Atoms Nuclear Decay

Atoms, nuclear decay, electronic structure, and atomic chemical behavior Atomic Nucleus (PHY, GC) • Atomic number, atomic weight o Atomic number = number of protons ▪ Defines an element ▪ When two things have same number of protons, they are the same element o Atomic weight = weighted average of atomic mass for all isotopes of a given atom ▪ Atomic mass = number of protons + neutrons ▪ Atomic mass used for an isotope ▪ Atomic weight used for an element • Neutrons, protons, isotopes o Neutrons = neutral particles in nucleus o Protons = positive particles in nucleus o Isotopes = things with same number of protons, but different number of neutrons o Nucleons = protons or neutrons • Nuclear forces, binding energy o Two forces are at work in the nucleus: strong force and electromagnetic force o Strong force binds nucleons together, and therefore contribute to binding energy o Electromagnetic force due to electrostatic repulsion between the positively charged protons in the nucleus o Nucleus stays together because the strong force is much stronger than electromagnetic repulsion o Strong force also called the “nuclear force” • Radioactive decay o α, β, γ decay ▪ α decay: ▪ β decay: Ejection of a helium nucleus at a relatively low speed Ejection of a high speed electron Release of a high energy electromagnetic wave ▪ γ decay: o Half-life, exponential decay, semi-log plots ▪ Half-life is the time it takes for the amount of something to half due to decay ▪ After 1 HL, amount decreases by half ▪ After 2 HL, amount decreases by a factor of 4 ▪ After 3 HL, amount decreases by a factor of 8 ▪ Exponential decay: ▪ • Semi-log plots: convert exponential curves into straight lines • Something that curves up becomes a straight line with positive slope • Something that curves down becomes a straight line with negative slope • For exponential decay, a semi-log plot graphs log of amount vs time • For exponential decay, a semi-log plot is a straight line with negative slope • Semi-log plot intercepts the x axis where the original y value is 1 Mass spectrometer o Mass spec is when you bombard a molecule with electrons o When electrons smash into your molecule, it is fragmented into ions ▪ The faster (higher energy) the bombarding electron, the more fragmentation ▪ The more fragmentation, the smaller the molecular ion peak o These ions have a characteristics mass to charge ratio (m/e or m/z) o A magnetic field resolves (separates) the different m/z ions so they can be individually detected and plotted on a spectrum o Resulting spectrum plots Relative abundance vs the m/z ratio o The parent peak, or molecular ion peak, is the peak that depicts the ion of the molecule without fragmentation. It has the highest m/z ratio o Peaks clustered really close to one another depicts isotopes o Base peak is the tallest peak (most abundant species) o Useful for: ▪ Measuring molecular weight of a molecule ▪ Identify molecule by fragmentation patterns ▪ Identify heteroatoms by their characteristic isotope ratios Electronic Structure (PHY, GC) • Orbital structure of hydrogen atom, principal quantum number n, number of electrons per orbital (GC) o In the Bohr model, the hydrogen electron orbits the nucleus. o In quantum mechanics, hydrogen electron exists in a spherical probability cloud around the nucleus o The principle quantum number, n, defines what shell the electron is in o n values start from one: 1,2,3 … etc o Higher n shells are higher in energy (if subshells are the same) o There are n2 orbitals per shell o There are 2 electrons per orbital o Thus, there are 2n2 electrons per shell • Ground state, excited states o Electrons are normally in their ground state o When they absorb energy, they get promoted to excited states o Excited states are higher in energy than ground states o Excited state electrons come back down to the ground state via release of energy • Absorption and emission line spectra o The absorption spectrum shows what wavelengths of light are absorbed o The absorption spectrum looks like black lines on a rainbow