CBSE Class 11 Chemistry Structure of Atom MCQs Set 08

Answer: (b) h/mv

Explanation:
1. de Broglie proposed that matter has wave properties, with λ = h/mv = h/p.
2. The wavelength is inversely proportional to momentum.

Teacher's Note:
1. A heavier or faster particle has a shorter wavelength.
2. h is Planck's constant.

Answer: (c) 1.3 × 10⁻³⁴ m

Explanation:
1. λ = h/mv = 6.626 × 10⁻³⁴ ÷ (0.1 × 50).
2. λ = 1.3 × 10⁻³⁴ m, far too small to detect.

Teacher's Note:
1. Convert the mass to kg (100 g = 0.1 kg).
2. Wave nature of large objects cannot be observed.

Answer: (a) 7.3 × 10⁻¹⁰ m

Explanation:
1. λ = h/mv = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 1 × 10⁶).
2. λ = 7.3 × 10⁻¹⁰ m, similar to atomic dimensions.

Teacher's Note:
1. The small mass of the electron gives a measurable wavelength.
2. Check the power of ten at the end.

Answer: (d) half

Explanation:
1. λ = h/mv, so λ is inversely proportional to v.
2. Doubling v halves λ.

Teacher's Note:
1. Inverse proportion: when one doubles, the other halves.
2. The mass is unchanged.

Answer: (a) Δx · Δp ≥ h/4π

Explanation:
1. The product of the uncertainties in position and momentum can never be smaller than h/4π.
2. Precise knowledge of one means imprecise knowledge of the other.

Teacher's Note:
1. For an electron this rules out fixed orbits.
2. The effect is negligible for large objects.

Answer: (b) 5.8 × 10⁵ m s⁻¹

Explanation:
1. Δv = h/(4π m Δx) = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 9.11 × 10⁻³¹ × 1.0 × 10⁻¹⁰).
2. Δv = 5.8 × 10⁵ m s⁻¹.

Teacher's Note:
1. Use Δx · m Δv = h/4π.
2. A small Δx gives a large Δv.

Answer: (c) the probability density of finding the electron at a point

Explanation:
1. ψ is the amplitude of the electron wave and has no direct physical meaning.
2. ψ² at a point is proportional to the probability of finding the electron there.

Teacher's Note:
1. ψ² is never negative.
2. Where ψ² is zero there is a node.

Answer: (d) the region of space around the nucleus where the probability of finding an electron is maximum

Explanation:
1. An orbital is a wave function with definite energy and shape; its boundary surface encloses about 90 per cent probability.
2. An orbit is a fixed circular path, which is not allowed by the uncertainty principle.

Teacher's Note:
1. Orbit: Bohr, circular. Orbital: quantum mechanics, three-dimensional.
2. Each orbital holds at most two electrons.

Answer: (a) 9

Explanation:
1. Number of orbitals in a shell = n² = 9.
2. These are one 3s, three 3p and five 3d orbitals.

Teacher's Note:
1. Number of electrons in a shell is 2n² = 18.
2. Do not confuse orbitals with electrons.

Answer: (c) 0, 1 and 2

Explanation:
1. l can take integer values from 0 to n − 1.
2. For n = 3: l = 0, 1, 2, which are the 3s, 3p and 3d subshells.

Teacher's Note:
1. There are n values of l for a shell.
2. l = 0 is always allowed.

Answer: (b) 1p

Explanation:
1. For n = 1 only l = 0 is allowed, so only a 1s orbital exists.
2. 2s, 2p and 3d are all allowed because l is less than n.

Teacher's Note:
1. l must be smaller than n.
2. p needs l = 1, so n must be at least 2.

Answer: (d) 5

Explanation:
1. For l = 2, m_l = −2, −1, 0, +1, +2.
2. The number of values is 2l + 1 = 5, so there are five d orbitals.

Teacher's Note:
1. Number of orbitals in a subshell = 2l + 1.
2. s: 1, p: 3, d: 5, f: 7.

Answer: (c) 32

Explanation:
1. Maximum electrons in a shell = 2n².
2. For n = 4: 2 × 16 = 32.

Teacher's Note:
1. It is 2 electrons per orbital and n² orbitals.
2. 4s² + 4p⁶ + 4d¹⁰ + 4f¹⁴ = 32.

Answer: (a) 1 and 1

Explanation:
1. Radial nodes = n − l − 1 = 3 − 1 − 1 = 1.
2. Angular nodes = l = 1.

Teacher's Note:
1. Total nodes = n − 1 = 2.
2. Add radial and angular nodes as a check.

Answer: (b) 2

Explanation:
1. Radial nodes = n − l − 1 = 3 − 0 − 1 = 2.
2. An s orbital has no angular node, so total nodes = 2 = n − 1.

Teacher's Note:
1. 1s has no node and 2s has one node.
2. Each step in n adds one radial node for s orbitals.

Answer: (d) d_z²

Explanation:
1. d_z² has two lobes along the z axis and a collar (ring) of electron density in the xy plane.
2. The other four d orbitals have four lobes in a cloverleaf shape.

Teacher's Note:
1. d_xy, d_yz, d_zx lie between the axes.
2. d_x²−y² has lobes along the x and y axes.

Answer: (c) 4s, because (n + l) = 4 for 4s and 5 for 3d

Explanation:
1. Orbitals with lower (n + l) have lower energy: 4s has n + l = 4 and 3d has n + l = 5.
2. So 4s is filled before 3d in neutral atoms.

Teacher's Note:
1. If (n + l) is equal, the one with lower n has lower energy.
2. Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s ...

Answer: (b) Hund's rule of maximum multiplicity

Explanation:
1. Hund's rule says electrons occupy degenerate orbitals singly, with parallel spins, before any pairing.
2. So the three 2p electrons go into 2p_x, 2p_y and 2p_z, one in each.

Teacher's Note:
1. Pairing costs repulsion energy.
2. Pauli limits an orbital to two electrons of opposite spin.

Answer: (c) [Ar] 3d⁵ 4s¹

Explanation:
1. A half-filled d subshell is extra stable because of symmetrical distribution and exchange energy.
2. One electron moves from 4s to 3d, giving 3d⁵ 4s¹.

Teacher's Note:
1. Copper is the other common exception: [Ar] 3d¹⁰ 4s¹.
2. Both d5 and d10 are stable arrangements.

Answer: (d) [Ar] 3d¹⁰ 4s¹

Explanation:
1. A completely filled 3d subshell is extra stable.
2. So one electron from 4s goes into 3d, giving 3d¹⁰ 4s¹.

Teacher's Note:
1. Exchange energy is largest for d5 and d10.
2. Cu has one unpaired electron.