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# Fall Final Study Guide! CHEM-1070-30

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This 18 page Study Guide was uploaded by Nina Kalkus on Wednesday December 9, 2015. The Study Guide belongs to CHEM-1070-30 at Tulane University taught by Schmehl, Russell in Fall 2015. Since its upload, it has received 280 views. For similar materials see General Chemistry I in Chemistry at Tulane University.

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If Nina isn't already a tutor, they should be. Haven't had any of this stuff explained to me as clearly as this was. I appreciate the help!

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Date Created: 12/09/15

Chemistry Fall Final 2015: Study Guide! Balancing Oxidation-Reduction Reduction: In acidic solution: Step 1: Split the reaction into 2 half reactions Step 2: Balance the elements that aren't oxygen or hydrogen Step 3: Balance oxygens by adding water Step 4: Balance hydrogens by adding protons (H ) + Step 5: Balance charges by adding electrons Step 6: Scale the equations so that electrons will cancel out when you combine half reactions Step 7: Combine half reactions and cancel out common terms In basic solution: *Follow steps 1-7 as in acidic solution + Step 8:Add OH- to cancel out H Step 9: Combine OH and H to form water Example 1) Balance MnO + I ===> Mn + I - 2+ in acidic solution 4 - 2+ 2 Step 1: MnO ===4 Mn - I =====> I 2 Step 2: MnO ===4 Mn 2+ - 2I ====> I 2 Step 3: MnO ===4 Mn + 4H O 2+ 2 - 2I ====> I 2 + - 2+ Step 4: 8H +MnO ===> M4 + 4H O 2 2I ====> I 2 - - 2+ Step 5: 5e + 8H O +2MnO ===> Mn 4 4H O 2 2I ====> I + 2e2 - - 2+ Step 6: 2*(5e + 8H O + 2nO ===> Mn + 4H O) 2 5*(2I ====> I + 2e-) 2 Step 7: 16H + 2MnO + 10I =4=> 2Mn - 2+ + 8H O 2 5I 2 Example 2) Balance ClO ===> ClO 2 Cl 3- - - Step 1: ClO ==2> ClO 3 ClO ===> Cl - 2 - Step 2: ClO ==2> ClO 3 - ClO ==2> Cl Step 3: H O2+ ClO ===>2ClO 3 - ClO ==2> Cl + 2H O 2 Step 4: H O + ClO ===> ClO + 2H - + 2+ 2 - 3 4H + ClO ===> 2l + 2H O 2 - + - Step 5: H O2+ ClO ===>2ClO + 2H 1e 3 5e- + 4H + ClO ===> C2 + 2H O - 2 - + - Step 6: 5*(H O 2 ClO ===> C2O + 2H + 1e) 3 5e- + 4H + ClO ===> C2 + 2H O - 2 - + - Step 7: 6ClO + 2H O ===2 5ClO + 6H + Cl 3 Step 8: 6OH + 6ClO + 3H O ===> 6OH + 5ClO + 6H + Cl - - + - 2 2 3 Step 9: 6OH + 6ClO ===> 52lO + 3H O + Cl 3- 2 - Limiting Reagents: Limiting reagents are the reactants in a reaction that stop the reaction first. This happens because the amount of this reactant will run out before the other reactant(s). To find the limiting reagent, you must pay close attention to the moles of each reactant that are required in a chemical equation. For example, in the equation 2AgI + Na S →Ag S +22 NaI, if2you have 3 moles ofAgI and 4 moles of Na S, 2heAgI would be the limiting reagent, because to have both reactants react completely, you would need 2 moles ofAgI to every 1 mole of Na S, and in thi2 case that would mean you'd need 8 moles ofAgI. Example 1) A0.696 mol sample of Cu is added to 136 mL of 6.0 M HNO .Assuming the foll3wing reaction is the only one that occurs, will the Cu react completely? 