#1 of 81
UHS 2024
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Which of the following electronic configuration is correct for carbon?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The electronic configuration maps the exact location of all electrons in an atom based on the Aufbau principle.
Formula: $$ Z = \sum \text{electrons} $$
Solution: - Carbon (C) is element number 6 on the periodic table, so it has \( Z = 6 \).
- A neutral atom of Carbon has exactly 6 electrons to place.
- First, the lowest energy level \( 1s \) gets 2 electrons: \( 1s^2 \).
- Next, the \( 2s \) orbital gets 2 electrons: \( 2s^2 \).
- There are \( 6 - 4 = 2 \) electrons remaining. These go into the \( 2p \) subshell: \( 2p^2 \).
- Final configuration: \( 1s^2, 2s^2, 2p^2 \).
Why other options are incorrect: Option A is Nitrogen (7 electrons). Option B is Oxygen (8 electrons). Option D is Boron (5 electrons).
#2 of 81
SZABMU 2024
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How many electrons will be accommodated in sub-shell with Azimuthal quantum number \( l = 2 \)?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The total electron capacity of a subshell depends on the number of spatial orbitals it contains, with each orbital holding 2 electrons.
Formula: $$ \text{Max Electrons} = 2(2l + 1) $$
Solution: - An azimuthal quantum number of \( l = 2 \) specifically denotes a \( d \)-subshell.
- First, calculate the number of orbitals: \( 2(2) + 1 = 5 \) orbitals.
- Multiply by 2 electrons per orbital: \( 5 \times 2 = 10 \) electrons.
Why other options are incorrect: 2 is the capacity for an \( s \)-subshell (\( l=0 \)). 6 is the capacity for a \( p \)-subshell (\( l=1 \)). 12 is mathematically invalid for standard quantum configurations.
#3 of 81
SZABMU 2024
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Which of the following element will show electronic configuration of outermost shell like \( ns^2, np^5 \)?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The general outermost electronic configuration \( ns^2, np^5 \) indicates there are 7 valence electrons. Elements with 7 valence electrons belong to Group 17 (Halogens).
Formula: $$ \text{Valence electrons} = 2 + 5 = 7 $$
Solution: - Let's check the valence configurations for the options:
- Carbon (C, Group 14): \( 2s^2, 2p^2 \) (4 valence electrons)
- Silicon (Si, Group 14): \( 3s^2, 3p^2 \) (4 valence electrons)
- Sulfur (S, Group 16): \( 3s^2, 3p^4 \) (6 valence electrons)
- Chlorine (Cl, Group 17): \( 3s^2, 3p^5 \) (7 valence electrons)
- Chlorine perfectly matches the \( ns^2, np^5 \) pattern.
Why other options are incorrect: They belong to different periodic groups and therefore have different numbers of electrons in their outermost \( p \)-subshell.
#4 of 81
SZABMU 2024
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(Deleted) Electronic configuration of \( _{11}\text{Na}^{23} \) is
D
\( \text{[Ne]} 3s^2 3p \dots \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: Noble gas notation simplifies electronic configurations by replacing inner, fully filled shells with the bracketed symbol of the preceding noble gas.
Formula: None required.
Solution: - Sodium (Na) has an atomic number \( Z = 11 \).
- The nearest preceding noble gas is Neon (Ne), which has \( Z = 10 \).
- Neon's configuration perfectly covers the core electrons: \( 1s^2, 2s^2, 2p^6 \).
- Sodium has one remaining valence electron (\( 11 - 10 = 1 \)).
- This final electron enters the next available energy level, which is the \( 3s \) orbital.
- The condensed configuration is therefore \( \text{[Ne]} 3s^1 \).
Why other options are incorrect: Option B represents Magnesium (\( Z=12 \)). Option C represents a Sodium ion (\( \text{Na}^+ \)). Option D implies too many electrons.
#5 of 81
DUHS 2024
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"In an orbital of an atom, no two electrons can have the same set of four quantum numbers, at least one quantum number must be different." This statement is:
B
Pauli's exclusion principle
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Electrons are fermions, and quantum mechanics dictates that fermions cannot occupy the exact same quantum state simultaneously.
Formula: None required.
Solution: - The Pauli Exclusion Principle explicitly formulates this rule for atoms.
- Even if two electrons share the exact same spatial orbital (meaning their \( n \), \( l \), and \( m \) quantum numbers are identical), they must have different spin quantum numbers (\( s \)).
- One must be spin up (\( +1/2 \)) and the other spin down (\( -1/2 \)).
Why other options are incorrect: The Aufbau principle is about filling order from lowest to highest energy. Hund's rule is about maximizing unpaired electrons in degenerate orbitals. Wiswesser's rule relates to line notation in chemical structures, entirely unrelated to quantum numbers.
#6 of 81
DUHS 2024
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Maximum number of electrons in a given sub-shell are calculated by:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The maximum number of electrons in a subshell is calculated by the formula 2(2l + 1) or 4l + 2. The formula (2l + 1) gives the number of orbitals.
Key: Option D (2(2l+1)).
#7 of 81
DUHS 2024
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The neutron was discovered by
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The neutron was the last of the three major subatomic particles to be discovered because it carries no electrical charge, making it difficult to detect with electromagnetic fields.
Formula: None required.
Solution: - In 1932, Sir James Chadwick bombarded Beryllium with alpha particles.
- He observed highly penetrating, uncharged radiation being emitted.
- He proved that this radiation consisted of neutral particles with a mass slightly greater than a proton. He named this particle the neutron.
Why other options are incorrect: J.J. Thomson discovered the electron. Eugen Goldstein discovered positive rays (protons/canal rays). Max Planck discovered quantum energy packets (photons).
#8 of 81
BUMHS 2024
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The charge of electron is:
A
\( 1.602 \times 10^{-16} \text{ C} \)
B
\( 1.602 \times 10^{-19} \text{ C} \)
C
\( 1.602 \times 10^{-17} \text{ C} \)
D
\( 1.602 \times 10^{-18} \text{ C} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The fundamental elementary charge is the electrical charge carried by a single proton or electron.
Formula: $$e = 1.602 \times 10^{-19}\text{ C}$$
Solution: - Robert Millikan measured the charge of a single electron in 1909 through his famous Oil Drop Experiment.
- He determined the elementary charge to be approximately \(1.602 \times 10^{-19}\) Coulombs (C).
- Because an electron is negatively charged, its exact value is \(-1.602 \times 10^{-19}\text{ C}\), but questions typically ask for the magnitude.
Why other options are incorrect: The exponents \(-16\), \(-17\), and \(-18\) represent orders of magnitude that are thousands of times too large.
#9 of 81
BUMHS 2024
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An electron in an atom is completely describe by its ____ quantum numbers
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Correct Key: Option D
Diagnostic Explanation
Concept: The state of any electron in a quantum mechanical atomic model is defined by a specific set of parameters called quantum numbers.
Formula: None required.
Solution: - To completely describe the unique address and state of an electron in an atom, exactly four quantum numbers are required:
- 1. Principal (\( n \)): Energy level / Shell.
- 2. Azimuthal (\( l \)): Orbital shape / Subshell.
- 3. Magnetic (\( m \)): 3D spatial orientation.
- 4. Spin (\( s \)): Direction of intrinsic electron spin.
- According to the Pauli Exclusion Principle, no two electrons can share the same set of these four numbers.
Why other options are incorrect: There are strictly four recognized quantum numbers used to define an electron's state.
#10 of 81
BUMHS 2024
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The only group of lines occurring in the visible region of the hydrogen spectrum is labelled as the
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: When excited hydrogen electrons fall back to lower energy levels, they emit photons. The destination level dictates the energy (and thus the spectrum band) of the emitted light.
Formula: $$ E = h\nu $$
Solution: - Lyman series: Electrons fall to \( n=1 \). The energy gap is massive, emitting high-energy ultraviolet (UV) light.
- Balmer series: Electrons fall to \( n=2 \). The energy gap matches the energy of visible light photons. The lines appear as specific colors (red, green, blue-violet) visible to the human eye.
- Paschen, Brackett, and Pfund series: Electrons fall to \( n=3, 4, 5 \) respectively. These gaps are very small, emitting low-energy infrared (IR) light.
Why other options are incorrect: Paschen and Pfund are strictly in the invisible infrared region. Lyman is strictly in the invisible ultraviolet region.
#11 of 81
NUMS 2024
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The value of "\( l \)" for \( 4p \) orbital is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The letter designation of an orbital directly corresponds to a specific integer value of the azimuthal quantum number (\( l \)).
Formula: $$ s=0, p=1, d=2, f=3 $$
Solution: - The orbital given is "\( 4p \)".
- The principal quantum number is the coefficient: \( n = 4 \).
- The letter "\( p \)" universally represents an azimuthal quantum number of \( l = 1 \).
- Therefore, the value of \( l \) is 1.
Why other options are incorrect: 0 is for an \( s \)-orbital. 2 is for a \( d \)-orbital. 3 is for an \( f \)-orbital.
