CBSE Class 11 Chemistry Structure of Atom MCQs Set 05

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. According to de Broglie's equation, λ = h/mv.
2. Because of the inverse relation of wavelength and mass, the wavelength associated with ordinary (macroscopic) objects is so short, owing to their large mass, that their wave properties cannot be detected.
3. For microscopic particles of very small mass, λ is measurable and significant.
4. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Electron diffraction confirms the wave nature of electrons; no such effect can be seen for a cricket ball.
2. Small mass means large λ, which is why only microscopic particles show observable wave behaviour.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Radius, \( r_n = \frac{n^2 h^2}{4\pi^2 m e^2 Z} = \frac{n^2}{Z} \times 0.529 \) Å.
2. For hydrogen, Z = 1 and n = 1, giving r₁ = 0.529 Å.
3. With increase of n, r_n also increases, indicating a greater separation between the orbit and the nucleus.
4. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. 0.529 Å (52.9 pm) is called the Bohr radius, a₀.
2. Radius increases as n² and decreases as 1/Z.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Energies of the orbitals in hydrogen or hydrogen-like species depend only on the principal quantum number n.
2. Energies of the orbitals in multi-electron atoms depend on the quantum numbers n and l, i.e. on more than one quantum number.
3. Both statements are correct, but the Reason (about multi-electron atoms) does not explain the Assertion (about one-electron species).

Teacher's Note:
1. In hydrogen, 2s and 2p have the same energy; in multi-electron atoms, 2s is lower than 2p.
2. The Reason describes a different system from the Assertion, so it cannot explain it.

Answer: (c) A is true, but R is false.

Explanation:
1. According to Heisenberg's uncertainty principle, the exact position and exact momentum of an electron cannot be determined simultaneously. The Assertion is correct.
2. As a consequence, the path (trajectory) of an electron in an atom is not clearly defined. The Reason is incorrect.
3. Thus, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. The uncertainty principle rules out definite paths, so it directly contradicts the Reason.
2. Quantum mechanics speaks only of the probability of finding an electron.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. With the help of the principal quantum number n, the average distance of the electron from the nucleus can be determined.
2. The distance of the electron from the nucleus, and the energy and size of the orbital, increase as n increases (K, L, M, N … for n = 1, 2, 3, 4 …).
3. The Reason is a true general statement about removing an electron, but it is not the correct explanation of the Assertion.
4. Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Teacher's Note:
1. Higher n means the electron is, on average, farther from the nucleus and less tightly held.
2. Two true statements are not always cause and effect; check whether R directly answers 'why' for A.

Answer: (c) A is true, but R is false.

Explanation:
1. Spin angular momentum of the electron, a vector quantity, can have two orientations (represented by + and − signs) relative to a chosen axis.
2. These two orientations are distinguished by the spin quantum number m_s, equal to +½ or −½. They are the two spin states of the electron, normally represented by ↑ (spin up) and ↓ (spin down).
3. The signs do not signify the wave function. Assertion is correct, but Reason is incorrect.

Teacher's Note:
1. The + and − signs in m_s refer to spin orientation, not to the sign (phase) of ψ.
2. The sign of the wave function is shown by shading or colour in orbital diagrams, a completely different idea.

Answer: (c) A is true, but R is false.

Explanation:
1. The magnetic quantum number m_l can have values −l, −(l − 1), …, 0, …, (l − 1), l, i.e. a total of (2l + 1) values. The Assertion is correct.
2. It gives information about the number of orbitals and their orientation, not their shape (the shape is given by l). The Reason is incorrect.
3. Thus, Assertion is correct, but Reason is incorrect.