background o The emission spectrum shows what wavelengths of light are emitted o The emission spectrum looks like colored lines on a black backgrou nd o The absorption spectrum corresponds to the emission spectrum in pattern o The emission spectrum shifts to a slightly longer wavelength • Use of Pauli Exclusion Principle o 2 electrons in the same orbital must be of different spins • Paramagnetism and diamagnetism o Paramagnetic = substance has unpaired electrons o Diamagnetic = all of the substance’s electrons are paired • Conventional notation for electronic structure (GC) • • • • Bohr atom o Electron orbiting the nucleus in a circular orbit o Larger n values have larger orbiting radii Heisenberg Uncertainty Principle o States that you can never simultaneously know the exact position and exact speed of an object because everything in the universe behaves both like a particle and wave at the same time Effective nuclear charge (GC) o Effective nuclear charge = nuclear charge - shielding electrons o Shielding electrons are those that stand between the nucleus and the electron we are interested in o Shielding electrons are those that are in subshells closer to the nucleus (lower in energy) than the electron we are interested in o MCAT questions usually give diagram of the Bohr model, in which case, the shield electrons are those that orbits at a smaller radius o The higher the effective nuclear charge for an electron, the more stable it is (higher ionization energy, not easily knocked off) o Effective nuclear charge increases for outer electrons as you go across (left to right) the periodic table Photoelectric effect o Incident light on metallic surfaces will cause the emission of electrons o The kinetic energy of the ejected electron does not depend on the intensity of the light o The kinetic energy depends on the energy of the individual photons, not how much total light there is. A high intensity radio source would not produce the photoelectric effect at all because no individual photons would have sufficient energy, even though there was high energy overall, while low intensity x-rays would lead to the photoelectric effect The Periodic Table - Classification of Elements into Groups by Electronic Structure (GC) • Alkali metals • • • • • • o Single valence electron - low ionization energy, very reactive o Wants to lose that electron to achieve empty valence shell o More reactive as you go down because of increasing radii o Reacts with oxygen to form oxides o Reacts with water to form hydroxides and releases hydrogen o Reacts with acids to form salts and releases hydrogen o Most commonly found in the +1 oxidation state Alkaline earth metals: their chemical characteristics o 2 valence electrons - relatively low ionization energy, quite reactive o Wants to lose both electrons to achieve empty valence shell o More reactive as you go down because of increasing radii o Reacts with oxygen to form oxides o Reacts with water to form hydroxides and releases hydrogen o Reacts with acids to form salts and releases hydrogen o Most commonly found in the +2 oxidation state Halogens: their chemical characteristics o 7 valence electrons (2 from s subshell and 5 from p subshell) - high electron affinity, very reactive o Wants to gain one electron to achieve full valence shell o More reactive as you go up because of decreasing radii o Reacts with alkali metals and alkaline earth metals to form salts o Most commonly found in the -1 oxidation state Noble gases: their physical and chemical characteristics o Full valence shell of 8 - high ionization energy couple with low electron affinity o Don’t react o Found in the oxidation state of 0 Transition metals o High conductivity due to free flowing (loosely bound) outer d electrons o In the presence of ligands (when in a chemical complex), the d orbitals become nondegenerate (different in energy) o Electron transitions between nondegenerate d orbitals gives transition metal complexes vivid colors o Varied oxidation states - but always + Representative elements o Representative elements