7 Molecular Structure questions

Questions

1. Draw Lewis structures for the following molecules and ions and use VSEPR to predict their shapes
   1. H₃O⁺,
   2. OH^–,
   3. CH₃OH,
   4. CH₂F₂,
   5. HOCl,
   6. CH₂O,
   7. PH₃,
   8. H₂S.
2. Draw Lewis structures for the sulfate ion and the nitrate ion, including resonance structures. Use VSEPR to predict their shapes.
3. Draw three different forms of phosphorus chloride entities found in natural crystals, Indicate if any of them has resonance structures and predict their shapes with VSEPR and find the formal charge on the P atom in each.
   1. PCl₅
   2. PCl₄⁺
   3. PCl₆^–
4. Draw Lewis structures and use VSEPR to predict the shapes of the following ions
   1. CLF₄⁺
   2. AsF₆^–
   3. NO₃^–
   4. SO₃^2-
5. Estimate the enthalpy change for the reaction using average bond energies
   CH₄(g) + 2 Cl₂(g) –> CH₂Cl₂(g) + 2 HCl(g)                                                                         [-202 kJ/mol]
6. Estimate the enthalpy change for the pyrolysis of ethane
   H₃C-CH₃ –> H₂C=CH₂ + H₂                                                                                                   [+140 kJ/mol]
7. Estimate the enthalpy of formation for trichloromethane in the gas phase using bond energies and any other necessary thermochemical data                                                                             [-93 kJ/mol]
8. Iodine has a normal valence of 1 and normal hyper valence states of 3,5,7. Draw orbital box diagrams for each of these four valence states.
9. Which diatomic molecules of the elements of the first two rows contain valence electrons in the * anti-bonding molecular orbitals
10. Which diatomic molecules of the elements of the first two rows contain valence electrons in the * antibonding orbitals.
11. What is the bond order of the following diatomic species?
    1. H₂
    2. N₂
    3. C₂
    4. He₂⁺
    5. N₂⁺
    6. O₂^–
12. What is the hybridization of all the atoms (except hydrogen) in the following molecules:
    1. C₂H₆
    2. CH₃NH₂
    3. CH₃OH
    4. CH₃CH=CH₂
    5. CH₃CCH
    6. CH₂O
13. Which of the first 10 elements do not form diatomic neutral molecules?
14. Draw the following species and predict their shape and the hybridization of the central atom.
    1. OH^–
    2. NH₂^–
    3. H₃O⁺
15. Draw a Lewis structure and a sketch showing the orbitals of the ethyne (aka acetylene) molecule, C₂H₂.  What is the hybridization of the carbon orbitals?  indicate where the sigma and pi orbitals are and indicate with + or -, the phases of the various parts.

Answers

1.

1. H₃O⁺,
2. OH^–,
3. CH₃OH,
4. CH₂F₂,
5. HOCl,
6. CH₂O,
7. PH₃,
8. H₂S.

2.

The sulfate ion has 6 resonance structures shown.  The S atom has 4 domains, so is tetrahedral geometry and since no lone pairs, also tetrahedral shape.

The nitrate ion has three resonance structures.  The N can have only 4 bonds maximum because it’s a row 2 element so only has four orbitals available (2s ,2p_x,2p_y,2p_z).  So it can only form one double bond at a time.  Since N normally has 3 bonds and one lone pair, this drawing shows N looking like C, so it must have a +1 charge.  There are three domains around the N so it’s geometry is trigonal planar. Since there are no lone pairs on the N, the shape is also trigonal planar.

3.

4.

a.	The Chlorine has 4 bonds and one lone pair, like a column 6 hypervalent atom. But Chlorine is column 7 so its charge is +1. Because the fluorine cannot have more than one bond, there is no resonance.
b.	The arsenic is a column 7 element but is drawn here looking like a nobel gas so charge is -1.  none of the Fluorine atoms can have more than one bond so no resonance.
c.	The N has 4 bonds, so looks like C, it is positive.  the double bonded O is neutral (looks normal) and the single bonded O are negative 1 (look like F).  There is resonance possible here as each O in turn can be double bonded; so, three resonance structures.
d.	The S has four bonds and one lone pair. It looks like a normal hypervalent column 6 atom so it is neutral. The double bonded O is neutral and the single bonded O is -1.  Since the O atoms can take turns being double bonded, three resonance structures are possible.  The S is tetrahedral geometry (4 domains) and trigonal pyramid shape (one of the domains is a lone pair.)

5.
CH₄(g) + 2 Cl₂(g) –> CH₂Cl₂(g) + 2 HCl(g)

draw the Lewis structures for each molecule and count the number and type of each bond formed and broken.

CH₄ ==> 4 C-H bonds broken
Cl₂ ==> 1 Cl-Cl bond per molecule so 2 bonds broken

CH₂Cl₂ ==> 2 C-H bonds formed and 2 C-Cl bonds formed
HCl ==> 2 H-Cl bonds formed.

let’s look up the average bond energy for each of these bonds and add them into the equation.

Notice that I added four C-H bond energies and then subtracted 2 of them.  I could have cancelled out two of those and just added two C-H bond energies to get the same answer.

