User:Hnedrfrieowjiahfguh/answers

User:FuzzyCatPotato | User:FuzzyCatPotato/practice | User:FuzzyCatPotato/answers

Lab equipment practice

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Laboratory spatula: Scoops stuff, moves it, applies it

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Erlenmeyer flask: Beakers with thin necks

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Test tube rack: Holds test tubes

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Graduated cylinder: Holds, measures liquids

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Watch glass: Holds liquid to evaporate it, holds solids for weighing, or covers beaker

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Crucible: Holds hot objects

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Laboratory iron rings: Holds items above work surface

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Evaporation dish: Evaporates

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Funnel:

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Test tube: Holds materials

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Ring stand (with attached titration device): Holds iron rings

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Spring clamps: Holds test tubes

Click!

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Wire gauze: Holds hot materials, diffuses heat

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Beakers: Holds liquids for stirring, heating, etc.

Click!

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Pipeclay triangle: Supports a crucible over a Bunsen burner

Click!

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Test tube brush: Cleans test tubes

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Florence flask: Beaker that ensures uniform heating, stirring, with thin neck

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Pasteur pipet (dropper): Moves liquid; large mouth; ungraduated

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Graduated pipet (dropper): Moves liquid, in measured amounts

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Bunsen burner: Emits sootless open gas flame

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Disposable weighing boat: Used for weighing

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Wash bottle: Used for dispensing fluid

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Wash bottle: Used for dispensing fluid

Click!

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Crucible tongs: Picks up hot stuff

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Forceps: Picks up small stuff

Click!

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Beaker tongs: Picks up beakers

Nuclear balancing practice

₈₈²²⁶Ra = ₂⁴α + X

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86222Rn

₀¹n + ₉₂²³⁵U = ₅₃¹³⁹I + 2₀¹n + X

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3995Y

Ionic equation balancing practice

Barium nitrate: Ba²⁺ + NO₃^-

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2+ + 1- (unbalanced) Ba2+(NO3-)2 2+ + 2- (balanced)

Ammonium sulfate: NH₄⁺ | SO₄^2-

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1+ | 2- (unbalanced) (NH4+)2SO42- 2+ | 2- (balanced)

Iron(III) chloride: Fe³⁺ | Cl^-

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3+ | 1- (unbalanced) Fe3+Cl3- 3+ | 3- (balanced)

Aluminum sulfide: Al³⁺ | S^2-

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3+ | 2- (unbalanced) Al3+2S32- 6+ | 6- (balanced)

Magnesium carbonate: Mg²⁺ | CO₃^2-

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2+ | 2- (balanced)

Zinc hydroxide: Zn²⁺ | OH^-

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2+ | 1- (unbalanced) Zn2+ | (OH-)2 2+ | 2- (balanced)

Magnesium carbonate: Mg³⁺ | CO₄^3-

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3+ | 3- (balanced)

Naming ionic compounds practice

Formula to words

H₂O:

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Hydrogen monoxide

P₂O₅:

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Diphosphorus pentoxide

CO₂:

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Carbon dioxide

P₂O₅:

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Carbon monoxide

P₂O₅:

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Dinitrogen monoxide

N₂O₄:

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Nitrogen tetroxide

SO₃:

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Sulfur trioxide

NO:

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Nitrogen monoxide

NO₂:

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Nitrogen dioxide

As₂O₅:

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Diarsenic pentoxide

PCl₃:

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Phosphorous trichloride

CCl₄:

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Carbon tetrachloride

SeF₆:

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Selenium hexafluoride

Words to formula

Diphosphorus pentoxide:

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P2O5

Carbon dioxide:

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CO2

Carbon monoxide:

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P2O5

Dinitrogen monoxide:

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P2O5

Silicon dioxide:

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SiO2

Carbon tetrabromide:

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CBr4

Sulfur dioxide:

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SO2

Phosphorus pentabromide:

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PBr5

Iodine trichloride:

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ICl3

Nitrogen triiodide:

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NI3

Dinitrogen trioxide:

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N2O3

Solution concentration practice

Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?

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1.5 L * (0.5 mol / 1 L) * (58.44 g / 1 mol) = 43.8 g NaCl

Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution?

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MstockVstock = MdiluteVdilute (11.6 mol / L)(x L) = (3.0 mol / L)(0.250 L) x = (3.0 mol) * (0.250 L) / (11.6 L) x = 0.065 L

How many grams of NaOH are contained in 270.0 mL of a 0.450 M sodium hydroxide solution?

