, Question 1
An element's ground-state electron configuration ends in 4d. Which set of
quantum numbers correctly describes the LAST electron added according to
the Aufbau principle?
A. n=4, =2, m=+2, ms=+½
B. n=4, =2, m=0, ms=½
C. n=4, =1, m=+1, ms=+½
D. n=5, =0, m=0, ms=+½
Correct Answer: A - n=4, =2, m=+2, ms=+½
RATIONALE
The 4d subshell fills after 5s; by Hund's rule the five 4d electrons
occupy all five m values singly with parallel spins. The last electron
enters the highest available m (+2) with ms=+½, giving option A.
Option D incorrectly invokes a 5p electron, and C misidentifies the
subshell as =1.
Question 2
Which of the following explains why the first ionization energy of nitrogen
(1402 kJ/mol) exceeds that of oxygen (1314 kJ/mol)?
A. Oxygen's larger nuclear charge increases electron shielding, lowering
its ionization energy.
B. Nitrogen's half-filled 2p subshell confers extra stability, while
oxygen's paired 2p electron experiences greater electron-electron
repulsion.
C. Oxygen has a smaller atomic radius, which paradoxically decreases the
energy required to remove an electron.
D. Nitrogen's 2p electrons are in a lower principal energy level than
oxygen's.
Correct Answer: B - Nitrogen's half-filled 2p subshell confers
extra stability, while oxygen's paired 2p electron experiences
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,greater electron-electron repulsion.
RATIONALE
Nitrogen's 2p³ half-filled configuration is exchange-energy stabilized,
whereas oxygen's 2p contains a paired electron whose repulsion makes
it easier to remove. This anomaly overrides the general left-to-right
increase in ionization energy. Options A, C, and D misapply shielding,
radius, or principal quantum number arguments.
Question 3
A 25.0 mL sample of 0.100 M HSO is titrated with 0.100 M NaOH. What is
the pH at the second equivalence point? (K for HSO = 1.2 × 10²)
A. 7.00
B. 8.10
C. 11.30
D. 12.60
Correct Answer: C - 11.30
RATIONALE
At the second equivalence point, all HSO is converted to SO², a weak
base (Kb = Kw/Ka = 8.3 × 10¹³). The resulting pH from SO²
hydrolysis is calculated to be approximately 11.3, making C correct.
Option A assumes a strong acid-strong base titration, and B and D
reflect miscalculations of the weak base hydrolysis.
Question 4
For the reaction 2 NO(g) + O(g) -> 2 NO(g), the rate law is rate = k[NO]²[O].
If the concentration of NO is doubled and O is halved, the initial rate will:
A. remain unchanged
B. double
C. quadruple
D. halve
Page 3
, Correct Answer: B - double
RATIONALE
Substituting [NO] -> 2[NO] and [O] -> ½[O] into the rate law gives
rate' = k(2[NO])²(½[O]) = 2k[NO]²[O], so the rate doubles. This is a
direct application of reaction order effects. The other options reflect
errors in exponent handling or arithmetic.
Question 5
A 2.50 g sample of a hydrocarbon is burned in excess oxygen, producing 7.90
g CO and 3.23 g HO. What is the empirical formula?
A. CH
B. CH
C. CH
D. CH
Correct Answer: A - CH
RATIONALE
Moles C = 7.90 g CO / 44.01 g/mol = 0.1795 mol; moles H = 2 ×
(3.23 g HO / 18.02 g/mol) = 0.3585 mol. The C:H ratio is
0.1795:0.3585 1:2, giving the empirical formula CH. The other
options require different C:H ratios that do not match the data.
Question 6
A 0.500 g sample of an unknown metal reacts with excess HCl to produce
0.0523 g of H gas. If the metal forms a +3 ion, what is its approximate molar
mass?