Teacher's Note:
1. Shape → l; orientation → m_l.
2. The number of m_l values equals the number of orbitals in the subshell.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a given principal quantum number, s, p, d, f … subshells all have different energies in a multi-electron atom.
2. This is because mutual repulsion exists among the electrons; inner electrons shield the outer ones differently for different l values.
3. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. In the hydrogen atom, by contrast, all subshells of a shell have the same energy because there is no electron–electron repulsion.
2. Shielding and penetration (s > p > d > f) are what split the subshell energies.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Half-filled and fully filled degenerate sets of orbitals acquire extra stability because of their symmetry (and larger exchange energy).
2. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹) are standard examples of this extra stability.
2. Exchange energy is the other key factor; it is maximum when a subshell is half-filled or full.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. An s-orbital electron is more tightly bound to the nucleus than a p-orbital electron because s electrons penetrate closer to the nucleus and are less shielded than p electrons.
2. As a result, the effective nuclear charge (Z_eff) experienced by the electron decreases with increase of the azimuthal quantum number (l).
3. Since a larger Z_eff means stronger binding, the Reason correctly explains the Assertion.
4. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Penetration order s > p > d > f leads to Z_eff order s > p > d > f for the same shell.
2. This is also why, in a multi-electron atom, a 2s electron has lower energy than a 2p electron.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a given orbital, only the boundary surface diagram of constant probability density |ψ|² is taken to be a good representation of its shape.
2. It encloses the region or volume in which the probability of finding the electron is very high, i.e. about 90%, and so helps to visualise the orbital.
3. Both Assertion and Reason are correct statements and Reason is the correct explanation of the Assertion.

Teacher's Note:
1. A boundary surface usually encloses about 90% probability; a 100% boundary would be infinitely large.
2. This is why orbitals are drawn as spheres, dumb-bells and cloverleaves.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. According to the (n + l) rule, the energy of an orbital depends on both n and l.
2. 5s (n + l = 5) has higher energy than 4s (n + l = 4) because it has a higher principal quantum number; both have l = 0.
3. The Reason is a correct statement (only for one-electron species such as H is l immaterial), but it does not explain the Assertion, since 4s and 5s have the same l.
4. Both Assertion and Reason are correct statements, but Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. When two orbitals have the same l, their energy order is decided by n.
2. Check whether the quantity named in the Reason actually differs between the two orbitals being compared.

Answer: (d) A is false, but R is true.

Explanation:
1. An atomic orbital is designated by three quantum numbers n, l and m_l; m_s is not needed for an orbital. The Assertion is incorrect.
2. The state of an electron in an atom is specified by four quantum numbers n, l, m_l and m_s. The Reason is correct.
3. Thus, Assertion is incorrect but Reason is a correct statement.

Teacher's Note:
1. Orbital: n, l, m_l (three numbers). Electron: n, l, m_l, m_s (four numbers).
2. Each orbital can hold two electrons that differ only in m_s.

Answer: (d) A is false, but R is true.

Explanation:
1. Isoelectronic ions have the same number of electrons but different nuclear charge; their size decreases with increase in atomic number.
2. Example: Na⁺ (Z = 11, 10 electrons) 0.95 Å; Mg²⁺ (Z = 12, 10 electrons) 0.65 Å; Al³⁺ (Z = 13, 10 electrons) 0.50 Å.
3. A greater nuclear attraction pulls the electrons closer and makes the ion smaller, so the Reason is correct.
4. Thus, Assertion is incorrect but Reason is a correct statement.

Teacher's Note:
1. Among isoelectronic species, more protons for the same electrons means a smaller size.
2. Order for the 10-electron series: N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.

Answer: (d) A is false, but R is true.

Explanation:
1. (2l + 1) gives the number of orbitals in a subshell, not the number of electrons; the maximum number of electrons is 2(2l + 1). The Assertion is incorrect.
2. For a given n, l can have only the values 0, 1, 2, …, (n − 1), i.e. 0, 1, 2, 3 for s, p, d and f. The Reason is correct.
3. Thus, Assertion is incorrect but Reason is a correct statement.

Teacher's Note:
1. Orbitals in a subshell = 2l + 1; electrons = 2(2l + 1).
2. For n = 3, l = 0, 1, 2 only; there is no 3f subshell.

Answer: (d) A is false, but R is true.

Explanation:
1. For a multi-electron atom such as sodium, the energy of an electron is determined by both n and l, so the Assertion is incorrect.
2. For a one-electron system, the energy expression obtained from quantum mechanics is the same as that obtained in Bohr's theory; the Reason is correct.
3. Thus, Assertion is incorrect but Reason is correct.

