MCAT General Chemistry Questions And
Answers | Verified
Which of the following increases with increasing atomic number within a family on the
periodic table?
A) electronegativity
B) electron affinity
C) atomic radius
D) ionization energy CORRECT ANSWERS C) atomic radius
Only atomic radius increases going down a column. Size of an atom is determined not
only by effective nuclear charge, but also by number of electron shells. Moving down a
column, the number of electron shells increases. Because the additional shells require
additional space, atomic radius increases as number of electron shells increases.
Which of the following most likely represents the correct order of ion size from greatest
to smallest?
A) O2-, F-, Na+, Mg2+
B) Mg2+, Na+, F-, O2-
C) Na+, Mg2+, O2-, F-
D) Mg2+, Na+, O2-, F- CORRECT ANSWERS A) O2-, F-, Na+, Mg2+
This is an isoelectronic series, which means that the number of electrons on each ion is
the same. In an isoelectronic series, the nuclear charge increases with increasing
atomic number and draws the electrons inward with greater force. The ion with the
fewest protons produces the weakest attractive force on the ions, so it has the largest
size. Oxygen has the smallest atomic number, so it has the fewest protons, and
therefore will have the greatest ion size.
Which of the following best explains why sulfur can make more bonds than oxygen?
A) sulfur is more electronegative than oxygen
B) oxygen is more electronegative than sulfur
C) sulfur has 3d orbitals not available to oxygen
D) sulfur has fewer valence electrons CORRECT ANSWERS C) sulfur has 3d orbitals
not available to oxygen
Only elements in the third and higher periods of the periodic table can form more than 4
bonds. Second period elements, including oxygen, have four valence orbitals (one 2s
and three 2p) with which to form bonds. By contrast, third period elements like sulfur
can form bonds with not only one 3s and three 3p, but also five 3d orbitals. Remember
that the second quantum number can have any integer value from 0 to n-1, so when n =
3, three subshells (s, p, and d) are available.
,Which of the following species has an unpaired electron in its ground-state electronic
configuration?
A) Ne
B) Ca+
C) Na+
D) O2- CORRECT ANSWERS B) Ca+
Atoms and ions with electron configurations identical to those of noble gases do not
have any unpaired electrons in their ground state. Ne, Na+, and O2- all have the same
electron configuration, [Ne], and therefore have no unpaired electrons in their ground
state. Ca+, by contrast, has a ground state configuration of [Ar]4s1, and has one
unpaired electron in its 4s subshell.
What is the electron configuration of chromium?
A) [Ar] 3d6
B) [Ar] 4s13d5
C) [Ar] 4s23d3
D) [Ar] 4s24d4 CORRECT ANSWERS B) [Ar] 4s13d5
Even without pre-existing knowledge of the electron configuration of Cr, strongest
answer could have been identified by eliminating improbable choices. Choice C can be
eliminated because it has the wrong number of electrons. Choice D can be eliminated
because Cr ground state electrons exist in the 3d rather than 4d subshell. Given Hund's
rule, which states that the most stable arrangement of electrons is the one with the most
unpaired electrons, choice B might have seemed more likely to be the best answer.
In reference to the photoelectric effect, which of the following will increase the kinetic
energy of a photoelectron?
A) increasing the work function
B) increasing the frequency of the incident light
C) increasing the number of photons in the incident light
D) increasing the mass of photons in the incident light CORRECT ANSWERS B)
increasing the frequency of the incident light
KE = hf - phi, in which h is Planck's constant, f is the frequency of incident light, and phi
is the work function of the metal. In other words, only the frequency of the incident light
and the work function of the metal affect the KE of ejected electrons, so C and D can be
eliminated. Note that photons are conventionally held to be massless, so this eliminates
D as well. The work function is the minimum energy required to eject an electron from a
solid state into a surrounding vacuum. Increasing the work function decreases the
kinetic energy of the ejected electron. Only increasing frequency increases KE.
When an electron moves from a 2p to a 3s orbital, the atom containing that electron:
A) becomes an new isotope
B) becomes a new element
C) absorbs energy
, D) releases energy CORRECT ANSWERS C) absorbs energy
3s orbitals are at higher energy than 2p orbitals. For an electron to move from 2p to 3s,
the atom containing that electron must absorb energy. In order to become a new
isotope, the atom must gain or lose a neutron. In order to become a new element, the
atom must gain or lose a proton. The atom would release energy if the electron moved
to a lower energy orbital.
Aluminum only has one oxidation state, while chromium has several. Which of the
following is the best explanation for this difference?
A) electrons in the d orbitals of Cr may or may not be used to form bonds
B) electrons in the p orbitals of Cr may or may not be used to form bonds
C) electrons in the d orbitals of Al may or may not be used to form bonds
D) electrons in the p orbitals of Al may or may not be used to form bonds CORRECT
ANSWERS A) electrons in the d orbitals of Cr may or may not be used to form bonds
Based on Cr's position in the periodic table, it can be inferred that it is a transition metal.
As such, the electrons in its d orbitals have the ability to move into valence orbitals and
form bonds. Depending on the number of electrons that move from the d orbitals, its
oxidation state can vary considerably. Note that aluminum, like all elements in the third
period and higher, has d orbitals. However, unlike chromium, Al does not, in its ground
state, have electrons in its d orbitals; its configuration is [Ne]3s23p1. Therefore, C can
be eliminated.
Which of the following molecules has the greatest dipole moment?
A) H2
B) O2
C) HF
D) HBr CORRECT ANSWERS C) HF
The dipole moment will be greatest for the atoms with the greatest difference in
electronegativity. Based upon periodic trends, H and F will have the greatest difference
in electronegativity, and therefore the greatest dipole moment.
A natural sample of carbon contains 99% of C-12. How many moles of C-12 are likely to
be found in a 48.5 gram sample of carbon obtained from nature?
A) 1
B) 4
C) 12
D) 49.5 CORRECT ANSWERS B) 4
Solving this problem does not require complicated calculations. Simply assume that
100% of the sample is 12-C. The molecular weight of 12-C is 12 g/mol. 48.5 g x 12
g/mol = ~4 mol.