#12 of 81
NUMS 2024
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The element having the electronic configuration of noble gas notation \( \text{(Kr)} 5s^2 4d^6 \) is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: To identify an element from its electron configuration, sum the total number of electrons, which equals its atomic number in a neutral state.
Formula: $$ Z = \text{Core electrons} + \text{Valence electrons} $$
Solution: - The core is represented by Krypton (Kr). Looking at the periodic table, Krypton has an atomic number of 36. So, there are 36 core electrons.
- The valence shell adds more: \( 5s^2 \) adds 2 electrons. \( 4d^6 \) adds 6 electrons.
- Total electrons = \( 36 + 2 + 6 = 44 \).
- An atomic number of 44 corresponds to Ruthenium (Ru).
Why other options are incorrect: Strontium (Sr) is 38. Palladium (Pd) is 46. Zirconium (Zr) is 40. Only Ruthenium possesses exactly 44 protons.
#13 of 81
NUMS 2024
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Choose the allowed sets of quantum numbers from the following:
A
\( n = 3, l = 2, m = 0, s = +1/2 \)
B
\( n = 3, l = 3, m = 0, s = -1/2 \)
C
\( n = 4, l = 3, m = 4, s = +1/2 \)
D
\( n = 4, l = 2, m = 4, s = -1/2 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The validity of a set of quantum numbers depends on strict boundary rules connecting \( n, l \), and \( m \).
Formula: $$ l = 0 \text{ to } (n-1) $$ $$ m = -l \text{ to } +l $$
Solution: - Let's evaluate each option to see if it breaks any rules:
- Option A: \( n=3 \). Max \( l = 2 \). So \( l=2 \) is allowed. If \( l=2 \), \( m \) can be from -2 to +2. So \( m=0 \) is allowed. Spin is +1/2. This set is perfectly valid.
- Option B: \( n=3 \). Max \( l = 3-1 = 2 \). Here \( l=3 \), which violates the rule (\( l \) cannot equal \( n \)). Invalid.
- Option C: \( n=4 \), \( l=3 \) is allowed. However, if \( l=3 \), the max value of \( m \) is +3. Here \( m=4 \) is given, which violates the rule. Invalid.
- Option D: \( n=4 \), \( l=2 \) is allowed. If \( l=2 \), max \( m \) is +2. Here \( m=4 \) is given, which violates the rule. Invalid.
Why other options are incorrect: Options B, C, and D mathematically violate the restrictive boundary rules of quantum mechanics.
#14 of 81
DUHS 2023
Permalink
Which of the following is different in \( \text{Na}^+ \) (Z=11), \( \text{Mg}^{+2} \) (Z=12) and \( \text{Al}^{+3} \) (Z=13)?
C
Electronic configuration
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:These three ions are
isoelectronic, meaning they possess the exact same number of electrons and electron shell configuration. The only thing separating them is the unchanging contents of their respective nuclei.
Formula:$$ \text{Electrons in ion} = Z - \text{Charge} $$
Solution:- Let's analyze their electron counts:
- \( \text{Na}^+ \): \( 11 \text{ (protons)} - 1 \text{ (lost electron)} = 10 \text{ electrons} \).
- \( \text{Mg}^{+2} \): \( 12 \text{ (protons)} - 2 \text{ (lost electrons)} = 10 \text{ electrons} \).
- \( \text{Al}^{+3} \): \( 13 \text{ (protons)} - 3 \text{ (lost electrons)} = 10 \text{ electrons} \).
- Because they all have 10 electrons, their electronic configurations and number of shells (2 shells: K=2, L=8) are perfectly identical.
- However, their atomic numbers (Z) are 11, 12, and 13 respectively. Atomic number dictates the Number of Protons, which remains entirely different and untouched by the ionization process.
Why other options are incorrect:- Options A, B, and C: As proven above, these parameters are identical (10 electrons, \( 1s^2 2s^2 2p^6 \), 2 shells) for all three species.
#15 of 81
UHS 2023
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When \( 6d \) orbital is filled, the entering electron goes into?
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Correct Key: Option C
Diagnostic Explanation
Concept: The order of orbital filling is strictly determined by the Aufbau principle, which uses the \( (n+l) \) rule to rank energy levels.
Formula: $$ \text{Energy} \propto (n+l) $$
Solution: - The \( 6d \) orbital has \( n=6 \) and \( l=2 \). Its \( (n+l) \) sum is \( 6 + 2 = 8 \).
- After \( 6d \), the next orbital must either have a higher \( (n+l) \) sum, or the same sum but a higher principal quantum number \( n \).
- Let's test \( 7p \): \( n=7, l=1 \implies n+l = 7 + 1 = 8 \).
- Since both \( 6d \) and \( 7p \) have a sum of 8, the orbital with the higher \( n \) is filled next. Thus, \( 7p \) follows \( 6d \).
Why other options are incorrect: \( 7s \) is filled much earlier (\( n+l = 7 \)). \( 7d \) and \( 7f \) have much higher energy levels and follow later in the sequence.
#16 of 81
SZABMU 2023
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Which of the following orbital will be filled first than \( 4p \)?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: Orbitals are filled in order of increasing energy according to the \( (n+l) \) rule.
Formula: $$ \text{Energy order: } 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p $$
Solution: - Note on question validity: Technically, all the listed orbitals (\( 1s, 2p, 3d, 4s \)) are filled before \( 4p \). However, standard exam questions of this type usually intend to ask which orbital is filled immediately before, or they are testing specific \( n+l \) boundary logic.
- The official key selects \( 4s \). Let's compare \( 4s \) and \( 4p \):
- \( 4s \): \( n=4, l=0 \implies n+l = 4 \)
- \( 4p \): \( n=4, l=1 \implies n+l = 5 \)
- Because \( 4s \) has a lower \( (n+l) \) value, it clearly fills before \( 4p \). (In the exact filling sequence, the order is \( 4s \rightarrow 3d \rightarrow 4p \)).
Why other options are incorrect: While \( 1s \), \( 2p \), and \( 3d \) are indeed filled before \( 4p \), \( 4s \) is the most recent principal shell companion and aligns with the provided answer key's logic emphasizing the \( n+l \) difference between the 4th shell subshells.
#17 of 81
SZABMU 2023
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If value of azimuthal quantum number is 1 then value of \( m \) will range from:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The magnetic quantum number (\( m \)) depends entirely on the azimuthal quantum number (\( l \)) and defines the possible 3D orientations of the subshell.
Formula: $$ m = -l, \dots, 0, \dots, +l $$
Solution: - The azimuthal quantum number is given as \( l = 1 \) (which corresponds to a \( p \)-subshell).
- The allowable values for \( m \) range from \( -l \) to \( +l \) in integer steps.
- Therefore, the values are \( -1, 0, +1 \).
- The range is strictly \( -1 \text{ to } 1 \).
Why other options are incorrect: The other ranges imply values of \( l \) greater than 1, violating the fundamental bounds of the magnetic quantum number.
#18 of 81
SZABMU 2023
Permalink
The dumbbell like orbitals can move in how many directions?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The "dumbbell" shape refers to the \( p \)-orbital. The number of directions (orientations) it can take in space is given by its magnetic quantum numbers.
Formula: $$ \text{Orientations} = 2l + 1 $$
Solution: - A dumbbell shape means it is a \( p \)-orbital, so its azimuthal quantum number is \( l = 1 \).
- The number of spatial orientations is \( 2(1) + 1 = 3 \).
- These three orientations align along the Cartesian axes: \( p_x \), \( p_y \), and \( p_z \).
Why other options are incorrect: 1 orientation is for the spherical \( s \)-orbital. 5 orientations are for the \( d \)-orbital. 7 orientations are for the \( f \)-orbital.
#19 of 81
SZABMU 2023
Permalink
The relationship between quantum number \( n \) and \( l \) is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The azimuthal quantum number (\( l \)) is strictly limited by the principal quantum number (\( n \)).
Formula: $$ l_{max} = n - 1 $$
Solution: - For any given principal shell \( n \), the allowed values for \( l \) start at 0 and go up to a maximum of \( n - 1 \).
- Therefore, the mathematical boundary defining the maximum possible value of \( l \) is \( l = n - 1 \).
Why other options are incorrect: Options A and D mathematically imply that \( l \) is larger than \( n \), which is physically impossible. Option B arbitrarily restricts the maximum limit too early.
#20 of 81
ETEA 2023
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The number of orientations of a sub-shell for which \( n = 4 \) and \( l = 3 \) will be:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The number of spatial orientations of any given subshell is independent of the principal quantum number (\( n \)) and depends entirely on the azimuthal quantum number (\( l \)).
Formula: $$ \text{Orientations} = 2l + 1 $$
Solution: - We are given \( l = 3 \) (which corresponds to an \( f \)-subshell).
- Substitute \( l=3 \) into the formula: \( 2(3) + 1 = 6 + 1 = 7 \).
- Therefore, there are 7 distinct spatial orientations (degenerate orbitals) for this subshell.
