Periodic Table: Organization, Blocks & Ionization

Periodic Table Organization

• Elements are arranged by increasing atomic number.
• Periods are the rows, and groups are the columns.
• Elements in the same group share similar physical and chemical properties due to having the same number of outer shell electrons.
• Periodicity is the repeating patterns of atomic, physical, and chemical properties that occur with increasing atomic number.

Blocks of the Periodic Table

• The periodic table is divided into s, p, d, and f blocks.
• Block designation depends on the orbital where the highest energy electron is located.
• S-block elements have outer shell electrons in the s subshell.
• D-block elements have outer shell electrons in the d subshell.
• P-block elements have outer shell electrons in the p subshell.

Ionization Energy

• Ionization energy measures the energy needed to remove electrons from atoms, creating positive ions.
• The first ionization energy is the energy to remove one electron from each atom in one mole of a gaseous element to form one mole of gaseous 1+ ions.
• Successive ionization energies refer to the removal of electrons after the first.
• Successive ionization energies increase, with larger 'jumps' between shells, providing evidence for electron shells.
• These jumps indicate when an electron is removed from a lower energy shell closer to the nucleus.
• For example, the jump between the third and fourth ionization energies of aluminum is large, indicating the start of electron removal from a lower energy shell.

Factors Affecting Ionization Energies (MASARA)

• Shielding (S): Lower number of shells, means less shielding, leading to greater nuclear attraction.
• Charge (C): Greater nuclear charge (more protons), means greater nuclear attraction.
• Atomic Radius (A): Smaller radius leads to a greater nuclear attraction because outer shell electrons are closer to the nucleus.
• Attraction (A): Greater nuclear attraction requires more energy to remove an outer shell electron.

Periodic Trends in Ionization Energies

• Across Periods: Shielding remains the same, nuclear charge increases, atomic radius decreases leading to increased ionization energy.
• Down Groups: Shielding and atomic radius increase, outweighing the increase in nuclear charge, leading to decreased ionization energy.
• Anomalies occur from group 2 to group 3 (ex: Beryllium to Boron) because electrons are being removed from the p subshell which is higher energy and further from the nucleus than the s subshell.
• Anomalies also occur from group 5 to group 6 (ex: Nitrogen to Oxygen) electrons pair in the p subshell, making them easier to remove due to repulsion.
• Orbitals fill singly before pairing.

Periodic Trends in Bonding and Structure

• Across period 2 and 3 melting point increases from group 1 to group 4 due to giant metallic and covalent structures.
• Melting points decrease rapidly from group 4 to 5, due to simple molecular structures with weak intermolecular forces.

Group 2: Alkaline Earth Metals

• Group 2 elements have two electrons in their outer shell (s subshell).
• Ionization energy decreases down the group due to increased atomic radius and shielding, outweighing increased nuclear charge.
• Melting points decrease down the group due to weakened metallic bonding.
• React with water in a redox reaction to form hydroxides and hydrogen.
• Solubility of group 2 hydroxides increases down the group, leading to higher pH (more alkaline).
• Magnesium hydroxide is used to neutralize stomach acid (antacid).
• Calcium hydroxide is used to neutralize acidic soils.

Sulfate Ion Test

• Acidified barium nitrate (Ba(NO3)2) is used to test for sulfate ions (SO42-).
• Barium sulfate (BaSO4), a white precipitate, forms.
• The solution is acidified to remove carbonate ions (CO32-) that would form a similar white precipitate.

Halogens

• Electronegativity decreases down the group due to increased shielding and atomic radius.
• Boiling point increases down the group due to stronger London forces caused by more electrons.
• Halogens have seven outer shell electrons, needing one more to achieve an octet (8 electrons).
• Reactivity decreases down the group due to decreased nuclear attraction, making them weaker oxidizing agents.

Halogen Displacement Reactions

• A halogen can displace a halide ion from a solution.
• Chlorine reacts with bromide ions to form chloride ions and bromine.

