Nov. 21 2003 NAME____________________________

Chemistry 101 Section 1

HOUR EXAM #3

12:10 – 1:30

IMPORTANT NOTE--For partial credit be sure to show your reasoning. Please attempt all parts of all problems--no partial credit can be given for blank spaces. If you need more paper, write on the back of the page and indicate clearly that you have done so.

(Possibly Useful )Physical Constants and Equations

NA = 6.022 x 1023 Mass of an electron 9.1 x 10-28 g

Mass of a proton 1.67 x 10-24 g h = 6.63 x 10 -34 J.s/photon

Mass of a neutron 1.67 x 10-24 g c = 3.0 x 10 8 m/s

E = hn E= hc/l 273 K = 0 °C. P V = n R T

Visible light 400 nm to 900 nm UV light 200 to 400 nm

Density of water 1 g/cm3 R = 0.082 L atm / K mol = 8.314 J / K mol

R = 0.082 L atm / K mol = 8.314 J / K mol

Specific heat of water = 4.18 J / g °C 1.00 atm = 760 mm Hg

Approximate Grading Scale

>84 4.0

>70 3's

>49 2's

>20 1's

The Chemistry 101/104 marathon is almost 50% complete. Don’t give up! Hard work, and perseverance will get you to your goals. Come to class, pay attention, read your text, do HW problems and more, attend office hours and PLI sessions, and seek connections between lab and lecture. Above all, relax so you can think clearly about each problem. Remember, Philadelphia is the city of Rocky and Allen Iverson, two “Marathon Men” who don’t give up.

Good Skill!

NAME_____Basic to HW level_________________

Problem 1 (20 points)

Fill in the blanks and circle the correct choices—the options are separated by “or”.

a) The ____O____ atom has a ground state electronic configuration of 1s² 2s² 2p⁴

b) The ___N_______anion has a charge of –1 and a ground state electronic configuration of 1s² 2s² 2p⁴

c) The _____F______cation has a charge of +1 and a ground state electronic configuration of 1s² 2s² 2p⁴

d) The bond order of a molecule having the electronic configuration 1 is _2_______ and the molecule is paramagnetic or diamagnetic.

e) The overlap of a 2s and an sp2 orbital produces a _sigma_____ bond or no bond

f) The overlap of a 2 pz and a 2 pz bond produces a _pi______bond along the x-axis or no bond

g) The overlap of a 2py and a 2pz orbital produce a _________bond along the x-axis or no bond

h) Phosphorus has a ___larger_______ atomic radius than nitrogen, but is __less__electronegative than N.

i) The bond angle around a central atom having two bonding electron pairs and two lone pairs is slightly _less______than ____109.5_______degrees.

j) A molecule whose central C atom makes two double bonds has a ____linear__shape.

NAMEHW level/synthesis 5 pts__________

Problem 2 (20 points)

Draw Lewis structures for the following molecules and ions.

For each one, indicate resonance structures, formal charges, the shape around the central atom and whether it is polar or non-polar.

The central atom is underlined.

CH₂F₂

Tetrahedral and polar with formal charges on each atom = 0.

You need to draw and think in 3D in order to see that this is polar with the negative end towards the Fs.

ClO₃^-1

Pyramidal with 2 Cl=O double bonds. There are 3 resonance structures and the O end of this polar ion is the negative end. Cl and 2 O’s have formal charges of 0 and one O has a FC of –1. If your structue has just one Cl=O then the Cl FC is +2 which is not as good.

NNO

Linear molecule slightly negative towards the O. The bottom structure is slightly better due to the FC distribution. The two structures can be considered resonance structures.

NAME_____________HW level_______________

Problem 3 (30 points)

a) Using valence bond theory and accurate drawings, describe the bonding of azide, N3 ^–1 as completely at you can.

N=N=N ^–1

Please remember that only hybridization around the central atoms is of interest and in this molecule we would count 2 electrion groups around the central N so it’s linear.

Problem 3 (continued)

b) In the epinephrine (or adrenaline) molecule below, indicate the geometry around each central atom. Before starting remember that you may need to add lone pairs, H or C to the drawing.

1) Indicate two atoms having sp³ geometry

Any non-ring C, O, or N

2) Indicate two atoms having sp² geometry

All the ring C’s

3) Is the ring planar? Explain your answer.

Yes—each C is trigonal planar

4) Find two carbon-carbon bonds having different lengths and state which one is shorter.

All the ring C-C bonds are the same length and shorter than the ones outside the ring.

5) What is the formal charge of N?

FC on N = 0

6) Are resonance structures possible?

Yes, in the ring—this makes each C-C equivalent to a 1.5 bond order.

7) Is this molecule soluble in water? Why or why not?

Probably, it is slightly polar due to the O-H groups on one end.

8) Indicate a bond that can rotate freely that changes the overall shape of the molecule.

Any sigma bond linking groups of atoms, but you must indicate a bond and not an atom.

Synthesis level

Problem 4 (20 points)

A mystery ion under study in the Bryn Mawr College Physics Department has 3 electrons and a snazzy new tunable laser excites the ground state ion to different excited states. These excited ions can be quickly trapped and analyzed. The ground state ion has two atoms and a bond length of 149 pm. As the laser scans various wavelengths starting from 900 nm, the first absorption occurs at 720 nm and produces an excited state ion whose bond length is 140 pm. The final absorption occurs at a wavelength of 210 nm and this excited state ion has a bond length of 149 pm. The Physics student running the experiment is perplexed about the changes in bond length and seeks help from Chemistry students who have recently studied chemical bonding.

a) Using the energy diagram for the ions orbitals below, explain to the physics student what is happening in the two experiments and why the bond length varies.

b) What is energy difference between the HOMO (highest occupied molecular orbital) and the LUMO (lowest unoccupied molecular orbital)? HOMO and LUMO both refer to the ground state electron configuration. Express you answer in kJ/mol.

This problem is meant to see if you can synthesize many of the ideas we’ve learned about in a completely different context. You should try to relax and think about solving it bit by bit. And don’t let this problem bother you on other problems.

The ground state has 3 electrons and a configuration of s 1s ² s *1s ¹ and a bond order of 1/2 that corresponds to a bond length of 149 pm. The first transition (red arrow) goes from the HOMO to the LUMO. The configuration of the excited state is s 1s ² s 2s ¹ and the bond order is 3/2 so the bond is shorter, 140 pm. The second transition (blue arrow) creates an excited state whose configuration is s 1s ² s * 2px ¹ and the bond order is 1/2 so we again have a bond length of 149 pm.