Why other options are incorrect: 1 is for \( l=0 \). 3 is for \( l=1 \). 5 is for \( l=2 \).
#21 of 81
ETEA 2023
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Photon of the lowest wavelength is related to:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The wavelength of an emitted photon is inversely proportional to the energy of the electron transition. Larger energy drops produce shorter (lower) wavelengths.
Formula: $$ E = \frac{hc}{\lambda} $$
Solution: - The energy of a transition depends on how close to the nucleus the electron falls.
- Transitions to \( n=1 \) (Lyman series) have the highest energy and lowest wavelength, but Lyman is not an option.
- Among the given options, the Balmer series drops to the lowest level (\( n=2 \)).
- This results in the largest energy gap compared to Paschen (\( n=3 \)), Brackett (\( n=4 \)), and Pfund (\( n=5 \)).
- Consequently, the Balmer series emits photons with the highest energy and the lowest wavelength among the choices provided.
Why other options are incorrect: The other series involve drops to higher principal quantum shells (smaller energy gaps), resulting in longer wavelengths (infrared region).
#22 of 81
ETEA 2023
Permalink
The maximum \( e/m \) ratio for positive rays is obtained when the discharge tube contains:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: Positive rays (canal rays) are ionized gas molecules. Their charge-to-mass ratio depends entirely on the mass of the gas used in the tube.
Formula: $$ \frac{e}{m} = \frac{\text{Charge}}{\text{Mass}} $$
Solution: - To maximize the fraction \( e/m \), the denominator (mass) must be as small as possible.
- Let's compare the atomic/molecular masses of the gases:
- Helium (\( \text{He} \)) \( \approx 4 \text{ amu} \)
- Nitrogen (\( \text{N}_2 \)) \( \approx 28 \text{ amu} \)
- Neon (\( \text{Ne} \)) \( \approx 20 \text{ amu} \)
- Hydrogen (\( \text{H}_2 \)) \( \approx 2 \text{ amu} \) (and a single ionized H atom is \( \approx 1 \text{ amu} \)).
- Hydrogen has the lowest mass of any element, so it yields the absolute maximum \( e/m \) ratio.
Why other options are incorrect: They are much heavier gases, which increases the denominator and significantly lowers the \( e/m \) ratio.
#23 of 81
ETEA 2023
Permalink
The correct sequence of electronic configuration is:
A
\( 4p, 5s, 4d, 5p, 6s, 4f, 5d \)
B
\( 4p, 4s, 4d, 5p, 6s, 4f, 6d \)
C
\( 4p, 4s, 4d, 5p, 5s, 4f, 5d \)
D
\( 4p, 3s, 4d, 5p, 6s, 4f, 5d \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The Aufbau principle states that orbitals are filled in order of their increasing \( (n+l) \) values.
Formula: $$ \text{Energy} \propto (n+l) $$
Solution: - Let's verify the \( (n+l) \) values for the correct sequence (Option A):
- \( 4p \): \( 4+1=5 \)
- \( 5s \): \( 5+0=5 \) (Higher \( n \) than \( 4p \), so it comes after)
- \( 4d \): \( 4+2=6 \)
- \( 5p \): \( 5+1=6 \) (Higher \( n \) than \( 4d \), so it comes after)
- \( 6s \): \( 6+0=6 \) (Higher \( n \) than \( 5p \), so it comes after)
- \( 4f \): \( 4+3=7 \)
- \( 5d \): \( 5+2=7 \) (Higher \( n \) than \( 4f \), so it comes after)
- The sequence strictly adheres to the \( (n+l) \) hierarchy.
Why other options are incorrect: They jumble the sequence, placing higher energy subshells before lower ones (e.g., jumping from \( 4p \) back down to \( 4s \) or \( 3s \)).
#24 of 81
DUHS 2023
Permalink
Radius of first orbit of hydrogen is \( 0.53\text{\AA} \). Which orbit has a radius of \( 4.77\text{\AA} \)?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: According to Bohr's model, the radius of an electron's orbit is directly proportional to the square of its principal quantum number (\( n \)).
Formula: $$ r_n = r_1 \times n^2 $$
Solution: - We are given the radius of the first orbit: \( r_1 = 0.53\text{\AA} \).
- We are given a target radius: \( r_n = 4.77\text{\AA} \).
- Substitute the values into the formula: \( 4.77 = 0.53 \times n^2 \).
- Divide both sides by 0.53: \( n^2 = \frac{4.77}{0.53} = 9 \).
- Take the square root: \( n = 3 \).
- Therefore, the orbit is the third orbit.
Why other options are incorrect: Second orbit would be \( 0.53 \times 4 = 2.12\text{\AA} \). Fourth orbit would be \( 0.53 \times 16 = 8.48\text{\AA} \).
#25 of 81
DUHS 2023
Permalink
Which of the following has only one orientation in space, in the magnetic field?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The spatial orientation of an orbital is determined by its magnetic quantum number (\( m \)). A purely spherical shape looks identical from every angle.
Formula: $$ \text{Orientations} = 2l + 1 $$
Solution: - The number of orientations depends on the azimuthal quantum number \( l \).
- For an \( s \)-orbital, \( l = 0 \).
- Applying the formula: \( 2(0) + 1 = 1 \).
- The \( s \)-orbital is perfectly spherical, meaning it has only one possible orientation in 3D space, regardless of the magnetic field applied.
Why other options are incorrect: The \( p \)-orbital has 3 orientations. The \( d \)-orbital has 5 orientations. The \( f \)-orbital has 7 orientations.
#26 of 81
DUHS 2023
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What's the mass of one proton?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The mass of subatomic particles is a fundamental constant. Protons and neutrons make up almost all the mass of an atom, while electrons are comparatively negligible.
Formula: $$ m_p \approx 1836 \times m_e $$
Solution: - The rest mass of a proton is \( 1.672 \times 10^{-27} \text{ kg} \).
- The rest mass of an electron is \( 9.109 \times 10^{-31} \text{ kg} \).
- Dividing the mass of the proton by the mass of the electron gives a ratio of approximately 1836.
- This means you would need about 1836 electrons to equal the mass of a single proton.
Why other options are incorrect: A positron is the antimatter equivalent of an electron, so it has the same tiny mass as an electron. Protons and neutrons have nearly identical masses, so a proton is definitely not equal to 1836 neutrons.
#27 of 81
BUMHS 2023
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What is the quantum number for the unpaired electron of chlorin atom \( (n, l, m) \)?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The last (valence) electron in a Halogen occupies a \( p \)-orbital.
Formula: None required.
Solution: - Chlorine has an atomic number \( Z=17 \).
- The full electronic configuration is \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^5 \).
- The valence subshell is \( 3p \). Expanding this using Hund's Rule gives: \( 3p_x^2, 3p_y^2, 3p_z^1 \).
- The single unpaired electron is in the \( 3p \) orbital.
- For a \( 3p \) orbital, the principal quantum number \( n = 3 \).
- The azimuthal quantum number for a \( p \)-orbital is \( l = 1 \).
- The magnetic quantum number \( m \) can be \( -1, 0, \) or \( +1 \). Based on the provided options, \( (3, 1, 1) \) is the only valid set that correctly describes a \( 3p \) orbital.
Why other options are incorrect: Options A and B point to the second shell (\( n=2 \)), which is fully filled in Chlorine. Option D points to a \( 3s \) orbital, which is fully paired (\( 3s^2 \)).
#28 of 81
NUMS 2023
Permalink
The amount of energy associated with quantum of radiation of directly proportional to:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Max Planck established the fundamental relationship between the energy of a quantum (photon) of electromagnetic radiation and its wave properties.
Formula: $$ E = h\nu $$
Solution: - In Planck's quantum equation, \( E \) is energy, \( h \) is Planck's constant, and \( \nu \) is the frequency of the radiation.
- Because \( h \) is a constant multiplier, energy (\( E \)) is mathematically directly proportional to frequency (\( \nu \)).
- Higher frequency radiation (like X-rays) carries more energy per quantum than lower frequency radiation (like radio waves).
Why other options are incorrect: Energy is inversely proportional to wavelength (\( \lambda \)). Velocity (\( c \)) is constant for all electromagnetic radiation in a vacuum, so it cannot act as a directly proportional variable for changing energy.
#29 of 81
NUMS 2023
Permalink
If value of azimuthal quantum number is 2 then total values of magnetic quantum number will be:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The total number of magnetic quantum number (\( m \)) values indicates how many spatial orientations (orbitals) exist within a given subshell.
Formula: $$ \text{Total values} = 2l + 1 $$
Solution: - The azimuthal quantum number is given as \( l = 2 \) (a \( d \)-subshell).
- Substitute \( l = 2 \) into the formula: \( 2(2) + 1 = 4 + 1 = 5 \).
- The specific values are: \( -2, -1, 0, +1, +2 \).
- There are exactly 5 possible values.
