Fiveable 5-Hour Cram Study Guide: Units 1-5

Atomic Structure and Properties

During the Fiveable 5-hour cram session, the primary focus for Unit 1 includes atomic structure and properties, encompassing concepts such as moles, molar mass, mass spectroscopy, electron configuration, photoelectron spectroscopy, and periodic trends. In terms of formatting and precision, students are advised not to box their answers and to look for parentheses which may provide helpful hints. All final calculations should be reported using three significant figures, and the value for $G$ should be considered at most.

At the fundamental level, the identity of an element is determined by the number of its protons. For instance, Chlorine has an atomic number of $17$, meaning a neutral Chlorine atom contains exactly $17$ electrons. Chlorine exists as different isotopes, which are defined as atoms of the same element that possess a different number of neutrons. Specifically, Chlorine has two common isotopes: ${^{35}\text{Cl}}$ and ${^{37}\text{Cl}}$. In ${^{37}\text{Cl}}$, there are $17$ protons and $20$ neutrons, while ${^{35}\text{Cl}}$ contains $18$ neutrons. The mass number is calculated as the sum of protons and neutrons. Atomic mass is measured in atomic mass units ($\text{amu}$). The average atomic mass of Chlorine is $35.45$, which represents a weighted average of its naturally occurring isotopes. Given this value, it is clear that ${^{35}\text{Cl}}$ is more abundant than ${^{37}\text{Cl}}$. For example, if ${^{35}\text{Cl}}$ has an abundance of $76\%$ and ${^{37}\text{Cl}}$ has an abundance of $24\%$, the calculation for average atomic mass would be $(35 \times 0.76) + (37 \times 0.24) = 35.4\,\text{amu}$.

Moles, Molar Mass, and Empirical Formulas

The fundamental unit for quantity in chemistry is the mole, where $1\,\text{mole} = 6.02 \times 10^{23}$ particles. For Chlorine, $1\,\text{mole}$ is equivalent to $35.45\,\text{g}$. To determine the mass of a specific number of molecules, such as $1.25 \times 10^{24}$ molecules of $\text{CO}_2$, one must first calculate the molar mass of the compound. For $\text{CO}_2$, this is calculated as $1 \times (12.01) + 2 \times (16) = 44.01\,\text{g/mol}$. The calculation follows: $\frac{1.25 \times 10^{24}\,\text{molecules}}{6.02 \times 10^{23}\,\text{molecules/mol}} \times 44.01\,\text{g/mol} = 91.4\,\text{g}$.

Empirical formulas represent the lowest whole-number ratio of atoms in a compound. To determine an empirical formula from percent composition, such as a compound containing $38.3\%$ Chlorine and $61.7\%$ Fluorine, a two-step process is followed. First, convert the mass (assuming a $100\,\text{g}$ sample) to moles: $\frac{38.3\,\text{g Cl}}{35.45\,\text{g/mol}} = 1.08\,\text{mol Cl}$ and $\frac{61.7\,\text{g F}}{19.00\,\text{g/mol}} = 3.24\,\text{mol F}$. Second, divide both values by the smallest number of moles: $\frac{1.08}{1.08} = 1$ for Cl and $\frac{3.24}{1.08} = 3$ for F. This results in the empirical formula $\text{ClF}_3$.

Electron Configuration and Orbital Mechanics

Electron configurations describe the location of electrons within an atom. For Chlorine, which has $17$ electrons, the configuration is written as $1s^2 2s^2 2p^6 3s^2 3p^5$. The energy level of an orbital indicates its distance from the nucleus. An orbital itself is defined as a region in space where an electron spends $90\%$ of its time. Different orbitals have distinct shapes and electron capacities. The $s$ orbital is spherical and can hold up to $2$ electrons. The $p$ orbital is shaped like a dumbbell and consists of $3$ orbitals, holding up to $6$ electrons. The $d$ orbital consists of $5$ orbitals holding up to $10$ electrons, and the $f$ orbital consists of $7$ orbitals holding up to $14$ electrons. All orbitals hold electrons in pairs with opposite spins.

