Question

identify the properties of the compou particle arrangement and the strength purtides? do now: identify bond type of the following compounds by finding number of valence electrons and applying octet rule. h₂o, co₂, nacl

Explanation:

To determine the bond types of \( \ce{H_2O} \), \( \ce{CO_2} \), and \( \ce{NaCl} \), we analyze each compound:

1. For \( \boldsymbol{\ce{H_2O}} \) (Water)

• Valence Electrons Calculation:

Hydrogen (\( \ce{H} \)) has 1 valence electron, and oxygen (\( \ce{O} \)) has 6 valence electrons.
Total valence electrons: \( 2 \times 1 + 6 = 8 \).

• Bond Type:

\( \ce{O} \) shares electrons with two \( \ce{H} \) atoms (each \( \ce{H} \) shares 1 electron, \( \ce{O} \) shares 2 electrons). This is a covalent bond (electron sharing between nonmetals, \( \ce{H} \) and \( \ce{O} \) are nonmetals).

2. For \( \boldsymbol{\ce{CO_2}} \) (Carbon Dioxide)

• Valence Electrons Calculation:

Carbon (\( \ce{C} \)) has 4 valence electrons, and oxygen (\( \ce{O} \)) has 6 valence electrons.
Total valence electrons: \( 4 + 2 \times 6 = 16 \).

• Bond Type:

\( \ce{C} \) shares electrons with two \( \ce{O} \) atoms (double bonds: \( \ce{O}=\ce{C}=\ce{O} \)). This is a covalent bond (electron sharing between nonmetals, \( \ce{C} \) and \( \ce{O} \) are nonmetals).

3. For \( \boldsymbol{\ce{NaCl}} \) (Sodium Chloride)

• Valence Electrons & Octet Rule:

Sodium (\( \ce{Na} \), a metal) has 1 valence electron (loses it to achieve octet), and chlorine (\( \ce{Cl} \), a nonmetal) has 7 valence electrons (gains 1 electron to achieve octet).

• Bond Type:

\( \ce{Na} \) transfers its valence electron to \( \ce{Cl} \), forming \( \ce{Na^+} \) and \( \ce{Cl^-} \) ions. The electrostatic attraction between these ions is an ionic bond (electron transfer between a metal and a nonmetal).

Summary of Bond Types:

• \( \ce{H_2O} \): Covalent bond (polar covalent, due to electronegativity difference).
• \( \ce{CO_2} \): Covalent bond (nonpolar covalent, symmetric structure).
• \( \ce{NaCl} \): Ionic bond.

Answer:

To determine the bond types of \( \ce{H_2O} \), \( \ce{CO_2} \), and \( \ce{NaCl} \), we analyze each compound:

1. For \( \boldsymbol{\ce{H_2O}} \) (Water)

• Valence Electrons Calculation:

Hydrogen (\( \ce{H} \)) has 1 valence electron, and oxygen (\( \ce{O} \)) has 6 valence electrons.
Total valence electrons: \( 2 \times 1 + 6 = 8 \).

• Bond Type:

\( \ce{O} \) shares electrons with two \( \ce{H} \) atoms (each \( \ce{H} \) shares 1 electron, \( \ce{O} \) shares 2 electrons). This is a covalent bond (electron sharing between nonmetals, \( \ce{H} \) and \( \ce{O} \) are nonmetals).

2. For \( \boldsymbol{\ce{CO_2}} \) (Carbon Dioxide)

• Valence Electrons Calculation:

Carbon (\( \ce{C} \)) has 4 valence electrons, and oxygen (\( \ce{O} \)) has 6 valence electrons.
Total valence electrons: \( 4 + 2 \times 6 = 16 \).

• Bond Type:

\( \ce{C} \) shares electrons with two \( \ce{O} \) atoms (double bonds: \( \ce{O}=\ce{C}=\ce{O} \)). This is a covalent bond (electron sharing between nonmetals, \( \ce{C} \) and \( \ce{O} \) are nonmetals).

3. For \( \boldsymbol{\ce{NaCl}} \) (Sodium Chloride)

• Valence Electrons & Octet Rule:

Sodium (\( \ce{Na} \), a metal) has 1 valence electron (loses it to achieve octet), and chlorine (\( \ce{Cl} \), a nonmetal) has 7 valence electrons (gains 1 electron to achieve octet).

• Bond Type:

\( \ce{Na} \) transfers its valence electron to \( \ce{Cl} \), forming \( \ce{Na^+} \) and \( \ce{Cl^-} \) ions. The electrostatic attraction between these ions is an ionic bond (electron transfer between a metal and a nonmetal).

Summary of Bond Types:

• \( \ce{H_2O} \): Covalent bond (polar covalent, due to electronegativity difference).
• \( \ce{CO_2} \): Covalent bond (nonpolar covalent, symmetric structure).
• \( \ce{NaCl} \): Ionic bond.