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CBSE Class 11 Chemistry · 10 questions · 24 marks
The tidy solar-system picture of the atom collapses once you look closely at light and electrons. This chapter walks you from Bohr's quantised orbits to the quantum mechanical model, where an electron is described by four quantum numbers and an orbital is a region of probability rather than a path.
Which set of quantum numbers is NOT permitted for an electron in an atom?
Answer
n = 2, l = 2, m_l = 0, m_s = −1/2 is not permitted. The azimuthal quantum number l can only take values from 0 to n − 1, so for n = 2 the allowed values of l are 0 and 1 only. An l value of 2 would mean a 2d subshell, which does not exist. The other three sets are valid: they describe a 3d, a 4s and a 3p electron respectively, and in each case m_l lies within the range −l to +l.
How many radial nodes does a 4p orbital possess?
Answer
2 radial nodes. The number of radial (spherical) nodes is given by n − l − 1. For a 4p orbital, n = 4 and l = 1, so the number of radial nodes = 4 − 1 − 1 = 2. It also has l = 1 angular node (a nodal plane), making 3 total nodes, consistent with the general result of n − 1 total nodes.
The ground-state electronic configuration of chromium (Z = 24) is [Ar] 3d⁵4s¹ rather than [Ar] 3d⁴4s². The best reason is:
Answer
A half-filled 3d⁵ set gives extra exchange energy and symmetrical stability. Once 3d and 4s are close in energy, promoting one 4s electron to 3d costs little but produces five singly occupied, parallel-spin d orbitals. This maximises exchange energy and gives a symmetrical charge distribution, so [Ar] 3d⁵4s¹ is the lower-energy arrangement. The first option is wrong because 4s lies below 3d in the neutral, unfilled atom; the Pauli principle allows up to ten 3d electrons; and chromium in its ground state is neutral, not ionised.
The energy of an electron in the n = 3 level of a hydrogen atom is closest to:
Answer
−1.51 eV. For hydrogen, E_n = −13.6 ÷ n² eV. With n = 3, E₃ = −13.6 ÷ 9 = −1.51 eV. The values −13.6 eV, −3.40 eV and −0.85 eV correspond to n = 1, n = 2 and n = 4 respectively. The negative sign indicates the electron is bound to the nucleus; zero energy corresponds to a free electron at n = ∞.
Assertion (A): The de Broglie wavelength of a moving cricket ball is far too small to be measured, while that of an electron of the same speed is measurable. Reason (R): The de Broglie wavelength is inversely proportional to the mass of the particle, and the mass of a cricket ball is enormous compared with that of an electron.
Answer
Both A and R are true and R is the correct explanation of A. From λ = h ÷ mv, at a fixed speed the wavelength varies as 1/m. Planck's constant is only 6.626 × 10⁻³⁴ J s, so for a ball of roughly 0.16 kg the wavelength comes out around 10⁻³³ m, unimaginably smaller than any object we could diffract it against. For an electron of mass 9.11 × 10⁻³¹ kg the wavelength is of the order of atomic spacings, which is exactly why electron diffraction by crystals is observable. The reason therefore explains the assertion.
Calculate the wavelength of the photon emitted when an electron in a hydrogen atom falls from n = 4 to n = 2. (Take 1.097 × 10⁷ m⁻¹ as the Rydberg constant)
Answer
Using 1/λ = R(1/n₁² − 1/n₂²) with n₁ = 2 and n₂ = 4: 1/λ = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × (0.2500 − 0.0625) = 1.097 × 10⁷ × 0.1875 = 2.057 × 10⁶ m⁻¹. So λ = 1 ÷ 2.057 × 10⁶ = 4.86 × 10⁻⁷ m, that is about 486 nm. This blue-green line belongs to the Balmer series, since the transition ends at n = 2 and lies in the visible region.
Write the ground-state electronic configuration of a manganese atom (Z = 25) and of the Mn²⁺ ion. Explain which electrons are lost on ionisation and state the number of unpaired electrons in the ion.
Answer
Manganese atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s², usually written [Ar] 3d⁵ 4s². On ionisation the electrons are removed from the orbital with the highest principal quantum number first, not simply from the last-filled orbital. Once 3d is occupied it drops below 4s in energy, so the two 4s electrons are removed first. Mn²⁺ is therefore [Ar] 3d⁵, with five electrons in five degenerate d orbitals. By Hund's rule each occupies a separate orbital with parallel spin, giving 5 unpaired electrons — which is why Mn²⁺ salts are strongly paramagnetic.