Periodic Trends and Free Response Applications

Atomic radius refers to the size of the atom, measured from the center (nucleus) to the outer orbitals. Ionization energy is defined as the amount of energy required to remove an electron from an atom. The further an electron is from the nucleus, the easier it is to remove, resulting in lower ionization energy. Conversely, electrons closer to the nucleus experience less shielding and are harder to remove, requiring higher ionization energy. For example, Lithium ($\text{Li}: 1s^2 2s^1$) has electrons closer to the nucleus with less shielding compared to Sodium ($\text{Na}: 1s^2 2s^2 2p^6 3s^1$), whose valence electron is further away. Thus, Na has a larger atomic radius and lower ionization energy than Li.

In a provided Free Response Question (FRQ #1), a graph of electron configuration corresponds to the sequence $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$, which can be abbreviated using noble gas notation as $[\text{Ar}] 4s^2$. This configuration identifies the element as Calcium ($\text{Ca}$).

Unit 2: Molecules and Ionic Compounds

Unit 2 covers ionic compounds, metals, alloys, molecular compounds, and Lewis structures. Ionic bonds involve the transfer of electrons between a metal and a nonmetal, such as in $\text{NaCl}$, where $\text{Cl}$ is more electronegative and extracts an electron from $\text{Na}$. Ionic compounds typically have high melting and boiling points and conduct electricity when dissolved in water or in a molten state, but not as solids. Metals and alloys are characterized by the "sea of electrons" model, where electrons are shared among metal atoms. This leads to properties such as malleability, ductility, luster, and high conductivity of heat and electricity. Covalent bonds occur between two nonmetals with large electronegativities where electrons are shared. Molecular compounds generally have low melting and boiling points and do not conduct electricity.

Common ionic charges across the periodic table include $+1$ for Group 1 ($\text{H}^+, \text{Li}^+, \text{Na}^+, \text{K}^+, \text{Rb}^+, \text{Cs}^+, \text{Fr}^+$), $+2$ for Group 2 ($\text{Be}^{2+}, \text{Mg}^{2+}, \text{Ca}^{2+}, \text{Sr}^{2+}, \text{Ba}^{2+}, \text{Ra}^{2+}$), and $+3$ for Group 13 ($\text{B}^{3+}, \text{Al}^{3+}, \text{In}^{3+}, \text{Tl}^{3+}$). Transition metals like $Zn$ typically form $2+$ ions. Nonmetals form anions such as $N^{3-}, P^{3-}, Sb^{3-}$ in Group 15; $O^{2-}, S^{2-}, Se^{2-}, Te^{2-}$ in Group 16; and $F^-, Cl^-, Br^-, I^-$ in Group 17. Noble gases in Group 18 ($\text{He}, \text{Ne}, \text{Ar}, \text{Kr}, \text{Xe}, \text{Rn}$) have an oxidation state of $0$.

Lewis Structures and Molecular Geometry

To draw a Lewis structure, follow these steps: 1) Count the total number of valence electrons. For $\text{SF}_4$, this is $6 + 7(4) = 34\,\text{valence electrons}$. 2) Determine the central atom (typically the least electronegative, Carbon is always central if present). 3) Connect all atoms with single bonds. 4) Add valence electrons to exterior atoms to satisfy the octet rule. 5) Place remaining electrons on the central atom. 6) If the central atom has fewer than 8 electrons, create double or triple bonds.

Hydrogen is an exception; it never has lone pairs and only needs $2$ electrons. In the case of $\text{HCN}$, the total valence electron count is $1+4+5=10$. The structure is \text{H-C\equiv N:}. Another exception allows central atoms from the 3rd period and below to have more than $8$ valence electrons. In FRQ #2, $\text{HNO}_2$ has $1+5+6(2)=18$ electrons; its Lewis structure is \text{H-\ddot{O}-\ddot{N}=\ddot{O}}. The nitrogen hybridization in this molecule is $sp^2$ because it has two bonding sets and one lone pair, resulting in a bent shape ($120^{\circ}$). Hybridization involves the mixing of atomic orbitals to create molecular orbitals.