3 Cu(s) + 8HNO (aq)3----> 3 Cu(NO ) (aq) +3 2 O(l) + 2NO2g) What we really need to find out: Is Cu or HNO the limi3ing reagent? What we know: We need 3 moles of Cu to every 8 moles of HNO 3 We have 0.696 moles of Cu and 136 mL of 6.0 M HNO 3 Step 1: Convert 136 mL of 6.0 M HNO into mole3 136 mL*(1L/1000mL)*(6 mole/1L)= 0.816 moles HNO 3 Step 2: See if ratio of moles Cu to moles HNO is equa3 to 3:8 3(x)= 0.696 moles Cu—-> x= 0.232 8(x)= # moles HNO we ne3d to complete reaction 8(0.232)= 1.856 moles HNO , which3is greater than 0.816 Answer: No, we do not have enough HNO for Cu to rea3t completely Example 2: Aside reaction in the manufacture of rayon from wood pulp is 3 CS +26 NaOH ----> 2 Na CS + Na2CO 3 3H O 2 3 2 How many grams of Na CS are 2rod3ced in the reaction of 92.5 mL of liquid CS (d=1.26 g/mL) 2 and 2.78 mol NaOH? What we really need to find out: Which is the limiting reagent? This will tell us how many moles of Na CS w2 can3produce with these amounts of reactants, and then we can convert that to grams. What we know: We need 3 moles of CS to every 62moles of NaOH We have 92.5 mL of liquid CS which 2as a density of 1.26 g/mL and 2.78 moles of NaOH The complete reaction will produce 2 moles of Na CS to eve2y 3 3 moles CS an2 every 6 moles NaOH Step 1: Convert 9.25 mL of CS into moles 2 92.5 mL*(1.26 g/mL)*(1 mole/76 g)= 1.533 mole CS Step 2: See if ratio of moles CS to 2oles NaOH is equal to 3:6 3(x)= 1.533 moles CS ----2 x= 0.511 6(0.511)= 3.066 moles NaOH, which is greater than 2.78 moles NaOH that we have, which means that NaOH is the limiting reagent 6(z)= 2.78 moles NaOH ----> z= 0.463 2(z)= # moles Na CS 2hat3can be produced 2(0.463)= 0.926 moles Na CS 2 3 0.926 moles Na CS *(154 g/1 mole)= 142.6 2 3 Answer: We will produce 142.6 grams of Na CS 2 3 Calorimetry: Important points to know: 1 calorie= 4.184 Joule Specific heat of water (heat capacity of water) = (4.18J/ (gram of water*degree Celsius)) quantity of heat= q = mass of substance*specific heat*temperature change = heat capacity*temperature change qsystem -qsurroundings qreaction-qcalorimeter qcalorimetereat capacity of calorimeter*temperature change Change in internal energy= ΔU= q (heat) + w (work)= U final U initial If reaction occurs at constant volume:ΔU= q rxn+ w= q rxn + 0= q v Enthalpy = H=ΔU +ΔPV If a process is carried out under constant temperature and pressure and with work limited to pressure-volume work, thenΔH=ΔU + PΔV and heat flow isΔH= q p Hess's Law: If a process occurs in stages or steps, the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps Translation: If you carry out a reaction in more than one steps, the sum of the enthalpy changes of those steps will equal the enthalpy change of the reaction if you did it in one step. Enthalpies of formation: The standard enthalpy of formation of a pure element in its reference form is 0. Standard enthalpy change: ΔH^o ΔH^o= (sum of ΔH^o(prfducts)*v ) – (sup of ΔH^o(reactanfs)*v ) r v= stoichiometric coefficients (number that comes before molecule in rxn equation) PV= nRT R= .08206 L atm/ mol K Practice Problems: 1) Calculate the quantity of heat, in kilojoules, required to raise the temperature 9.25 L water from 22.0 to 29.4 degrees Celsius What we need to know: How many grams of water do we have? What is the temperature change? What we know: We have 9.25 L of water Water is going from 22.0 to 29.4 degrees Celsius Specific heat of water is 4.18 Joules/(g*degree C) Step 1: Convert water to grams 9.25 L*(1000 mL/1 L)*(1 g/1 mL)= 9250 g Step 2: Find temperature change 29.4-22.0= 7.4 degrees Celsius Step 3: Use specific heat of water to find quantity of heat in Joules 9250 g*7.4 degrees C*(4.18 Joules/(g*degrees C))= 