How much of each starting material would you use to prepare 2.00 L of each of the following solutions? (a) 0.250 M NaOH from solid NaOH (b) 0.160 M NaOH from 1.00 M NaOH stock solution (c) 0.150 M K2CrO4 from solid K2CrO4 (d) 0.210 M K2CrO4 from 1.75 M K2CrO4 stock solution

Moles practice

How many grams of lithium are in 3.50 moles of lithium?

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(3.50 moles Li / 1) * (6.94 g Li / 1 mol Li) = 45.1 g Li

How many moles of lithium are in 18.2 grams of lithium?

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(18.2 grams Li / 1) * (1 mol Li / 6.94 g Li) = 2.62 mol Li

How many atoms of lithium are in 3.50 moles of lithium?

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(3.50 moles Li / 1) * (6.022e23 atoms Li / 1 mol Li) = 2.11e24 atoms Li

How many atoms of lithium are in 18.2 grams of lithium?

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(18.2 grams Li / 1) * (1 mol Li / 6.94 g Li) * (6.022e23 atoms Li / 1 mol Li) = 1.58e24 atoms Li

Calculate each of the following quantities. (a) Mass (g) of 0.59 mol of MnSO4 (b) Amount (mol) of compound in 13.3 kg of Fe(ClO4)3 (c) Number of N atoms in 78.2 mg of NH4NO2

Calculate each of the following quantities. (a) Mass (g) of 0.66 mol of KMnO4 (b) Amount (mol) of O atoms in 8.85 g of Ba(NO3)2 (c) Number of O atoms in 7.8 10-3 g of CaSO4 · 2 H2O

Empirical formula practice

Calculate the formula mass and percentage composition of magnesium carbonate.

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magnesium carbonate = MgCO3 Formula mass: 24.31 g Mg + 12.01 g C + 3(16.00 g O) = 84.32 g MgCO3 Percentage composition: Magnesium: (24.31 g/84.32 g) * 100% = 28.83% Carbon: (12.01 g/84.32 g) * 100% = 14.24% Oxygen: (48 g/84.32 g) * 100% = 56.93% Total: 28.83% + 14.24% + 56.93% = 100%

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

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Pretend 100g. Find mass: Carbon: 49.32% C * 100g = 49.32g C Oxygen: 43.84% C * 100g = 43.84g O Hydrogen: 6.85% C * 100g = 6.85g H Find moles: Carbon: 49.32g C/(12.01g C/mol C) = 4.107 mol C Oxygen: 43.84g O/(16.00g O/mol O) = 2.740 mol O Hydrogen: 6.85g H/(1.01g H/mol H) = 6.782 mol H Divide moles: Carbon: 4.107 mol C/2.740 mol O = 1.499 C/O Oxygen: 2.740 mol O/2.740 mol O = 1 O/O Hydrogen: 6.782 mol H/2.740 mol O = 2.475 H/O Multiply: Carbon: 1.499 C/O * 2 O ≈ 3 Oxygen: 1 O/O * 2 O ≈ 2 Hydrogen: 2.475 H/O * 2 O ≈ 5 Consolidate: C3H5O2 The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? Answer Add masses: Carbon: (12.01g C/mol C) * 3 mol C = 36.03 g C Oxygen: (16.00g O/mol O) * 2 mol O = 32.00 g O Hydrogen: (1.01g H/mol H) * 5 mol H = 5.05 g H Total: 73.08 g Divide given mass by total: 146 g/73.08 g = 2 Multiply empirical formula by number: (C3H5O2) * 2 = (C3H5O2)2 = C6H10O4

A chloride of silicon contains 79.1 mass % Cl. If the molar mass is 269 g/mol, what is the molecular formula?

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Pretend 100g. Find mass: 79.1% Cl * 100g = 79.1 g Cl 79.1 g / (35.453 g/mol) = 2.23 mol Cl 100 g (Cl + Si) - 79.1 g Cl = 20.9 g Si 20.9 g / (28.09 g/mol) = 0.744 mol Si Find ratios: 2.23 Cl / 0.744 = ca. 3 Cl 0.744 Cl / 0.744 = ca. 1 Si Empirical formula: SiCl3 Molar mass: 134.449 g/mol 269 g/mol / 134.449 g/mol = 2 Molecular formula: Si2Cl6

A compound contains only carbon, hydrogen, and oxygen. Combustion of 8.544 mg of the compound yields 12.81 mg CO2 and 3.50 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

