Chemistry Unit 1 Review PDF

S1.1.1 Matter and energy Pure substances and mixtures Diatomic Elements: HOFBrINCl Here are concise definitions and examples for each term: Matter: Anything that has mass and occupies space. ○ Example: Water, air, and a rock are all matter. Pure Substance: A mate...

S1.1.1 Matter and energy Pure substances and mixtures Diatomic Elements: HOFBrINCl Here are concise definitions and examples for each term: Matter: Anything that has mass and occupies space. ○ Example: Water, air, and a rock are all matter. Pure Substance: A material with a constant composition and distinct chemical properties. ○ Example: Pure distilled water (H₂O). Element: A substance made of only one type of atom, which cannot be broken down into simpler substances. ○ Example: Oxygen (O₂). Compound: A substance formed when two or more elements chemically bond together in a fixed ratio. ○ Example: Carbon dioxide (CO₂). Mixture: A combination of two or more substances that are not chemically bonded and retain their individual properties. ○ Example: Salt and sand mixed together. Homogeneous Mixture: A mixture with a uniform composition throughout, where the different components are not distinguishable. ○ Example: Salt dissolved in water (saltwater). Heterogeneous Mixture: A mixture where the components are not uniformly distributed, and individual substances are easily distinguishable. ○ Example: A salad with lettuce, tomatoes, and cucumbers.  Ways to separate matter: filtration, distillation, paper chromatography Filtration: Used to separate a heterogeneous mixture of a solid and liquid. ○ Ex: CaCO3 and water Distillation: Used to separate a homogeneous mixture of a liquid and liquid using difference in boiling point. ○ Ex: ethanol and water Chromatography (know the difference): Used to separate complex mixtures using columns, paper or special plates. Column chromatography: depending on how fast or slow fluids go through, you can separate the layers. Paper chromatography: When you drop the paper in and the water sinks though, and it depends on how fast or slow the ink bleeds. You can measure and see where the colors are. Thin-layer chromatography: You dip the whole paper into the water, and you can see the clump be separated by the water. S1.1.2/1.1.3 States of Matter Kinetic molecular theory All matter is made up of small particles. These particles all have kinetic energy (the energy of motion) which causes the particles to constantly move. The amount of kinetic energy is proportional to the temperature of the substance; therefore, the particles have greater motion at higher temperatures (straight line motion) and lesser motion at lower temperatures (vibrational motion). Collisions between particles are elastic, which means no loss in kinetic energy. KMT can explain phase changes ○ The faster the particles move, the less they can be contained. When substances are heated, the KMT and energy increases, which can change the state of matter. If a substance cools, KMT reduces and therefore it could turn into a solid. Kinetic energy is the energy of motion. ○ Dependent on temperature and measured in Joules. Temperature is proportional to the average amount of kinetic energy a substance has. ○ Independent of mass and measured in degrees. ○ The amount of KE each particle has in a given sample will vary, therefore temperature is the average KE Heat is the amount of thermal energy transferred from one object to another. ○ Dependent on mass and measured in Joules Inter-particle force Inter-particle force refers to the forces of attraction or repulsion between particles (atoms, molecules, or ions) in a substance. These forces determine the physical properties of a material, such as its phase (solid, liquid, or gas), melting and boiling points, and solubility. Types of Interparticle Forces: Van der Waals forces (weakest): Attraction between molecules due to temporary dipoles. ○ Example: Between oxygen molecules (O₂). Dipole-dipole interactions: Attraction between polar molecules with permanent dipoles. ○ Example: Between hydrogen chloride (HCl) molecules. Hydrogen bonding: Strong dipole-dipole interaction involving hydrogen bonded to nitrogen, oxygen, or fluorine. ○ Example: Between water molecules (H₂O). Ionic bonds (strongest): Attraction between positively and negatively charged ions. ○ Example: Between sodium (Na⁺) and chloride (Cl⁻) ions in NaCl. Solid, liquid, gas, aqueous states within reactions 1. Solid (s): A substance in its solid state has a definite shape and volume, with its particles closely packed in a fixed arrangement. Solids are rigid and do not flow. In chemical equations, the solid state is represented by (s). ○ Example: NaCl(s) → Sodium chloride in solid form. 2. Liquid (l): A