Why other options are incorrect: 3 is for \( l=1 \). 7 is for \( l=3 \). 10 is the number of
electrons that the subshell can hold, not the number of magnetic orientations.
#30 of 81
NUMS 2023
Permalink
Total number of directions of f-orbitals in space are:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The number of spatial directions (orientations) of an orbital type is determined by its azimuthal quantum number (\( l \)).
Formula: $$ \text{Directions} = 2l + 1 $$
Solution: - For an \( f \)-orbital, the azimuthal quantum number is \( l = 3 \).
- Substitute into the orientation formula: \( 2(3) + 1 = 6 + 1 = 7 \).
- Therefore, an \( f \)-subshell splits into 7 distinct spatial orientations in a magnetic field.
Why other options are incorrect: 3 is for \( p \)-orbitals. 5 is for \( d \)-orbitals. 6 is the maximum electron capacity of a \( p \)-subshell, not a number of directions.
#31 of 81
NUMS 2023
Permalink
Which of the following quantum number is not obtained from Schrodinger Wave equation?
A
Principal Quantum Number
C
Azimuthal Quantum Number
D
Magnetic Quantum Number
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The Schrödinger wave equation relies on three spatial coordinates (e.g., x, y, z or r, \( \theta, \phi \)) to map the probability density of an electron.
Formula: $$ \hat{H}\psi = E\psi $$
Solution: - Solving the Schrödinger equation inherently yields three constants of integration, which correspond to the principal (\( n \)), azimuthal (\( l \)), and magnetic (\( m \)) quantum numbers.
- These three numbers entirely describe the 3D spatial properties of the orbital.
- The spin quantum number (\( s \)) is a relativistic, intrinsic property of the electron itself, independent of spatial coordinates. It was proposed separately by Uhlenbeck and Goudsmit and later derived from Dirac's relativistic wave equation, not Schrödinger's.
Why other options are incorrect: Principal, Azimuthal, and Magnetic are all direct mathematical outcomes of solving the Schrödinger equation.
#32 of 81
NUMS 2023
Permalink
The electronic configuration for degenerate orbitals is explained by:
D
Pauli exclusion principle
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: When electrons enter a subshell with multiple orbitals of identical energy (degenerate orbitals), they must follow a rule that minimizes electron-electron repulsion.
Formula: None required.
Solution: - Hund's Rule explicitly states that electrons must fill degenerate orbitals (like \( p_x, p_y, p_z \)) singly first, with parallel spins, before pairing up.
- This maximizes the total spin state and reduces electrostatic repulsion, leading to the most stable ground state configuration.
Why other options are incorrect: The Aufbau principle explains vertical filling across different energy shells. The Pauli exclusion principle limits each orbital to a maximum of two electrons. The "\( n+1 \)" rule is a typo for the \( n+l \) rule, which is part of Aufbau.
#33 of 81
UHS 2022
Permalink
Which gas in the discharge tube produces lightest canal rays particles?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Canal rays (positive rays) are the positively charged ions left behind when cathode rays knock electrons off the gas atoms in a discharge tube.
Formula: $$ \text{Mass of canal ray} = \text{Mass of gas atom} - \text{Mass of lost electron} $$
Solution: - The mass of the positive ion directly depends on the mass of the gas used in the tube.
- Hydrogen (\(\text{H}_2\)) is the lightest element in the periodic table.
- When a hydrogen atom loses an electron, it becomes a single proton (\(\text{H}^+\)), which is the lightest possible canal ray particle.
- Consequently, hydrogen yields the highest \(e/m\) ratio for positive rays.
Why other options are incorrect: Helium, Neon, and Argon are all significantly heavier inert gases, so their corresponding positive ions are much more massive.
#34 of 81
SZABMU 2022
Permalink
Principal quantum number is represented by the symbol:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Quantum mechanics utilizes specific, universally accepted variables to denote the four quantum numbers.
Formula: None required.
Solution: - The principal quantum number, which denotes the main energy shell and distance from the nucleus, is represented by the letter \( n \).
- The azimuthal (orbital shape) is \( l \).
- The magnetic (spatial orientation) is \( m \).
- The spin is \( s \).
Why other options are incorrect: They correspond to the other three quantum numbers.
#35 of 81
SZABMU 2022
Permalink
Shape of the sub shell is explained by which quantum number:
A
Principal quantum number
B
Magnetic quantum number
C
Azimuthal quantum number
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The 3D geometric shape of an electron's probability cloud is determined by its angular momentum.
Formula: None required.
Solution: - The azimuthal quantum number (also called the angular momentum quantum number), represented by \( l \), dictates the shape of the orbital.
- If \( l=0 \), the shape is a sphere (\(s\)-orbital).
- If \( l=1 \), the shape is a dumbbell (\(p\)-orbital).
- If \( l=2 \), the shape is generally a double-dumbbell (\(d\)-orbital).
Why other options are incorrect: Principal handles size, Magnetic handles spatial orientation (tilt/direction), and Spin handles internal angular momentum.
#36 of 81
SZABMU 2022
Permalink
The electronic configuration for degenerated orbitals is explained by:
C
Pauli's exclusion principle
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: Degenerate orbitals are orbitals that possess the exact same energy level (e.g., \( p_x, p_y, p_z \)). A specific rule dictates how electrons fill these equal-energy slots.
Formula: None required.
Solution: - Hund's Rule of Maximum Multiplicity states that when filling degenerate orbitals, electrons will first fill them singly, with parallel spins, before they begin to pair up.
- This minimizes electron-electron repulsion, resulting in a more stable, lower-energy atomic state.
Why other options are incorrect: Aufbau explains filling across different energy levels. Pauli's principle dictates that paired electrons must have opposite spins. The \( (n+l) \) rule determines the overall energy order of subshells.
#37 of 81
SZABMU 2022
Permalink
Maximum number of electrons which can be placed in one orbital is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Pauli's Exclusion Principle puts a hard mathematical limit on the capacity of a single spatial orbital.
Formula: None required.
Solution: - According to Pauli's Exclusion Principle, no two electrons can have the exact same set of four quantum numbers.
- If two electrons are in the same exact orbital, they share identical \( n, l \), and \( m \) values.
- The only quantum number left to differentiate them is the spin quantum number (\(s\)), which only has two possible states: \( +1/2 \) (spin up) and \( -1/2 \) (spin down).
- Therefore, a maximum of 2 electrons can occupy a single orbital.
Why other options are incorrect: 1 is an incomplete orbital. 3 or more violates the Pauli Exclusion Principle, as it would require two electrons to have identical spins.
#38 of 81
ETEA 2022
Permalink
Greater the wavelength associated with the photon:
B
Its energy will be variable
D
Its energy will remains constant
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The relationship between the energy of a photon and its wavelength is governed by Planck's equation, showing an inverse proportionality.
Formula: $$ E = \frac{hc}{\lambda} $$
Solution: - In the equation, \( E \) is energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is wavelength.
- Because \( \lambda \) is in the denominator, as wavelength increases (gets greater), the total energy of the photon decreases (gets smaller).
- Long-wavelength radiation (like radio waves) has very low energy, while short-wavelength radiation (like gamma rays) has massive energy.
Why other options are incorrect: Energy is not directly proportional, nor is it independent (constant or random variable) with respect to wavelength.
#39 of 81
ETEA 2022
Permalink
Photon of which of the following series will have largest wavelength?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Wavelength is inversely proportional to energy. Therefore, the spectral series involving transitions in the highest energy levels (smallest energy gaps) produces photons with the lowest energy and largest wavelength.
Formula: $$ \Delta E = \frac{hc}{\lambda} $$
Solution: - The Hydrogen emission spectrum series fall to different base levels:
- Lyman: falls to \( n=1 \) (UV region, huge energy, shortest wavelength)
- Balmer: falls to \( n=2 \) (Visible)
- Paschen: falls to \( n=3 \) (IR)
- Brackett: falls to \( n=4 \) (IR)
- Pfund: falls to \( n=5 \) (Far IR, smallest energy transitions, largest wavelengths).
- Because the energy gap between higher shells (e.g., \( n=6 \) to \( n=5 \)) is extremely small, the emitted photon has minimal energy and maximum wavelength.
Why other options are incorrect: The other series all drop to lower principal quantum numbers, meaning their electron transitions cover larger energy gaps, yielding shorter wavelengths.
#40 of 81
DUHS 2022
Permalink
The equation \( \bar{\nu} = R_h \left[ \frac{1}{2^2} - \frac{1}{n^2} \right] \) where \( n > 2 \) is the equation of H-spectrum for:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The Rydberg formula calculates the wave number of spectral lines based on the energy levels involved in the electron transition.
Formula: $$ \bar{\nu} = R_h \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] $$
Solution: - In the given equation, the lower energy level (destination shell) is locked at \( n_1 = 2 \).
- The upper energy levels are \( n = 3, 4, 5\dots \) (since \( n > 2 \)).
- Transitions that end at the \( n=2 \) shell exclusively generate the Balmer series, which lies predominantly in the visible light spectrum.