6.             H₃C-CH₃ –> H₂C=CH₂ + H₂

# bonds

H₃C-CH₃ | C-H ==> 6  C-C ==> 1

H₂C=CH₂ | C-H ==> 4  C=C ==> 1

net two C-H bonds broken

H₂          |   H-H ==> 1

This time, I’m going to take make note that I only break 2 of the C-H bonds.

7.  The formation reaction is a defined type.  one mole of the named material is the only product and all reactants are elements in their standard state.

Formation of trichloromethane (CHCl₃) is

C(s,graphite) + 1/2 H₂(g) + 3/2Cl₂(g) CHCl₃         

So, we have a problem here.  Bond energies are only reliable if all reactants and products are gases. So, we need to add in the energy needed to raise the solid carbon to gaseous state.  That data can be found in the thermodynamic tables.

C(s) C(g)  kJ/mol

So, our gas-phase reaction (not quite the formation of CHCl₃) will be

C(g) + 1/2 H₂(g) + 3/2Cl₂(g) CHCl₃
This is the equation we can get enthalpy using bond energies.

looking at the gas phase reaction, we see that we broke 1/2 of an H-H and 2/3 of a Cl-Cl bond and formed a C-H, and 3 C-Cl bonds.

kJ/mol

so

8.  Iodine valence states

9 and 10.

Draw the energy levels for the atoms (see below) and then fill them up from bottom to top (Aufbau process) with the correct number of electrons.  Any molecule with more than 10 electrons will have at least one in an antibonding orbital, but it will take more than 14 electrons to reach the σ* orbital.
(remember single-single-double single-double single)

11.  Use the same manifold of energy levels for the molecules we used in 9 and 10 and count bonds-antibonds.  a bond is a bonding orbital with 2 electrons in it; similar for antibonds.  For n=2 molecules, I will ignore the 1s energy levels since they will always cancel out.

1. H₂   2 electrons, so 1 bond → BO = 1
2. N₂   10 electrons, so 4 bonds – 1 antibond → BO = 3
3. C₂   8 electrons, so 3 bonds – 1 antibond → BO = 2
4. He₂⁺  3 electrons, so 1 bonds – .5 antibond → BO = .5
5. N₂⁺   9 electrons, so 4 bonds – .5 antibond → BO = 3.5
6. O₂^–    13 electrons, so 4 bonds – 2.5 antibond → BO = 1.5

Note that the half bonds and/or antibonds are the result of an orbital with only 1 electron.

12. What is the hybridization of all the atoms (except hydrogen) in the following molecules:

1. 1. C₂H₆ rewrite structurally H₃C-CH₃ each C has 4 domains (3 bonds to H and 1 to the neighboring C. So both C atoms are sp³ hybridized.
   2. CH₃NH₂ the C and the N both have 4 domains (C: 4 bonds, N: 3 bonds & 1 lone pair) so both are sp³ hybridized.
   3. CH₃OH  The C has 4 domains, so sp³. The O has 2 bond domains and 2 lone pairs, so also sp³.
   4. CH₃CH=CH₂ The first C has four domains (3 C-H bonds and one C-C bond) so sp³.  The two double bonded C atoms have only 3 domains (2 single bonds and 1 double bond) so sp² hybridized.
   5. CH₃CCH  The first C is sp³, like the ones in the previous questions.  The two C atoms that have a triple bond on them have only 2 domains (1 single bond and 1 triple bond), so they are sp hybridized.
   6. CH₂O  redraw structurally.  H₂C=O. C has 3 bonding domains (2 single bonds and 1 double bond) and O has 1 double bond and 2 lone pairs for a total of 3 domains.   Both the C and the O are sp² hybridized.

13.  Use the same energy level diagrams as for 9,10,11.  Fill up the orbitals according to Aufbau, look for molecules with unpaired electrons.  Those will be paramagnetic, also look for molecules with bond order of zero.  Those simply won’t form a molecule (diatomic or paramagnetic).  I’ll give the answers but leave it to you to draw all those diagrams ;).

1. H₂    BO = 1, diamagnetic ♦
2. He₂ BO = 0 so no molecule
3. Li₂   BO = 1, diamagnetic ♦
4. Be₂ BO = 0 so no molecule
5. B₂   BO = 1 (two half-filled anti bonds, so paramagnetic)
6. C₂    BO = 2, diamagnetic ♦
7. N₂    BO = 3, diamagnetic ♦
8. O₂    BO = 2, (two half filled π* orbitals, so paramagnetic)
9. F₂     BO = 1, diamagnetic ♦
10. Ne₂   BO = 0, so no molecule

14.

1. OH^–  pretty simple.  3 lone pairs on O and one O-H bond.  O is sp³ hybridized shape is moot, there’s only 1 bond.
2. NH₂^–  Also simple,  N has 2 bonds and 2 lone pairs (the extra electron due to the charge), so sp³hybridized, tetrahedral geometry and ‘bent’ shape.
3. H₃O⁺  O has 3bonds and 1 lone pair; so 3 domains, sp³ hybridized.

15.  Here is a sketch of the orbitals, done two ways.  the first has just sticks coloured to represent the type of the bond.  the second is an attempt to show the 3d shape of the same colour coded orbitals.  note that the blue π bond is perpendicular to the screen so I represented the part behind the page with dots.