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8.544 mg CxHyOz + ∞ mg O2 = 12.81 mg CO2 + 3.50 mg H2O Molar masses: CxHyOz: 176.1 g/mol CO2: 44.01 g/mol H2O: 18.016 g/mol Product milligrams: 12.81 mg CO2 = 12.01 (element molar mass) / 44.01 (compound molar mass) * 12.81 = 3.496 mg C 3.50 mg H2O = 2.016 (element molar mass) / 18.016 (compound molar mass) * 3.50 = 0.3917 mg H O: 8.544 - (3.496 + 0.3917) = 4.656 mg O Moles: C: 3.496 / 12.011 = 0.2911 H: 0.3917 / 1.008 = 0.3886 O: 4.656 / 16.00 = 0.2911 Ratio: C: 0.2911 / 0.2911 = 1 ; *3 = 3 H: 0.3886 / 0.2911 = 1.33 ; * 3 = 4 O: 0.2911 / 0.2911 = 1 ; * 3 = 3 Empirical formula: C3H4O3 Empirical formula mass: 88.06 g 171.6 / 88.06 = 2 Chemical formula: C6H8O6

A chloride of silicon contains 79.1 mass % Cl. (a) What is the empirical formula of the chloride? (b) If the molar mass is 269 g/mol, what is the molecular formula?

A 0.370-mol sample of a metal oxide (M2O3) weighs 55.4 g. (a) How many moles of O are in the sample? (b) How many grams of M are in the sample? (c) What element is represented by the symbol M?

Menthol (script M = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g menthol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol's molecular formula?

Mass spectrometry charts practice

See: http://www.sciencegeek.net/APchemistry/APtaters/MassSpec.htm

Methylene Chloride (CH₃Br). Br is 50.69% ⁷⁹Br and 49.31% ⁸¹Br.

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Methylene Chloride (CH₂Cl₂). Cl is 75.77% ³⁵Cl and 24.23% ³⁷Cl.

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Vinyl Chloride (CH₂CHCl). Cl is 75.77% ³⁵Cl and 24.23% ³⁷Cl.

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Stoichiometry practice

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

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Aluminum oxide: Alx+3Oy-2 Balance: Al2O3 Equation: XAl + YO2 = ZAl2O3 Balance: 4Al + 3O2 = 2Al2O3 Al2O3 molar mass = 26.98 * 2 + 16.00 * 3 = 101.96 g/mol Convert: 6.50 g Al * (1 mol Al / 26.98 g Al) * (2 mol Al2O3 / 4 mol Al) * (101.96 g / 1 mol Al2O3) 6.50 / 26.98 x 2 / 4 x 101.96 = 12.3 g

Bornite (Cu3FeS3) is a copper ore used in the production of copper. When heated, the following reaction occurs.

2 Cu₃FeS₃(s) + 7 O₂(g) = 6 Cu(s) + 2 FeO(s) + 6 SO₂(g)

If 3.98 metric tons of bornite is reacted with excess O2 and the process has an 76.9% yield of copper, how much copper is produced?

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3.98 metric tons = 3.98e6g 3.98e6g (2 Cu3FeS3(s)) + ∞(7 O2(g)) = x(6 Cu(s)) Molar mass Cu3FeS3: 342.681 g/mol (3.98e6) * (1/342.681) * (6/2) * (63.546) * (1/1e6) * (0.769) = 1.70 metric tons Cu

Limiting reactant practice

What is the limiting reactant in the synthesis of ammonia (NH₃) if you have 5.0 moles of N₂ and 10.0 moles of H₂(g) in the reaction vessel?

N₂(g) + 3H₂(g) = 2NH₃(g)

Answer
N2 as limiting: (N2)5.0 + (3H2(g))∞ = (2NH3(g))5/2 H2 as limiting: (N2)∞ + (3H2(g))10/3 = (2NH3(g))10/6 H2 is limiting.

Photoelectric effect practice

• http://www.sciencegeek.net/Activities/Ehf.html

If a beam of light with energy 4.0 eV (1 eV = 1.602e-19 J) strikes a gold surface (φ = 5.1 eV), what is the maximum kinetic energy of the ejected electrons?

