $\text{Na}^+$ and $\text{Cl}^-$
d. $\text{NaCl}$

2. Ca and Cl

a. $\text{Ca} :$; $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{Cl}}} \cdot$ (two Cl diagrams)
b. Ca donates 2 electrons (1 to each Cl).
c. $\text{Ca}^{2+}$ and $\text{Cl}^-$ (two $\text{Cl}^-$)
d. $\text{CaCl}_2$

3. K and F

a. $\text{K} \cdot$; $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{F}}} \cdot$
b. K donates 1 electron to F.
c. $\text{K}^+$ and $\text{F}^-$
d. $\text{KF}$

4. Mg and I

a. $\text{Mg} :$; $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{I}}} \cdot$ (two I diagrams)
b. Mg donates 2 electrons (1 to each I).
c. $\text{Mg}^{2+}$ and $\text{I}^-$ (two $\text{I}^-$)
d. $\text{MgI}_2$