286121 Joules Step 4: Convert to kilojoules 286121 Joules*(1 kJ/1000 J)= 286.121 kJ Answer: 286.121 kJ are required Calorimetry Practice Continued: 2)A74.8 g sample of copper at 143.2 degrees Celsius is added to an insulated vessel containing 165 mL of glycerol, C H3O 8l)3(d= 1.26 g/mL), 24.8 degrees C. The final temperature is 31.1 degrees Celsius. The specific heat of copper is 0.385 Joules/(g*degree C). What is the specific heat of glycerol in Joules/(mole*degree C)? What we know: We have 74.8 g sample of copper and 165 mL of glycerol The specific heat of copper is 0.385 Joules/(g*degree C) The temperature goes from 143.2 to 31.1 degrees Celsius for copper, and from 24.8 to 31.1 degrees Celsius for glycerol q copperq glycerol density of glycerol is 1.26 grams per mL There are 420 grams per mole of glycerol What we need to know:Amount of glycerol in grams Temperature change for each substance Step 1: Convert glycerol to grams 165 mL*(1.26 g/mL)= 207.9 g glycerol Step 2: Find temperature change for each substance Copper: 31.1-143.2= -112.1 degrees Celsius---> ΔT= 112.1 Glycerol: 31.1- 24.8= 6.3 degrees Celsius ---> ΔT= 6.3 Step 3: Solve for heat capacity of glycerol using equations for q q copperq glycerol mass of copper*specific heat of copper*temperature change of copper= mass of glycerol*specific heat of glycerol*temperature change of glycerol 74.8 g*0.385 Joules/(g*degree C)*112.1 degrees C= 207.9 g glycerol*specific heat of glycerol*6.3 degrees Specific heat of glycerol= 2.645 Joules/(g*degree C) Step 4: Convert from Joules/(g*degree C) to Joules/(mole*degree C) 2.645 Joules/(g*degree C)*(420 g/1mole)= 226.78 Joules/(mole*degree C) Answer: The specific heat of glycerol is 226.78 Joules/(mole*degree C) 3) The combustion of methane gas, the principal constituent of natural gas, is represented by the equation CH (4) + 2 O (g2 ------> CO (g)2+ 2 H O(l)2 ΔH = -890.3 kJ What mas of methane, in kilograms, must be burned to liberate 2.80*10 kJ of heat? What we know: Burning 1 mole of methane will produce 890.3 kJ of heat There are 16 grams in every 1 mole of methane What we need to know: The ratio of 2.80*10 kJ to 890.3 kJ 7 Step 1: Find out h7w many times 890.3 goes into 2.80*10 kJ 2.80*10 kJ/ 890.3= 31450.07 Step 2: Convert moles CH to g4ams 31450.07 moles methane*(16 grams/1 mole)= 503201.17 g methane Step 3: Convert grams to kilograms 503201.17 g methane*(1 kg/1000 g)= 503.20117 kg Answer: Must burn 503 kg of methane to liberate 2.80*10 kJ of heat 7 Light-energy-wave/particle problems Important things to remember: 1 meter= 10 micrometers, 10 nm (nanometer), 10 pm (picometer) 8 C= speed of light= 3*10 meters/sec C= υ*λ υ= frequency--> units are Hz=1/sec λ= wavelength--> units are meters Planck's constant: h=6.626*10 -34Joules*sec E= Energy per photon= h*υ p= momentum= (h*υ)/C= h/λ -----> λ= (h)/(m*u) (m=mass, u= velocity) -31 mass of electron= 9.109*10 kg Molecular mass=M= (m/V) (RT/P) 15 2 2 Balmer's equation: υ= 3.2881*10 s^-1((1/2 )- (1/n )) Bohr atom description equations: rn= n a o a = 53 picometers o 2 -18 E n Energy of an electron at a level= (-R )/(nH) R H 2.179*10 ΔE= R ((H/n )-(i/n )) f where n= initial level and n = final level *only works in hydrogen-like species with only 1 electron; for other species: E = (-Z R )/(n )2 n H Uncertainty principle: Δx= uncertainty in position, Δp= uncertainty in momentum ΔxΔp≥ h/(4π) Δp= m*Δu----->Δu= uncertainty of velocity