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KEmax = Ei - φ Ei = 4.0 eV φ = 5.1 eV KEmax = 4.0 eV - 5.1 eV KEmax = -1.1 eV = 0

Zinc experiment

1. φ Zn = 6.9e-19 J
2. λ UV lamp = 254 nm
3. λ red laser pointer = 700 nm
4. Calculate energy:
   1. Equation (for light):
      1. E = hν
      2. ν = c/λ
      3. E = hc/λ
   2. UV lamp: E = (6.626e-34)(2.998e8)/(254e-9) Js*m/s/m = 7.82e10-19 J
5. Red laser pointer: E = (6.626e-34)(2.998e8)/(700e-9) Js*m/s/m = 2.84e10-19 J
6. Ejects electrons?
   1. UVL: yes
   2. RLP: no
7. Calculate photons:
   1. UVL: (1e-3J/s) * (photon/2.84e-19 J) * 60s = 2.1e17 photons
8. It works!

Single-electron atoms

1. Schroedinger equation: ĤΨ = EΨ
   1. Ĥ = Hamiltonian operator
   2. Ψ = wavefunction (description of the orbital)
   3. E = binding energy

Hydrogen orbital change

Wavefunctions: http://quantum.phys.cmu.edu/CQT/chaps/cqt02.pdf

1. For a hydrogen atom:
   1. ĤΨ = ΨE
   2. ĤΨ = Ψ * (-1/n²) * (me⁴/8ε₀²h²)
      1. m = m_e = mass of electron
      2. e = charge on the e-
      3. ε₀ = permittivity constant of a vacuum = 8.854e-12 C²J^-1m^-1
      4. h = 6.626e-34 Js
      5. n = principle quantum number (interger)

Rydberg constant in binding energy in general

1. Rydberg constant (R):
   1. ĤΨ = Ψ * -R_H/n²
   2. R_H = (me⁴/8ε₀²h²) = 2.18e-18 J
2. Binding energy (E): E = -R_H/n² (Always negative)
3. Ionization energy (IE): IE_n = -E_n (Always positive)

States of excitement in general

1. "Excited state"
   1. n₁ = ground
   2. n₂ = first excited state
   3. n₃ = second excited state
   4. n₄ = third excited state
2. E_n = -Z²R_H / n² (ONLY IF 1 ELECTRON; EITHER HYDROGEN OR 1+ IONS)
   1. He¹⁺ 1e^- atom: Z = 2
   2. Li²⁺ 1e^- atom: Z = 3
   3. Tb⁶⁴⁺ 1e^- atom: Z = 65

Series of hydrogen orbital changes

1. Series:
   1. n_f = 1: Lyman series, UV range
   2. n_f = 2: Balmer series, UV range
   3. n_f = 3: Paschen series, UV range
   4. n_f = 4: Brackett series, UV range

Photon emission in hydrogen orbital change

1. Setup
   1. Evacuated glass tube, filled with H₂
   2. Negative electrode, positive electrode
   3. Disperse emission
   4. Analyse wavelength
2. ν = ΔE/h = (E_i - E_f)/h
3. Four wavelengths:
   1. Red: n=3 to n=2; 656 nm; longest wavelength, lowest frequency, lowest energy
   2. Green: n=4 to n=2; 486 nm
   3. Purple: n=5 to n=2; 434 nm
   4. Violet: n=6 to n=2 (shortest wavelength, highest frequency, highest energy); 410 nm

Frequency in hydrogen orbital change

1. General:
   1. Combine
      1. ν = (E_i - E_f)/h
      2. E_n = -R_H / n²
      3. ν = ((-R_H / n_i²) - (-R_H / n_f²))/h
   2. Simplify
      1. ν = (-R_H/h) * ((1/n_i²) - (1/n_f²))/h
      2. ν = (R_H/h) * ((1/n_f²) - (1/n_i²))
2. For n_f=2:
   1. ν = (R_H/h) * (1/4 - (1/n_i²))

Rydberg constant in frequency in general

1. Rydberg constant Mk. II:
   1. R_H/h = ℜ = 3.29e15 s^-1
   2. R∞ = 1.097373157e7 s^-1

Energy levels practice in general

Calculate the wavelength of radiation emitted by a hydrogen atom when an electron makes a transition from the n=3 to the n=2 energy level. (E_n=3 to E_n=2).