A. 27 g/mol
B. 52 g/mol
C. 58 g/mol
D. 112 g/mol
Correct Answer: A - 27 g/mol
Page 4
An element's ground-state electron configuration ends in 4d. Which set of
quantum numbers correctly describes the LAST electron added according to
the Aufbau principle?
A. n=4, =2, m=+2, ms=+½
B. n=4, =2, m=0, ms=½
C. n=4, =1, m=+1, ms=+½
D. n=5, =0, m=0, ms=+½
Correct Answer: A - n=4, =2, m=+2, ms=+½
RATIONALE
The 4d subshell fills after 5s; by Hund's rule the five 4d electrons
occupy all five m values singly with parallel spins. The last electron
enters the highest available m (+2) with ms=+½, giving option A.
Option D incorrectly invokes a 5p electron, and C misidentifies the
subshell as =1.
Question 2
Which of the following explains why the first ionization energy of nitrogen
(1402 kJ/mol) exceeds that of oxygen (1314 kJ/mol)?
A. Oxygen's larger nuclear charge increases electron shielding, lowering
its ionization energy.
B. Nitrogen's half-filled 2p subshell confers extra stability, while
oxygen's paired 2p electron experiences greater electron-electron
repulsion.
C. Oxygen has a smaller atomic radius, which paradoxically decreases the
energy required to remove an electron.
D. Nitrogen's 2p electrons are in a lower principal energy level than
oxygen's.
Correct Answer: B - Nitrogen's half-filled 2p subshell confers
extra stability, while oxygen's paired 2p electron experiences
Page 2
,greater electron-electron repulsion.
RATIONALE
Nitrogen's 2p³ half-filled configuration is exchange-energy stabilized,
whereas oxygen's 2p contains a paired electron whose repulsion makes
it easier to remove. This anomaly overrides the general left-to-right
increase in ionization energy. Options A, C, and D misapply shielding,
radius, or principal quantum number arguments.
Question 3
A 25.0 mL sample of 0.100 M HSO is titrated with 0.100 M NaOH. What is
the pH at the second equivalence point? (K for HSO = 1.2 × 10²)
A. 7.00
B. 8.10
C. 11.30
D. 12.60
Correct Answer: C - 11.30
RATIONALE
At the second equivalence point, all HSO is converted to SO², a weak
base (Kb = Kw/Ka = 8.3 × 10¹³). The resulting pH from SO²
hydrolysis is calculated to be approximately 11.3, making C correct.
Option A assumes a strong acid-strong base titration, and B and D
reflect miscalculations of the weak base hydrolysis.
Question 4
For the reaction 2 NO(g) + O(g) -> 2 NO(g), the rate law is rate = k[NO]²[O].
If the concentration of NO is doubled and O is halved, the initial rate will:
A. remain unchanged
B. double
C. quadruple
D. halve
Page 3
, Correct Answer: B - double
RATIONALE
Substituting [NO] -> 2[NO] and [O] -> ½[O] into the rate law gives
rate' = k(2[NO])²(½[O]) = 2k[NO]²[O], so the rate doubles. This is a
direct application of reaction order effects. The other options reflect
errors in exponent handling or arithmetic.
Question 5
A 2.50 g sample of a hydrocarbon is burned in excess oxygen, producing 7.90
g CO and 3.23 g HO. What is the empirical formula?
A. CH
B. CH
C. CH
D. CH
Correct Answer: A - CH
RATIONALE
Moles C = 7.90 g CO / 44.01 g/mol = 0.1795 mol; moles H = 2 ×
(3.23 g HO / 18.02 g/mol) = 0.3585 mol. The C:H ratio is
0.1795:0.3585 1:2, giving the empirical formula CH. The other
options require different C:H ratios that do not match the data.
Question 6
A 0.500 g sample of an unknown metal reacts with excess HCl to produce
0.0523 g of H gas. If the metal forms a +3 ion, what is its approximate molar
mass?
A. 27 g/mol
B. 52 g/mol
C. 58 g/mol
D. 112 g/mol
Correct Answer: A - 27 g/mol
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