2 P Electron example essay topic
680 words
Chemistry Study Guide Oct 2nd 1 hour Exam Chapter 9- Thermodynamics KE = 1/2 mv 2 w = F xw = force x distance A state function refers to a property of the system that depends only on its present state. Internal Energy = heat + work E = q + w Pressure = Force / Area = P = F / A Work = - external pressure x change in volume = - P VEnthalpyH = E + Pvp = E + P V H = qp H = H products - H reactants Ideal Gas Law PV = nRT Energy "heat" = 3/2 R Cv = 3/2 R = "heat" required to change the temp of 1 mol of gas by 1 K at constant volume Energy required = "heat" - energy needed - energy needed to do to change the translational the PV workenergyCp = 3/2 + R = 5/2 R = Cv + R = CpE = 3/2 RT (per mole) E = Cv T (per mole) E = n Cv T"Heat" required = qp = n Cp T = n (Cv + R) T = n Cv T + nR T ( E) (PV) H = E + (PV) H = E + (nRT) = E + nR T H = n Cp Energy released = energy absorbed = m x Cp x T (mass) (specific heat) (change in Temperature) E = q + w = q = qv Bomb calorimeter = H = TCp Ho reaction = ∑ Hof (products) - ∑ Hof (reactants) Chapter 12 c = λ ν speed of light (3.0 E 8 m) = (wavelength) (frequency) E = nh vs. or E = hv if n = 1 Planck's constant = h = 6.626 x 10-34 J sE photon = hv = hc/λ KE electron = 1/2 mv 2 = hv - h voE = mc 2 De Broglieλ = h / mv PE = -Z e 2/rZ = atomic number = distance between nucleusΗ ψ = Ε ψ Ψ 2 = Probability Density = energy level = shape ml = orientation ms = spin Effective nuclear charge = Ziff = Z actual - effect of electron-electron repulsion 1 eV = 1.602 x 10-19 Ionization Energy (g) X+ (g) + e- First Ionization energy increases as we go across a period from left to right due to shielding effects Shielding occurs because electrons repel each other First ionization energy decreases as we go down a group. As n increases, the size of the orbital increases, and the electron is easier to remove. Exceptions: Be to B - decrease in IE shows that the electrons in 2's orbital effectively shield the 2 p electron.
N to O- drop in IE because of addition of electron in first p orbital that results in a pair that repel each other and make either of them easier to remove. Electron Affinity X (g) + e- X- (g) Down a group - more positive since electron is added at increasing distances from the nucleus. Increase across period because effective nuclear charge is increasing. Electro negativity = (Electron Affinity + Ionization Potential) /2 Lattice Energy Li+ (g) + Cl- (g) Li Cl (s) Lattice Energy = k (Q 1 Q 2) /r H = ∑ D (Bonds broken) - ∑ D (Bonds formed) Formal charge = (# of valence electrons on a free atom) - (# of valence electrons assigned to the atom in molecule) (Valence electron) assigned = (# of lone pair electrons) + 1/2 (# of shared electrons) 1.
Lone pair electrons belong entirely to the atom in question 2.2. Shared electrons are divided equally between the 2 sharing atoms Chapter 14 Bond Order = (# of bonding electrons - # of anti bonding electrons) /2 Chap 20 CO 2 CN- NO 2- en NH 3 H 2 O OH- F- Cl- Br- I- (strong field) (weak field) (diamagnetic) (paramagnetic) (Low spin) (High spin) Magnitude of for ligand increases as charge on metal ion increases.
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