General Chemistry Exam & Worksheet Study Guidance

Q1. Use Lewis theory to determine the chemical formula for the compound formed between Al and F.

Background

Topic: Ionic Bonding & Lewis Theory

This question tests your understanding of how to use Lewis theory to predict the formula of an ionic compound formed between a metal (aluminum) and a nonmetal (fluorine).

Key Terms and Concepts:

• Lewis theory: Atoms gain, lose, or share electrons to achieve a noble gas configuration.

• Ionic compound: Formed when electrons are transferred from a metal to a nonmetal.

• Aluminum (Al): Group 13, typically forms a $3^+$ ion.

• Fluorine (F): Group 17, typically forms a $1^-$ ion.

Step-by-Step Guidance

1. Determine the charge on each ion: Al forms $\mathrm{Al^{3+}}$, F forms $\mathrm{F^-}$.

2. Use the charges to find the ratio of ions needed for a neutral compound. The total positive and negative charges must balance.

3. Set up the formula so that the total positive charge from Al equals the total negative charge from F.

4. Write the empirical formula using the smallest whole-number ratio of ions.

Try solving on your own before revealing the answer!

Final Answer: AlF3

Aluminum forms $\mathrm{Al^{3+}}$ and fluorine forms $\mathrm{F^-}$. Three F- ions are needed to balance one Al3+ ion, so the formula is AlF3.

Q2. Choose the compound below that should have the highest melting point according to the ionic bonding model.

Background

Topic: Ionic Bonding & Melting Points

This question tests your understanding of how lattice energy and ionic charges affect the melting points of ionic compounds.

Key Terms and Concepts:

• Lattice energy: The energy required to separate one mole of an ionic solid into gaseous ions.

• Melting point: Higher lattice energy generally means a higher melting point.

• Lattice energy increases with higher ionic charges and smaller ionic radii.

Step-by-Step Guidance

1. Identify the charges on the cations and anions in each compound.

2. Recall that lattice energy increases with the product of the charges ($|q_1 q_2|$) and decreases with increasing ionic radius.

3. Compare the charges and sizes for each compound to predict which has the highest lattice energy.

4. Relate the highest lattice energy to the highest melting point.

Try solving on your own before revealing the answer!

Final Answer: MgF2

MgF2 has small, highly charged ions, leading to a high lattice energy and thus a high melting point.

Q3. How many of the following molecules are polar? XeO2, SiCl2F2, C2Br2, SeF6

Background

Topic: Molecular Polarity

This question tests your ability to determine molecular polarity based on molecular geometry and the presence of polar bonds.

Key Terms and Concepts:

• Polarity: A molecule is polar if it has a net dipole moment.

• Symmetry: Symmetrical molecules with identical surrounding atoms are often nonpolar.

• Electronegativity: Differences in electronegativity between atoms can create polar bonds.

Step-by-Step Guidance

1. Draw the Lewis structure for each molecule to determine its shape.

2. Assess the symmetry of each molecule and the direction of bond dipoles.

3. Decide if the dipoles cancel (nonpolar) or if there is a net dipole (polar).

4. Count how many of the four molecules are polar.

Try solving on your own before revealing the answer!

Final Answer: 2

SiCl2F2 and XeO2 are polar; C2Br2 and SeF6 are nonpolar due to symmetry.

Q4. A molecule containing a central atom with sp3 hybridization has a(n) ________ electron geometry.

Background

Topic: Hybridization and Molecular Geometry

This question tests your understanding of how the hybridization of the central atom relates to the electron geometry of a molecule.

Key Terms and Concepts:

• sp3 hybridization: Mixing one s and three p orbitals to form four equivalent orbitals.

• Electron geometry: The arrangement of electron domains (bonding and lone pairs) around the central atom.

• sp3 hybridization corresponds to four electron domains.

Step-by-Step Guidance

1. Recall the number of electron domains associated with sp3 hybridization.

2. Identify the electron geometry that results from four electron domains.

3. Match this geometry to the correct name (e.g., tetrahedral, trigonal planar, etc.).

Try solving on your own before revealing the answer!

Final Answer: Tetrahedral

sp3 hybridization leads to a tetrahedral electron geometry because there are four electron domains.

Q5. What are the F-Te-F bond angles in TeF6?

Background

Topic: Molecular Geometry & Bond Angles

This question tests your knowledge of the geometry and bond angles in molecules with octahedral geometry.

Key Terms and Concepts:

• TeF6: Tellurium hexafluoride, central atom with six bonding pairs.

• Octahedral geometry: Six regions of electron density around the central atom.

• Bond angles in an octahedral geometry are 90°.

Step-by-Step Guidance

1. Determine the number of electron domains around Te in TeF6.

2. Identify the electron geometry (octahedral for six domains).

3. Recall the bond angles associated with an octahedral geometry.

Try solving on your own before revealing the answer!

Final Answer: 90°

In an octahedral geometry, all F-Te-F bond angles are 90°.