The following table summarizes molecular geometry and hybridization based on the number of electron domains (bonding sets + lone pairs):

• 2 bonds, 0 lone pairs: Linear ($sp$)
• 3 bonds, 0 lone pairs: Trigonal Planar ($sp^2$)
• 2 bonds, 1 lone pair: Bent $120^{\circ}$ ($sp^2$)
• 4 bonds, 0 lone pairs: Tetrahedral ($sp^3$)
• 3 bonds, 1 lone pair: Trigonal Pyramidal ($sp^3$)
• 2 bonds, 2 lone pairs: Bent $109^{\circ}$ ($sp^3$)
• 5 bonds, 0 lone pairs: Trigonal Bipyramidal ($sp^3d$)
• 4 bonds, 1 lone pair: See-saw ($sp^3d$)
• 3 bonds, 2 lone pairs: T-shape ($sp^3d$)
• 6 bonds, 0 lone pairs: Octahedral ($sp^3d^2$)

Resonance, Formal Charge, and Coulomb's Law

Resonance structures occur when there is more than one acceptable Lewis structure for a molecule, such as in $\text{SO}_2$. Choosing the "best" structure involves calculating formal charge ($\text{FC}$) using the formula: $\text{FC} = \text{Valence Electrons} - \text{Dots} - \text{Lines}$. The goal is to minimize formal charges toward zero. In the $\text{:}\,\ddot{\text{O}}\text{=}\ddot{\text{S}}\text{-}\ddot{\text{O}}\text{:}$ resonance structure, the $\text{FC}$ for Sulfur is $6 - 2 - 3 = +1$, one Oxygen is $6 - 4 - 2 = 0$, and the other Oxygen is $6 - 6 - 1 = -1$.

Coulomb's Law states that opposite charges attract and that attraction is stronger when charges are closer together. In an FRQ comparing lattice energy, it was noted that $\text{Sr}(\text{OH})_2$ has less lattice energy than $\text{Mg}(\text{OH})_2$. This is because the $\text{Sr}^{2+}$ ion is larger than the $\text{Mg}^{2+}$ ion due to additional occupied energy levels. Per Coulomb's Law, the distance between the nuclei of $\text{Mg}$ and $\text{O}$ in the hydroxide is shorter, resulting in stronger attraction and higher lattice energy.

Unit 3: Intermolecular Forces (IMF)

Intermolecular forces are the forces of attraction between particles. Hydrogen bonding is not a true chemical bond but a strong IMF occurring when Hydrogen is covalently bonded to highly electronegative elements Nitrogen ($\text{N}$), Oxygen ($\text{O}$), or Fluorine ($\text{F}$), such as in $\text{H-F}$. Dipole-dipole forces are attractions between polar molecules like $\text{HCl}$. London Dispersion Forces ($\text{LDF}$) exist between nonpolar molecules, including diatomic molecules ($Br_2, I_2, N_2, Cl_2, H_2, O_2, F_2$), noble gases, and hydrocarbons. LDF results from the random motion of electrons; molecules with more electrons are more polarizable and thus have stronger LDF.

The strength of the IMF determines several physical properties: 1) Stronger IMF leads to higher melting and boiling points. 2) Stronger IMF leads to higher viscosity and resistance to flow. 3) Stronger IMF leads to higher surface tension. 4) Stronger IMF leads to lower vapor pressure, as it is harder for the liquid to evaporate into a gas. In terms of states of matter, solids are densely packed with the strongest IMF, liquids allow for flow with intermediate IMF, and gases have huge distances between particles with very low IMF. Pressure in a gas is caused by collisions. Real gases like $\text{CO}_2, \text{O}_2,\, \text{He}$ behave most ideally at high temperatures and low pressures.

Gas Laws and Solution Chemistry

The Ideal Gas Law is expressed as $PV = nRT$, where $R = 0.0821\,\text{dm}^3\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1}$. Related individual laws include Boyle's Law ($P_1V_1 = P_2V_2$, an inverse relationship), Charles's Law ($\frac{V_1}{T_1} = \frac{V_2}{T_2}$), and Avogadro's Law ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$). Temperature must always be converted to Kelvin: $T_{\text{K}} = T_{\text{C}} + 273$.

Solubility is based on the principle "like dissolves like." Polar solvents (water, acids, bases, sugars, alcohols, ionic compounds) dissolve polar solutes. Nonpolar solvents (fats, oils, waxes, diatomics, noble gases) dissolve nonpolar solutes. Quantitatively, increasing temperature increases the solubility of solids and liquids but decreases the solubility of gases. Molarity ($M$) is the primary measure of concentration, defined as $\frac{\text{moles solute}}{\text{dm}^3\,\text{solution}}$. For example, the molarity of $75\,\text{g}$ of $\text{NaOH}$ in a $215\,\text{cm}^3$ solution is calculated by converting grams to moles ($\frac{75}{40.0} = 1.875\,\text{mol}$) and dividing by volume in $dm^3$ ($0.215\,\text{dm}^3$), yielding $8.72\,M$.