substance in its liquid state has a definite volume but takes the shape of its container. The particles in a liquid are close together but can move past each other, allowing liquids to flow. Liquids are represented by (l) in reactions. ○ Example: H₂O(l) → Water in its liquid form. 3. Gas (g): In the gaseous state, a substance has neither a definite shape nor volume. Gas particles are far apart and move freely, expanding to fill any container. Gases are denoted by (g) in chemical reactions. ○ Example: O₂(g) → Oxygen gas. 4. Aqueous (aq): An aqueous solution is a substance dissolved in water. When a substance is in its aqueous state, it is dissolved in water, forming a homogeneous mixture. The (aq) symbol is used to represent aqueous solutions in reactions. ○ Example: NaCl(aq) → Sodium chloride dissolved in water. Phase change graphs *understand how to read this Names of changes of state Endothermic: A reaction or process that absorbs heat from its surroundings, resulting in a decrease in temperature around it. Example: Melting ice absorbs heat to transition from solid to liquid. Exothermic: A reaction or process that releases heat into its surroundings, causing an increase in temperature. Example: Combustion of wood releases heat and light as energy. Convert between values of Celsius and Kelvin temperature scales S1.2.1 Structure of the atom Looking at the periodic table, does the amount of IE follow an overall trend for elements in the same group or period? Are there any exceptions? First IE values decrease going down a group. First IE values increase overall going across a period. Exception 1: Be → B ○ Losing the single e- in 2p (B) requires less energy to remove than a paired e- in 2s (Be) Exception 2: N → O ○ Losing the paired e- in 2p (O) requires less energy to remove than the half-filled 2p sublevel (N) Nuclear Models of the Atom DEMOCRITUS: Solid indivisible particles Democritus and Leucippus Matter is made up of solid indivisible particles DALTON: Billiard balls Dalton’s model (1808) John Dalton - one type of atom for each element THOMSON: Plum pudding Thomson’s model (1904) Discovered electrons Atoms were made of positive stuff Negative electron floating around “Plum-Pudding” model RUTHERFORD/BOHR: Planetary system Rutherford’s model (1911) Discovered dense positive piece at the center of the atom- nucleus Electrons would surround it Mostly empty space “Nuclear model” Bohr’s model (1913) The electrons must move like planets around the sun, in circular orbits at different levels. Different amounts of energy separate one level from another. This became the “Planetary model” Electric repulsion from each other and attraction means electrons cannot leave the orbit but do not stick to the nucleus.. TODAY'S MODEL: Best described mathematically, attributing both wave-like properties and particle-like properties to electrons Quantum Mechanical Model (Present) Relative masses and charges of subatomic particles. Use nuclear symbol notation to deduce the number of protons, neutrons, electrons in atoms and ions. S1.2.2 Isotopes  Perform calculations involving relative atomic masses and abundance of isotopes from given data (solve for average atomic mass or isotopic abundance) Relative abundance: how much actually exists in nature. Just like your mark is a “weighted” average in this course, the relative atomic mass is also a weighted average of the isotopes present. This is known as the relative atomic mass, Ar. Consider chlorine: SAMPLE PROBLEM Cl-35 is 75.77% Cl-37 is 24.23% (as you can see, they add up to 100%) Calculate the relative atomic mass of Cl. AVERAGE ATOMIC MASS FORMULA Average atomic mass = % abund. iso 1 (mass of iso 1) + % abund. iso 2 (mass of iso 2) + …. AAMCl = (% of Cl-35 × mass of Cl-35) + (% of Cl-37×mass of Cl-37) AAMCl = (0.7577 x 35) + (0.2433 x 37) AAMCl = 26.5195+8.9651 AAMCl = 35.4846 u Where have you seen this value before? (atomic mass of chlorine on the periodic table) 1. Silver has two naturally occurring isotopes; 107Ag and 109Ag. The relative atomic mass of silver is 107.87. Determine the percent abundance of each isotope. Round your answer to two decimal places. SAMPLE PROBLEM Before solving anything, define your variables: Let x = Ag-107 Let 1-x = y = Ag-109 Step 1: set up what you know into the average atomic mass equation x = % of 107Ag y = % of 109Ag Step 2: set up a 2nd equation using the knowledge that all the % abundances should add up to 1 (100%) x+y=1 y=1-x Step 3: solve using substitution or elimination 107.87=(x × 107)+[y × 109] 107.87 = (x×107)+[(1−x)×109] 107.87=107x + 109(1−x) 107.87=107x + 109 −109x 107.87=109 − 2x 2x=109 − 107.87 2x=1.13 x = 1.13/2 x = 0.565 (56.5%) Therefore, y = 1-x y = 0.435 (43.5%) Step 4: Write your statement Therefore, the