Chemistry 65.100 A and V

Second Test - November 8, 1996 5:00 - 6:30 PM

ANSWERS

1a. COCl₂ O(-2), Cl(-1), thus C(+4)

1b. CrO₄^-2 O(-2), thus Cr(+6)

1c. NH₂OH H(+1), O(-2), thus N(-1)

2a. H₅IO₆ is the oxidizing agent and I₂ is the reducing agent

I₂ --> IO₃^- is one unbalanced half reaction

H₅IO₆ --> IO₃^- is the other

I₂ + 6 H₂O --> 2 IO₃^- + 12 H⁺ + 10 e^- is balanced

H₅IO₆ + H⁺ + 2 e^- --> IO₃^- + 3 H₂O is balanced

thus, I₂ + 6 H₂O --> 2 IO₃^- + 12 H⁺ + 10 e^-

and 5 H₅IO₆ + 5 H⁺ + 10 e^- --> 5 IO₃^- + 15 H₂O

can be added to give the balanced reaction:

I₂ + 5 H₅IO₆ --> 7 IO₃^- + 7 H⁺ + 9 H₂O

2b. CrO₄^-2 is the oxidizing agent and AsH₃ is the reducing agent

CrO₄^-2 --> Cr⁺³ is one unbalanced half reaction

AsH₃ --> As is the other

CrO₄^-2 + 8 H⁺ + 3 e^- --> Cr⁺³ + 4 H₂O is balanced

AsH₃ + 3 OH^- --> As + 3 H₂O + 3 e^- is balanced

these can be added to give the balanced reaction:

CrO₄^-2 + AsH₃ + 8 H⁺ + 3 OH^- --> Cr⁺³ + As + 7 H₂O

combining the H⁺ and OH^- on the LHS gives:

CrO₄^-2 + AsH₃ + 5 H⁺ + 3 H₂O --> Cr⁺³ + As + 7 H₂O

cancelling the waters gives:

CrO₄^-2 + AsH₃ + 5 H⁺ --> Cr⁺³ + As + 4 H₂O

to balance the equation in basic solution, add 5 OH^- ions to each side, combine those on the LHS with the H⁺ ions, and cancel the waters:

CrO₄^-2 + AsH₃ + H₂O --> Cr⁺³ + As + 5 OH^-

3. ‘’Formal charge’’ is the difference between the number of valence electrons in the free, unbound atom, and the number associated with the atom in a compound.

Note that this year, I want the shape of the molecule, not the shape of the electron clouds about the central atom for any VSEPR question:

4a. CrO₄^- The number of valence electrons is equal to 6 + (4 x 6) + 1 = 24. The Lewis structure for this compound is thus:

We see that there are no lone pairs around the central atom, thus the VSEPR notation is AX₄, and the shape must be TETRAHEDRAL.

4b. SiCl₄ The number of valence electrons is equal to 4 + (4 x 7) = 32. The Lewis structure for this compound is thus:

We see that there are no lone pairs around the central atom, thus the VSEPR notation is AX₄, and the shape must be TETRAHEDRAL.

4c. I₃^- The number of valence electrons is equal to (3 x 7) + 1 = 22. The Lewis structure for this compound is thus:

We see that there must be three lone pairs around the central atom, thus the VSEPR notation is AX₂E₃, and the geometry is TRIGONAL BIPYRAMIDAL. However, the three I atoms are LINEAR (either answer is OK)

5. DH^o = DH^o_f(SO_3(g)) - DH^o_f(SO_2(g)) (note that DH^o_f(O_2(g)) = 0)

= -395.7 - (-296.8)

= -98.9 kJ/mol

1 kg SO₂ x (1000 g / 1 kg) x (1 mol SO₂ / 62.1 g SO₂) x (98.9 kJ released / 1 mol SO₂)

= 1,593 kJ released

6a. Dissolving NaCl in water results in a POSITIVE DS, since the ions go from a very ordered state in the crystal to a relatively disordered state in solution

6b. Evaporating Cl₂ liquid to gas results in a POSITIVE DS, since a gas is much more disordered than a liquid.

7a. bond order of C₂ = (6 - 2)/2 = 2

7b. C₂⁺ bond order = (5 - 2)/2 = 1.5

7c. C₂⁺⁴ has lost all four p_2p electrons. Its bond order is thus (2 - 2)/2 = 0, and so could not exist

7d. The bond order of C₂ is greater than that of C₂⁺. The bond energy of C₂ is therefore greater.

7e. C₂⁺ is paramagnetic, since it has one unpaired electron in the p_2p orbital.