Answer
Frequency: ν = (RH/h) * ((1/nf2) - (1/ni2)) ν = 2.18e-18/6.626e-34 * ((1/(2^2)) - (1/(3^2)) ν = 4.5695409e+14 s-1 Wavelength: λ = c/ν = 2.998e8/4.5695409e+14 = 6.56e-7 m = 656 nm

1e^- atom

1. Frequency:
   1. For n_i > n_f: ν = (R_HZ²/h) * ((1/n_f²) - (1/n_i²))
   2. For n_f > n_i: ν = (R_HZ²/h) * ((1/n_i²) - (1/n_f²))
2. Binding energy:
   1. Hydrogen: E_n = -R_H/h
   2. All 1-electron: E_n = -Z²R_H/h

Quantum numbers

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

1. Ψ_nlm(r,θ,φ)
2. n
   1. Name: Principle quantum number
   2. Describes: Total binding energy of electron (potential + kinetic)
   3. Describes: Shell (and number of subshells)
   4. Rules: Any positive integer above 1
3. l
   1. Name: Angular momentum quantum number
   2. Describes: Angular energy of the electron
   3. Describes: Subshell (l = 0 (s), l = 1 (p), l = 2 (d), l = 3 (f))
   4. Rules: Any positive integer above 0 up to l = n-1
4. m (or m_l)
   1. Name: Magnetic quantum number
   2. Describes: Shape of orbital / how electron behaves in magnetic field / z component of angular momentum
   3. Describes: Subshell description (m = -1 (x), m = 0 (z), m = +1 (y))
   4. Rules: Any integer above 0 from m = -l to m = l
5. Alternate description:
   1. Ψ₁₀₀(r,θ,φ) = 1s orbital
6. Degenerate orbitals: for n orbitals, n² orbitals are degenerate (have the same energy)
7. For 1-electron atoms, all subshells have the same energy

	State label	Wavefunction	Orbital	Es	Es[J]
n = 1 l = 0 m = 0	100	Ψ100	1s	-RH/12	-2.9e-18 J
n = 2 l = 0 m = 0	200	Ψ200	2s	-RH/22	-5.45e-19 J
n = 2 l = 1 m = +1	210	Ψ211	2px or 2py (opp of 5th)	-RH/22	-5.45e-19 J
n = 2 l = 1 m = 0	210	Ψ210	2pz	-RH/22	-5.45e-19 J
n = 2 l = 1 m = -1	21-1	Ψ21-1	2px or 2py (opp of 3rd)	-RH/22	-5.45e-19 J

Wavefunction physical interpretation

https://youtu.be/Pj2fkkZ6Gto?t=1630

1. Max Born
   1. [Ψ_nlm(r,θ,φ)]²
   2. Probability/Volume
   3. Probability of finding an electron in a given volume
   4. Never reaches 0 through space
2. Components
   1. Ψ_nlm(r,θ,φ) = R_nl(r) x Y_lm(θ,φ)
   2. Wavefunction = radial wavefunction x angular wavefunction
   3. For all s orbitals, Y is a constant
3. Areas of 0 probability: Nodes
   1. Only occurs between orbitals
   2. Number of nodes = n - l - 1

Radial probability distribution

1. For s orbitals: RPD = Ψ² * 4πr²Ψ²dr
   1. Probability = Probability/Volume * Volume
   2. Most probable : r_mp = a₀ = Bohr radius = 0.529 Ångstroms

p orbitals

1. l = 1
2. 3 orbitals
   1. for m = +/-1, p_x/p_y
3. θ and φ (angular) dependence
4. Total of 6 lobes
   1. Each plane: Two lobes, separated by nodal plane
5. Ψ²_2py =
   1. Highest probability: Along Y axis
   2. Positive Ψ: Where y is positive
   3. Nodal plane: xz plane (φ = 0 degrees)
6. Nodes:
   1. Total: n - 1
   2. Angular: l
   3. Radial: n - l - 1
7. As l increases, r_mp decreases

Spin magnetic quantum number

1. m_s = +1/2 (spin up) or -1/2 (spin down)
2. Independent of orbital, only describes electron
3. Discovery:
   1. Emission spectrum of sodium
   2. Would expect 1 line at given frequency
   3. 2 lines very slightly above and below given frequency (doublet)

Pauli exclusion principle

1. No two electrons in same atom can have same quantum numbers
2. Distinction: orbital has 3 quantum numbers, electron 4
3. Limits to 2 electrons per orbital

Multi-electron atoms

How many electrons can exist in a 2p orbital? 6. 2p has 3 complete orbitals, 2px, 2py, 2pz; each complete orbital can have 2 electrons.

Shroedinger equation for multiple electrons

Base: ĤΨ = EΨ

1. H (1 electron): ĤΨ(r₁θ₁φ₁) = EΨ(r₁θ₁φ₁)
2. He (2 electrons): ĤΨ(r₁θ₁φ₁r₂θ₂φ₂) = EΨ(r₁θ₁φ₁r₂θ₂φ₂)
3. Li (3 electrons): ĤΨ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃) = EΨ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃)
4. At high enough levels, it is mathematically impossible to solve the Shroedinger equation; instead, an approximation is used.

