Chemistry I Module 2 Exam 2026
Question 1
An isotope of an element is parsed analytically and shown to contain 24 protons, 28
neutrons, and 24 electrons. Identify the correct atomic symbol.
A) \(_{24}^{28}\text{Cr}\)
B) \(_{24}^{52}\text{Cr}\)
C) \(_{28}^{52}\text{Ni}\)
D) \(_{52}^{24}\text{Cr}\)
Answer: B) \(_{24}^{52}\text{Cr}\)
Rationale: The atomic number (subscript) is determined entirely by the number of
protons, which is 24 (Chromium, Cr). The mass number (superscript) represents the
sum of the protons and neutrons within the nucleus: \(24 + 28 = 52\).
Question 2
Albert Einstein explained the Photoelectric Effect by treating light as stream packets of
energy. This physical phenomenon validates which property of light?
A) Continuous wave nature
B) Pure mathematical refraction
C) Particulate (photon) nature
D) Total destructive interference
Answer: C) Particulate (photon) nature
Rationale: The photoelectric effect demonstrates that incoming light behaves as
discrete packets of localized kinetic energy called photons. When these particle packets
strike a metallic plate, they transfer energy instantly to eject a specific electron, a
behavior that classical wave theory cannot explain.
Question 3
Calculate the exact frequency (in \(\text{s}^{-1}\)) of an electromagnetic wave that
possesses a wavelength (\(\lambda \)) of \(4.25 \times 10^{-7}\text{ m}\). \([c = 3.00
\times 10^8\text{ m/s}]\).
A) \(1.28 \times 10^2\text{ s}^{-1}\)
B) \(7.06 \times 10^{14}\text{ s}^{-1}\)
, C) \(1.42 \times 10^{-15}\text{ s}^{-1}\)
D) \(7.06 \times 10^{15}\text{ s}^{-1}\)
Answer: B) \(7.06 \times 10^{14}\text{ s}^{-1}\)
Rationale: Use the standard light wave equation: \(c = \lambda\nu\). Isolate frequency
(\(\nu \)): \(\nu = c / \lambda\). Substituting the given values yields: \(\nu = (3.00 \times
10^8\text{ m/s}) / (4.25 \times 10^{-7}\text{ m}) = 7.0588 \times 10^{14}\text{ s}^{-1}\),
which rounds to \(7.06 \times 10^{14}\text{ s}^{-1}\).
Question 4
According to the modern Quantum Mechanical model of the atom, which parameter
describes the regional space mapped by an atomic orbital?
A) The exact trajectory path an electron follows around the nucleus.
B) A 3D probability contour map showing where an electron is likely to be found 90% of
the time.
C) A completely solid, unyielding shell filled with static negative charges.
D) The path along which neutrons migrate between electron shells.
Answer: B) A 3D probability contour map showing where an electron is likely to
be found 90% of the time.
Rationale: The Heisenberg Uncertainty Principle states that we cannot know an
electron's exact path. Therefore, the modern quantum mechanical model replaces
Bohr's circular paths with orbital probability regions derived from wave equations.
Question 5
What is the correct ground-state electron configuration for a neutral Manganese (Mn)
atom? (Atomic number = 25)
A) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}4p^{5}\)
B) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\)
C) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}\)
D) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{6}\)
Answer: B) \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\)
Rationale: Manganese contains 25 electrons. Following the Aufbau principle, the
subshells fill in order of increasing energy: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\) accounts
for the first 18 electrons (Argon core), followed by filling the \(4s^{2}\) subshell, and
placing the remaining 5 electrons into the \(3d\) subshell.
Question 6
Which noble gas configuration represents the correct abbreviated ground-state
arrangement for a neutral Selenium (Se) atom? (Atomic number = 34)
A) \([\text{Ne}] 3s^2 3p^6 3d^{10} 4s^2 4p^4\)
B) \([\text{Ar}] 4s^2 3d^{10} 4p^4\)