percent abundance of Ag-107 is 56.5% and the percent abundance of Ag-109 is 43.5% S1.2.3 Mass Spectra Understand how a Mass spectrometry machine works and is able to detect isotopes of an element A mass spectrometer is used to determine the relative atomic masses of elements using their isotopic composition. A mass spectrometer works by injecting a sample of an element: vaporizing the sample to separate the atoms ionizing the atoms to have a positive charge accelerating the ions to sort them by mass detecting the mass of the charged particles The ions produced are sorted and detected by their mass/charge (m/z) ratio. How does this spectrum relate to the abundance of each H isotope? The relative height of each peak corresponds to the abundance of each isotope in the sample tested, based on its m/z ratio. The ratio of each peak, relative to the total, can be used to calculate the percentage abundance of that isotope. Ex: 1H is [47.6 / (47.6+0.50+1.67)] The % abundance of 1H is 95.64% in this particular sample. Perform calculations involving relative atomic masses and abundance of isotopes from given data (solve for average atomic mass or isotopic abundance) POSSIBILITY 1: Given the abundance and mass/charge ratio, determine the percentage abundance of each isotope for an element, possible combinations of isotopes, and the mass of each molecule (and in what proportions?) Step 1: Calculate the percentage abundance of each isotope Relative abundance/sum of all relative abundances = percentage abundance Step 2: Determine the possible combinations of isotopes in the element Usually: a-a: mass a + mass a = corresponding mass a-b: mass a + mass b = corresponding mass b-b: mass b + mass b = corresponding mass Step 3: Proportions of each​molecule For each possible combination Percent abundance x percent abundance = relative proportion of combination Final result: mass of a-a: corresponding mass u(relative proportion of combination percentage %) mass of a-b mass of b-b EXAMPLE 1: STATION 1: Determine the percentage abundance of each isotope for the sample of Chlorine shown in the mass spectra. If 2 chlorine atoms bond to form a molecule of Cl2. What possible combinations of isotopes can bond together and what would the mass of each molecule be (and in what proportions!)? Step 1: Calculate the percentage abundance of each isotope The relative abundance is already given: Chlorine-35 (35Cl): 100.00% Chlorine-37 (37Cl): 31.96% Total relative abundance=100+31.96=131.96 Percentage abundance 35Cl 100/131.96 x 100 = 75.79% Percentage abundance 37Cl 31.96/131.96 x 100 = 24.21% Step 2: Determine the possible combinations of isotopes in Cl2​ When two chlorine atoms bond to form Cl2​, the possible combinations of isotopes and their corresponding masses would be: 35 1. Cl−35Cl ○ Mass = 34.969+34.969=69.938 35 37 2. Cl− Cl ○ Mass = 34.969+36.966=71.935 37 3. Cl−37Cl ○ Mass = 36.966+36.966=73.932 Step 3: Proportions of each Cl2​molecule Using the percentage abundances of the isotopes, we can determine the relative proportion of each Cl2\text{Cl}_2Cl2​molecule: 35 1. Cl−35Cl Probability = 0.7579×0.7579=0.574 (or 57.4%) 35 2. Cl−37Cl (which is the same as 37Cl−35Cl) Probability = 2×(0.7579×0.2421)=0.367 (or 36.7%) 37 3. Cl−37Cl Probability = 0.2421×0.2421=0.0586 (or 5.86%) Final result: Mass of 35Cl−35Cl: 69.938 u (57.4%) Mass of 35Cl−37Cl: 71.935 u (36.7%) Mass of 37Cl−37Cl : 73.932 u (5.86%) POSSIBILITY 2: Given a graph without percentage abundance but mass/charge ratio, determine the percentage abundance and relative atomic mass (AAM), and identify the unknown element using the periodic table. Step 1: Measure the height of each bar for each isotope, then add them all and divide the individual heights for percent abundance. a + b + c = total Isotope 1 = a/total = percent abundance (in decimals) Step 2: use AAM to calculate the relative atomic mass and identify with a periodic table AAM = #1(percent abundance (in decimals) x mass/charge ratio) + #2 (percent abundance (in decimals) x mass/charge ratio) + … EXAMPLE 2: STATION 4: Use a ruler to measure the height of each peak. Determine the percentage abundance of each isotope and the relative atomic mass. Identify the unknown element using the Periodic Table. 