In FRQ #3, a gas at $P = 0.40\,\text{atm}$, $V = 25.0\,\text{dm}^3$, and $T = 393\,\text{K}$ results in $n = \frac{(0.40 \times 25.0)}{(0.0821 \times 393)} = 0.309\,\text{moles}$, which rounds to $0.31\,\text{moles}$. Higher temperatures increase the average kinetic energy of particles, leading to more frequent and more energetic collisions that can overcome activation energy.

Spectroscopy and Light Interaction

Light affects matter according to the equations $E = h\nu$ and $c = \lambda\nu$. Here, $\lambda$ is wavelength (in $m$), $\nu$ is frequency (in $Hz$ or $s^{-1}$), and $E$ is energy. Constants include $c = 3.00 \times 10^8\,\text{m/s}$, $h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s}$, and $1\,\text{mol} = 6.02 \times 10^{23}\,\text{photons}$. Beer’s Law relates absorbance to concentration: $A = abc$ (or $A = \epsilon b C$), where $A$ is absorbance, $b = 1\,\text{cm}$ is path length, and $c$ is concentration. As concentration increases, absorbance increases linearly.

Unit 4: Chemical Reactions and Stoichiometry

Chemical reactions involve forming or breaking bonds, resulting in the creation of new matter, whereas physical processes (like state changes) do not form new bonds. Stoichiometry uses mole ratios from balanced equations to convert between substances. For the reaction $\text{CH}_4 + 2\text{O}_2 \rightarrow 2\text{H}_2 ext{O} + \text{CO}_2$, the ratio of $\text{CH}_4$ to $\text{O}_2$ is $1:2$. Limiting reactants determine when a reaction stops based on which reactant is consumed first. For the reaction $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$, if starting with $25\,\text{g SO}_2$ ($0.39\,\text{mol}$) and $43\,\text{g O}_2$ ($1.34\,\text{mol}$), $\text{SO}_2$ is the limiting reactant because \frac{0.39}{2} < \frac{1.34}{1}.

Reaction types include precipitation reactions, where a solid is produced. Solubility rules (SNAP: $\text{Na}^+, \text{NO}_3^-, \text{NH}_4^+, \text{K}^+$) indicate these ions are always aqueous. Redox reactions involve electron transfer and changes in oxidation states. Rules for oxidation states include: 1) Elements in their standard state are $0$ ($ext{Na}(s), ext{O}_2(g)$). 2) Monatomic ions equal their charge ($ext{Mg}^{2+} = +2, ext{Cl}^- = -1$). 3) Oxygen is usually $-2$ and Hydrogen is $+1$. 4) The sum of oxidation numbers must equal the overall charge. In FRQ #4, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$ as its oxidation state changes from $+7$ to $+2$, indicating a gain of $5$ electrons.

Unit 5: Kinetics

Kinetics is the study of reaction rates. The Collision Model dictates that a reaction only occurs if particles collide with sufficient energy and the correct orientation. Activation energy ($E_a$) is the minimum energy required to initiate a reaction. Heating a reaction increases the number of particles with enough energy to overcome this barrier.

Rate laws relate concentration to rate: $\text{Rate} = k[A]^x[B]^y$. A zero-order reaction ($x=0$) means concentration does not affect rate. A first-order reaction ($x=1$) means the rate is directly proportional to concentration. A second-order reaction ($x=2$) means the rate is proportional to the square of the concentration. The units for the rate constant $k$ vary: $0^{th}$ order is $\text{M/s}$, $1^{st}$ order is $1/s$, $2^{nd}$ order is $1/( ext{M}\cdot ext{s})$, and $3^{rd}$ order is $1/( ext{M}^2\cdot ext{s})$. For half-life, only first-order reactions are typically discussed in this context. Integrated rate laws provide linear graphs for different orders: concentration vs. time for $0^{th}$, $\ln(\text{concentration})$ vs. time for $1^{st}$, and $1/(\text{concentration})$ vs. time for $2^{nd}$. Reaction mechanisms consist of steps, where the slowest step is the rate-determining step and must match the experimental rate law. Potential energy (PE) diagrams illustrate exothermic reactions where the energy of the products is lower than the reactants.