IndietroGeneral Chemistry Exam 2 Study Guide – Step-by-Step Guidance
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Q1. What is a cation?
Background
Topic: Ions and Ionic Compounds
This question tests your understanding of basic terminology in chemistry, specifically the definition of a cation.
Key Terms
Cation: An ion with a positive charge, formed when an atom loses one or more electrons.
Anion: An ion with a negative charge, formed when an atom gains one or more electrons.
Step-by-Step Guidance
Recall that atoms can gain or lose electrons to form ions.
Think about what happens to the charge of an atom when it loses electrons versus when it gains electrons.
Consider which type of ion (cation or anion) results from the loss of electrons.
Try solving on your own before revealing the answer!
Final Answer:
A cation is an ion with a positive charge, formed when an atom loses one or more electrons.
Q2. How does a covalent bond form?
Background
Topic: Chemical Bonding
This question tests your understanding of how atoms share electrons to form covalent bonds.
Key Terms
Covalent bond: A chemical bond formed when two atoms share one or more pairs of electrons.
Electron pair: Two electrons shared between atoms in a bond.
Step-by-Step Guidance
Recall that atoms form bonds to achieve a stable electron configuration, often resembling the nearest noble gas.
Think about how nonmetals interact with each other to achieve this stability.
Consider what happens to the electrons during the formation of a covalent bond.
Try solving on your own before revealing the answer!
Final Answer:
A covalent bond forms when two atoms share one or more pairs of electrons to achieve a stable electron configuration.
Q3. How many pairs of electrons are in a covalent single bond and a covalent triple bond?
Background
Topic: Covalent Bonding and Electron Pairs
This question tests your knowledge of the number of shared electron pairs in different types of covalent bonds.
Key Terms
Single bond: A covalent bond involving one pair of shared electrons.
Triple bond: A covalent bond involving three pairs of shared electrons.
Step-by-Step Guidance
Recall the definition of a single covalent bond and how many electrons are shared.
Think about how many electron pairs are involved in a triple bond.
Remember that each pair consists of two electrons.
Try solving on your own before revealing the answer!
Final Answer:
A covalent single bond has 1 pair of electrons (2 electrons), and a covalent triple bond has 3 pairs of electrons (6 electrons).
Q4. How does an ionic bond form?
Background
Topic: Ionic Bonding
This question tests your understanding of the process by which atoms transfer electrons to form ionic bonds.
Key Terms
Ionic bond: A chemical bond formed by the electrostatic attraction between oppositely charged ions.
Cation: Positively charged ion.
Anion: Negatively charged ion.
Step-by-Step Guidance
Recall that ionic bonds typically form between metals and nonmetals.
Think about what happens to the electrons during the formation of an ionic bond.
Consider the resulting charges on the atoms after electron transfer.
Try solving on your own before revealing the answer!
Final Answer:
An ionic bond forms when one atom transfers electrons to another, resulting in the formation of oppositely charged ions that are held together by electrostatic attraction.
Q5. Calculate the percent by mass of Be in beryllium phosphate, beryllium nitrate, and beryllium phosphide.
Background
Topic: Percent Composition by Mass
This question tests your ability to calculate the percent by mass of an element in a compound using molar masses.
Key Formula:
Step-by-Step Guidance
Write the chemical formulas for beryllium phosphate, beryllium nitrate, and beryllium phosphide.
Calculate the molar mass of each compound by adding the atomic masses of all atoms in the formula.
Determine the total mass of beryllium atoms in one mole of each compound.
Set up the percent by mass formula for Be in each compound, but do not calculate the final value yet.
Try solving on your own before revealing the answer!
Final Answer:
Beryllium phosphate (Be3(PO4)2): %Be = 13.6% Beryllium nitrate (Be(NO3)2): %Be = 7.2% Beryllium phosphide (Be3P2): %Be = 36.1%
Each value is calculated by dividing the total mass of Be in one mole of the compound by the molar mass of the compound, then multiplying by 100%.
Q6. What is the ground state electron configuration of Zn2+, Fe3+, S2-? Are any of these paramagnetic?
Background
Topic: Electron Configuration and Magnetism
This question tests your ability to write electron configurations for ions and determine if they are paramagnetic (contain unpaired electrons).
Key Terms and Concepts:
Electron configuration: The arrangement of electrons in an atom or ion.
Paramagnetic: Species with unpaired electrons.