Halide Ion Tests

• Silver ions (Ag+) are used to test for halide ions (Cl-, Br-, I-).
• Silver halides form precipitates with different colors:
  • Silver chloride (AgCl): white precipitate
  • Silver bromide (AgBr): cream precipitate
  • Silver iodide (AgI): yellow precipitate
• Solubility in ammonia differentiates the precipitates:
  • AgCl: soluble in dilute ammonia
  • AgBr: soluble in concentrated ammonia
  • AgI: insoluble in concentrated ammonia

Uses of Chlorine

• Chlorine dissolves in water to form hydrochloric acid (HCl) and chloric(I) acid (HClO).
• This is a disproportionation reaction where chlorine is both oxidized and reduced.
• HClO is an oxidizing agent that disinfects water.
• Chlorine reacts with cold dilute sodium hydroxide (NaOH) to form bleach (sodium chlorate(I), NaClO).
• This is also a disproportionation reaction.

Anion Testing Order

• Carbonate ions (CO32-) are tested first.
• Sulfate ions (SO42-) are tested second.
• Halide ions (X-) are tested last.

Carbonate Ion Test

• Add dilute strong acid (e.g., nitric acid) to the sample.
• Carbon dioxide gas bubbles (effervescence) form.
• Bubble the CO2 gas through limewater, which will turn cloudy due to calcium carbonate precipitate.

Sulfate Ion Test

• Add barium ions (Ba2+), typically from barium nitrate.
• White precipitate of barium sulfate (BaSO4) forms.

Halide Ion Test

• Add silver ions (Ag+), typically from silver nitrate.
• A precipitate of silver halide (AgX) forms.

Cation Testing (Ammonium Ions)

• Add aqueous sodium hydroxide (NaOH) and heat gently.
• Ammonia gas (NH3) is released.
• Test the gas with damp litmus paper, which will turn blue due to the basic ammonia.

Electron Configuration

• Electron configuration of a bromide ion (Br-) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6.

Enthalpy

• Enthalpy (H) is the thermal energy stored in a system.
• Enthalpy change is measured as heat energy change under standard conditions.
  • One atmosphere pressure
  • 298 Kelvin (25 degrees Celsius) temperature
  • One mole per decimeter cubed concentration
• Standard states are the state of a substance under standard conditions.
• Enthalpy change of reaction defined as when a reaction takes place in the molar quantity shown in a chemical equation.
• Enthalpy change of formation defined as when one mole of a compound is formed from it's constituent elements.
• Enthalpy change of combustion defined as when one mole of a substance is completely burnt in oxygen.
• Enthalpy change of neutralization defined as when an acid reacts with a base to form one mole of H2O.

Exothermic and Endothermic Reactions

• Exothermic reactions release energy, resulting in a negative enthalpy change.
• Endothermic reactions absorb energy, resulting in a positive enthalpy change.
• Activation energy is the minimum energy required for a reaction to occur (always positive).
• Catalysts lower activation energy.
• Enthalpy change of reaction is calculated by subtracting the sum of bond enthalpies of reactants from the products.
• Bond breaking is endothermic, and bond making is exothermic.

Bond Enthalpies

• Bond enthalpies are always positive because they are endothermic, requiring energy from the surroundings.

Calorimetry

• Calorimetry measures the amount of heat given off or taken in during a chemical reaction.
• The formula Q = mcΔT is used to find the heat change in a reaction, which can then be used to find the enthalpy change.
  • Q represents the heat change in joules.
  • m represents the mass of the substance in grams.
  • c is the specific heat capacity (4.18 J/g°C for water).
  • ΔT is the temperature change in degrees Celsius or Kelvin.

Example Calculation: Magnesium Reacts with Aqueous Silver Nitrate

• Magnesium reacts with silver nitrate according to the equation: Mg + 2AgNO3 → 2Ag + Mg(NO3)2.
• To determine the enthalpy change of this reaction, a student adds 25 cm³ of 0.512 mol/dm³ silver nitrate to magnesium powder.
• The initial temperature is 19.5°C, and the maximum temperature reached is 47.5°C, indicating an exothermic reaction.
• Calculate the enthalpy change in kilojoules per mole for the reaction.

