Question

name: ______
pd: ____
directions: for each of the following ionic compounds: draw the lewis structure

1. li₂o\t2. mgs\t3. k₃n
2. cacl₂\t5. bao\t6. alf₃
3. srbr₂\t8. na₂se\t9. al₂s₃
4. mg₃n₂

Answer:

To draw the Lewis structures for these ionic compounds, we follow these general steps for ionic compounds (composed of cations and anions):

Step 1: Identify the ions

For an ionic compound \( M_xA_y \), the cation is \( M^{n+} \) and the anion is \( A^{m-} \), where \( n \) and \( m \) are the charges (determined by valence electrons or group number).

Step 2: Determine charges

• Group 1 metals (Li, Na, K, etc.): \( +1 \) charge.
• Group 2 metals (Mg, Ca, Ba, Sr, etc.): \( +2 \) charge.
• Group 13 metals (Al): \( +3 \) charge.
• Group 15 nonmetals (N, P, etc.): \( -3 \) charge (gain 3 electrons to fill octet).
• Group 16 nonmetals (O, S, Se, etc.): \( -2 \) charge (gain 2 electrons).
• Group 17 nonmetals (F, Cl, Br, etc.): \( -1 \) charge (gain 1 electron).

Example: \( \boldsymbol{\ce{Li2O}} \) (1)

• Li (Group 1): \( \ce{Li^+} \) (loses 1 electron).
• O (Group 16): \( \ce{O^{2-}} \) (gains 2 electrons).
• Charge balance: \( 2 \times (+1) + 1 \times (-2) = 0 \), so 2 \( \ce{Li^+} \) and 1 \( \ce{O^{2-}} \).
• Lewis structure: \( [\ce{Li^+}]_2 \: \: \ce{O^{2-}} \) (or draw each ion with valence electrons: Li has 0 valence electrons (lost 1), O has 8 (gained 2), so \( \ce{Li^+} \) (no dots) and \( \ce{O^{2-}} \) (8 dots, 2- charge)).

Example: \( \boldsymbol{\ce{MgS}} \) (2)

• Mg (Group 2): \( \ce{Mg^{2+}} \) (loses 2 electrons).
• S (Group 16): \( \ce{S^{2-}} \) (gains 2 electrons).
• Charge balance: \( 1 \times (+2) + 1 \times (-2) = 0 \), so 1 \( \ce{Mg^{2+}} \) and 1 \( \ce{S^{2-}} \).
• Lewis structure: \( \ce{Mg^{2+}} \: \: \ce{S^{2-}} \) (Mg has 0 valence electrons, S has 8 dots, 2- charge).

Example: \( \boldsymbol{\ce{K3N}} \) (3)

• K (Group 1): \( \ce{K^+} \) (loses 1 electron).
• N (Group 15): \( \ce{N^{3-}} \) (gains 3 electrons).
• Charge balance: \( 3 \times (+1) + 1 \times (-3) = 0 \), so 3 \( \ce{K^+} \) and 1 \( \ce{N^{3-}} \).
• Lewis structure: \( [\ce{K^+}]_3 \: \: \ce{N^{3-}} \) (K has 0 valence electrons, N has 8 dots, 3- charge).

Example: \( \boldsymbol{\ce{CaCl2}} \) (4)

• Ca (Group 2): \( \ce{Ca^{2+}} \) (loses 2 electrons).
• Cl (Group 17): \( \ce{Cl^-} \) (gains 1 electron).
• Charge balance: \( 1 \times (+2) + 2 \times (-1) = 0 \), so 1 \( \ce{Ca^{2+}} \) and 2 \( \ce{Cl^-} \).
• Lewis structure: \( \ce{Ca^{2+}} \: \: [\ce{Cl^-}]_2 \) (Ca has 0 valence electrons, each Cl has 8 dots, 1- charge).

Example: \( \boldsymbol{\ce{AlF3}} \) (6)

• Al (Group 13): \( \ce{Al^{3+}} \) (loses 3 electrons).
• F (Group 17): \( \ce{F^-} \) (gains 1 electron).
• Charge balance: \( 1 \times (+3) + 3 \times (-1) = 0 \), so 1 \( \ce{Al^{3+}} \) and 3 \( \ce{F^-} \).
• Lewis structure: \( \ce{Al^{3+}} \: \: [\ce{F^-}]_3 \) (Al has 0 valence electrons, each F has 8 dots, 1- charge).

Example: \( \boldsymbol{\ce{Al2S3}} \) (9)

• Al (Group 13): \( \ce{Al^{3+}} \) (loses 3 electrons).
• S (Group 16): \( \ce{S^{2-}} \) (gains 2 electrons).
• Charge balance: \( 2 \times (+3) + 3 \times (-2) = 0 \), so 2 \( \ce{Al^{3+}} \) and 3 \( \ce{S^{2-}} \).
• Lewis structure: \( [\ce{Al^{3+}}]_2 \: \: [\ce{S^{2-}}]_3 \) (Al has 0 valence electrons, each S has 8 dots, 2- charge).

Example: \( \boldsymbol{\ce{Mg3N2}} \) (10)

• Mg (Group 2): \( \ce{Mg^{2+}} \) (loses 2 electrons).
• N (Group 15): \( \ce{N^{3-}} \) (gains 3 electrons).
• Charge balance: \( 3 \times (+2) + 2 \times (-3) = 0 \), so 3 \( \ce{Mg^{2+}} \) and 2 \( \ce{N^{3-}} \).
• Lewis structure: \( [\ce{Mg^{2+}}]_3 \: \: [\ce{N^{3-}}]_2 \) (Mg has 0 valence electrons, each N has 8 dots, 3- charge).

General Format for Ionic Lewis Structures

For \( \ce{M^+_x A^-_{y}} \), draw the cation (no valence electrons, or \( \ce{M^+} \)) and the anion (with 8 valence electrons, \( \ce{A^-} \)) in the ratio given by the formula.

Final Answer (Example for \( \boldsymbol{\ce{Li2O}} \))

The Lewis structure for \( \ce{Li2O} \) is:
\( [\ce{Li^+}] \: [\ce{Li^+}] \: [\overset{..}{\underset{..}{\ce{O}}}]^{2-} \) (or simplified as \( [\ce{Li^+}]_2 \: \ce{O^{2-}} \) with the oxide ion showing 8 valence electrons).

For other compounds, follow the same logic (identify ions, balance charges, draw cations and anions with appropriate charges and valence electrons).