Teacher's Note:
1. Sodium has 11 electrons, so its orbital energies depend on n and l.
2. Only H-like species have energies that depend on n alone.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. An s-orbital electron is more tightly bound to the nucleus than a p-orbital electron because s electrons penetrate closer to the nucleus and are less shielded than p electrons.
2. As a result, the effective nuclear charge (Z_eff) experienced by the electron decreases with increase of the azimuthal quantum number (l).
3. Since a larger Z_eff means stronger binding, the Reason correctly explains the Assertion.
4. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. s electrons penetrate closer to the nucleus than p electrons of the same shell.
2. Larger Z_eff means stronger attraction and tighter binding.

Answer: (c) A is true, but R is false.

Explanation:
1. Orbit and orbital are different concepts, so the Assertion is correct.
2. The Reason has the definitions interchanged. Its correct form is: an orbit represents the circular path along which the electron moves, but an orbital represents the three-dimensional space around the nucleus within which the probability of finding the electron is maximum.
3. Thus, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. Orbit is a Bohr-model idea (definite path); orbital is a quantum-mechanical idea (probability region).
2. Watch for swapped definitions in Assertion–Reason questions.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The principal quantum number determines the size of the orbital, i.e. the most probable distance of the electron from the nucleus.
2. The farther the electron is from the nucleus, the less tightly it is held and the higher its energy; hence the energy is largely determined by n.
3. Assertion and Reason both are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. n fixes both the size and, to a large extent, the energy of an orbital.
2. In H-like species n alone fixes the energy; in other atoms l makes a smaller correction.

Answer: (c) A is true, but R is false.

Explanation:
1. The spin quantum number describes the spin of the electron (clockwise or anticlockwise), so the Assertion is correct.
2. The value of the spin quantum number can be either +½ or −½ and it is independent of the value of the principal quantum number. The Reason is incorrect.
3. Thus, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. m_s is the only quantum number that does not depend on the others.
2. Every orbital, whatever its n, holds one +½ and one −½ electron.

Answer: (c) 3.2 × 10⁻¹⁹ J

Explanation:
1. W₀ = 4.2 eV = 4.2 × 1.60 × 10⁻¹⁹ J = 6.72 × 10⁻¹⁹ J.
2. Energy of the photon, E = hc/λ = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) ÷ (2000 × 10⁻¹⁰ m) = 9.94 × 10⁻¹⁹ J.
3. Kinetic energy of the fastest electron = E − W₀ = (9.94 − 6.72) × 10⁻¹⁹ J = 3.22 × 10⁻¹⁹ J ≈ 3.2 × 10⁻¹⁹ J.

Teacher's Note:
1. Convert the work function from eV to J (1 eV = 1.6 × 10⁻¹⁹ J) before subtracting.
2. 1 Å = 10⁻¹⁰ m; 2000 Å is UV light, energetic enough to cause emission here.

Answer: (c) Both (a) and (b)

Explanation:
1. The number of electrons ejected depends upon the intensity or brightness of light (above the threshold frequency).
2. Brightness and intensity mean the same thing here, so both (a) and (b) are correct.

Teacher's Note:
1. Intensity controls how many electrons are emitted; frequency controls their kinetic energy.
2. Below the threshold frequency, no electrons are emitted however bright the light is.

Answer: (b) proportional to the frequency of the electromagnetic radiation

Explanation:
1. From hν = hν₀ + ½m_ev², KE = h(ν − ν₀).
2. Hence the kinetic energy of the ejected electron increases linearly with the frequency of the radiation, as stated in the passage, i.e. option (b).

Teacher's Note:
1. Strictly, KE varies linearly with ν (a straight line of slope h), and is zero at ν = ν₀.
2. Energy and frequency have different units, so options (a) and (c) cannot even be compared.