include the s block and p block of the periodic table o No free flowing (loosely bound) outer d electrons o Valence shell fills from left (1 electron) to right (8 electrons) o Standard nomenclature: I A, II A, III A, IV A, V A, VI A, VII A, VIII A Metals and non-metals o Metals are to the left of metalloids o Non-metals are to the right of metalloids o Metalloids: diagonal line from Boron to Polonium: B, Si, As, Te, Ge, Sb, (Po) • Oxygen group o The group (column) that contains oxygen o Oxygen and sulfur are chemically similar (if question asks you what element you can substitute for oxygen and keep same chemical reactivity, choose sulfur) o Se - Te - Po - non-metal - metalloid - metal (or metalloid) The Periodic Table - Variations of Chemical Properties with Group and Row (GC) • Valence electrons o Electrons in the outer shell o Ranges from 1 to 8 from left to right of the representative elements o The valence electron rule does not apply to transition metals • First and second ionization energy o Definition of first IE: the energy needed to knock off the first valence electron o Definition of second IE: energy needed to knock off the second valence electron o Prediction from electronic structure for elements in different groups or rows ▪ IE decreases as you go down because of increasing radii ▪ IE increases as you go right because of decreasing radii ▪ Highest peaks are noble gases ▪ Lowest troughs are alkali metals ▪ Local maxima occurs for filled subshells and half-filled p subshells ▪ Second IE is always higher than the first IE (usually a lot higher) ▪ Alkali metals and hydrogen: first ionization energy very low. Second ionization much higher. ▪ Alkaline earth metals: first ionization energy low. Second IE also low • Electron affinity o Definition: electron affinity is the amount of energy released when something gains an electron (how easily it can gain an electron) o Variation with group and row ▪ As you go down a group, EA decreases because of larger radii ▪ As you go across (left to right) a row, EA increases ▪ Highest peaks are for the halogens ▪ Lowest for noble gases • Electronegativity o Definition: electronegativity is how much something hoards electrons in a covalent bond • • o Comparative values for some representative elements and important groups ▪ Electronegativity increases toward the top right ▪ Fluorine is the most EN element ▪ Things around fluorine are highly EN: N, O, Cl, Br ▪ Halogens are EN, especially toward the top of the group ▪ Noble gases can be very EN if they participate in bond formation (Kr and Xe) ▪ Non-metals are more EN than metals ▪ Covalent bond is a sharing of electrons between elements ▪ The more EN element in a covalent bond gets a larger share of electrons and has a partial negative charge ▪ The less EN (more electropositive) element in a covalent bond gets a smaller share of the electrons and has a partial positive charge ▪ If the EN difference is too great, an ionic bond occurs instead of a covalent one ▪ Ionic bonds result from a complete transfer of electrons from the electropositive element to the electronegative element Electron shells and the sizes of atoms o Electron shells ▪ Defined by the principle quantum number - the n value ▪ Going down the periodic table means jumping to the next shell ▪ As you fill to the next shell (Ne to Na), the effective nuclear charge decreases because the old shell stands in between the nucleus and the new shell ▪ Filling to the next shell causes a jump in atom size because of decreased effective nuclear charge ▪ As you go down a group (Na to K0, the atomic size increases even though the effective nuclear charge stays the same, because higher shells have a larger radius than lower shells ▪ Going across the periodic table means filling up the same shell (by going through subshells) ▪ As you fill up a shell, the effective nuclear charge increases because the atomic number (protons) is increasing while the same-shell electrons you add do not shield one another ▪ With increasing effective nuclear charge, the electrostatic attraction (F=kQq/r2) between the nucleus and the electrons increases, so the atom becomes more compact ▪ The increasing effective nuclear charge and electrostatic attraction is why going across a periodic table means decreasing atomic size o Sizes of atoms ▪ Size increases as you go down a column ▪ Size decreases as you go across (to the right of) a row ▪ Atomic sizes may overlap if you zigzag on the periodic table Electron shells and the sizes of ions o Sizes of ions ▪ Number of protons affects how attracted the electrons ▪ Higher # of protons pulls electrons closer, reducing the radius ▪ Pay attention to whether the molecule has lost electrons in its outer subshell Stoichiometry (GC) • Molecular weight o Molecular weight is numerically equal to molecular mass (amu) o 1 amu = 1 g/mol o 12Carbon has 12 amu and weighs 12g/mol • Empirical versus molecular formula • • • • • o Empirical formula is what you get after dividing everything in the molecular formula by the highest common factor Metric units commonly used in the context of chemistry o Molarity = M = mol/L o Molality = m = mol/kg o Mass = kg. Molar mass = g/mol Description of composition by percent mass o %mass = mass of species of interest / total mass * 100 Mole concept, Avogadro’s number NA o 1 mole = 1 mol = 1 Avogadro’s number = 6.02E23 molecules Definition of density o Density = mass/volume = kg/m3 o Often in chemistry, specific gravity is used o S.G. = number of times the density of water = density of substance / density of water o Density of water = 1 g/mL = 1g/cm3 o S.G. of water = 1 g/cm3 / 1 g/cm3 = 1 o Density of lead = 11 g/cm3 o S.G. of lead = 11 g/cm3 / 1 g/cm3 = 11 o S.G. is unitless Oxidation number o Common oxidizing and reducing agents o Disproportionation reactions ▪ An element in a single oxidation state reacts to form 2 different oxidation states ▪ Disproportionation can occur when a species undergo both oxidation and reduction ▪ For example: 2Cu+ → Cu + Cu2+ • Here, the Cu+ acts as both oxidizing and reducing agent and simultaneously reduce and oxidize itself • The oxidized Cu+ becomes Cu2+ • • The reduced Cu+ becomes Cu Description of reactions by chemical equations o Conventions for writing chemical equations ▪ Phases • (s) = solid • (l) = liquid • (g) = gas • (aq) = aqueous (dissolved in water) ▪ Coefficient • An equation with coefficients is a balanced equation ▪ Direction • A single head arrow denotes the reaction goes to completion in the direction of the arrow • A double-sided arrow denotes a reaction in equilibrium • A double-sided arrow with one side larger than the other denotes an equilibrium in favor of the side of the larger arrow ▪ Charge • Denotes charge and magnitude, for example +, -, 2+, 5-, … etc • Neutral charges are not denoted o Balancing equations, including redox equations ▪ Balance the combustion of propanol: C3H8O + O2 → CO2 + H2O ▪ Pick out atom (or group) that is easiest to balance (usually represented in only 1 term on both sides of the equation. In this case it is carbon ▪ C3H8O + O2 → 3CO2 + H2O ▪ The next easiest to balance is hydrogen ▪ C3H8O + O2 → 3CO2 + 4H2O ▪ Leave the hardest to last, oxygen. O is present in every term of the equation, so if we tried to balance O first, we’d have a hard time. ▪ C3H8O + 9/2 O2 → 3CO2 + 4H2O ▪ Even though we balanced every term, we’re not done yet. We need to get rid of any fractions, so multiply every term by 2 ▪ 2C3H8O + 9O2 → 6CO2 + 8H2O ▪ Balancing oxidation-reduction (redox) equations • 1. Separate into half reactions o There will be 2 half equations: one will be oxidation, the other reduction o Half equations contain only species of interest - those containing the atom that undergoes a change in oxidation state o Anything that is not covalently attached to the atom is not part of the species of interest o Anything that does not undergo a change in oxidation state is a spectator ion/species • 2. Balance each of the half reactions o Balance both charge and atoms o To balance one oxygen atom: ▪ ▪ Under acidic conditions: add H2O to the side that needs the oxygen atom, then add