Why other options are incorrect: Lyman ends at \( n_1 = 1 \). Paschen ends at \( n_1 = 3 \). Pfund ends at \( n_1 = 5 \).
#41 of 81
DUHS 2022
Permalink
The quantum number allows following orbitals in N-shell.
B
\( s \) & \( p \) orbital
C
\( s, p, d \) & \( f \) orbital
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The principal quantum number \( n \) restricts the number of subshells available within a given electron shell to exactly \( n \) subshells.
Formula: $$ l = 0, 1, 2 \dots (n-1) $$
Solution: - The shells are designated K, L, M, N corresponding to \( n = 1, 2, 3, 4 \).
- For the N-shell, \( n = 4 \).
- The possible azimuthal quantum numbers are \( l = 0, 1, 2, 3 \).
- These correspond directly to the \( s, p, d \), and \( f \) subshells.
- Therefore, the N-shell fully supports \( s, p, d \), and \( f \) orbitals.
Why other options are incorrect: Options A and B are incomplete subsets of the orbitals present in the N-shell.
#42 of 81
BUMHS 2022
Permalink
Which concept was introduced by the scientist Bohr?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: Niels Bohr revolutionized atomic theory by introducing quantization to classical planetary models.
Formula: $$ mvr = \frac{nh}{2\pi} $$
Solution: - Prior to Bohr, Rutherford's model couldn't explain why orbiting electrons didn't spiral into the nucleus due to continuous energy radiation.
- Bohr proposed that electrons move in fixed, circular orbits (shells) around the nucleus without radiating energy.
- He defined these stationary states as specific "energy levels" (K, L, M, etc.).
Why other options are incorrect: The concept of 3D "orbitals" and "sub-energy levels" (subshells) was developed later by quantum mechanics pioneers like Schrödinger and Heisenberg.
#43 of 81
BUMHS 2022
Permalink
What is the electronic configuration of C?
C
\( 1s^2, 2s^2, 2p^6, 3s^2 \)
D
\( 1s^2, 2s^2, 2p^6, 3s^2, 3p^3 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: The electronic configuration is built by sequentially filling atomic orbitals according to the Aufbau principle until all electrons are placed.
Formula: None required.
Solution: - Carbon (C) has an atomic number \( Z = 6 \).
- Therefore, a neutral Carbon atom possesses exactly 6 electrons.
- Fill the \( 1s \) orbital: 2 electrons \( \implies 1s^2 \) (4 remaining).
- Fill the \( 2s \) orbital: 2 electrons \( \implies 2s^2 \) (2 remaining).
- Place the remaining 2 electrons into the \( 2p \) orbital \( \implies 2p^2 \).
- Total configuration: \( 1s^2, 2s^2, 2p^2 \).
Why other options are incorrect: Option B is Oxygen (8). Option C is Magnesium (12). Option D is Phosphorus (15).
#44 of 81
BUMHS 2022
Permalink
With the increase in the value of principal quantum number "\(n\)" the size of the s-orbitals?
C
May or may not remain the same
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The principal quantum number \(n\) determines the main energy level and the average distance of the electron cloud from the nucleus.
Formula: $$ r \propto n^2 $$
Solution: - The radius of an orbital is directly tied to the square of the principal quantum number.
- As \( n \) steps up from 1 to 2 to 3 (e.g., \( 1s \rightarrow 2s \rightarrow 3s \)), the electron occupies higher energy states that pull it further away from the attractive pull of the nucleus.
- Consequently, the physical volume and size of the s-orbital consistently increases.
Why other options are incorrect: The size of an orbital cannot decrease or remain static as higher energy shells inherently encompass a larger spatial volume.
#45 of 81
NUMS 2022
Permalink
Maximum electrons that can be placed in p subshell are:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Subshell capacity is determined by the number of degenerate orbitals it contains, multiplied by 2 (Pauli's limit).
Formula: $$ \text{Max Electrons} = 2(2l + 1) $$
Solution: - For a p-subshell, the azimuthal quantum number \( l = 1 \).
- Number of orbitals = \( (2 \times 1) + 1 = 3 \) distinct spatial orientations (\(p_x, p_y, p_z\)).
- Since each orbital holds a maximum of 2 electrons: \( 3 \times 2 = 6 \) electrons.
Why other options are incorrect: 2 is the capacity for an s-subshell. 10 is for a d-subshell. 14 is for an f-subshell.
#46 of 81
NUMS 2022
Permalink
Apply \( (n+l) \) rule and calculate which of the following orbital has maximum energy?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The \( (n+l) \) rule dictates that orbitals with higher \( (n+l) \) sums exist at higher energy levels.
Formula: $$ \text{Energy} \propto (n+l) $$
Solution: - Let's calculate the values for the valid options:
- \( 4s \): \( n=4, l=0 \implies n+l = 4 \)
- \( 3s \): \( n=3, l=0 \implies n+l = 3 \)
- \( 2p \): \( n=2, l=1 \implies n+l = 3 \)
- Since \( 4s \) yields the highest sum (4), it represents the maximum energy among the valid choices.
- Note: The \( 2d \) orbital is physically impossible because if \( n=2 \), \( l \) can only be 0 or 1.
Why other options are incorrect: Their \( (n+l) \) sums are lower, indicating they are filled earlier because they are lower in energy.
#47 of 81
NUMS 2022
Permalink
The mass of electron is:
A
\( 1836 \) times more than mass of proton
B
\( 1836 \) times less than mass of proton
C
\( 1836 \) times more than mass of neutron
D
\( 1836 \) time more than mass of hydrogen atom
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The electron is a fundamental lepton, carrying an incredibly tiny mass compared to the nucleons (protons and neutrons) that make up the atomic core.
Formula: $$ \frac{m_p}{m_e} \approx 1836 $$
Solution: - The absolute rest mass of a proton is approximately \( 1.672 \times 10^{-27} \text{ kg} \).
- The rest mass of an electron is approximately \( 9.109 \times 10^{-31} \text{ kg} \).
- Dividing the two yields a ratio of roughly 1836.
- This means a single proton is 1836 times heavier than an electron, or conversely, an electron is 1836 times lighter (less massive) than a proton.
Why other options are incorrect: Saying the electron is "more" massive than a proton, neutron, or whole hydrogen atom is factually backwards.
#48 of 81
NUMS 2022
Permalink
The shape of an orbital is determined by which quantum number?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Quantum numbers collectively describe the state and probabilistic location of an electron. Each number corresponds to a specific physical property of the orbital.
Formula: None required.
Solution: - The principal quantum number (\( n \)) determines the size and energy level.
- The azimuthal quantum number (\( l \)), also known as the angular momentum quantum number, dictates the 3D geometric shape of the electron cloud (e.g., spherical for \( l=0 \), dumbbell for \( l=1 \)).
- The magnetic quantum number (\( m \)) describes spatial orientation.
- The spin quantum number (\( s \)) describes the electron's intrinsic spin.
Why other options are incorrect: \( n \), \( m \), and \( s \) relate to size, orientation, and spin respectively, not the shape of the orbital.
#49 of 81
NUMS 2022
Permalink
Proton number of \( _{11}\text{Na}^{23} \) is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The standard atomic symbol notation \( ^A_Z\text{X} \) clearly defines both the mass number and the atomic number of an element.
Formula: $$ \text{Proton Number} = Z $$
Solution: - In the notation \( _{11}\text{Na}^{23} \), the lower left subscript is the atomic number (\( Z \)).
- The atomic number is exactly equal to the number of protons in the nucleus.
- Therefore, the proton number is 11.
Why other options are incorrect: 23 is the mass number (protons + neutrons). 12 is the number of neutrons (\( 23 - 11 \)). 34 is the sum of the mass and atomic numbers, which has no physical meaning.
#50 of 81
PMC 2021
Permalink
Which of the following has the lowest \(e/m\) ratio?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The charge-to-mass ratio (\(e/m\)) is directly proportional to the charge of the particle and inversely proportional to its mass.
Formula: $$ \frac{e}{m} = \frac{\text{Charge}}{\text{Mass}} $$
Solution: - For \( \text{H}^+ \): Charge = +1, Mass \( \approx 1 \) amu. \( e/m \propto 1/1 = 1 \).
- For \( \text{He}^+ \): Charge = +1, Mass \( \approx 4 \) amu. \( e/m \propto 1/4 = 0.25 \).
- For \( \text{Li}^{2+} \): Charge = +2, Mass \( \approx 7 \) amu. \( e/m \propto 2/7 \approx 0.28 \).
- For \( \text{Be} \) (assuming neutral atom as written): Charge = 0. \( e/m = 0 \). Even if inferred as a canal ray ion like \( \text{Be}^+ \) (Mass \( \approx 9 \) amu), \( e/m \propto 1/9 \approx 0.11 \).
- In either case, Be (or its single positive ion) provides the mathematically lowest value due to having the highest mass relative to charge.
Why other options are incorrect: Hydrogen has the highest \(e/m\) ratio due to its minimal mass, and the others are intermediate.
