Question

practicing ionic bonding
i. writing lewis dot structures
a. for each of the elements below, completes the following steps.

1. write the electron configuration of the element
2. circle the valence shells of the element
3. write the lewis dot structure for the element.

	electron configuration	lewis dot structure
al
o
ar
n

b. demonstrate with lewis dot structures and electron configurations how the following elements would ionically bond.
li + f

be + o

mg + cl

Answer:

Part I.a: Writing Lewis Dot Structures for Elements

Element: Li (Lithium)

Step 1: Electron Configuration

Lithium has an atomic number of 3. The electron configuration is built by filling orbitals: $1s^2 2s^1$.

Step 2: Valence Shell

The outermost (valence) shell is the $2s$ subshell (circled: $1s^2 \boldsymbol{\circ 2s^1}$).

Step 3: Lewis Dot Structure

Valence electrons = 1 (from $2s^1$). The Lewis dot structure is $\text{Li} \cdot$ (one dot representing the valence electron).

Element: Al (Aluminum)

Step 1: Electron Configuration

Aluminum (atomic number 13) has the configuration: $1s^2 2s^2 2p^6 3s^2 3p^1$.

Step 2: Valence Shell

The valence shell is the $3s^2 3p^1$ subshell (circled: $1s^2 2s^2 2p^6 \boldsymbol{\circ 3s^2 3p^1}$).

Step 3: Lewis Dot Structure

Valence electrons = 3 (from $3s^2 3p^1$). The Lewis dot structure is $\text{Al} \cdot \cdot \cdot$ (three dots).

Element: O (Oxygen)

Step 1: Electron Configuration

Oxygen (atomic number 8) has the configuration: $1s^2 2s^2 2p^4$.

Step 2: Valence Shell

The valence shell is the $2s^2 2p^4$ subshell (circled: $1s^2 \boldsymbol{\circ 2s^2 2p^4}$).

Step 3: Lewis Dot Structure

Valence electrons = 6 (from $2s^2 2p^4$). The Lewis dot structure is $\ddot{\text{O}} \cdot \cdot$ (two lone pairs + two unpaired electrons, total 6 dots).

Element: Ar (Argon)

Step 1: Electron Configuration

Argon (atomic number 18) has the configuration: $1s^2 2s^2 2p^6 3s^2 3p^6$.

Step 2: Valence Shell

The valence shell is the $3s^2 3p^6$ subshell (circled: $1s^2 2s^2 2p^6 \boldsymbol{\circ 3s^2 3p^6}$).

Step 3: Lewis Dot Structure

Valence electrons = 8 (full octet). The Lewis dot structure is $\ddot{\text{Ar}} \ddot{}$ (four lone pairs, total 8 dots).

Element: N (Nitrogen)

Step 1: Electron Configuration

Nitrogen (atomic number 7) has the configuration: $1s^2 2s^2 2p^3$.

Step 2: Valence Shell

The valence shell is the $2s^2 2p^3$ subshell (circled: $1s^2 \boldsymbol{\circ 2s^2 2p^3}$).

Step 3: Lewis Dot Structure

Valence electrons = 5 (from $2s^2 2p^3$). The Lewis dot structure is $\ddot{\text{N}} \cdot \cdot \cdot$ (one lone pair + three unpaired electrons, total 5 dots).

Part I.b: Ionic Bonding with Lewis Dot Structures and Electron Configurations

1. $\text{Li} + \text{F}$

• Li (Lithium):

Electron configuration: $1s^2 2s^1$ → Loses 1 electron to form $\text{Li}^+$ (configuration: $1s^2$, same as He).
Lewis dot: $\text{Li} \cdot
ightarrow \text{Li}^+$ (no valence electrons).

• F (Fluorine):

Electron configuration: $1s^2 2s^2 2p^5$ → Gains 1 electron to form $\text{F}^-$ (configuration: $1s^2 2s^2 2p^6$, same as Ne).
Lewis dot: $\ddot{\text{F}} \cdot \cdot \cdot \cdot \cdot
ightarrow \ddot{\text{F}}^- \ddot{}$ (full octet).

• Ionic Bond: $\text{Li}^+$ and $\text{F}^-$ attract via electrostatic forces. The compound is $\text{LiF}$.

2. $\text{Be} + \text{O}$

• Be (Beryllium):

Electron configuration: $1s^2 2s^2$ → Loses 2 electrons to form $\text{Be}^{2+}$ (configuration: $1s^2$, same as He).
Lewis dot: $\text{Be} \cdot \cdot
ightarrow \text{Be}^{2+}$ (no valence electrons).

• O (Oxygen):

Electron configuration: $1s^2 2s^2 2p^4$ → Gains 2 electrons to form $\text{O}^{2-}$ (configuration: $1s^2 2s^2 2p^6$, same as Ne).
Lewis dot: $\ddot{\text{O}} \cdot \cdot
ightarrow \ddot{\text{O}}^{2-} \ddot{}$ (full octet).

• Ionic Bond: $\text{Be}^{2+}$ and $\text{O}^{2-}$ attract. The compound is $\text{BeO}$.

3. $\text{Mg} + \text{Cl}$

• Mg (Magnesium):

Electron configuration: $1s^2 2s^2 2p^6 3s^2$ → Loses 2 electrons to form $\text{Mg}^{2+}$ (configuration: $1s^2 2s^2 2p^6$, same as Ne).
Lewis dot: $\text{Mg} \cdot \cdot
ightarrow \text{Mg}^{2+}$ (no valence electrons).

• Cl (Chlorine):

Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^5$ → Gains 1 electron to form $\text{Cl}^-$ (configuration: $1s^2 2s^2 2p^6 3s^2 3p^6$, same as Ar).
Lewis dot: $\ddot{\text{Cl}} \cdot \cdot \cdot \cdot \cdot
ightarrow \ddot{\text{Cl}}^- \ddot{}$ (full octet).

• Ionic Bond: $\text{Mg}^{2+}$ needs two $\text{Cl}^-$ ions (since Mg loses 2, Cl gains 1). The compound is $\text{MgCl}_2$.

Summary Table (Part I.a)

Element	Electron Configuration	Lewis Dot Structure
Al	$1s^2 2s^2 2p^6 3s^2 3p^1$	$\text{Al} \cdot \cdot \cdot$
O	$1s^2 2s^2 2p^4$	$\ddot{\text{O}} \cdot \cdot$
Ar	$1s^2 2s^2 2p^6 3s^2 3p^6$	$\ddot{\text{Ar}} \ddot{}$
N	$1s^2 2s^2 2p^3$	$\ddot{\text{N}} \cdot \cdot \cdot$

Final Answer (Key Takeaways)

• Lewis Dot Structures represent valence electrons as dots.
• Ionic Bonding involves electron transfer to achieve stable octets (or duets for H/He-like ions).

For specific sub-questions (e.g., electron configurations, Lewis dots, or ionic bonding), refer to the detailed steps above.