Chemistry 65

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                                      CHEM 1000 A, V Midterm Test #1
                                            October 17, 2003

Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.     Ammonia, NH3, has a tetrahedral arrangement of charge clouds around the nitrogen atom. Why is the
       H-N-H bond angle (107.3°) less than the ideal tetrahedral angle (109.5°)?

       In NH3, there is a lone pair of electrons on the N atom. This pair repels the N-H bonding pairs more
       strongly than they repel one another, resulting in the H-atoms being pushed together slightly, reducing
       the bond angle.

2.     What two concepts are brought together in the Planck equation, E = h?

       This equation brings together the ideas that light is wavelike, hence has a wavelength, , and has
       particle-like properties, i.e. is composed of photons having energy E.

3.     Name three physical properties of molecules that can be explained using molecular orbital theory

       MOs correctly predict the magnetic properties of molecules (i.e. para- or diamagnetism), as well as the
       variation of bond length and bond energies when the molecules are ionized. Also, by calculating the
       bond order, we can preedict whether the molecule has a single, double or triple bond (any 3 of these 4).

4.     Which of CaBr2 or SrCl2 is more ionic? Why?

       The difference in electronegativity between Sr and Cl is greater than that between Ca and Br. Thus,
       SrCl2 is more ionic.

5.     Explain how the electron affinity changes going from left to right across a period, and why.

       Electron affinity increases (i.e. becomes more negative) going from left to right across a period. This is
       because the effective nuclear charge increases in this direction due to an increasing number of protons in
       the nucleus, i.e. less shielding of the nucleus by the valence electrons going L to R. The energy released
       upon addition of an electron is therefore greater.

6.     In the molecule C2H2, there is a triple bond between the carbon atoms. What kinds of orbitals does one
       carbon atom use to bond with the other?

       Each carbon uses an sp hybrid orbital and two p-orbitals to bond with the other.

 Part B. Answer any two questions. If you answer all three, the best two answers will count.
(20 marks each)

B1.    (a) Use VSEPR theory to predict the shape of XeCl4.

       8 + (4 x 7) = 36 electrons. Making single bonds from the central Xe atom to each terminal Cl atom uses
       8 of these. Completing the octets around each Cl atom uses another 24, for a total of 32 electrons used.
       This leaves two pairs to assign to the Xe atom. The shorthand notation of XeCl4 is therefore AX4E2,
       which is a square planar molecule.

(b) What kind of hybrid orbitals is the S atom using to bond with the O atoms in a molecule of SO2?.

Using VSEPR theory, SO2 is of the form AX2E, and there there are three electron clouds around the
central S atom. The S atom must therefore be using sp2 hybrid orbitals to bond with the O atoms.

(c) A molecular orbital (MO) diagram for fluorine (F2) is shown below. How can you deduce from the
MO diagram that F2 has a single bond?

The bond order = (8 – 6) / 2 = 1, i.e. a single bond.

(d) What will happen to the bond length and bond energy if we ionize F2 to F2+ and why?

The electron lost upon ionization comes from the 2p orbital. the bond order of F2+ is thus (8 – 5) / 2 =
1.5. In other words, loss of an antibonding electron causes the bond order to increase and so the bond
energy of F2+ will be greater than that of F2, and the bond length of F2+ will be less than that of F2.

(e) Is F2 paramagnetic, diamagnetic, both or neither? Why?

F2 contains paired electrons, but no unpaired electrons. It is therefore diamagnetic, but not paramagnetic.






B2.    (a) An electron falls from the n = 5 level in a hydrogen atom and emits a photon having a wavelength of
       1281 nm. To which level, m, does it fall?

       1     1  1 R R
         R 2  2 2  2
           m n  m n
           R 1 R
       or,     
           m2  n 2
                1   1
       or, m      2
                R n 
                   1          1
                   1
                              2 3
          0.01097nm (1281nm) 5 

       (b) What is the energy of the photons having a wavelength of 1281 nm? Express in kJ mol-1.

        E  h
                           3.00  108 m s 1
         6.63  1034 J s
                            1281 10 9 m
         1.55  10 19 J (per photon)

        x6.02  10 23 photon mol 1  93519 J mol 1
         93.52 kJ mol 1

B3. (a) Predict the formulas of compounds formed from:
       (i)    Mg and S     MgS
       (ii)   Na and Se Na2Se
       (iii)  V with S     V2S5
       (iv)   Y with O     Y2O3
       (v)    Al with N AlN

(b) Why is the ionization potential of Zn greater than those of its two neighbours, Cu and Ga?

       Zinc’s electronic structure is [Ar] 4s2 3d10. Since this is two filled sub-shells, the electrons are
       energetically stable, i.e. at lower energies than those of Cu ([Ar] 4d13d10) or Ga ([Ar]4d23d104p1), and
       are therefore more difficult to remove, i.e. require more energy to remove.

(c) Which of As or Se has the larger atomic radius? Why?

       As is a larger atom since Se has more protons, which draw the electrons inwards.

(d) Why is the second ionization potential of each element greater than the first?
       The first electron is being removed from a singly charged positive ion. The second is being removed
       from a doubly charged ion, which requires more energy.

(e) Which of LiF or KBr would have the higher lattice energy? Why?

       Li and F are smaller than K and Br respectively. Hence, the Li is closer to the F than the K is to the Br,
       resulting in a larger lattice energy in LiF.

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