#51 of 81
PMC 2021
Permalink
M-shell contain:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The principal quantum number \(n\) determines the number of subshells within that shell.
Formula: $$ \text{Number of subshells} = n $$
Solution: - The shells are lettered K, L, M, N... corresponding to \( n = 1, 2, 3, 4\dots \)
- For the M-shell, \( n = 3 \).
- An energy level \(n\) contains exactly \(n\) subshells.
- Therefore, the M-shell has 3 subshells: \(s\) (\(l=0\)), \(p\) (\(l=1\)), and \(d\) (\(l=2\)).
Why other options are incorrect: Option A describes the K-shell. Option C describes the L-shell. Option D describes the N-shell.
#52 of 81
PMC 2021
Permalink
Total no. of electrons that can be accommodated in f-subshell:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The maximum number of electrons in any subshell is derived from its azimuthal quantum number \(l\).
Formula: $$ \text{Max Electrons} = 2(2l + 1) $$
Solution: - For an f-subshell, the azimuthal quantum number is \( l = 3 \).
- Number of degenerate orbitals = \( 2l + 1 = 2(3) + 1 = 7 \) orbitals.
- Following Pauli's Exclusion Principle, each orbital holds up to 2 electrons.
- Total electrons = \( 7 \times 2 = 14 \).
Why other options are incorrect: 6 is for a p-subshell. 10 is for a d-subshell.
#53 of 81
PMC 2021
Permalink
\( n+l \) value for \( 5d \) is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The \( (n+l) \) rule is used to determine the relative energy levels of atomic orbitals.
Formula: $$ \text{Value} = n + l $$
Solution: - The term "\(5d\)" directly gives us the principal quantum number \( n = 5 \).
- The letter "\(d\)" corresponds to an azimuthal quantum number \( l = 2 \).
- Adding them together: \( 5 + 2 = 7 \).
Why other options are incorrect: They result from incorrect memorization of the \(l\) value for the d-subshell (e.g., using \(l=1\) yields 6, which is wrong).
#54 of 81
PMC 2021
Permalink
Each electron in an atom must have its own unique set of quantum number is a statement of ____
C
Pauli exclusion principle
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The foundational rules of quantum mechanics govern how electrons populate atoms to prevent the physical impossibility of two identical fermions occupying the exact same space and state.
Formula: None required.
Solution: - The Pauli Exclusion Principle explicitly states that no two electrons in a single atom can have an identical set of all four quantum numbers (\(n, l, m, s\)).
- Even if two electrons share the same orbital (meaning \(n, l\), and \(m\) are identical), they must have opposite spins (one \(s = +1/2\), the other \(s = -1/2\)).
Why other options are incorrect: The Aufbau principle dictates the order of energy filling (lowest energy first). Hund's rule dictates that electrons singly occupy degenerate orbitals before pairing up.
#55 of 81
PMDC 2020
Permalink
The relationship between quantum number \(n\) and \(l\) is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The azimuthal quantum number (\(l\)) is bounded by the principal quantum number (\(n\)).
Formula: $$ l = 0, 1, 2, \dots, (n - 1) $$
Solution: - For a given principal shell \(n\), the possible values of \(l\) start from \(0\) and go up to a maximum of \(n-1\).
- Therefore, the maximum possible value (and the defining mathematical boundary) linking the two is \( l = n - 1 \).
Why other options are incorrect: \(l\) can never be equal to or greater than \(n\), which immediately disqualifies expressions like \(n = l - 1\).
#56 of 81
PMDC 2020
Permalink
Quantum number values for "\(2p\)" orbitals are:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Orbital nomenclature directly states the principal quantum number and uses a letter code for the azimuthal quantum number.
Formula: None required.
Solution: - The number "2" in "\(2p\)" denotes the principal quantum number: \( n = 2 \).
- The letter "\(p\)" denotes the subshell type. The mapping is \( s=0, p=1, d=2, f=3 \).
- Therefore, for a \(p\)-orbital, \( l = 1 \).
- Result: \( n=2, l=1 \).
Why other options are incorrect: \(n=2, l=0\) is the \(2s\) orbital. The others denote \(1s\) or invalid states.
#57 of 81
PMDC 2020
Permalink
Which pair has 1 electron in its outer most '\(s\)' orbital?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Elements in Group 1 (Alkali metals) naturally have \( ns^1 \) configurations. Certain transition metals exhibit anomalous configurations resulting in an \( ns^1 \) outer shell.
Formula: None required.
Solution: - Sodium (Na, \(Z=11\)): \( [Ne] 3s^1 \) (1 electron in outermost \(s\)).
- Chromium (Cr, \(Z=24\)): \( [Ar] 4s^1 3d^5 \) (anomalous config to achieve half-filled stability, leaving 1 electron in the outermost \(s\)).
- Therefore, the pair Na & Cr both fit the condition.
Why other options are incorrect: Iron (Fe) is \( 4s^2 \). Manganese (Mn) is \( 4s^2 \). Helium (He) is \( 1s^2 \).
#58 of 81
NUMS 2019
Permalink
Maximum numbers of electron in a sub-shell is given by:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The number of orbitals in a sub-shell is dictated by the magnetic quantum number \(m\), and Pauli's Exclusion Principle states each orbital can hold 2 electrons.
Formula: $$ \text{Orbitals} = 2l + 1 $$ $$ \text{Max Electrons} = 2(2l + 1) $$
Solution: - For any azimuthal quantum number \(l\), there are \( (2l + 1) \) degenerate orbitals.
- Since each orbital holds a maximum of 2 electrons (with opposite spins), we multiply the number of orbitals by 2.
- The final formula is \( 2(2l+1) \).
Why other options are incorrect: \( 2l+1 \) calculates the number of orbitals, not electrons. The others are mathematically invalid expressions for quantum capacity.
#59 of 81
NUMS 2019
Permalink
The order of energy level in \( _{19}\text{K} \) is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The filling of electrons into orbitals follows the Aufbau principle and the \( (n+l) \) rule.
Formula: $$ \text{Energy} \propto (n+l) $$
Solution: - Potassium (K) has \( Z = 19 \).
- The electron configuration is: \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1 \).
- After filling the \( 3p \) subshell, the next available subshells are \( 4s \) and \( 3d \).
- Calculate \( n+l \): For \( 4s \), \( 4+0=4 \). For \( 3d \), \( 3+2=5 \).
- Since \( 4s \) has lower energy than \( 3d \), the electron enters \( 4s \). Thus, the progressive filling order at the valence boundary is \( 3p \rightarrow 4s \).
Why other options are incorrect: They do not represent the consecutive energy levels being filled for the 19th electron of Potassium.
#60 of 81
SET 2019
Permalink
Which of the following element in its atomic state has electrons fully occupying first two spherically symmetrical orbitals?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: Spherically symmetrical orbitals correspond to \(s\)-orbitals (\(l=0\)).
Formula: None required.
Solution: - The first two spherically symmetrical orbitals are the \(1s\) and \(2s\) orbitals.
- Fully occupying both means filling them with a total of 4 electrons: \( 1s^2, 2s^2 \).
- The element with an atomic number of 4 is Beryllium (Be).
Why other options are incorrect: Helium only fully occupies the first \(s\)-orbital (\(1s^2\)). Oxygen and Carbon have electrons in \(p\)-orbitals, going beyond just the first two \(s\)-orbitals.
#61 of 81
ETEA 2019
Permalink
Quantum number which describes the orientation of orbitals in three dimensional space is:
B
Azimuthal quantum number
C
Principal quantum number
D
Magnetic quantum number
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: Each of the four quantum numbers defines a unique aspect of an electron's state in an atom.
Formula: None required.
Solution: - Principal (\(n\)): Distance from nucleus / Energy level.
- Azimuthal (\(l\)): Shape of the orbital (e.g., sphere, dumbbell).
- Magnetic (\(m\)): Spatial orientation of the orbital along the x, y, and z axes in a magnetic field.
- Spin (\(s\)): Intrinsic angular momentum (spin) of the electron.
Why other options are incorrect: Only the magnetic quantum number corresponds to 3D spatial orientation.
#62 of 81
MDCAT 2019
Permalink
Which of the following is the electronic configuration of Cr?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Chromium (\(Z = 24\)) exhibits an anomalous electronic configuration to achieve the extra stability associated with a symmetrical, half-filled \(3d\)-subshell.
Formula / Rule:$$\text{Aufbau principle with half-filled } d^5 \text{ stability exception: } [\text{Ar}]\, 3d^5\, 4s^1$$
Solution:- Chromium has 24 electrons. The noble gas core Argon (\([\text{Ar}]\)) accounts for 18 electrons.
- The standard Aufbau sequence predicts \([\text{Ar}]\, 3d^4\, 4s^2\). However, transferring one electron from the \(4s\) orbital to the \(3d\) subshell yields \([\text{Ar}]\, 3d^5\, 4s^1\).
