MCAT General Chemistry Exam 100 Questions
and Verified Correct Answers latest 2025–2026
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 - answer>>>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- - answer>>>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 - answer>>>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- - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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.
I-131 is a radionuclide used in the treatment of thyroid carcinomas. If a sample's activity dropped
from 100 mCi to 12.5 mCi in 12 days, what is the half-life of I-131?
A) 1.5 days
B) 3.0 days
C) 4.0 days
D) 6.0 days - answer>>>C) 4.0 days
One reliable strategy to solve half-life problems is to write out the remaining activity above each
half-life. Knowing that 12 days represents 3 half-lives allows the problem to be solved with
division. 12 days/3 hl = 4 days/1 hl.
23/12-Mg decays into 23/11-Na with the emission of:
A) a proton
B) an electron
C) a positron and a neutrino
D) a positron - answer>>>C) a positron and a neutrino
Remember that charge and mass are always conserved during nuclear decay on the MCAT. This
means that the sum of the reactant masses and the sum of the reactant charges are equal to the
sum of the product masses and product charges. 23 = 23, so there is nothing to worry about to
balance the masses in this equation. because the charge number is not balanced, a new charge
needs to be introduced into the decay equation to balance it. 12 = 11 + x, so x = +1. A particle with
a positive charge and no mass is a positron. A positron must be emitted alongside the decay
product of this reaction in order for the reaction to be balanced. during positron emission,
neutrinos, which are particles lacking charge and mass are also released, so C is the best answer
choice.
A PET scan uses a derivative of glucose with a radioactive fluoride-18 atom in place of the 2'
oxygen. The radioactive fluoride emits a positron which allows physicians to measure metabolic
activity, a key marker for identifying cancerous tumors. Which decay process is exhibited by F-18?
A) alpha decay
B) beta decay
C) gamma decay
D) glycolytic decay - answer>>>B) beta decay
and Verified Correct Answers latest 2025–2026
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 - answer>>>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- - answer>>>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 - answer>>>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- - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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 - answer>>>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.
I-131 is a radionuclide used in the treatment of thyroid carcinomas. If a sample's activity dropped
from 100 mCi to 12.5 mCi in 12 days, what is the half-life of I-131?
A) 1.5 days
B) 3.0 days
C) 4.0 days
D) 6.0 days - answer>>>C) 4.0 days
One reliable strategy to solve half-life problems is to write out the remaining activity above each
half-life. Knowing that 12 days represents 3 half-lives allows the problem to be solved with
division. 12 days/3 hl = 4 days/1 hl.
23/12-Mg decays into 23/11-Na with the emission of:
A) a proton
B) an electron
C) a positron and a neutrino
D) a positron - answer>>>C) a positron and a neutrino
Remember that charge and mass are always conserved during nuclear decay on the MCAT. This
means that the sum of the reactant masses and the sum of the reactant charges are equal to the
sum of the product masses and product charges. 23 = 23, so there is nothing to worry about to
balance the masses in this equation. because the charge number is not balanced, a new charge
needs to be introduced into the decay equation to balance it. 12 = 11 + x, so x = +1. A particle with
a positive charge and no mass is a positron. A positron must be emitted alongside the decay
product of this reaction in order for the reaction to be balanced. during positron emission,
neutrinos, which are particles lacking charge and mass are also released, so C is the best answer
choice.
A PET scan uses a derivative of glucose with a radioactive fluoride-18 atom in place of the 2'
oxygen. The radioactive fluoride emits a positron which allows physicians to measure metabolic
activity, a key marker for identifying cancerous tumors. Which decay process is exhibited by F-18?
A) alpha decay
B) beta decay
C) gamma decay
D) glycolytic decay - answer>>>B) beta decay