Hartree orbitals

Approximation: Treat the wavefunction of the many orbitals as the product of a one electron approximation for each.

1. He (2 electrons):
   1. Ψ(r₁θ₁φ₁r₂θ₂φ₂) = Ψ(r₁θ₁φ₁) * Ψ(r₂θ₂φ₂)
   2. Ψ(e^-#1, e^-#2) = Ψ(e^-#1) * Ψ(e^-#2)
   3. Ψ_100-1/2 * Ψ_100+1/2
   4. 1s(1) * 1s(2)
2. Li (3 electrons):
   1. Ψ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃) = Ψ(r₁θ₁φ₁) * Ψ(r₂θ₂φ₂) * Ψ(r₃θ₃φ₃)
   2. Ψ(e^-#1, e^-#2, e^-#3) = Ψ(e^-#1) * Ψ(e^-#2) * Ψ(e^-#3)
   3. Ψ_100-1/2 * Ψ_100+1/2 * Ψ_200-1/2
   4. 1s(1) * 1s(2) * 2s(1)

Shorthand

1. H (1 electron): 1s¹
2. He (2 electrons): 1s²
3. Li (3 electrons): 1s²2s²
4. Be (4 electrons): 1s²2s¹
5. B (5 electrons): 1s²2s²2p¹

Multi-electron vs hydrogen atom wavefunctions

Argon: 1s²2s²2p⁶3s²3p⁶

1. Similarities:
   1. Similar in shape
   2. Identical Radial Probability Distribution and nodes
2. Differences:
   1. Each multi-electron orbital is smaller than the corresponding hydrogen atom orbital, because the nucleus is more positively charged.
   2. Orbital energy depends on both the shell (n) and the subshell (l) or angular momentum quantum number.
   3. Binding orbital energy in multi-electron atoms is lower (more negative) than in corresponding H-atom orbitals.
   4. In hydrogen, all orbitals for a given n have same energy level; in a multi-electron atom, orbitals increase in energy for both n and l
      1. For 1-electron: E_n = -IE_n = -Z²R_H/n²
      2. For multi-electron: E_n = -IE_n = -(Z^nl_eff)²R_H/n²
      3. Z != Z_effective
3. Shielding and binding energy:
   1. Scenario: Helium (He) nucleus has 2 protons (z = 2). Electron 1 is very close to nucleus. Electron 2 is far enough out that it doesn't affect electron 1. Electron 1 is thus "no shielded" and electron 2 is "totally shielded".
   2. E_He = -IE_He = -(Z_eff)²R_H/n² = -(+2)²R_H/1² = 8.72 x 10^-18 J, Z_eff = 1
   3. E_He = -IE_He = -(Z_eff)²R_H/n² = -(+1)²R_H/1² = 2.18 x 10^-18 J, Z_eff = 2
   4. Reality: IE_He = 3.94 x 10^-18 J
4. Finding Z_eff:
   1. Z_eff = [n²(IE/R_H)]^1/2
   2. He: [1²(3.94 x 10^-18 J/2.18 x 10^-18 J)]^1/2 = 1.34
5. Shielding
   1. S is less shielded than P, because over RPD, average Z_eff of 2p is less than Z_eff of 2s (etc.)
   2. RPD of orbitals

Electron configurations

1. Aufbau principle:
   1. Pauli Exclusion Principle: Only 1 spin in any given suborbital
   2. Hund's rule: At the same energy level, a single e^- enters state before a second enters said state
2. Oxygen: https://youtu.be/f7RRqxv2pzg?t=2198
   1. O: 1s²2s²2p⁴
   2. O_ml: 1s²2s²2px²2px¹2px¹
3. Sodium:
   1. Na: [Ne]3s¹
   2. Na_ml: 1s²2s²2px²2px²2px²3s¹

Exceptions

1. 1: Half-filled d orbitals are more stable than half-filled s
   1. V: [Ar]4s²3d³
   2. Cr: [Ar]4s¹3d⁵
   3. Mn: [Ar]4s²3d⁵
2. 2: Filled d orbitals are more stable than half-filled s
   1. Ni: [Ar]4s²3d⁸
   2. Cu: [Ar]4s¹3d¹⁰
   3. Zn: [Ar]4s²3d¹⁰

