JEE Chemistry Structure of Atom MCQs Set 21

Chemistry Objective Questions and Answers: Structure Of Atom

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Multiple Choice Questions

Question 1: The uncertainty in the momentum of an electron is 1 × 10-5 kg m/s. The uncertainty in its position will be (h = 6.62 × 10-34 kg m2/s):
(a) 2.36 × 10-28 m
(b) 5.25 × 10-28 m
(c) 2.27 × 10-30 m
(d) 5.27 × 10-30 m
Show Answer & Explanation

Answer: (d) 5.27 × 10-30 m

Explanation:
1. Δx = h/(4πΔp) = 6.62 × 10-34 / (4 × 3.14 × 1 × 10-5).
2. Δx = 5.27 × 10-30 m.

Teacher's Note:
1. Δx is inversely proportional to Δp.
2. The factor 4π is easy to forget.

Question 2: If the kinetic energy of an electron is increased four times, the wavelength of the de Broglie wave associated with it becomes:
(a) half
(b) one-fourth
(c) four times
(d) two times
Show Answer & Explanation

Answer: (a) half

Explanation:
1. λ = h/√(2mKE), so λ ∝ 1/√KE.
2. When KE becomes 4 times, λ becomes 1/2 of its initial value.

Teacher's Note:
1. Wavelength is inversely proportional to the square root of kinetic energy.
2. Momentum doubles when KE increases four times.

Question 3: The two electrons in the K shell of an atom differ in:
(a) principal quantum number
(b) azimuthal quantum number
(c) magnetic quantum number
(d) spin quantum number
Show Answer & Explanation

Answer: (d) spin quantum number

Explanation:
1. Both electrons have n = 1, l = 0 and m = 0.
2. By the Pauli exclusion principle they must have opposite spins, +1/2 and -1/2.

Teacher's Note:
1. The 1s orbital holds a maximum of two electrons.
2. Spin is the fourth quantum number.

Question 4: What is the lowest value of n that allows g orbitals to exist?
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. g orbitals have l = 4.
2. Since l can take values from 0 to n - 1, we need n - 1 ≥ 4, so the lowest n is 5.

Teacher's Note:
1. s, p, d and f correspond to l = 0, 1, 2 and 3.
2. Then g corresponds to l = 4.

Question 5: The number of spherical (radial) nodes in a 3p orbital is:
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Number of radial nodes = n - l - 1 = 3 - 1 - 1 = 1.
2. Total nodes are n - 1 = 2, of which one is angular (l = 1).

Teacher's Note:
1. Radial nodes = n - l - 1.
2. Angular nodes = l.

Question 6: Which of the following sets of quantum numbers (n, l, ml, ms) are not possible? I: n = 0, l = 0, ml = 0, ms = +1/2; II: n = 1, l = 0, ml = 0, ms = -1/2; III: n = 1, l = 1, ml = 0, ms = +1/2; IV: n = 2, l = 1, ml = 0, ms = -1/2; V: n = 3, l = 3, ml = -3, ms = +1/2; VI: n = 3, l = 1, ml = 0, ms = +1/2
(a) I, II and III
(b) I, III and V
(c) I, IV and VI
(d) I, II, III and V
Show Answer & Explanation

Answer: (b) I, III and V

Explanation:
1. n cannot be zero (set I), l must be less than n (sets III and V have l = n).
2. Sets II, IV and VI obey all the rules.

Teacher's Note:
1. Rules: n ≥ 1; l = 0 to n - 1; ml = -l to +l.
2. Check each set one by one.

Question 7: The total number of orbitals associated with the third shell is:
(a) 2
(b) 4
(c) 9
(d) 3
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. The number of orbitals in a shell is n2.
2. For n = 3 it is 9 (one 3s, three 3p and five 3d).

Teacher's Note:
1. Maximum electrons in a shell = 2n2.
2. For n = 3 this is 18.

Question 8: The number of angular nodes for a 4d orbital is:
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. The number of angular nodes equals the azimuthal quantum number l.
2. For d orbitals l = 2, so there are 2 angular nodes.

Teacher's Note:
1. Radial nodes for 4d = 4 - 2 - 1 = 1.
2. Total nodes = 3.