4cm + 0.3cm + 0.2cm = 4.5cm X-28: 4/4.5 = 0.889 X-29: 0.3 / 4.5 = 0.067 X-30: 0.2 /4.5 = 0.044 AAM = (0.889 x 28) + (0.067 x 29) + (0.044 x 30) AAM = 28.1555 Looking at the periodic table, the element that has the closest atomic mass is Silicon POSSIBILITY 3: Given a couple of spectrums, determine which spectra best represents a certain element. Step 1: Using the periodic table, identify which isotope is predominant, and select the mass spectra that shows the highest bar for that isotope. EXAMPLE 3: STATION 5 Determine which spectra best represents atoms of iodine. Justify your answer. Looking at the Periodic Table, Iodine has an atomic mass of 126.90 meaning that it is closer to 127 than it is 126 which suggests that there is a higher abundance of I-127 which spectrum C illustrates. POSSIBILITY 4: Use a stability chart to determine which isotopes are stabler. EXAMPLE 4: STATION 6: Some isotopes are stable and others aren’t. Using the data in the table, predict the stability of the following isotopes: Carbon-13 vs carbon-12 Helium-4 vs helium-3 Iodine-127 and Iodine-124 C-12 6 Even 6 Even Super Stable  C-13 6 Even 7 Odd Stable He-3 2 Even 1 Odd Stable He-4 2 Even 2 Even Super Stable I-124 53 Odd 71 Odd Not Stable I-127 53 Odd 74 Even Stable POSSIBILITY 5: Analyze mass spectra diagrams to draw conclusions/explanations. EXAMPLE 5: STATION 7: Using the periodic table, state the Ar for iodine and tellurium. What do you notice about these masses and the placement of each element on the periodic table? In nature, 127I is the most abundant isotope. Describe how the mass spectrum of Iodine would differ from Tellurium. Iodine: 126.90 u Tellurium: 127.60 u Iodine (atomic number 53) and tellurium (atomic number 52) are neighboring elements in the periodic table. However, despite iodine having a higher atomic number, its atomic mass is slightly lower than that of tellurium. This is an unusual observation since atomic mass generally increases with atomic number. The reason for this anomaly lies in the isotopic composition of each element. Tellurium has multiple isotopes of varying masses (as seen in the mass spectrum), while iodine primarily exists as a single isotope, 127I. As a result, the average atomic mass of tellurium is slightly higher than iodine. Tellurium: As shown in the mass spectrum provided, tellurium has multiple isotopes with varying abundances, resulting in multiple peaks at different mass numbers. The most prominent peaks correspond to 128Te and 130Te, which are the most abundant isotopes. Iodine: In contrast, iodine in nature consists almost entirely of the isotope 127I. Therefore, the mass spectrum of iodine would be significantly simpler than that of tellurium. It would show: A single large peak at mass number 127, representing 127I. There would be no other peaks unless extremely rare isotopes or ion fragments were present. S1.3.1 Emission and absorption Describe relationship between color, wavelength, frequency, and energy across the electromagnetic spectrum Electromagnetic Spectrum ○ Light, as a form of electromagnetic radiation, exhibits wave-particle duality, meaning it has both wave-like (energy and wavelength) and particle-like (mass, photon) properties. The electromagnetic spectrum includes a broad range of wavelengths, frequencies, and energies, from radio waves with long wavelengths to gamma rays with very short wavelengths. The visible light spectrum is just a small part of this vast range. Wavelength, Frequency, and Energy Relationship: ○ Wavelength (λ): refers to the distance between two consecutive peaks of a light wave and is measured in units such as meters, nanometers (nm), or micrometers (𝛍m). ○ Frequency (f):s the number of waves that pass a given point per second, measured in (Hz). ○ Energy (E): The energy of a light wave is directly proportional to its frequency and inversely proportional to its wavelength. This means that shorter wavelengths (like ultraviolet light) have higher frequencies and more energy, while longer wavelengths (like infrared light) have lower frequencies and less energy. Wavelength and frequency are inversely related: shorter wavelengths mean higher frequency. Frequency and energy are directly related: higher frequency means more energy. Wavelength and energy are inversely related: shorter wavelengths mean more energy. Wave-Particle Duality and Photon Energy ○ Light behaves both as a wave and a particle, meaning it can be described as energy traveling in wave form or as discrete particles called photons. ○ A photon carries a specific amount of energy, calculated using the equation E=hf, where E is energy, h is Planck’s constant (6.626 x 10⁻³⁴ J·s), and f is the frequency of the light.  