Wave mechanics: λ= 2L/n L= length of “string”, n+1= total number of nodes 2 2 2 E k Energy of wave= (n h )/8mL Quantum numbers: n= principal quantum number= 1,2,3,4,.... (integers starting at 1 and increasing) tells you what the orbital is: orbital number=n l = orbital angular momentum quantum number= 0,1,2,3,...n-l (can't be larger than n-l) tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m l magnetic quantum number= -l, (-l+1),...,(l-1),+l (starting at -l and going to positive l up by 1) corresponds to number of orbitals in each subshell (s:n=1,l=0, ml=0-->only 1 orbital) m s electron spin quantum number ( + ½, - ½) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) can't have [↑↑] or [↓↓] Ahelpful table: Energy level Sublevels # of orbitals # of electrons 1 s 1 2 2 s,p 1,3= 4 8 3 s,p,d 1,3,5=9 18 4 s,p,d,f 1,3,5,7=16 32 Aufbau process: *fill in the s subshell first *can't have electrons of the same spin in an orbital Ground state: if all orbitals have at least one electron, they do [↑ ][↑ ][↑ ] Excited state: if all orbitals could have at least one electron, one has moved [↑↓][↑ ][ ] Atomic and Radial Nodes: helpful shortcuts Total number of nodes= n-1 # radial nodes= n-(l+1) # angular nodes= l Practice Problems: 1) Use the Balmer equation to determine a) the frequency, in s , of the radiation corresponding to n=5 b) the wavelength, in nanometers, of the line in the Balmer series corresponding to n=7 c) the value of n corresponding to the Balmer series line at 380 nm -1 a) The frequency, in s , of the radiation corresponding to n=5 Step 1: plug n=5 into Balmer's equation υ= 3.2881*10 s ((1/2 ) – (1/5 ))= 6.90501*10 s 14 -1 Answer= 6.90501*10^14 s^-1 b) The wavelength, in nanometers, of the line in the Balmer series corresponding to n=7 What we need to know: The frequency of the radiation corresponding to n=7 What we know: C= λυ---> λ= C8υ C= 3*10 meters/sec Step 1: Plug in the Balmer equation to λ= C/υ 15 -1 2 2 λ= C/(3.2881*10 s ((1/2 )15 (-1n )))2 2 λ= (3*10 )/(3.2881*10 s ((1/2 ) – (1/n ))) Step 2: Plug n=7 into the new equation 8 15 -1 2 2 -7 λ= (3*10 )/(3.2881*10 s ((1/2 ) – (1/7 )))= 3.97*10 Step 3: Convert the wavelength from meters to nanometers 3.97*10 *(10 nm/1m)= 397.4 nanometers Answer= 397.4 nanometers c) The value of n corresponding to the Balmer series line at 380 nm What we need to know: What frequency corresponds to the wavelength of 380nm? What we know: C= λυ---> υ= C/λ 8 C= 3*10 meters/sec Step 1: Convert frequency from nanometers to meters 9 -7 380 nm*(1m/ 10 nm)= 3.8*10 m Step 2: Use the equation υ= C/λ to find frequency υ= (3*10 )/(3.8*10 )= 7.89*10 14 Step 3: Use Balmer's equation to solve for n υ= 7.89*10 = 3.2881*10 ((1/2 ) – (1/n )) 2 n=10 2)Acertain radiation has a wavelength of 574 nm. What is the energy, in Joules, of a) one photon b) a mole of photons of this radiation a)Acertain radiation has a wavelength of 574 nm. What is the energy, in Joules, of one photon? What we need to know: The frequency corresponding to this wavelength What we know: E= hυ υ= C/λ E= (hC)/λ λ= 574 nm Step 1: Convert wavelength from nanometers to meters λ= 574 nm*(1 m/10 nm)= 5.74*10 meters -7 Step 2: Plug in wavelength to E= (hC)/λ equation -34 8 -7 -19 E= (6.626*10 *3*10 )/ 5.74*10 = 3.463*10 Joules/photon b) Acertain radiation has a wavelength of 574 nm. What is the energy, in Joules, of a mole of photons of this radiation? What we know: 1 mole= 