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Chemistry 1010
Fall (Loader) Version 2
Test # 3
[50 Marks Total]
Name: _____________________
November, 17th 2003
MUN #: ______________________
Marking Scheme
QUESTIONS
Part A
Part B
Part C
VALUE
20
15
15
MARK
Total
Part A: Multiple choice questions. Circle the letter for the one correct answer.
[Each Question = 2 marks]
1.
Sulfur can form compounds with various oxidation numbers. In which of the
compounds
2−
A: S2Cl2, B: H2 S 2 O 7 , C: SO 2−
3 , D: SO 4 , E: SO 2 , F: Na 2 S
does sulfur have the oxidation number of +6?
(a) B and D
(b) C and E
(c) A and F
(d) B and C
(e) B and E
2.
2 H+(aq) + Zn(s) → Zn2+(aq) + H2(g)
2 H+(aq) + Pb(s) → no reaction
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Pb2+(aq) + Zn(s) → Pb(s) + Zn2+(aq)
Pb2+(aq) + Cu(s) → no reaction
Based on these results, the activity series for the elements used in the
reactions is
(a) Cu > Pb > H2 > Zn
(b) H2 < Zn < Pb < Cu
(c) H2 > Zn > Cu > Pb
(d) Zn > H2 > Pb > Cu
(e) Zn > Cu > H2 > Pb
3.
Calculate the frequency of visible light having a wavelength of 486.1 nm.
(a) 6.17 x 1014 Hz
(b) 4.86 x 10–7 Hz
(c) 2.06 x 1014 Hz
(d) 2.06 x 106 Hz
(e) 1.20 x 10–15 Hz
4.
How many unpaired electrons are present in the ground state of a chlorine
atom?
(a) 5
(b) 4
(c) 3
(d) 2
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(e) 1
CHEMISTRY 1010
5.
According to the octet rule (law), what ionic charges are necessary for the
element aluminum, Al, and sulfur, S, so that each has a full valence octet of
electrons?
(a) Al3+ and S3(b) Al3+ and S2(c) Al3+ and S6(d) Al2+ and S2(e) Al+ and S3-
6.
Choose the general electron configuration (where n = principal quantum
number) that best describes the nitrogen family of elements as a group.
(a) ns1 np4
(b) ns2 np4
(c) ns2 np5
(d) ns2 np6
(e) ns2 np3
7.
For the elements of the Third Period of the Periodic Table, going from left to
right, which of the following trends is true?
(a) electronegativity increases
(b) atomic size increases
(c) the attractive force between the electrons and the nucleus generally
decreases because the nuclear charge increases
(d) the outer electrons are less strongly held by the atom
(e) the elements become more metallic in their properties
8.
Which of the following bond types is likely to be the least polar?
(a) N − H in NH3
(b) O − H in H2O
(c) F − H in HF
(d) Cl − H in HCl
(e) C − H in CH4
9.
Select the best choice to complete the equation Li2O(s) + H2O(l) → ?
(a) Li2(OH)2(s)
(b) 2 LiO(aq) + H2(g)
(c) 2 Li+(aq) + 2 OH-(aq) (In any reaction where O2-(aq) might form the
oxide ion reacts with the water to give 2 OH-(aq) )
(d) 2 LiH(s) + O2(g)
2−
(e) Li 2+
2 (aq) + O (aq)
10.
Which of the following species has the largest radius
(a) Sr 2+
(b) Br −
(c) Rb +
(d) Kr
(e) Se 2− (Note: All of these ions are isoelectronic)
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CHEMISTRY 1010
Part B
Short Answer Questions
(20 Marks)
Show your working in the space provided.
B1.
[2]
(a)
Copper metal dissolves in nitric acid by reaction with the nitric acid
or nitrate ion in acid solution rather than by the reaction directly with
hydrogen ion from the acid. The unbalanced equation is:
Cu(s) + HNO3 (aq) → Cu2+(aq) + NO2(g) + H2O(l)
(i)
Assign the oxidation number to each element in the reaction
before and after the reaction.
(−2)
Cu(0)(s) + H (+1) N (+6) O 3 (aq) → Cu(+2)2+(aq) + N(+4)O(-2)2(g) + H(+1)2O(-2)(l)
The numbers in parenthesis are the oxidation numbers
for the elements.
(ii)
[2]
(b)
[2]
Indicate which species is the reducing agent (reductant) and
which is the oxidizing agent (oxidant) in the reaction.
The reducing agent is the Cu(s) (it is oxidized)
The oxidizing agent is the HNO3 (aq) (it is reduced)
Draw the shapes and show the relative orientation of the 4p
orbitals.
A picture of these can be fount on page 177 in your textbook.