H+ to the other side ▪ Under basic conditions: add 2OH- to the side that needs the oxygen atom, then add H2O to the other side o The Ion-Electron Method: you balance out the atoms first, then charge o The Oxidation-State Method: treat the species of interest as a single atom (those that undergo a change in oxidation number) then balance it • 3. Recombine the half reactions o Multiply each half reaction by a factor, such that when you add them together, the electrons cancel out o It’s like you’re trying to solve a simultaneous equation and you want to eliminate the electron term • 4. Finishing touches o Combine any identical species on the same side of the equation o Cancel out any identical species on opposite sides of the equation o Add back in the spectator ions o For the oxidation-state method, now is also the time to balance out the oxygens and hydrogens o Check to make sure that both sides of the equation have equal number of atoms and neutral net charge Example using ion-electron method: K2Cr2O7 (aq) + HCl (aq) → KCl (aq) + CrCl3 (aq) + H2O (l) + Cl2 (g) • 1. Separate into half reactions o Reduction: Cr2O72- → Cr3+ o Oxidation: Cl- → Cl2 o Species of interest for the oxidation reaction is Cl-, not HCl, because the H+ is not covalently attached to our atom of interest, and the hydrogen proton breaks off in aqueous solution o Similarly, we use Cr2O72- and not K2Cr2O7 o K+ is the spectator ion • 2. Balance each of the half reactions o The Ion-Electron Method: you balance out the atoms first, then charge ▪ Balancing atoms for the reduction half reaction (Ion-electron method): • 1. Cr2O72- → Cr3+ • 2. Cr2O72- → 2Cr3+ • 3. Cr2O72- + 14H+ → 2Cr3+ + 7H2O ▪ Balancing charge for the reduction half reaction (Ion-electron method): • 1. Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O ▪ Do the same thing for the oxidation half reaction (Ion-electron method): • 1. Cl- → Cl2 • 2. 2Cl- → Cl2 • 3. 2Cl- → Cl2 + 2e• 3. Recombine the half reactions o Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O o 2Cl- → Cl2 + 2eo Multiply everything in the second equation by 3 o 6Cl- → 3Cl2 + 6eo Add the two equations together o Cr2O72- + 14H+ +6e- + 6Cl- → 2Cr3+ + 7H2O + 3Cl2 + 6eFinishing touches o Except for the electrons, there are no like terms to combine or cancel at this time o Cr2O72- + 14H+ + 6Cl- → 2Cr3+ + 7H2O + 3Cl2 o For the ion-electron method, the equation is already balanced. However, you need to add back in the spectator ions. When adding back the spectator ions, whatever you do to the left side, you do to the right o To the left side: the dichromate came in counter-ioned with K+, so add 2K+ o To the right side: do the same o K2Cr2O7 + 14H+ + 6Cl- → 2Cr3+ + 7H2O + 3Cl2 + 2K+ o Referring back to the original equation, the H’s and Cl’s on the left came in as HCl, so in order to balance the extra 14 - 6 = 8 H’s, you add 8 Cl’s. As always, if you add 8 Cl’s to the left, go ahead and add the same to the right o Kr2Cr2O7 + 14HCl → 2Cr3+ + 7H2O + 3Cl2 + 2K+ + 8Clo Done focusing on the left side. A quick look at the right side shows that we need to combine 2 of the Cl- with the 2 K+, and the remaining 6 Clgoes with the Cr. Thus, the final balanced redox equation is: o K2Cr2O7 (aq) + 14HCl (aq) → 2CrCl3 (aq) + 7H2O (l) + 3Cl2 (g) + 2KCl (aq) Example using oxidation state-method: K2Cr2O7 (aq) + HCl (aq) → KCl (aq) + CrCl3 (aq) + H2O (l) + Cl2 (g) • 1. Separate into half reactions (same as the ion-electron method) o Reduction: Cr2O72- → Cr3+ o Oxidation: Cl- → Cl2 • 2. Balance each of the half reactions o The Oxidation-State Method: focus on atom of interest ▪ Balancing the atom of interest for the reduction half reaction (Oxidation-state method) • 1. Cr2O72- → Cr3+ • 2. Cr2O72- → 2Cr3+ • 3. Each oxygen is 2- so the 2 Cr on the left must be 6+ • 2Cr6+ → 2Cr3+ ▪ Balancing charge for the atom of interest in the reduction half reaction (Oxidation-state method) • 1. 2Cr6+ + 6e- → 2Cr3+ ▪ Do the same thing for the oxidation half reaction (Oxidation-state method) • 1. Cl- → Cl2 • 2. 2Cl- → Cl2 • 3. 2Cl- → 2Cl0 • 4. 