Calculation Steps

• Calculate the energy released in the reaction using Q = mcΔT.
  • Volume of silver nitrate = 25 cm³ which is equivalent to 25g
  • m = 25 grams (assuming density of solution is the same as water).
  • c = 4.18 J/g°C (specific heat capacity of water).
  • ΔT = 47.5°C - 19.5°C = 28°C.
  • Q = 25g * 4.18 J/g°C * 28°C = 2926 joules.
• Convert joules to kilojoules: 2926 J / 1000 = 2.926 kJ.
• Calculate the moles of silver nitrate using n = CV:
  • n = (25 / 1000 dm³) * 0.512 mol/dm³ = 0.0128 moles.
• Divide the kilojoules by the moles to get the enthalpy change value: 2.926 kJ / 0.0128 mol = 228.5 kJ/mol.
• Since the equation involves two moles of silver nitrate, multiply by 2: 228.5 kJ/mol * 2 = 457.1 kJ/mol.
• Express the answer to three significant figures and include a negative sign for the exothermic reaction: -457 kJ/mol.

Tips for Calculations

• Carefully read the questions and decipher the information provided, looking for clues.
• Always check the molar quantities in the equation and make sure to define the type of enthalpy change.
• Ensure the correct sign is used: negative for exothermic (temperature increase) and positive for endothermic (temperature decrease reactions).

Hess's Law

• Hess's Law states that the enthalpy change for a chemical reaction is independent of the route taken.
• If a reaction can occur by two routes, the enthalpy change will be the same for each route (Root 1 = Root 2).

Drawing Hess Cycles

• Use the data given to determine which type of cycle to draw (combustion or formation).
• Combustion Cycles: Arrows point downwards, with carbon dioxide and H2O at the bottom of the cycle. Enthalpy changes of combustion values are used.
• Formation Cycles: Arrows point upwards, with elements at the bottom. Enthalpy changes of formation values are used.

Calculation Example: Enthalpy Changes of Formation

• Given enthalpy changes of combustion for carbon, hydrogen, and nonane, calculate the standard enthalpy change of formation of nonane.

Steps in calculation

• Given C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
• Given H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
• Given C9H20(l) + 14O2 -> 9CO2(g) + 10H2O(l) ΔH = -6125 kJ/mol
• Given Carbon and Hydrogen -> C9H20(l) -> Nonane combustion products
• The equation for the formation of nonane is: 9C(s) + 10H2(g) → C9H20(l).
• Draw a combustion cycle with downward arrows.
• Multiply the enthalpy change of combustion of carbon by 9: -394 kJ/mol * 9 = -3546 kJ.
• Multiply the enthalpy change of combustion of hydrogen by 10: -286 kJ/mol * 10 = -2860 kJ.
• Adding the values for carbon and hydrogen gives -6406.
• The enthalpy change of combustion of nonane is -6125 kJ/mol.
• To find the enthalpy change of formation, change the sign of the nonane combustion value and add it to the value of the combustion of carbon and hydrogen.
• −6406 + 6125 = −281 kJ/mol.

Calculation Example: Formation Values

• Reaction: 4NH3 + 5O2 → 4NO + 6H2O ΔH = −908 kJ/mol
• Given the enthalpy changes of formation for ammonia and water, calculate the standard enthalpy change of formation for NO.

Steps in calculation

• Given the enthalpy change of formation of ammonia NH3(g) ΔH = -46 kj/mol
• Given the enthalpy change of formation of water H2O(l) ΔH = -242 kj/mol
• Draw a cycle with upward arrows.
• Multiply the enthalpy change of formation of ammonia by 4: -46 kJ/mol * 4 = -184 kJ.
• Multiply the enthalpy change of formation of water by 6: -242 kJ/mol * 6 = -1452 kJ.
• Write out the equation to find the value of the N3 change of formation of nitrous oxide
• Find the value of the arrow to complete the cycle by using −908 = −1452 + 4X + 184 where X is the unknown ΔH.
• Re-arrange the equation to find the value of 4X by simplifying the numbers −908 = − 1268 + 4X
• Find the value of X 360 = 4X
• Therefore X = 90
• Simplify 360 / 4 = 90

Collision Theory

• For a reaction to occur, particles must collide with sufficient energy to overcome the activation energy (Ea).
• Particles must collide with the correct orientation for a reaction to take place.
• Example: Ammonia (NH3) reacting with H+ ions requires the lone pair on the nitrogen atom to align with the H+ ion.