- The half-filled \(3d^5\) subshell possesses greater exchange energy and maximum symmetry, making \([\text{Ar}]\, 3d^5\, 4s^1\) the actual, ground-state electronic configuration.
Why other options are incorrect:- Opt_A: \([\text{Ar}]\, 3d^5\, 4s^2\) totals 25 electrons (Manganese, \(\text{Mn}\)), not Chromium.
- Opt_B: \([\text{Ar}]\, 3d^4\, 4s^2\) is the unshifted hypothetical Aufbau configuration, which is less stable than the actual \(3d^5\, 4s^1\) ground state.
- Opt_C: \([\text{Ar}]\, 3d^6\, 4s^0\) has an excited/unstable configuration with paired \(d\) electrons.
#63 of 81
MDCAT 2019
Permalink
Nitrogen has the atomic number 7. Which of the following electronic configurations is of a Nitrogen atom in ground state?
A
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^1, 2\text{py}^1, 2\text{pz}^1 \)
B
\( 1\text{s}^2, 2\text{s}^2, 2\text{py}^2, 2\text{pz}^1 \)
C
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^2, 2\text{py}^1 \)
D
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^2, 2\text{pz}^1 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to Hund's Rule, electrons occupy degenerate orbitals singly, with parallel spins, before pairing up.
Formula:$$ \text{Total Electrons} = 7 $$
Solution:- The first two electrons fill the \( 1s \) orbital: \( 1s^2 \).
- The next two electrons fill the \( 2s \) orbital: \( 2s^2 \).
- The remaining three electrons must be distributed among the three degenerate \( 2p \) orbitals (\( p_x, p_y, p_z \)).
- By Hund's Rule, they each take one empty orbital rather than pairing up: \( 2p_x^1, 2p_y^1, 2p_z^1 \).
Why other options are incorrect:- Option B, Option C, Option D: These configurations show electrons pairing up in one of the p-orbitals while leaving another empty, which violently violates Hund's Rule of maximum multiplicity.
#64 of 81
MDCAT 2018
Permalink
Which is the correct electronic configuration of chromium (24Cr)?
A
\( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^4 \)
B
\( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1, 3d^5 \)
C
\( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6 \)
D
\( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Half-filled (\(d^5\)) and fully-filled (\(d^{10}\)) subshells offer extra thermodynamic exchange energy and spherical symmetry, making them exceptionally stable.
Formula: None required.
Solution: - Chromium has an atomic number \( Z = 24 \).
- The expected Aufbau configuration is \( [Ar] 4s^2 3d^4 \).
- However, to achieve the highly stable half-filled \(d\)-subshell, one electron is promoted from the \(4s\) orbital to the \(3d\) orbital.
- The actual configuration is anomalous: \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1, 3d^5 \).
Why other options are incorrect: Option A assumes rigid adherence to Aufbau without accounting for half-filled stability. Options C and D have incorrect total electron counts.
#65 of 81
MDCAT 2017
Permalink
Number of electrons in \( ^{71}_{31}\text{Ga}^{+3} \) will be
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: A positive ion (cation) is formed when a neutral atom loses electrons. The net positive charge equals the number of electrons lost.
Formula: $$ \text{Electrons} = Z - \text{Charge} $$
Solution: - The atomic number of Gallium (Ga) is given as \( Z = 31 \).
- A neutral atom of Ga has 31 electrons.
- The \( +3 \) charge indicates the loss of 3 electrons.
- Total electrons = \( 31 - 3 = 28 \).
Why other options are incorrect: 34 would mean it gained 3 electrons (anion). 29 and 30 miscalculate the subtraction.
#66 of 81
MDCAT 2017
Permalink
Isotopic symbol of ion of Sulphur-33 is \( ^{33}_{16}\text{S}^{-2} \). How many numbers of protons and neutrons are present if number of electrons are 18?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Protons are strictly determined by the atomic number (\(Z\)), and neutrons are determined by mass number (\(A\)) minus atomic number (\(Z\)), regardless of the electron count.
Formula: $$ P = Z $$ $$ N = A - Z $$
Solution: - The symbol \( ^{33}_{16}\text{S}^{-2} \) gives \( Z = 16 \) (subscript) and \( A = 33 \) (superscript).
- Number of protons \( (P) = Z = 16 \).
- Number of neutrons \( (n) = A - Z = 33 - 16 = 17 \).
Why other options are incorrect: They confuse the electron count (18) with the proton or neutron counts, which are fixed by the nucleus.
#67 of 81
MDCAT 2017
Permalink
Among the following, which contains same no. of electrons & proton but different no. of neutron:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Isotopes are atoms of the same chemical element that have the same atomic number but a different mass number.
Formula: None required.
Solution: - Because isotopes are the same element, their atomic number (\(Z\)) is identical, meaning they have the same number of protons and electrons (in a neutral state).
- The difference in their mass numbers is purely due to a different number of neutrons in the nucleus.
Why other options are incorrect: Isobars have the same mass number but different atomic numbers. Isotones have the same number of neutrons but different numbers of protons.
#68 of 81
MDCAT 2017
Permalink
Identify the correct option associated with the shape of p-orbital:
A
Spherical shape (s-orbital)
B
Four-lobed cloverleaf shape (d_{xy}-orbital)
C
Dumbbell shape with donut collar (d_{z^2}-orbital)
D
Two-lobed dumbbell shape (p_z-orbital)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The azimuthal quantum number \(l\) dictates the shape of the electron probability density (orbital).
Formula: $$ l=1 \implies \text{p-orbital} $$
Solution: - For an s-orbital (\(l=0\)), the shape is spherical.
- For a p-orbital (\(l=1\)), the shape is a dumbbell (two lobes separated by a single nodal plane).
- For a d-orbital (\(l=2\)), the shapes are typically double-dumbbells or a dumbbell with a collar (like \( d_{z^2} \)).
- Option D represents the classic dumbbell shape associated with p-orbitals.
Why other options are incorrect: Option A is an s-orbital. Options B and C represent d-orbitals.
#69 of 81
MDCAT 2016
Permalink
Number of neutrons in \( ^{66}_{30}\text{Zn} \) will be:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The number of neutrons in an isotope is the difference between its mass number (\(A\)) and atomic number (\(Z\)).
Formula: $$ N = A - Z $$
Solution: - From the symbol \( ^{66}_{30}\text{Zn} \), the mass number \( A = 66 \).
- The atomic number \( Z = 30 \).
- Number of neutrons \( N = 66 - 30 = 36 \).
Why other options are incorrect: 30 is the number of protons. 35 and 38 are mathematically incorrect subtractions.
#70 of 81
MDCAT 2016
Permalink
The maximum number of electrons in electronic configuration can be calculated by using formula:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: The total number of electrons that can be accommodated in a principal energy level (shell) depends solely on its principal quantum number \(n\).
Formula: $$ \text{Max Electrons per Shell} = 2n^2 $$
Solution: - For \( n=1 \) (K shell): \( 2(1)^2 = 2 \)
- For \( n=2 \) (L shell): \( 2(2)^2 = 8 \)
- For \( n=3 \) (M shell): \( 2(3)^2 = 18 \)
- This standard formula aggregates the capacities of all subshells within the \( n^{th} \) shell.
Why other options are incorrect: \( 2l+1 \) calculates the number of orbitals in a subshell, not the maximum electrons in a principal shell.
#71 of 81
ETEA 2016
Permalink
What are the values of principal quantum number and azimuthal quantum number for the last electron in chlorine atom?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The quantum numbers \(n\) (principal) and \(l\) (azimuthal) define the specific shell and subshell where the valence electron resides.
Formula: None required.
Solution: - Chlorine has an atomic number \( Z = 17 \).
- Its electronic configuration is \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^5 \).
- The last (valence) electron enters the \( 3p \) orbital.
- For a \( 3p \) orbital, the principal quantum number \( n = 3 \).
- The azimuthal quantum number for a \( p \)-orbital is \( l = 1 \).
- Therefore, \( (n, l) = (3, 1) \).
Why other options are incorrect: They do not match the quantum numbers of the \( 3p \) subshell.
#72 of 81
ETEA 2016
Permalink
Which of the following electronic configuration is / are correct?
(i) \( _{29}\text{Cu} = 1s^2, 2s^2, 3p^6, 3s^1 \)
(ii) \( _{29}\text{Cu} = [Ar] 4s^1 3d^{10} \)
(iii) \( _{24}\text{Cr} = [Ar] 4s^1 3d^{10} \)
(iv) \( _{24}\text{Cr} = [Ar] 4s^2 3d^4 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept: Anomalous electronic configurations occur in transition metals to maximize stability via fully-filled (\(d^{10}\)) or half-filled (\(d^5\)) subshells.
Formula: None required.
Solution: - (i) is completely missing the \(2p\), \(4s\), and \(3d\) orbitals. It is blatantly incorrect.
- (ii) correctly shows the anomalous configuration for Copper (\(Z=29\)), where an electron is promoted to achieve a stable, fully-filled \(3d\) subshell: \( [Ar] 4s^1 3d^{10} \).