6*10 23 Step1: Convert Joules/photon to Joules/mole 3.463*10 -19Joules/photon*(6*10 photons/1 mole)= 207,783.9721 Joules/mole 3) Use the description of the Bohr atom in the text to determine a) the radius, in nanometers, of the sixth Bohr orbit for hydrogen b) the energy, in Joules, of the electron when it is in this orbit a) Use the description of the Bohr atom in the text to determine the radius, in nanometers, of the sixth Bohr orbit for hydrogen. What we know: r = n a 2 n o n=6 ao= 53 picometers 12 9 1 m= 10 pm= 10 nm Step 1: Convert picometers to nanometers 53 pm*(1m/10 pm)*(10 nm/1 m)= 0.053 nm Step 2: Plug in a ,onow in nanometers, into r equanion 6 = 6 *0.053= 1.098 nm Answer: Radius of the sixth Bohr orbit for hydrogen equals 1.098 nm b) Use the description of the Bohr atom in the text to determine the energy, in Joules, of the electron when it is in this orbit. 2 What we know: E = (-n )/(n H -18 R H 2.179*10 Joules n=6 Step 1: Plug in the given n and R to tHe energy equation -18 2 -20 E 6 (-2.179*10 )/(6 )= -6.05*10 Joules Answer: The energy of an electron in the 6 Bohr orbit is -6.05*10 -20Joules 5) What must be the velocity, in meters per second, of a beam of electrons if they are to display a de Broglie wavelength of 1micrometer? What we know: 1 meter= 10 micrometers λ= (h)/(m*u) velocity of beam of electrons=velocity of one electron in that beam mass of electron= 9.109*10 -31kg -34 h= 6.626*10 Step 1: Convert wavelength from micrometers to meters λ= 1 micrometer*(1 m/10 micrometers)= 10 meters -6 Step 2: Plug in wavelength in meters, h, and mass of electron to the equation λ= (h)/(m*u) and solve for u 10 = (6.626*10 )/(9.109*10 *u)---> u= 727 m/s Answer: The velocity of a beam of electrons that display a de Broglie wavelength of 1 micrometer is 727 m/s 6)Aproton is accelerated to one-tenth the velocity of light, and this velocity can be measured with a precision of 1%. What is the uncertainty in the position of this proton? What we know: ΔxΔp≥ h/(4π)-->Δp=uncertainty of momentum, Δx= uncertainty of position Δp= m*Δu----->Δu= uncertainty of velocity C= speed of light= 3*10 m/s 8 uncertainty of something= value*(% precision/100) mass of proton= 1.67*10 kg 27 h= 6.626*10 -34 Step 1: Find velocity of proton u= (1/10*C)= (1/10)*(3*10 )= 30,000,000 m/s Step 2: Find the uncertainty of the velocity Δu= 30,000,000*(1/100)= 30,000 Step 3: Find uncertainty of momentum -27 -22 Δp= m*Δu= (1.67*10 )*(30,000)= 5.01*10 Step 4: Plug in uncertainty of momentum and h into ΔxΔp≥ h/(4π) and solve for Δx Δx*(5.01*10 )= (6.636*10 )/(4π)---> Δx= 1.05*10 -13m -13 Answer: The uncertainty in the position of the proton is 1.05*10 meters 7) What type of orbital (ie, 3s, 4p,...) is designated by these quantum numbers? a) n=5, l= 1, m= l b) n=4, l= 2, m= -l c) n= 2, l= 0, m= 0 l a) What type of orbital (ie, 3s, 4p,...) is designated by n=5, l= 1, m= 0? l What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f Answer: n=5, l= 1, m= 0 islthe orbital 5p b) What type of orbital (ie, 3s, 4p,...) is designated by n=4, l= 2, m= -2? l What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f Answer: n=4, l= 2, m= -2 is tle orbital 4d c) What type of orbital (ie, 3s, 4p,...) is designated by n= 2, l= 0, m= 0? l What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f Answer: n= 2, l= 0, m= 0 is thl orbital 2s 8) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have a) n=4, l=2, m= 1, l = + ½ s b) n=4, l= 2, m= 1 l c) n=4 d) n=4, l=2, m = + ½s a) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have n=4, l=2, m= 1, l = + ½? s What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m =lmagnetic quantum number= -l, (-l+1),...,(l-1),+l corresponds to number of orbitals in each subshell (s:n=1,l=0, m=0l->only 1 orbital) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) Step 1: Determine orbital number orbital #=n=4 Step 2:Add subshell type subshell type==> l=2==> subshell d==> 4d Step 3: Find orbital number in subshell th l= 2==> ml could be: -2,-1,0,1,2---> m= 1= 4 magnltic quantum #---> 4d^4 Step 4: Check electron spin m =s+ ½ -----> only 1 Answer: Only 1 can have n=4, l=2, m= 1, m = + ½ l s b) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have n=4, l=2, m= 1, l = + ½? s What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m =lmagnetic quantum number= -l, (-l+1),...,(l-1),+l corresponds to number of orbitals in each subshell (s:n=1,l=0, m=0l->only 1 orbital) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) Step 1: Determine orbital number orbital #=n=4 Step 2:Add subshell type subshell type==> l=2==> subshell d==> 4d Step 3: Find orbital number in subshell l= 2==> m coull be: -2,-1,0,1,2---> m= 1= 4 magnelic quantum #---> 4d^4 Can have 2 in each orbital slot Answer: 2 can have n=4, l=2, m= 1 l c) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have n=4, l=2? What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m l magnetic quantum number= -l, (-l+1),...,(l-1),+l corresponds to number of orbitals in each subshell (s:n=1,l=0, m=l-->only 1 orbital) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) Step 1: Determine orbital number orbital #=n=4 Step 2:Add subshell type subshell type==> l=2==> subshell d==> 4d d subshell has 5 orbitals and can therefore have 10 electrons Answer: 10 can have n=4, l=2 d) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have n=4? What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m = magnetic quantum number= -l, (-l+1),...,(l-1),+l l corresponds to number of orbitals in each subshell (s:n=1,l=0, m=l-->only 1 orbital) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) Step 1: Determine orbital number orbital #=n=4 4 can have subshells s,p,d,f--> respectively, 1,3,5,7 orbitals= total of 32 electrons Answer: 32 can have n=4 e) Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have n=4, l=2, m = s ½? What we know: n tells you what the orbital is: orbital number=n tells you what subshell type is: l=0=s, l=1=p, l=2=d, l=3=f m l magnetic quantum number= -l, (-l+1),...,(l-1),+l corresponds to number of orbitals in each subshell (s:n=1,l=0, m=l-->only 1 orbital) each orbital can have up to 1 electron that is + ½ (↑) and one that is – ½ (↓) Step 1: Determine orbital number orbital #=n=4 Step 2:Add subshell type subshell type==> l=2==> subshell d==> 4d--> 5 orbitals Step 3: Check electron spin m s + ½ -----> only one electron per orbital Answer: 5 can have n=4, l=2, m = + ½ s 9) Identify the orbital that has a) 1 radial node, 1 angular node b) no radial nodes, 2 angular nodes c) 2 radial nodes, 3 angular nodes a) Identify the orbital that has 1 radial node, 1 angular node What we know: Total number of nodes= n-1 # radial nodes= n-(l+1) # angular nodes= l Step 1: Find n n-1= total number of nodes= 1+1= 2 ----> n= 3 Step 2: Find l l= number of angular nodes= 1 Step 3: Find subshell type l=1 ----> subshell type p Answer: 3p b) Identify the orbital that has no radial nodes, 2 angular node What we know: Total number of nodes= n-1 # radial nodes= n-(l+1) # angular nodes= l Step 1: Find n n-1= total number of nodes= 0+2= 2 ----> n= 3 Step 2: Find l l= number of angular nodes= 2 Step 3: Find subshell type l=2 ----> subshell type d Answer: 3d c) Identify the orbital that has no radial nodes, 2 angular node What we know: Total number of nodes= n-1 # radial nodes= n-(l+1) # angular nodes= l Step 1: Find n n-1= total number of nodes= 2+3= 5 ----> n= 6 Step 2: Find l l= number of angular nodes= 3 Step 3: Find subshell type l=3 ----> subshell type f Answer: 6f 10) What is the expected ground-state electron configuration for each of the following elements? a) mercury b) calcium c) polonium a) What is the expected ground-state electron configuration for mercury? Step 1: Identify the noble gas that precedes the element Xenon---> [Xe] Step 2: Identify first subshell of electrons that should be added athays s subshell 6 period---> 6s goes past s subshell--> 6s 2 2 [Xe] 6s Step 3: Identify the next subshell of electrons that should be added 6 period--> 4f subshell, which has 14 electrons 14 goes pas2 f 14bshell--> 4f [Xe] 6s 4f Step 4: Identify the next subshell of electrons that should be added th 6 period--> 5d subshell, which has 10 electr10s Hg is in the last spot of d orbital--> 5d Answer: [Xe] 6s 4f 5d 14 10 b) What is the expected ground-state electron configuration for calcium? Step 1: Identify the noble gas that precedes the element Argon---> [Ar] Step 2: Identify first subshell of electrons that should be added always s subshell 4 period---> 4s s subshell has 2 electrons Ca at end of 4s subshell--> 4s 2 Answer: [Ar] 4s 2 c) What is the expected ground-state electron configuration for polonium? Step 1: Identify the noble gas that precedes the element Xenon---> [Xe] Step 2: Identify first subshell of electrons that should be added always s subshell th 6 period---> 6s goes past s subshell--> 6s 2 2 [Xe] 6s Step 3: Identify the next subshell of electrons that should be added th 6 period--> 4f subshell, which has 14 electrons goes past f subshell--> 4f 14 2 14 [Xe] 6s 4f Step 4: Identify the next subshell of electrons that should be added th 6 period--> 5d subshell, which has 10 electrons goes past 5d subshell--> [Xe] 6s 4f 5d 2 14 10 Step 5: Identify the next subshell of electrons that should be added 6 period--> 6p subshell, which has 6 electrons Po is in 4 spot in 6p subshell--> 6p 4 2 14 10 4 Answer: [Xe] 6s 4f 5d 6p 11) Between which two orbits of the Bohr hydrogen atom must an electron fall to produce light of wavelength 1876 nm? 2 2 What we know: ΔE= R ((1/n )-(1Hn )) i f R =H2.179*10 -18 E= hC/λ h= 6.626*10 -34 8 C= 3*10 n's must be positive integer values greater than 0 Step 1: Convert wavelength from nanometers to meters 1876 nm*(1 m/10 nm)= 1.876*10 meters -6 Step 2: Plug in C, h, and wavelength to E= hC/λ E= (6.626*10 *3*10 )/(1.876*10 )= 1.06*10 -6 -19Joules 2 2 Step 3: Plug in E and R into ΔH= R ((1/n )-(1/H )) i f 1.06*10 -19Joules= 2.179*10 *((1/n )-(1/n )i2 f2 2 2 (1/n i-(1/n )f 0.0486276 Step 4: Guess and check values for n and n i f 2 2 (1/2 )-(1/3 )==> too large of a value (1/3 )-(1/4 )= 0.0486276 Answer: n= 3,in= 2 f Lewis Structures, resonance, VSEPR Important things to remember: Atomic radius increases from right to left and top to bottom on periodic