Ion electron configurations

A d orbital with 2+ electrons has less energy (is more stable) than a s orbital. When removing the highest-energy atoms (making something ionic) an S

1. Vi: [Ar]4s²3d³
   1. Vi (reordered from lowest to highest energy orbital): [Ar]3d³4s²
2. Vi^-: [Ar]4s¹3d³

Practice

http://www.sciencegeek.net/Chemistry/taters/Unit2ElectronNotations.htm http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/electronconfigpractice.html

Photoelectric spectroscopy (PES)

https://en.wikipedia.org/wiki/Spectroscopy

1. X-rays are commonly used because they have sufficient energy; UV rays may be, but are sometimes insufficient.
2. Neon:
   1. 1s²2s²2p⁶ -> 1s²2s²2p⁵ + e^- with most KE
   2. 1s²2s²2p⁶ -> 1s²2s¹2p⁶ + e^- with middle KE
   3. 1s²2s²2p⁶ -> 1s¹2s²2p⁶ + e^- with least KE
3. IE = E_i + KE
   1. IE = Ionization energy
   2. E_i = Energy of photon
   3. KE = Kinetic energy of electron
4. Neon IE (with E_i = 1254)
   1. 1s²2s²2p⁶ ; KE = 1232 eV ; IE_2p = 22 eV (1254 - 1232)
   2. 1s²2s²2p⁶ ; KE = 1206 eV ; IE_2p = 48 eV (1254 - 1206)
   3. 1s²2s²2p⁶ ; KE = 384 eV ; IE_2p = 870 eV (1254 - 384)
   4. Each would be represented as a bar with x = KE in order 1s, 2s, 2p
5. Practice: If PES produces 5 lines, which elements could it be?
   1. 5 lines, 5 orbitals: 1s2s2p3s3p: Al, Si, P, S, Cl, Ar
6. Practice: PES would reveal how many lines in hafnium (z = 72)?
   1. 1s2s2p3s3p3d4s4p4d4f5s5p5d6s (or 1s2s2p3s3p4s3d4p5s4d5p6s4f5d)
   2. 14

https://youtu.be/LPh2Ut7D4WA?t=746

Periodic trends

Ionization energy

1. First (smallest) ionization energy
   1. Implied whenever "ionization energy" is stated
   2. IE = -E_nl
   3. Boron:
      1. B(1s²2s²2p¹) -> B⁺(1s²2s²) + e^-
      2. ΔE = E_p - E_r = IE = -E_2p
2. Second ionization energy
   1. FILLER
   2. Boron:
      1. B⁺(1s²2s²) -> B²⁺(1s²2s¹) + 2e^-
      2. ΔE = IE₂ = -E_2s in B₊
3. Third ionization energy
   1. FILLER
   2. Boron:
      1. B²⁺(1s²2s¹) -> B³⁺(1s²) + 3e^-
      2. ΔE = IE₃ = -E_2s in B₊
4. Periodic table:
   1. [1], [2]
   2. Moving right in a row, Z (charge) while n (shell) remains constant; thus, Z_eff (effective charge) increases, so IE increases
   3. Moving down in column, Z (charge) and n (shell) increases; increasing n overpowers increasing Z, so Z_eff decreases and IE decreases
   4. Exceptions in Lithium - Neon row: http://staff.norman.k12.ok.us/~cyohn/index_files/ionizationenergynotes_files/image001.gif
      1. Beryllium (1s²2s² to Boron 1s²2s²2p¹: Energy required to add new orbital is greater than the increased distance from the nucleus
      2. Nitrogen (1s²2s²2p³ to Boron 1s²2s²2p⁴: Oxygen has 4 electrons in a 6 electron-orbital (3 subshells of 2 each) which forces 2 electrons to pair up

• Which element has smaller IE (and why): Al or P?

1. Al (lower Z_eff)
2. P (lower Z_eff)
3. Al (higher Z_eff)
4. P (higher Z_eff)

Electron affinity

1. Electron affinity (EA or E_ea) = -ΔE
2. Chlorine:
   1. Cl + e^- -> Cl^- ; ΔE = -349 kJ/mol ; EA = 349 kJ/mol
   2. Energy is released, so negative ion is more stable than atom.
3. Nitrogen:
   1. N + e^- -> N^- ; ΔE = +7 kJ/mol ; EA = -7 kJ/mol
   2. Energy is added, so atom is more stable than negative ion.
4. Periodic table:
   1. Moving right in a row, EA increases.
   2. Moving down in a column, EA decreases.