- (iii) incorrectly assigns 35 electrons to Chromium (\(Z=24\)). Its correct config is \( [Ar] 4s^1 3d^5 \).
- (iv) is the predicted (but incorrect) Aufbau configuration for Chromium.
- Hence, ONLY (ii) is correct.
Why other options are incorrect: They include structurally flawed or experimentally incorrect configurations.
#73 of 81
MDCAT 2015
Permalink
There are four orbitals \(s, p, d\) and \(f\). Which order is correct with respect to the increasing energy of the orbitals?
A
\( 4s < 4p < 4d < 4f \)
B
\( 4s < 4f < 4p < 4d \)
C
\( 4p < 4s < 4f < 4d \)
D
\( 4f < 4s < 4d < 4p \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept: For orbitals within the same principal shell (same \(n\)), energy increases as the azimuthal quantum number (\(l\)) increases.
Formula: $$ \text{Energy sequence for same } n: s < p < d < f $$
Solution: - Given \( n=4 \) for all options.
- The \( l \) values are: \( s=0, p=1, d=2, f=3 \).
- Applying the \( n+l \) rule:
- \( 4s = 4+0 = 4 \)
- \( 4p = 4+1 = 5 \)
- \( 4d = 4+2 = 6 \)
- \( 4f = 4+3 = 7 \)
- Therefore, the increasing order is \( 4s < 4p < 4d < 4f \).
Why other options are incorrect: They jumble the well-established \( s, p, d, f \) subshell hierarchy.
#74 of 81
MDCAT 2015
Permalink
Which one of the following pairs has the same electronic configuration as possessed by neon (Ne-10)?
A
\( \text{Na}^+, \text{Cl}^- \)
B
\( \text{Na}^+, \text{Mg}^+ \)
C
\( \text{K}^+, \text{Cl}^- \)
D
\( \text{Na}^+, \text{F}^- \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: Isoelectronic species have the exact same number of electrons.
Formula: $$ \text{Total Electrons} = Z - \text{Charge} $$
Solution: - Neon (Ne) has an atomic number \( Z=10 \), meaning it has 10 electrons.
- Sodium ion (\( \text{Na}^+ \)): \( Z=11 \), charge = +1 \(\implies 11 - 1 = 10\) electrons.
- Fluoride ion (\( \text{F}^- \)): \( Z=9 \), charge = -1 \(\implies 9 - (-1) = 10\) electrons.
- Since both \( \text{Na}^+ \) and \( \text{F}^- \) have 10 electrons, they are isoelectronic with Neon.
Why other options are incorrect: \( \text{Cl}^- \) and \( \text{K}^+ \) have 18 electrons (isoelectronic with Argon), and \( \text{Mg}^+ \) has 11 electrons.
#75 of 81
MDCAT 2014
Permalink
According to the number of protons, neutrons and electrons given in the table, which one of the following options is correct?
| Species | Proton | Neutron | Electron |
|---|
| As | 33 | 42 | 30 |
| Ga | 31 | 39 | 28 |
| Ca | 20 | 20 | 20 |
A
\( \text{As}^{+3}, \text{Ga}, \text{Ca}^{-2} \)
B
\( \text{As}^{+3}, \text{Ga}^{+3}, \text{Ca}^{+2} \)
C
\( \text{As}^{+3}, \text{Ga}^{+3}, \text{Ca} \)
D
\( \text{As}^{+3}, \text{Ga}^{+2}, \text{Ca} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The charge of an ion is determined by the difference between its protons (positive charge) and electrons (negative charge).
Formula: $$ \text{Charge} = \text{Protons} - \text{Electrons} $$
Solution: - For As: \( 33 - 30 = +3 \implies \text{As}^{+3} \)
- For Ga: \( 31 - 28 = +3 \implies \text{Ga}^{+3} \)
- For Ca: \( 20 - 20 = 0 \implies \text{Ca}^0 \) (Neutral atom)
Why other options are incorrect: They assign incorrect charges, such as giving Calcium a charge when its protons and electrons are equal.
#76 of 81
MDCAT 2013
Permalink
Number of electrons in the outermost shell of chloride ion (\( \text{Cl}^- \)) is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Formation of an anion involves the addition of electrons to the valence shell of a neutral atom to achieve a stable noble gas configuration.
Formula: $$ \text{Electrons} = Z - \text{Charge} $$
Solution: - A neutral Chlorine (Cl) atom has atomic number \( Z = 17 \).
- Its electronic configuration is \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^5 \) (or 2, 8, 7).
- The outermost shell (\( n=3 \)) has \( 2 + 5 = 7 \) electrons.
- When it gains 1 electron to form the chloride ion (\( \text{Cl}^- \)), the configuration becomes \( 3s^2, 3p^6 \).
- The outermost shell now contains \( 2 + 6 = 8 \) electrons.
Why other options are incorrect: 17 is the total number of protons. 7 is the valence electrons in a neutral Cl atom, not the ion.
#77 of 81
MDCAT 2013
Permalink
Correct order of energy in the given sub-shells is:
A
\( 5s > 3d > 3p > 4s \)
B
\( 3p > 3d > 5s > 4s \)
C
\( 5s > 3d > 4s > 3p \)
D
\( 3p > 3d > 4s > 5s \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Orbital energy is governed by the \( (n+l) \) rule.
Formula: $$ \text{Energy order} \propto (n+l) $$
Solution: - Calculate \( (n+l) \) for each:
- \( 5s: 5 + 0 = 5 \)
- \( 3d: 3 + 2 = 5 \) (Lower \( n \) than \( 5s \), so lower energy)
- \( 4s: 4 + 0 = 4 \)
- \( 3p: 3 + 1 = 4 \) (Lower \( n \) than \( 4s \), so lower energy)
- Descending order of energy: \( 5s > 3d > 4s > 3p \).
Why other options are incorrect: They incorrectly sequence the \( n+l \) values or misapply the tie-breaking rule for identical \( n+l \) sums.
#78 of 81
MDCAT 2012
Permalink
The relative energies of \( 4s \), \( 4p \) and \( 3d \) orbitals are in the order:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept: The energy of an orbital is directly proportional to its \( (n + l) \) value (Aufbau Principle).
Formula: $$ \text{Energy} \propto (n + l) $$
Solution: - For \( 4s \): \( n=4, l=0 \implies n+l = 4 \)
- For \( 3d \): \( n=3, l=2 \implies n+l = 5 \)
- For \( 4p \): \( n=4, l=1 \implies n+l = 5 \)
- Since \( 4s \) has the lowest value, it has the lowest energy.
- For \( 3d \) and \( 4p \), both have \( n+l=5 \). When \( (n+l) \) values are identical, the orbital with the lower \( n \) value has lower energy. Thus, \( 3d < 4p \).
- Overall order: \( 4s < 3d < 4p \).
Why other options are incorrect: They violate the \( (n+l) \) rule for filling order.
#79 of 81
MDCAT 2012
Permalink
With increase in the value of principal quantum number "\(n\)", the shape of the \(s\) orbitals remains same although their sizes:
D
May or may not remain the same
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: The principal quantum number \(n\) determines the main energy level (shell) and the distance of the electron from the nucleus.
Formula: $$ \text{Size} \propto n^2 $$
Solution: - As the value of \( n \) increases (e.g., \( 1s \rightarrow 2s \rightarrow 3s \)), the electron cloud expands outward.
- The orbital maintains its spherical shape, but its volume and radius increase significantly.
Why other options are incorrect: Size cannot decrease or remain the same because higher energy shells are physically located further from the nucleus.
#80 of 81
MDCAT 2010
Permalink
Which quantum number tells us about orientation of orbitals?
A
Principal quantum number
C
Magnetic quantum number
D
Azimuthal quantum number
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept: Quantum numbers describe the properties of atomic orbitals and the properties of electrons in orbitals.
Formula: None required.
Solution: - The principal quantum number (\(n\)) describes energy and size.
- The azimuthal quantum number (\(l\)) describes the shape of the orbital.
- The magnetic quantum number (\(m\)) describes the orientation of the orbital in three-dimensional space.
- The spin quantum number (\(s\)) describes the spin of the electron.
Why other options are incorrect: Principal, spin, and azimuthal describe size/energy, spin, and shape respectively, not 3D spatial orientation.
#81 of 81
MDCAT 2010
Permalink
The ionization energy of hydrogen atom is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Ionization energy (IE) is the exact amount of energy required to remove an electron from a gaseous atom in its ground state.
Formula:$$ \text{IE of Hydrogen} = 1312.36 \text{ kJ/mol} \approx 1313 \text{ kJ/mol} $$
Solution:- The single electron in a Hydrogen atom is tightly held in the 1s orbital close to the nucleus.
- Experimental data establishes the required energy to overcome this electrostatic attraction is roughly 1313 kJ per mole of Hydrogen atoms.
Why other options are incorrect:- Option A, Option B, Option C: These values are orders of magnitude too low to ionize an atom.
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