Chemistry 65.100 A and V
Midterm Test #2
November 3, 2000

ANSWERS

A1. In the dichromate ion, Cr₂O₇^-2, the Cr atoms each have an oxidation number of +6.

A2. The d-orbitals are close in energy to the p-orbitals of phosphorus. Thus, bonds can be made between phosphorus and other atoms if these other atoms donate the bonding electrons into these d-orbitals.

A3. This reaction has 3 moles of gas on the left and only 2 on the right. Since the number of moles decreases, we expect the value of DS to be negative.

A4. If we remove an electron and the bond length decreases, that electron must have been in an antibonding molecular orbital.

A5. Since the B atom is making three bonds with other atoms, it must be using sp² hybrid orbitals.

A6. Enthalpy is a state function. Its value says NOTHING about the rate of a reaction.

B1. In all cases below, we put the least electronegative element at the centre!

(a) AsH₂⁺: Valence electrons = 5 + (2 x 1) - 1 = 6.
The skeletal structure is thus:

The other two electrons go on the As atom:

And the form is thus AX₂E. This is a bent molecule.

(b) AsCl₅: Valence electrons = 5 + (5 x 7)  = 40.
The skeletal structure is thus:

Completing the octets around the terminal atoms gives:

which uses all 40 electrons. This molecule is thus of the form AX₅. The shape is therefore trigonal bipyramidal.

(c) CBr₄: Valence electrons = 4 + (4 x 7) = 32
The skeletal structure is thus:

Completing the octets around the terminal atoms gives:

which uses all 32 electrons. The molecule is thus of the form AX₄, which is tetrahedral.

(d) BrCl₂^-: Valence electrons = 7 + (2 x 7) +1 = 22
The skeletal structure is thus:

Completing the octets around the terminal atoms gives:

which uses only 16 electrons. The last 6 go as lone pairs on the central Br atom:

This is of the form AX₂E₃, which is linear.

B2. First, we split the reaction into the oxidation and reduction half-reactions and balance each separately:
Oxidation: Bi⁺³ Ž BiO₃^-_(aq)   (here, Bi goes from an oxidation state of +3 to +5)
balancing for O: Bi⁺³ + 3 H₂O_(l) Ž BiO₃^-_(aq)
balancing for H: Bi⁺³ + 3 H₂O_(l) Ž BiO₃^-_(aq) + 6 H⁺_(aq)
balancing for charge: Bi⁺³ + 3 H₂O_(l) Ž BiO₃^-_(aq) + 6 H⁺_(aq) + 2 e^-

Reduction: Cr₂O₇^-2_(aq) Ž Cr⁺³_(aq) (Cr is reducred from +6 to +3)
balancing for Cr: Cr₂O₇^-2_(aq) Ž 2 Cr⁺³_(aq)
balancing for O: Cr₂O₇^-2_(aq) Ž 2 Cr⁺³_(aq) + 7 H₂O_(l)
balancing for H: Cr₂O₇^-2_(aq) + 14 H⁺_(aq) Ž 2 Cr⁺³_(aq) + 7 H₂O_(l)
balancing for charge: Cr₂O₇^-2_(aq) + 14 H⁺_(aq) + 6 e^-  Ž 2 Cr⁺³_(aq) + 7 H₂O_(l)

To get the electrons to cancel, multiply the oxidation reaction by 3, and add to the reduction reaction:

3 Bi⁺³ + 9 H₂O_(l) Ž 3 BiO₃^-_(aq) + 18 H⁺_(aq) + 6 e^-
Cr₂O₇^-2_(aq) + 14 H⁺_(aq) + 6 e^-  Ž 2 Cr⁺³_(aq) + 7 H₂O_(l)
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3 Bi⁺³ + 2 H₂O_(l) + Cr₂O₇^-2_(aq) Ž 3 BiO₃^-_(aq) + 2 Cr⁺³_(aq) + 4 H⁺_(aq)  (This is balanced in acidic solution)

To convert to a basic solution, add 4 OH^-_(aq) to each side:
3 Bi⁺³ + 2 H₂O_(l) + Cr₂O₇^-2_(aq) + 4 OH^-_(aq) Ž 3 BiO₃^-_(aq) + 2 Cr⁺³_(aq) + 4 H⁺_(aq) + 4 OH^-_(aq)

The 4 H⁺_(aq) + 4 OH^-_(aq) make 4 H₂O_(l):
3 Bi⁺³ + 2 H₂O_(l) + Cr₂O₇^-2_(aq) + 4 OH^-_(aq) Ž 3 BiO₃^-_(aq) + 2 Cr⁺³_(aq) +  4 H₂O_(l)

And finally, the waters cancel out giving:
3 Bi⁺³ + Cr₂O₇^-2_(aq) + 4 OH^-_(aq) Ž 3 BiO₃^-_(aq) + 2 Cr⁺³_(aq) +  2 H₂O_(l)

B3. The desired reaction is:

2 N_2(g) + 5 O_2(g) ® 2 N₂O_5(g)

and we have as data:

N₂O_5(g) + H₂O_(l) ® 2 HNO_3(l) DH = -77 kJ ("reaction 1")
1/2 N_2(g) + 3/2 O_2(g) + 1/2 H_2(g) ® HNO_3(l) DH = -174 kJ ("reaction 2")
H_2(g) + 1/2 O_2(g) ® H₂O_(l) DH = -286 kJ ("reaction 3")

Since we need 2 N_2(g) on the left side, start by using 4 x reaction 2:

2 N_2(g) + 6 O_2(g) + 2 H_2(g) ® 4 HNO_3(l) DH = 4(-174) = -696 kJ

And since we need only 5 O_2(g) on the left side and no H_2(g), take -2 times reaction 3:

2 H₂O_(l) ® 2 H_2(g) + O_2(g) DH = -2(-286) = +572 kJ

Then use -2 times reaction 1:

4 HNO_3(l) ® 2 N₂O_5(g) + 2 H₂O_(l)  DH = -2(-77) = +154 kJ

Adding these three reactions gives the desired reaction:

2 N_2(g) + 5 O_2(g) ® 2 N₂O_5(g)

and its enthalpy change is the sum of the above: DH = -696 + 572 + 154 = +30 kJ