chemistry : class11 Subshell Penetration Effect : Concept of subshells : Structure of Atom MCQ Questions and Answers
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Subshell Penetration Effect
MCQ questions and answers.
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MCQ Questions
1. For orbitals having the same principal quantum number n = 4, which order correctly represents their penetration toward the nucleus?
A.
4f > 4d > 4p > 4s
B.
4s > 4p > 4d > 4f
C.
4p > 4s > 4f > 4d
D.
4d > 4f > 4s > 4p
Correct Answer:
4s > 4p > 4d > 4f
Explanation:
For orbitals in the same shell, penetration decreases as the azimuthal quantum number increases. Thus, s orbitals penetrate most, followed by p, d, and f orbitals.
2. Which observation most directly demonstrates that a 2s electron penetrates the nuclear region more effectively than a 2p electron?
A.
The 2s orbital has a lower principal quantum number than the 2p orbital
B.
The radial probability distribution of 2s has electron density close to the nucleus, whereas 2p has zero density at the nucleus
C.
The 2p orbital has a greater number of angular nodes than the 2s orbital
D.
The 2s orbital contains fewer electrons than the 2p orbital
Correct Answer:
The radial probability distribution of 2s has electron density close to the nucleus, whereas 2p has zero density at the nucleus
Explanation:
An s orbital has nonzero probability density at the nucleus, while a p orbital has a node at the nucleus. This allows a 2s electron to spend more time near the nucleus and indicates greater penetration.
3. In a multi-electron atom, why does a 3s electron generally experience a greater effective nuclear attraction than a 3d electron in the same principal shell?
A.
The 3s electron has a larger principal quantum number
B.
The 3s electron penetrates closer to the nucleus and is therefore less shielded by inner electrons
C.
The 3d electron has no angular momentum
D.
The 3d electron always lies entirely outside the 4s orbital
Correct Answer:
The 3s electron penetrates closer to the nucleus and is therefore less shielded by inner electrons
Explanation:
Greater penetration allows a 3s electron to enter regions closer to the nucleus, where shielding is lower. Consequently, it experiences a larger effective nuclear charge than a 3d electron.
4. Which sequence correctly ranks the penetration of the following orbitals in a many-electron atom?
A.
3p > 2s > 3s > 3d
B.
3s > 2s > 3p > 3d
C.
2s > 3s > 3d > 3p
D.
3d > 3p > 3s > 2s
Correct Answer:
3s > 2s > 3p > 3d
Explanation:
For orbitals of the same n, 3s penetrates more than 3p and 3d. Between 3s and 2s, the 3s orbital has an additional radial region that allows appreciable density near the nucleus, and the standard penetration order used for subshell comparison is 3s > 2s > 3p > 3d.
5. Two electrons occupy the 4s and 3d subshells of the same atom. Which statement best explains why the 4s subshell can show greater penetration than 3d?
A.
Penetration depends only on the principal quantum number, so 4s must penetrate less
B.
The zero angular momentum of 4s permits electron density close to the nucleus, unlike the 3d subshell
C.
The 3d subshell has no radial nodes and therefore cannot approach the nucleus
D.
The 4s subshell is always closer to the nucleus at every radial distance
Correct Answer:
The zero angular momentum of 4s permits electron density close to the nucleus, unlike the 3d subshell
Explanation:
An s subshell has l = 0 and can have nonzero electron density at the nucleus. A d subshell has l = 2 and is excluded from the nuclear region by its angular behavior, so 4s has greater penetration.
6. If an electron is promoted from a 3p orbital to a 3d orbital without changing its principal quantum number, what happens to its penetration toward the nucleus?
A.
It increases because the angular momentum increases
B.
It decreases because the 3d orbital has greater angular momentum and stronger exclusion from the nuclear region
C.
It remains exactly unchanged because both orbitals belong to n = 3
D.
It becomes equal to the penetration of a 3s orbital
Correct Answer:
It decreases because the 3d orbital has greater angular momentum and stronger exclusion from the nuclear region
Explanation:
Within the same shell, increasing l reduces penetration. Therefore, 3d penetrates less than 3p, even though both have n = 3.
7. Which set of orbitals contains the orbital with the greatest penetration and the orbital with the least penetration, respectively?
A.
5s and 4f
B.
4p and 5d
C.
3d and 5s
D.
4f and 3s
Correct Answer:
5s and 4f
Explanation:
Among the listed orbitals, the s orbital has the greatest penetration because l = 0, while the f orbital has the least because l = 3. Hence, 5s is most penetrating and 4f is least penetrating in this comparison.
8. Which change would most strongly increase the penetration of an electron while keeping the principal quantum number constant?
A.
Changing from an ns orbital to an np orbital
B.
Changing from an nd orbital to an nf orbital
C.
Changing from an np orbital to an ns orbital
D.
Changing the electron spin from up to down
Correct Answer:
Changing from an np orbital to an ns orbital
Explanation:
At constant n, penetration increases when the azimuthal quantum number decreases. Changing from p, with l = 1, to s, with l = 0, produces a substantial increase in penetration.
9. Why can the energy ordering of subshells in a multi-electron atom differ from the hydrogen-like ordering based only on n?
A.
Subshell penetration changes the extent of shielding and therefore changes the effective nuclear charge experienced by electrons
B.
Electrons in multi-electron atoms have no principal quantum number
C.
All subshells with the same n necessarily have identical penetration
D.
The nucleus changes its charge whenever an electron enters a p or d subshell
Correct Answer:
Subshell penetration changes the extent of shielding and therefore changes the effective nuclear charge experienced by electrons
Explanation:
In multi-electron atoms, subshells with the same n have different penetration. More penetrating orbitals are less shielded and experience greater effective nuclear attraction, so their energies are not determined by n alone.
10. Which statement about the radial distribution of a 2s orbital best distinguishes penetration from the location of its most probable radial distance?
A.
A 2s electron cannot penetrate the nucleus because its radial probability is greatest at a finite distance
B.
The maximum radial probability identifies the only region where the electron exists
C.
Although 2s may have a radial-probability maximum away from the nucleus, its nonzero density near the nucleus demonstrates penetration
D.
Penetration is determined only by the number of radial nodes
Correct Answer:
Although 2s may have a radial-probability maximum away from the nucleus, its nonzero density near the nucleus demonstrates penetration
Explanation:
Radial probability and penetration are related but not identical. An s orbital can have substantial probability at a finite radius while still possessing nonzero electron density near the nucleus, which is the key feature of penetration.
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