Electronegativity

1. Electronegativity (χ) is proportional to (0.5)(EA + IE)
2. Periodic table:
   1. Upper right: high χ (electron expector)
   2. Bottom left: low χ (electron donor)

Atomic radius

1. Atomic radius: value of r which encompasses 90% of electron density
2. Atomic radius: value of r which is 0.5 of distance between two atoms in a compound
3. Both are very similar values
4. Periodic table:
   1. Moving right in a row, r decreases (because Z_Eff decreases).
   2. Moving down in a column, r increases (because new orbitals are larger).
5. Ions:
   1. F^- is larger than F (more shielding, less Z_Eff))
   2. Na⁺ is smaller than Na (less shielding, more Z_Eff))

Isoelectronic

1. Neon: 1s²2s²2p⁶
2. Flourine:
   1. F: 1s²2s²2p⁵
   2. F^-: 1s²2s²2p⁶
3. Sodium:
   1. Na: 1s²2s²2p⁶3s¹
   2. Na⁺: 1s²2s²2p⁶
4. Which atom is isoelectronic with Krypton (Z=36)? Selenium^2- (Z=34).

Covalent bonds

1. Chemical bonds: rearrangement of the nuclei and electrons of the bonded atoms results in a lower energy than separate atoms.
2. Covalent bond: a pair of electrons shared between 2 atoms.
3. Terms:
   1. Internuclear distance: r
   2. Energy: E = nuclear-nuclear repulsion + electron-nuclear attraction + electron-electron repulsion
   3. Disassociation energy: ΔE_d = energy necessary to break covalent bond = bond strength = distance from E_{H + H} to E_H2
   4. Energy vs interatomic distance [3][4]
4. Example with H:
   1. H⁺ + e^- = 0 kJ/mol
   2. 2H (unbonded): E = 2 * -1312 kJ/mol = -2624 kJ/mol
   3. H₂ (bonded): E = -3048 kJ/mol
   4. H₂ is lower energy, so 2H bonds
   5. ΔE_d = -2624 kJ/mol - (-3048 kJ/mol) = 424 kJ/mol
5. N₂ vs H₂: N₂ is stronger bond, because its ΔE_d is lower than H₂'s, which means that the bond is more stable.

Quantitative chemical shit

Finding the stuff you've got in a pile of shit

Gravimetric analysis

• http://www.dynamicscience.com.au/tester/solutions1/chemistry/analytical%20chem/2007nswgrv1.gif
• https://en.wikipedia.org/wiki/Gravimetric_analysis#Procedure

1. Obtain sample (analyte) with some
2. Form precipitate: Add solute to form insoluble precipitate containing sample and some known (eg, AgNO₃
3. Isolate precipitate:
   1. Filter the precipitate from the solvent (eg, using sintered glass filter and vacuum pump)
   2. Precipitate has water removed (eg, drying/heating/etc.)
4. Calculate mass of sample
   1. Precipitate is weighed
   2. Proportion of mass attributable to the precipitating agent is subtracted
   3. Remainder is mass of analyte

Later

• molecular geometry https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=molcular%20geometry&oq=molcular%20geometry&aqs=chrome..69i57j0l5.2521j0j7
  • http://intro.chem.okstate.edu/1314F00/Lecture/Chapter10/VSEPR.html
  • https://en.wikipedia.org/wiki/Molecular_geometry
  • http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Molecular_Geometry
  • http://www.chemmybear.com/shapes.html
  • http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/section_13/26009d78c259f1040f28320caaf413ba.jpg

• Shroedinger: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.html

• TITRATION!!!

• Reorganize

• General naming rules, move ionic naming rules

• Practice photoelectric effect.

• Gas laws

• Solubility rules (NO3, NH4, and Group I)

• http://ocw.mit.edu/courses/chemistry/5-111-principles-of-chemical-science-fall-2008/exams/

Dtown

1. http://www.webassign.net/web/Student/Assignment-Responses/last?dep=12315235
2. http://www.webassign.net/web/Student/Assignment-Responses/last?dep=12370788
3. http://www.webassign.net/web/Student/Assignment-Responses/last?dep=12466037
4. http://www.webassign.net/web/Student/Assignment-Responses/last?dep=12518626

Redox

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions http://www.webqc.org/balance.php?reaction=Ag%2BHNO3%3DAgNO3%2BH2O%2BNO2 http://www.chemteam.info/Redox/Balance-Redox-Acid.html http://www.sciencegeek.net/APchemistry/APtaters/Redox/redox1.htm