Diamagnetic: Species with all electrons paired.
Step-by-Step Guidance
Write the ground state electron configuration for the neutral atom.
Adjust the configuration for the ion by adding or removing electrons as appropriate.
Check for unpaired electrons to determine if the ion is paramagnetic or diamagnetic.
Repeat for each ion: Zn2+, Fe3+, S2-.
Try solving on your own before revealing the answer!
Final Answer:
Zn2+: [Ar] 3d10 (diamagnetic) Fe3+: [Ar] 3d5 (paramagnetic) S2-: [Ne] 3s2 3p6 (diamagnetic)
Fe3+ is paramagnetic because it has unpaired electrons.
Q7. What is the empirical formula for a compound of nitrogen and oxygen with 36.86% nitrogen?
Background
Topic: Empirical Formulas
This question tests your ability to determine the simplest whole-number ratio of atoms in a compound from percent composition.
Key Steps:
Assume 100 g of the compound to convert percentages to grams.
Convert grams to moles for each element.
Divide by the smallest number of moles to get the simplest ratio.
Step-by-Step Guidance
Assume you have 100 g of the compound: 36.86 g N and 63.14 g O.
Convert grams of N and O to moles using their molar masses ( g/mol, g/mol).
Divide each mole value by the smallest number of moles to get the ratio.
If necessary, multiply by a whole number to get integer subscripts for the empirical formula.
Try solving on your own before revealing the answer!
Final Answer:
The empirical formula is NO2.
This is based on the mole ratio of N to O after converting from percent composition.
Q8. If a compound has a molar mass of 180 g/mol and its empirical formula is C9H8O4, what is its molecular formula?
Background
Topic: Empirical and Molecular Formulas
This question tests your ability to determine the molecular formula from the empirical formula and molar mass.
Key Formula:
Step-by-Step Guidance
Calculate the empirical formula mass of C9H8O4.
Divide the given molar mass (180 g/mol) by the empirical formula mass to find n.
Multiply the subscripts in the empirical formula by n to get the molecular formula.
Try solving on your own before revealing the answer!
Final Answer:
The molecular formula is C9H8O4.
The empirical formula mass matches the molar mass, so the empirical and molecular formulas are the same.
Q9. If a compound is made of C (60%), H (4.48%), O (35.52%), what is its empirical formula? If this compound has a molar mass of 180 g/mol, what is its molecular formula?
Background
Topic: Empirical and Molecular Formulas from Percent Composition
This question tests your ability to determine both empirical and molecular formulas from percent composition and molar mass.
Key Steps:
Assume 100 g of the compound to convert percentages to grams.
Convert grams to moles for each element.
Divide by the smallest number of moles to get the simplest ratio.
Calculate the empirical formula mass and use the molar mass to find the molecular formula.
Step-by-Step Guidance
Assume 100 g: 60 g C, 4.48 g H, 35.52 g O.
Convert each to moles: for C, for H, for O.
Divide each by the smallest number of moles to get the ratio.
Write the empirical formula based on these ratios.
Calculate the empirical formula mass and determine n by dividing 180 g/mol by the empirical formula mass.
Multiply the empirical formula by n to get the molecular formula.
Try solving on your own before revealing the answer!
Final Answer:
Empirical formula: C9H8O4 Molecular formula: C9H8O4
The empirical formula mass matches the molar mass, so the empirical and molecular formulas are the same.
Q10. Draw the electron configuration of S2-, Br-, Fe3+, Zn2+, Mn, Sr.
Background
Topic: Electron Configurations
This question tests your ability to write electron configurations for both neutral atoms and ions.
Key Concepts:
Use the Aufbau principle, Hund's rule, and the Pauli exclusion principle to fill orbitals.
For ions, add electrons for anions and remove electrons for cations, starting from the outermost shell.
Step-by-Step Guidance
Write the electron configuration for the neutral atom.
Adjust for the ion by adding or removing electrons as needed.
Repeat for each species: S2-, Br-, Fe3+, Zn2+, Mn, Sr.
Use noble gas notation for brevity where appropriate.
Try solving on your own before revealing the answer!
Final Answer:
S2-: [Ne] 3s2 3p6 Br-: [Kr] Fe3+: [Ar] 3d5 Zn2+: [Ar] 3d10 Mn: [Ar] 4s2 3d5 Sr: [Kr] 5s2