table Cations (positive ions) are smaller than the atoms from which they are formed Anions are larger than the atoms from which they are formed Isoelectronic ions have identical electron configurations Ionization energy and ElectronAffinity increase from left to right and from bottom to top on periodic table Diamagnetic:All electrons are paired Paramagnetic: Unpaired electrons exist Practice Problems: 1) For each of the following pairs, indicate the atom that has the larger size: a) Te or Br b) K or Ca c) Ca or Cs d) N or O e) O or P f)Al orAu Answer: a) Te b) K c) Cs d) N e) P f)Au 2)Among the following ions, several pairs are isoelectronic. Identify these pairs: Fe , Sc , Ca , F, Co , Co , Sr , Cu^+, Zn ,Al 2+ 3+ Answer: Fe and Co (both Cr), Sc and Ca (bothAr), F andAl (both Ne), Cu and Zn (both Ni) + 2+ 3) Unpaired electrons are found in only one of the following species. Indicate which one: - 2+ 2+ 2- F, Ca , Fe , S F: 2s 2p + 1e= 8 e ====> [↑↓] [↑↓][↑↓][↑↓] 2+ 2 - - Ca : 4s – 2e= 0 e ====> [ ] Fe : 4s 3d – 2e= 6 e ====> [↑↓] [↑ ][↑ ][↑ ][↑ ][ ] S : 3s 3p + 2e = 8e ====> [↑↓] [↑↓][↑↓][↑↓] *Another helpful table Number of Electron- Number of VSEPR Molecular Ideal Bond Example electron group lone pairs Notation geometry Angles groups geometry (A=central) (degrees) 2 Linear 0 AX 2 Linear 180 BeCl 2 3 Trigonal 0 AX 3 Trigonal 120 BF 3 planar planar 3 Trigonal 1 AX 2 Bent 120 SO 2 planar 4 Tetrahedral 0 AX Tetrahedral 109.5 CH 4 4 4 Tetrahedral 1 AX 3 Trigonal 109.5 NH 3 pyramidal 4 Tetrahedral 2 AX 2 2 Bent 109.5 OH 2 5 Trigonal 0 AX 5 Trigonal 90, 120 PCl5 bipyramidal bipyramidal 5 Trigonal 1 AX 4 Seesaw 90, 120 SF4 bipyramidal 5 Trigonal 2 AX 3 2 T-shaped 90 ClF3 bipyramidal 5 Trigonal 3 AX 2 3 Linear 180 XeF 2 bipyramidal 6 Octahedral 0 AX 6 Octahedral 90 SF6 6 Octahedral 1 AX 5 Square 90 BrF5 pyramidal 6 Octahedral 2 AX 4 2 Square 90 XeF 4 planar 4) Each of the following molecules contains at least one multiple covalent bond. Give a plausible Lewis Structure for a) OCS b) CH CHO 3 c) F2CO d) Cl 2O e) C H 2 2 Answers: 5) The following polyatomic anions involve covalent bonds between O atoms and the central nonmetal atom. Propose an acceptable Lewis Structure for each: a) SO 32- - b) NO 2 Answers: Orbital hybridization: Important things to remember: The number of hybrid orbitals equal the total number of atomic orbitals that are combined' sp: linear (2 electron groups) sp : trigonal planar (3 electron groups) 3 sp : tetrahedral (4 electron groups) sp d: trigonal bipyramidal (5 electron groups) sp d : octahedral (6 electron groups) sigma vs pi bonds: first bond between two atoms is always sigma preceding bonds are pi if two atoms are triple bonded, there is 1 sigma bond and 2 pi bonds *pay attention to positive and negative signs on orbitals Molecular Orbitals: The number of molecular orbitals formed is equal to the number of atomic orbitals combined When two atomic orbitals are combined, one of the molecular orbitals is a bonding orbital at a lower energy, and one is an anti-bonding orbital at a higher energy In ground state configurations, electrons enter the lowest energy molecular orbit available The maximum number of electrons in a molecular orbital is two In ground-state configurations, electrons enter molecular orbits of identical energies singly before they pair up s orbital from top to right: Px, Pz, Py

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