Answer: (a) half of the potential energy

Explanation:
1. Kinetic energy in an orbit = \( \frac{Ze^2}{8\pi\varepsilon_0 r} \) … (i)
2. Potential energy in an orbit = \( -\frac{Ze^2}{4\pi\varepsilon_0 r} \) … (ii)
3. Comparing (i) and (ii), KE = −½ PE, i.e. |KE| = ½ |PE|.
4. Hence the magnitude of kinetic energy in an orbit is equal to half of the (magnitude of the) potential energy.

Teacher's Note:
1. In a Bohr orbit, E = KE + PE = −KE = ½PE.
2. So for H in n = 1: KE = +13.6 eV, PE = −27.2 eV, E = −13.6 eV.

Answer: (d) All of the above

Explanation:
1. (a) is correct: the energy of a quantum is E = hν.
2. (b) is correct: Planck's quantum theory explained the intensity distribution of black body radiation at different temperatures.
3. (c) is correct: photoelectrons are ejected instantly, with no time lag.
4. Since all three statements are correct, option (d) is the answer.

Teacher's Note:
1. The instant emission in the photoelectric effect could not be explained by the wave theory; it supports the photon idea.
2. Black body radiation and the photoelectric effect were the two key results that led to quantum theory.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Energies of the orbitals in hydrogen or hydrogen-like species depend only on the principal quantum number n.
2. Energies of the orbitals in multi-electron atoms depend on the quantum numbers n and l, i.e. on more than one quantum number.
3. Both statements are correct, but the Reason (about multi-electron atoms) does not explain the Assertion (about one-electron species).

Teacher's Note:
1. In H, 3s, 3p and 3d all have the same energy.
2. The Reason talks about a different kind of atom, so it cannot explain the Assertion.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. With the help of the principal quantum number n, the average distance of the electron from the nucleus can be determined.
2. The distance of the electron from the nucleus, and the energy and size of the orbital, increase as n increases (K, L, M, N … for n = 1, 2, 3, 4 …).
3. The Reason is a true general statement about removing an electron, but it is not the correct explanation of the Assertion.
4. Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Teacher's Note:
1. n fixes the average distance of the electron from the nucleus.
2. Ask whether R answers the 'why' of A before choosing option (a).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a given principal quantum number, s, p, d, f … subshells all have different energies in a multi-electron atom.
2. This is because mutual repulsion exists among the electrons; inner electrons shield the outer ones differently for different l values.
3. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Shielding by inner electrons is a result of electron–electron repulsion.
2. This splitting gives the order ns < np < nd < nf in multi-electron atoms.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Number of radial (spherical) nodes = n − l − 1. For p-orbitals, l = 1, so radial nodes = n − 2.
2. For 2p, n − 2 = 0; hence 2p-orbitals have no spherical node.
3. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. 3p has one spherical node and 4p has two.
2. Every p orbital still has one angular node (a nodal plane through the nucleus).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Half-filled and fully filled degenerate sets of orbitals acquire extra stability because of their symmetry (and larger exchange energy).
2. Thus both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Symmetrical distribution of electrons lowers the energy of the subshell.
2. Exchange energy is maximum for half-filled and fully filled subshells.

Answer: (d) infinite

Explanation:
1. Δx·Δp ≥ h/4π, so Δp ≥ h/(4πΔx).
2. When Δx = 0, Δp becomes infinite.

Teacher's Note:
1. Perfect knowledge of position means complete ignorance of momentum.
2. The product Δx·Δp can never be smaller than h/4π.

Answer: (b) 1.92 cm

Explanation:
1. Uncertainty in velocity, Δv = 3 × 10⁴ × 0.001/100 = 0.3 cm s⁻¹.
2. Δx = h/(4πmΔv) = 6.6 × 10⁻²⁷ erg s ÷ (4 × 3.14 × 9.1 × 10⁻²⁸ g × 0.3 cm s⁻¹)
3. Δx = 1.92 cm.

Teacher's Note:
1. Use h = 6.6 × 10⁻²⁷ erg s when the mass is in grams and velocity in cm s⁻¹.
2. An uncertainty of about 2 cm for an electron shows why definite orbits are meaningless.