State the Heisenberg uncertainty principle. The position of an electron is measured with an uncertainty of 2.0 × 10⁻¹¹ m. (i) Calculate the minimum uncertainty in its momentum. (ii) Calculate the corresponding minimum uncertainty in its velocity (mass of electron = 9.1 × 10⁻³¹ kg). (iii) Explain what this result implies for the idea of a definite electron orbit. (Take h = 6.626 × 10⁻³⁴ J s)
Answer
The Heisenberg uncertainty principle states that it is impossible to determine simultaneously the exact position and the exact momentum of a microscopic particle; the product of the uncertainties satisfies Δx × Δp ≥ h ÷ 4π. (i) Δp = h ÷ (4π Δx) = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 2.0 × 10⁻¹¹) = 6.626 × 10⁻³⁴ ÷ (2.51 × 10⁻¹⁰) = 2.64 × 10⁻²⁴ kg m s⁻¹. (ii) Δv = Δp ÷ m = 2.64 × 10⁻²⁴ ÷ 9.1 × 10⁻³¹ = 2.9 × 10⁶ m s⁻¹. (iii) The uncertainty in velocity is of the same order as the electron's own speed in an atom, so its velocity is essentially unknown once its position is pinned down to atomic dimensions. A Bohr orbit demands that position and velocity both be known exactly at every instant, which this result forbids. Hence the concept of a fixed circular path is abandoned and replaced by an orbital — a three-dimensional region where the probability of finding the electron, given by the square of the wave function, is high (conventionally 90%).
Explain the Aufbau principle, the Pauli exclusion principle and Hund's rule. Use them to build the electronic configuration of sulphur (Z = 16), showing the filling of the 3p subshell in orbital-box form and stating the number of unpaired electrons.
Answer
Aufbau principle: in the ground state, electrons occupy the available orbitals in order of increasing energy. The order is decided by the (n + l) rule — lower (n + l) fills first, and for equal (n + l) the orbital with the smaller n fills first. This gives the sequence 1s, 2s, 2p, 3s, 3p, 4s, 3d, and so on. Pauli exclusion principle: no two electrons in the same atom can have all four quantum numbers identical. Since an orbital fixes n, l and m_l, the only remaining difference is spin, so an orbital can hold a maximum of two electrons and they must have opposite spins. Hund's rule of maximum multiplicity: within a set of degenerate orbitals, pairing does not begin until every orbital of that set has one electron, and the singly occupied orbitals carry parallel spins. This minimises electron–electron repulsion and maximises exchange energy. For sulphur, 16 electrons fill as 1s² 2s² 2p⁶ 3s² 3p⁴. In the 3p subshell the four electrons distribute as 3p_x (↑↓), 3p_y (↑), 3p_z (↑): the first three enter separate orbitals with parallel spins by Hund's rule, and only the fourth is forced to pair up in an already occupied orbital. Sulphur therefore has 2 unpaired electrons in its ground state, which accounts for its paramagnetic behaviour and its common valency of two in compounds such as H₂S.
A school astronomy club photographs the spectrum of a distant star through a diffraction grating. The image shows a bright continuous band crossed by dark lines, and one prominent dark line sits at 656 nm, exactly where hydrogen is known to emit in the laboratory. Answer: (a) What type of spectrum is the club observing, and how does it arise? (b) To which spectral series of hydrogen does the 656 nm line belong, and between which levels does it arise? (c) Calculate the energy of a 656 nm photon in joules (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹). (d) What does the presence of this line tell the club about the star?
Answer
(a) It is an absorption spectrum. The hot, dense interior of the star emits a continuous spectrum, and as that light passes outward through the cooler gaseous atmosphere, atoms absorb photons of exactly the energies matching their own electronic transitions. Those wavelengths are removed from the beam and appear as dark lines against the bright background. (b) The 656 nm line lies in the visible red region and belongs to the Balmer series, which comprises transitions ending at (or, in absorption, starting from) n = 2. Specifically it is the H-alpha line, the n = 2 to n = 3 transition. (c) E = hc ÷ λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) ÷ (656 × 10⁻⁹) = (1.988 × 10⁻²⁵) ÷ (6.56 × 10⁻⁷) = 3.03 × 10⁻¹⁹ J, roughly 1.9 eV. (d) It shows that hydrogen is present in the star's outer atmosphere, and that some of those hydrogen atoms are already in the n = 2 excited state, which requires a high temperature. Because every element has a unique set of line positions, spectra like this let astronomers determine stellar composition without ever collecting a sample.
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