chemistry : class11 s-Subshell Characteristics : Concept of subshells : Structure of Atom MCQ Questions and Answers
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s-Subshell Characteristics
MCQ questions and answers.
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MCQ Questions
1. For an electron in an ns subshell, which combination of quantum-number restrictions is correct?
A.
l = 0 and m_l = 0
B.
l = 1 and m_l = 0, ±1
C.
l = n − 1 and m_l = ±l
D.
l = 0 and m_l = ±1
Correct Answer:
l = 0 and m_l = 0
Explanation:
An s subshell is defined by l = 0. Since m_l ranges from −l to +l, l = 0 permits only m_l = 0, so every s subshell contains one orbital.
2. Which statement correctly distinguishes the 1s and 2s subshells while preserving their common s-subshell characteristic?
A.
Both have zero angular nodes, but 2s has one radial node whereas 1s has none.
B.
Both have one angular node, but 2s has two radial nodes whereas 1s has one.
C.
1s has one angular node and 2s has zero angular nodes.
D.
Both have zero radial nodes, but 2s has one angular node.
Correct Answer:
Both have zero angular nodes, but 2s has one radial node whereas 1s has none.
Explanation:
For an ns orbital, the angular-node count is l = 0, so it is zero. The radial-node count is n − l − 1 = n − 1; therefore, 1s has 0 and 2s has 1 radial node.
3. An orbital has a spherically symmetric angular distribution and two radial nodes. Which designation is consistent with these observations?
A.
2s
B.
3s
C.
4s
D.
3p
Correct Answer:
3s
Explanation:
Spherical symmetry indicates l = 0, identifying an s orbital. For an ns orbital, radial nodes equal n − 1; two radial nodes therefore require n = 3, giving 3s.
4. What is the maximum number of electrons that can occupy a single s subshell, and why?
A.
1, because an s subshell contains only one value of m_l
B.
2, because an s subshell contains one orbital that can hold two opposite-spin electrons
C.
4, because l = 0 permits four spin states
D.
6, because every subshell can contain three orbitals
Correct Answer:
2, because an s subshell contains one orbital that can hold two opposite-spin electrons
Explanation:
For an s subshell, l = 0 gives only one possible m_l value, so it contains one orbital. That orbital accommodates at most two electrons with opposite spins.
5. Which statement about the angular part of every s-orbital wavefunction is correct?
A.
It depends on direction and has one nodal plane.
B.
It is independent of direction and has no angular node.
C.
It changes sign across the xy-plane and has one angular node.
D.
It has two angular nodes because m_l = 0.
Correct Answer:
It is independent of direction and has no angular node.
Explanation:
For an s orbital, l = 0. The angular wavefunction is therefore spherically symmetric, and the number of angular nodes equals l, which is zero.
6. A radial probability distribution for an s orbital becomes zero at a finite nonnuclear distance. What does this point represent?
A.
An angular node caused by l = 0
B.
A radial node at which the probability of finding the electron is zero
C.
The outer boundary of the s subshell
D.
The point of maximum electron density in every s orbital
Correct Answer:
A radial node at which the probability of finding the electron is zero
Explanation:
An s orbital has no angular nodes, but an ns orbital can have radial nodes. At a radial node, the radial wavefunction and corresponding probability of finding the electron are zero.
7. Which set gives the numbers of radial nodes in the 1s, 2s, and 4s orbitals, respectively?
A.
0, 1, and 3
B.
0, 2, and 4
C.
1, 2, and 4
D.
0, 1, and 4
Correct Answer:
0, 1, and 3
Explanation:
For an ns orbital, the number of radial nodes is n − 1 because l = 0. Thus 1s has 0, 2s has 1, and 4s has 3 radial nodes.
8. Which comparison of 2s and 2p orbitals is correct specifically with respect to angular nodes and radial nodes?
A.
2s has zero angular nodes and one radial node; 2p has one angular node and zero radial nodes.
B.
2s has one angular node and zero radial nodes; 2p has zero angular nodes and one radial node.
C.
Both have zero angular nodes and one radial node.
D.
Both have one angular node and zero radial nodes.
Correct Answer:
2s has zero angular nodes and one radial node; 2p has one angular node and zero radial nodes.
Explanation:
For 2s, l = 0, so angular nodes are zero and radial nodes are n − l − 1 = 1. For 2p, l = 1, so it has one angular node and zero radial nodes.
9. If the principal quantum number is increased while the orbital remains an s orbital, which change necessarily occurs?
A.
The number of angular nodes increases by one.
B.
The number of possible m_l values increases.
C.
The number of radial nodes increases by one for each unit increase in n.
D.
The orbital loses spherical symmetry.
Correct Answer:
The number of radial nodes increases by one for each unit increase in n.
Explanation:
All s orbitals retain l = 0, so they remain spherically symmetric with m_l = 0 and zero angular nodes. Their radial-node count is n − 1, which increases by one whenever n increases by one.
10. Which statement best explains why an s electron can have appreciable probability density at the nucleus, unlike an electron in a non-s orbital with the same principal quantum number?
A.
An s orbital has l = 0 and no angular node at the nucleus.
B.
An s orbital has the largest possible value of l.
C.
An s orbital has a nodal plane passing through the nucleus.
D.
An s orbital contains more orbitals than every non-s subshell.
Correct Answer:
An s orbital has l = 0 and no angular node at the nucleus.
Explanation:
The s subshell has l = 0, giving spherical symmetry and no angular node. Non-s orbitals have l greater than zero and an angular node that includes the nucleus, so their probability density is zero there.
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