Answer: (d) equal to or greater than λ

Explanation:
1. Light cannot locate an object more precisely than about its own wavelength.
2. To locate an electron with an uncertainty Δx, light of wavelength not longer than Δx must be used (λ ≤ Δx).
3. Hence, if light of wavelength λ is used, the uncertainty in position would be equal to or greater than λ.

Teacher's Note:
1. Shorter wavelength light gives better position resolution, but its photons carry more momentum and disturb the electron more.
2. This trade-off is the physical picture behind Heisenberg's principle.

Answer: (a) 8.0 × 10¹² m s⁻¹

Explanation:
1. Given Δx = Δp = mΔv.
2. From Δx·Δp = h/4π: (mΔv)² = h/4π ⇒ \( \Delta v = \frac{1}{m}\sqrt{\frac{h}{4\pi}} = \frac{1}{2m}\sqrt{\frac{h}{\pi}} \)
3. Δv = [1/(2 × 9.1 × 10⁻³¹)] × √(6.63 × 10⁻³⁴/3.14)
4. Δv = 7.98 × 10¹² m s⁻¹ ≈ 8 × 10¹² m s⁻¹.

Teacher's Note:
1. Set Δx = Δp, so Δp² = h/4π; find Δp first and then divide by m.
2. Such an enormous Δv shows the uncertainty principle is decisive for electrons.

Answer: (d) 5.27 × 10⁻³⁰ m

Explanation:
1. Δx·Δp = h/4π
2. Δx = 6.62 × 10⁻³⁴ ÷ (4 × 3.14 × 10⁻⁵) = 5.27 × 10⁻³⁵ ÷ 10⁻⁵
3. Δx = 5.27 × 10⁻³⁰ m.

Teacher's Note:
1. h/4π ≈ 5.27 × 10⁻³⁵ J s is worth remembering for quick calculations.
2. Mass is not needed here because Δp is given directly.

Answer: (c) 9

Explanation:
1. Number of orbitals = n² = 3² = 9 orbitals.
2. The 9 orbitals are 3s, 3p_x, 3p_y, 3p_z, 3d_x²−y², 3d_z², 3d_xy, 3d_yz and 3d_xz.

Teacher's Note:
1. Orbitals per shell = n²; 1 + 3 + 5 = 9 for n = 3.
2. Do not confuse with electrons per shell, 2n² = 18.

Answer: (b) 2p

Explanation:
1. n = 2 gives the shell number 2 and l = 1 is a p subshell.
2. Hence n = 2, l = 1 ⇒ 2p.

Teacher's Note:
1. l = 0, 1, 2, 3 correspond to s, p, d, f.
2. The subshell label is written as the value of n followed by the letter for l.

Answer: (b) 2

Explanation:
1. Any orbital can have a maximum of two electrons, which must be of opposite spin (Pauli exclusion principle).
2. Hence a 2s orbital can hold 2 electrons.

Teacher's Note:
1. Every orbital, whatever its type, holds at most two electrons.
2. The s subshell has only one orbital, so it too holds a maximum of 2 electrons.

Answer: (d)

Explanation:
1. A p orbital has two lobes on opposite sides of the nucleus along one axis, i.e. it is dumb-bell shaped.
2. Option (d) shows such a dumb-bell along the x-axis, i.e. the p_x orbital.
3. Option (a) is a four-lobed d orbital (d_xy), (b) is a four-lobed d orbital along the axes and (c) is a spherical s orbital.

Teacher's Note:
1. p orbitals have one nodal plane through the nucleus between the two lobes.
2. Four lobes indicate a d orbital; a sphere indicates an s orbital.

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. In the arrangement 1s ⇅, 2s ⇅, 2p ⇅ ↑, the spins of the paired electrons cancel (+½ − ½ = 0).
2. Number of unpaired electrons = 1.
3. Total spin = 1 × ½ = ½.

Teacher's Note:
1. Only unpaired electrons contribute to the total spin; total spin = (number of unpaired electrons) × ½.
2. Note that the arrangement shown violates Hund's rule; the correct 2p³ arrangement would give a total spin of 3/2.