2Cl- → 2Cl0 + 2e• 3. Recombine the half reactions o 2Cr6+ + 6e- → 2Cr3+ o 2Cl- → 2Cl0 + 2eo Multiply everything in the second equation by 3: o 2Cr6+ + 6e- → 2Cr3+ o 6Cl- → 6Cl0 + 6eo Add the two equations together • ▪ o 2Cr6+ + 6e- + 6Cl- → 2Cr3+ + 6Cl0 + 6e4. Finishing touches o Except for the electrons, there are no like terms to combine or cancel at this time… o 2 Cr6+ + 6Cl- → 2Cr3+ + 6Cl0 o Convert the atoms of interest into species of interest by referring back to original equation o K2Cr2O7 + 6HCl → 2CrCl3 + 3Cl2 o Now unlike the ion-electron method, where the equation is balanced and you only add back spectator ions at this stage of the game, the oxidation state method requires you to balance the equation again. This is because after you convert the atoms of interest back to their species of interest, the equation is no longer balanced. o Start with the oxygens. On the left you have 7 O, so add 7 H 2O to the right o K2Cr2O7 + 6HCl → 2CrCl3 + 3Cl2 + 7H2O o Now take care of the hydrogens. You have 6H on the left, but 14H on the right. That means you should add 8 more H’s to the left to make a total of 14. All 14 H’s on the left should be in the form of HCl (refer back to the original equation. Note, HCl here is both the species of interest and spectator species. Some of the HCl contributes to the Cl- → Cl2 oxidation, but the other portion of the HCl doesn’t undergo redox. It merely provides the H+ for the water and the Cl- for the KCl and CrCl 3) o K2Cr2O7 + 14HCl → 2CrCl3 + 3Cl2 + 7H2O o Now you see there’s 14 Cl to the left, and 12 Cl to the right. You need 2 more Cl’s on the right. Referring back to the original equation, all the right-sided cl’s come in the form of KCl (don’t modify the Cl2 since you’ve already correctly balanced it by the oxidation state method. when balancing equations at this stage, only play around with water and the spectator species) o K2Cr2O7 + 14HCl → 2CrCl3 + 3Cl2 + 7H2O + 2KCl o Upon examination of the equation, every atom is balanced. So the final balanced redox equation is: o K2Cr2O7 (aq) + 14HCl (aq) → 2CrCl3 (aq) + 3Cl2 (g) + 7H2O (l) + 2KCl (aq) o Limiting reactants ▪ Limiting reactant is reactant that will get all used up first ▪ What is the limiting reactant for the following reaction? • 3Xox + Ared → 3Xred + Aox • Given: you use 60g of Xox and 63g of Ared • Given: the molecular weight of Xox is 2 amu, and Ared is 7 amu • The first thing you do is convert everything in moles. 1 amu = 1 g/mol • Xox: 60 g/ 2 amu = 30 mols • Ared: 63 g/ 7 amu = 9 mols • Now here’s where stoichiometry comes in: divide the mols by the stoichiometric coefficient of the species: • 30 mols / 3 = 10 for Xox • 9 mols / 1 = 9 for Ared • Now compare the values. 9 is the smallest, so Ared is the limiting reactant. o Theoretical yields ▪ The theoretical yield is how much of the product will be made based on the stoichiometry • ▪ ▪ ▪ ▪ ▪ ▪ In calculating the theoretical yield, first find out what your limiting reactant is. Then, use your limiting reactant as the stoichiometric basis to calculate how much product you will get. In real life, experimental yield less than theoretical yield because of loss during steps of the reaction What is theoretical yield for 3Xred 3Xox + Ared → 3Xred + Aox If you react 60g of Xox with 63g of Ared given that the molecular weight of Xox is 2 amu, Ared is 7 amu, and Xred is 10 amu? • First, find who’s the limiting reagent • Using the method above, limiting reactant is Ared • Next, take the amount in mols of the limiting reactant (9 mols according to the above calculation) and do the stoichiometry to get to how many mols of 3Xred this will yield • 9 mols of Ared * 3 mols of Xred per 1 mol of Ared = 27 mols • Lastly, convert mols to grams: 27 mols * 10 g/mol = 270g • The theoretical yield for the above reaction is 270g of Xred Say you did an actual experiment of the above reaction and you managed to obtain 243g Xred, then the experimental yield is 243g Percent yield = experimental yield / theoretical yield x 100 For the above experiment, the percent yield would be 243 / 270 x 100 = 90%