Color and Wavelength: ○ Visible light, the part of the electromagnetic spectrum that we can perceive, consists of different colors that correspond to specific wavelengths. For instance, red light has a longer wavelength (~700 nm) and lower energy, while blue light has a shorter wavelength (~450 nm) and higher energy. Different colors of light are produced when light interacts with matter (e.g., through absorption or emission), which allows us to identify elements and compounds. Continuous Spectrum and Line Spectrum: ○ Continuous Spectrum: When white light, which contains all visible wavelengths, is passed through a prism, it separates into all the colors of the visible spectrum. This continuous range of colors shows that white light contains all the different wavelengths. ○ Line Spectrum (Emission Spectrum): When a sample of gas (like hydrogen) absorbs energy (such as from an external light source), the electrons in the atoms can become excited, jumping to higher energy levels. As these electrons return to their lower, ground states, they emit photons at specific wavelengths. This produces a line spectrum with distinct colors, each corresponding to a specific energy transition. These emission lines are unique to each element, making them useful in identifying the composition of the gas, similar to a fingerprint. Atomic Structure and Spectra: Each atom has distinct energy levels. When light (energy) is applied to a gas sample, electrons absorb energy and "jump" to higher energy levels, a process called excitation. However, electrons eventually return to their ground state, releasing the absorbed energy in the form of photons at specific wavelengths. The emitted colors form an emission spectrum that is characteristic of the atom or molecule. In summary: The visible light we see is just a small part of the electromagnetic spectrum, and light's interactions with atoms (like excitation and emission) allow us to use light as a tool to study atomic structure. The unique spectral lines produced by different elements provide insight into their electronic structure and behavior. Distinguish between a continuous and a line spectrum ○ Continuous Spectrum: When white light, which contains all visible wavelengths, is passed through a prism, it separates into all the colors of the visible spectrum. This continuous range of colors shows that white light contains all the different wavelengths. ○ Line Spectrum (Emission Spectrum): When a sample of gas (like hydrogen) absorbs energy (such as from an external light source), the electrons in the atoms can become excited, jumping to higher energy levels. As these electrons return to their lower, ground states, they emit photons at specific wavelengths. This produces a line spectrum with distinct colors, each corresponding to a specific energy transition. These emission lines are unique to each element, making them useful in identifying the composition of the gas, similar to a fingerprint. Don’t have to know details about EM spectrum: given in data booklet. S1.3.2 Hydrogen emission spectrum  Describe the emission spectrum of the hydrogen atom, including the relationships between the lines and energy transitions to the first, second and third energy levels. The emission spectrum of hydrogen is a series of lines that appear when the hydrogen atom’s single electron transitions between different energy levels. These transitions release photons with specific energies, which correspond to particular wavelengths of light. Since hydrogen only has one electron, the energy level transitions produce distinct, well-defined spectral lines. The energy of the 4 visible lines correspond to transitions ○ 6→2 ○ 5→2 ○ 4→2 ○ 3→2 All other transitions occur outside of the visible range Transitions back to the 1st energy level are in the UV range (more energy than visible light) Transitions back to the 3rd energy level are in the IR range (less energy than visible light) At lower energy levels, there are greater gaps/energy differences. At higher energy levels, the difference in energy is less. This is known as “convergence” S1.3.3-4 Main energy levels and sublevels Deduce the maximum number of electrons that can occupy each energy level. The number of electrons that each energy level can accommodate follows a pattern: 2n^2 Recognize the shape and orientation of an s atomic orbital and the three p atomic orbitals.  s orbitals are spherical in shape and can hold 2 e- p orbitals have a dumbbell shape - there are 3 for each coordinate (x, y and z) - for a total of 2 e- x 3 orbitals S1.3.5 Electron configuration Apply the Aufbau principle, Hund’s rule and the Pauli exclusion principle to deduce electron configurations for atoms and ions up to Z = 36. Aufbau principle: electrons are added from low to high energy example: You cannot place an electron in a 3s before filling up the 2p sublevel This relates to what you learned previously about filling in shells from the inside out Hund’s Rule: an electron must be placed in each orbital of a sublevel before electrons can be paired. Same idea as when drawing Bohr Rutherford Diagrams. example: you cannot have 2 electrons paired and an empty orbital in the same sublevel This relates to what you learned previously about placing electrons in the 4 corners before pairing them Pauli exclusion