(d)
[2]
Each of the electron box diagrams violates one of the rules of
aufbau for ground-state electron configurations. In each case name
(i)
4s ↑
Ar
4d ↑ ↑ ↑ ↑ ↑
4p ↑ ↑ ↑
Ar 4s ↑ 3d ↑ ↑ ↑ ↑ ↑ 4p ↑ ↑ ↑
The aufbau rules state that the obitals fill in order of
increasing energy so that the 3d orbitals fill before the
4d.
(ii)
[2]
1s ↑↓
2s ↑↓
2p ↑
↑
↑
The corrected version is as below. The original violates the
Pauli Exclusion Principle which requires that each
orbital can hold only two electrons which must have
opposite spins. "No two electrons may have the same
four quantum numbers"
1s ↑↓
[2]
(iii)
1s ↑↓
2s ↓↑
2s ↑
2p ↑
↑
↑
2p ↑ ↑ ↑
The corrected version is as below. The aufbau rules state that
the obitals fill in order of increasing energy so that the
2s orbitals fills completely before the 3p.
(e)
1s ↑↓
2s ↑
2p ↑ ↑
Write complete ground state electron configurations (eg. P: 1s2
2s22p6 3s23p3) for the following species:
1s2 2s22p6 3s23p6 4s1 3d10
[1]
(i)
Cu
[2]
.
(ii)
As 3− 1s2 2s22p6 3s23p63d10 4s24p6
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CHEMISTRY 1010
Section C
Long Answer Questions
(15 Marks)
Show your working in the space provided.
[3]
C1.
The laser light used in some compact disc players has a frequency,
ν = 3.94 x 1014 Hz. Calculate the energy of the light in kJ mol-1.
Reminder: Energy E = hν per photon.
E = (6.626 % 10 −34 J s % 3.94 % 10 14 s −1 ) per photon = 2.61 1 % 10 −19 J per photon
Now the Avogadro constant gives the number of photons per mole
so
E = 2.611 % 10 −19 J % 6.022 % 1023 mol −1 = 1.57 2 % 10 5 J mol −1
There are 1000 J kJ-1
E = 1.572 % 10 5 J mol −1 %
1
1000 J kJ −1
= 157. 1 kJ mol −1
Ans: 1.57 x 102 kJ mol-1
[7]
C2.
Balance the equation for the following redox in BASIC solution.
−
Cr2O3(s) + ClO − (aq) → CrO 2−
4 (aq) + Cl (aq)
Oxidation:
Cr2O3(s) → CrO 2−
4 (aq)
Cr2O3(s) → 2 CrO 2−
4 (aq)
5 H2O(l) + Cr2O3(s) → 2 CrO 2−
4 (aq)
+
5 H2O(l) + Cr2O3(s) → 2 CrO 2−
4 (aq) + 10 H (aq)
+
5 H2O(l) + Cr2O3(s) → 2 CrO 2−
4 (aq) + 10 H (aq) + 6 e
Reduction: ClO − (aq) → Cl − (aq)
ClO − (aq) → Cl − (aq) + H2O(l)
2 H+(aq) + ClO − (aq) → Cl − (aq) + H2O(l)
2 H+(aq) + ClO − (aq) + 2 e- → Cl − (aq) + H2O(l)
Combine and balance electron transfer.
+
−
2 H2O(l) + Cr2O3(s) + 3 ClO − (aq) → 2 CrO 2−
4 (aq) + 4 H (aq) + 3 Cl (aq)
Convert the reaction to that in base by adding 4 OH- to both sides.
−
4 OH-(aq) + Cr2O3(s) + 3 ClO − (aq) → 2 CrO 2−
4 (aq) + + 2 H2O(l)(aq) + 3 Cl (aq)
Check for balance
−
Ans: 4 OH-(aq) + Cr2O3(s) + 3 ClO − (aq) → 2 CrO 2−
4 (aq) + 2 H2O(l)(aq) + 3 Cl (aq)
C3.
[2]
25.00 mL of a solution of sulfuric acid requires 26.25 mL of 0.1550 mol L-1
sodium hydroxide solution for neutralization.
(a)
Write both the molecular and net-ionic equation for the
neutralization reaction.
H2SO4(aq) + 2 NaOH(aq) → 2 NaCl(aq) + 2 H2O(l)
H+(aq) + OH-(aq) → H2O(l)
[3]
(b)
Calculate the concentration of the solution of sulfuric acid in mol L-1.
Moles of NaOH used
26.26 mL
= 1000 mL L−1 % 0.1550 mol L −1 = 4.073 4 % 10 −3 mol
If we divide the equation by two we can easily see that 1 mole
of NaOH reacts with ½ mole of H2SO4.
½ H2SO4(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Moles of H2SO4 present in the 25.00 mL of solution
= ½ x 4.073 4 % 10 −3 mol
= 2.0344 x 10-3 mol
So the concentration of the solution was
2.0344 % 10 −3 mol
= 25.00%10 −3 L = 0.081375 mol L-1
Ans: 0.08138 mol L-1
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