Boltzmann Distribution

• Represents the energy of molecules in a reaction system.
• Y-axis: Number of molecules.
• X-axis: Energy of molecules.
• The area under the curve represents the total number of molecules.
• The area to the right of the activation energy (Ea) line represents molecules that can react if they collide with the correct orientation.
• The distribution starts at the origin (0,0) because zero molecules have zero energy.
• The curve never touches the x-axis after the origin, as molecules cannot have zero energy.
• Higher temperature: a larger proportion of molecules have energy higher than the activation energy, increasing the likelihood of reaction upon correct collision.
• Increased concentration: there are more molecules with energy higher than the activation energy, increasing the likelihood of reaction upon correct collision.
• Catalyst: the activation energy shifts to the left, meaning a larger proportion of reactant molecules have the required activation energy.

Rate of Reaction

• Rate = Change in concentration / Time.
• Units for rate: mol dm-3 s-1.
• Catalysts decrease the activation energy by providing an alternative reaction pathway without being used up.
• Homogeneous catalysts are in the same physical state as reactants, reacting to form an intermediate and are regenerated (e.g., phosphoric acid in the hydration of alkenes).
• Heterogeneous catalysts are in a different physical state from the reactants, usually solid, and reactants adsorb (weakly bond) to the surface, then desorb (e.g., platinum in catalytic converters).
• Benefits of catalysts in industry: lower demand for fossil fuels due to reduced energy use, leading to lower costs.

Continuous Monitoring

• Involves recording the amount of a reactant or product at regular time intervals to calculate the rate of reaction.
• Gas collection: measures the volume of gas formed using a gas syringe.
• Mass loss: measures the decrease in mass as gas evolves using a mass balance.
• Colorimetry: monitors concentration changes by measuring the absorbance of light through a sample; higher concentration results in higher absorbance.
• Rate graphs: tangents are used to calculate the rate of a curved graph.
• The gradient of the graph (change in y / change in x) equals the rate (change in concentration / time).

Chemical Equilibria

• Reversible reactions are represented by the equilibrium symbol.
• Dynamic equilibrium conditions: constant concentrations of reactants and products, the rate of the forward reaction equals the rate of the reverse reaction, and a closed system.
• Le Chatelier's principle: if reaction conditions of a dynamic equilibrium change, the equilibrium will shift to favor either the forward or reverse reaction to oppose the change.
• Effect of temperature: depends on whether the forward reaction is exothermic (releases energy) or endothermic (absorbs energy).
• Exothermic reactions increase the temperature of their surroundings, while endothermic reactions decrease it.
• Increasing the temperature of an exothermic reaction shifts the equilibrium in the endothermic (reverse) direction.
• Decreasing the temperature shifts the equilibrium in the exothermic (forward) direction.
• Increasing reactant concentration shifts the equilibrium towards the products.
• Increasing product concentration shifts the equilibrium towards the reactants.
• Pressure (only applies to gases): increasing pressure shifts the equilibrium to the side with the fewest gaseous moles.
• Decreasing pressure shifts the equilibrium to the side with more gaseous moles.
• Catalysts do not affect the position of equilibrium but increase the rate of reaction in both directions, establishing equilibrium faster.

Equilibrium Constant (Kc)

• Numerically indicates the position of equilibrium, representing the ratio of products to reactants.
• In Kc expressions, concentrations of reactants and products are raised to the power of their stoichiometric coefficients (moles).
• Units of Kc vary depending on the reaction's stoichiometry and the number of species involved.
• If Kc=1, equilibrium lies exactly halfway between reactants and products.
• If Kc

Learn about the periodic table's organization by atomic number, periods, and groups. Explore the s, p, d, and f blocks based on electron configuration. Understand ionization energy and its trends.