principle: electrons in the same orbital must have opposite spin, maximum of 2 electrons per orbital example: you cannot place 2 electrons with the same spin in the same orbital OR have 3 electrons in one orbital This explains why the electrons do not repel each other in the same orbital. Opposite spin allows for electrons to pair Full electron configurations and condensed electron configurations using the noble gas core Another way to communicate electron arrangement is by using electron configuration notation: For Ca: 1s22s22p63s23p64s2 Short form notation, Ca: [Ar] 4s2 Each electron is expressed using the format: Short form notation is sometimes used for longer configurations by including the previous noble gas in [ ] and then adding the rest. Orbital diagrams, i.e. arrow-in-box diagrams, should be used to represent the filling and relative energy of orbitals. EX: Can you map out the 20 electrons in Calcium? Each sublevel has a different amount of energy. Only the electrons in s and p orbitals have the right amount of energy to be lost as valence electrons Arrows are representing the number of electrons You can use lines or boxes  Using arrows to represent electrons, we can fill up the orbitals with 20 electrons The electron configurations of Cr and Cu as exceptions Chromium (Z=24) expected: [Ar] 4s23d4 observed: [Ar] 4s13d5 Copper (Z=29) expected: [Ar] 4s23d9 observed: [Ar] 4s13d10 S1.3.6 Calculations for Light, IE  In an emission spectrum, the limit of convergence at a higher frequency corresponds to the first ionization energy If you have a graph or chart be able to understand Ionization is said to have occurred when the electron is excited to the limit of convergence (n=∞) on an emission spectrum ○ For Hydrogen, the electron would transition from the ground state (n=1) to n=∞ Ionization Energy: The strength of attraction between the outermost electron and the nucleus will determine the amount of ionization energy. Atoms want to get rid of electrons to be stable. Electrons in higher energy levels are weakly attracted to the nucleus, therefore lower IE. The strength of attraction between the outermost electron and the nucleus will determine the amount of ionization energy. More protons in the nucleus mean a greater attraction, therefore greater IE. If enough energy is given to the electron, it will jump so high, it leaves the attraction of the nucleus at the limit of convergence One or more electrons can be removed when enough energy is provided for this transition - ionization energy We can calculate the amount of energy required for the first ionization energy from spectral data The greatest energy transition, measured by wavelength/frequency, is used to determine the IE.  Trends in first ionization energy across periods account for the existence of main energy levels and sublevels in atoms Looking at the periodic table, does the amount of IE follow an overall trend for elements in the same group or period? Are there any exceptions? First IE values decrease going down a group. First IE values increase overall going across a period. Exception 1: Be → B ○ Losing the single e- in 2p (B) requires less energy to remove than a paired e- in 2s (Be) Exception 2: N → O ○ Losing the paired e- in 2p (O) requires less energy to remove than the half-filled 2p sublevel (N) Calculating Ionization Energy ○ E = hf ○ c = fλ Using these 2 equations, we can determine the minimum amount of energy required to remove an electron from an atom This will calculate the energy of the convergence limit - the point at which the energy levels converge and the electron is no longer attracted to the nucleus ○ Convergence - the pathway a specific electron is taking S1.3.7 Successive IE’s Successive ionization energy data for an element give information that shows relations to electron configurations EXAMPLE:  Deduce the group of an element from its successive energy data Example: A graph of IE vs number of electrons removed can give information about the group an element is in, by indicating the likely number of valence electrons You can kind of tell this is argon because the jumps are in between the shells, with 2, 8, and 8 that remain steady Explanations of the trends and discontinuities in data on first ionization energy across a period Looking at the periodic table, does the amount of IE follow an overall trend for elements in the same group or period? Are there any exceptions? First IE values decrease going down a group. First IE values increase overall going across a period. Exception 1: Be → B ○ Losing the single e- in 2p (B) requires less energy to remove than a paired e- in 2s (Be) Exception 2: N → O ○ Losing the paired e- in 2p (O) requires less energy to remove than the half-filled 2p sublevel (N)