Question 9: Which of the following sets of quantum numbers is incorrect for a 4d electron?
(a) 4, 3, 2, +1/2
(b) 4, 2, 1, +1/2
(c) 4, 2, -2, +1/2
(d) 4, 2, 1, -1/2
Show Answer & Explanation

Answer: (a) 4, 3, 2, +1/2

Explanation:
1. A 4d electron has n = 4 and l = 2.
2. The set (4, 3, 2, +1/2) has l = 3, which belongs to a 4f electron.

Teacher's Note:
1. d orbitals have l = 2.
2. Check n first, then l, then ml within -l to +l.

Question 10: The electronic configuration of the O2- ion is:
(a) 1s2 2s2 2p4
(b) 1s2 2s2 2p6
(c) 1s2 2s2 2p5
(d) 1s2 2s2
Show Answer & Explanation

Answer: (b) 1s2 2s2 2p6

Explanation:
1. The oxygen atom has 8 electrons and O2- has 10 electrons.
2. The 10 electrons fill 1s2 2s2 2p6, the neon configuration.

Teacher's Note:
1. Add as many electrons as the negative charge.
2. O2- is isoelectronic with Ne.

Question 11: The pair of ions having the same electronic configuration is:
(a) Cr3+, Fe3+
(b) Fe3+, Mn2+
(c) Fe3+, Co3+
(d) Sc3+, Cr3+
Show Answer & Explanation

Answer: (b) Fe3+, Mn2+

Explanation:
1. Fe3+ (26 - 3 = 23 electrons) is [Ar]3d5 and Mn2+ (25 - 2 = 23 electrons) is also [Ar]3d5.
2. The other pairs have different numbers of d electrons.

Teacher's Note:
1. Remove 4s electrons before 3d electrons when forming cations.
2. d5 is half filled and particularly stable.

Question 12: The number of unpaired electrons in Ni2+ (atomic number 28) is:
(a) 0
(b) 2
(c) 4
(d) 6
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Ni is [Ar]3d84s2 and Ni2+ is [Ar]3d8.
2. Eight d electrons fill five orbitals with three pairs and two unpaired electrons.

Teacher's Note:
1. Use Hund's rule to place electrons singly first.
2. Magnetic moment = √(n(n + 2)) = 2.83 BM for n = 2.

Question 13: The atomic number of an element is 35 and its mass number is 81. The number of electrons in its outermost shell is:
(a) 3
(b) 5
(c) 7
(d) 9
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. The element is bromine with configuration [Ar]3d104s24p5.
2. The outermost (fourth) shell holds 2 + 5 = 7 electrons.

Teacher's Note:
1. Mass number is not needed to find the electronic configuration.
2. Bromine is a halogen with seven valence electrons.

Question 14: In a set of degenerate orbitals, electrons distribute themselves to retain like spins as far as possible. This statement belongs to:
(a) Pauli's exclusion principle
(b) the Aufbau principle
(c) Hund's rule of maximum multiplicity
(d) Slater's rule
Show Answer & Explanation

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

Explanation:
1. Hund's rule says that pairing starts only after every degenerate orbital has one electron, all with parallel spin.
2. This gives maximum multiplicity and lower energy.

Teacher's Note:
1. Pauli deals with identical sets of four quantum numbers.
2. Aufbau gives the order of filling of orbitals.

Question 15: An f-shell containing 6 unpaired electrons can exchange:
(a) 6 electrons
(b) 9 electrons
(c) 12 electrons
(d) 15 electrons
Show Answer & Explanation

Answer: (d) 15 electrons

Explanation:
1. Exchanges possible among n parallel-spin electrons = n(n - 1)/2.
2. For 6 electrons, 6 × 5/2 = 15.

Teacher's Note:
1. Exchange energy is highest for half-filled subshells.
2. Cr and Cu have extra stability for this reason.

Question 16: The number of metals out of Li (2.4 eV), Na (2.3), K (2.2), Mg (3.7), Cu (4.8), Ag (4.3), Fe (4.7), Pt (6.3) and W (4.75) which show the photoelectric effect with light of wavelength 300 nm (work functions in brackets) is:
(a) 2
(b) 4
(c) 6
(d) 8
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Photon energy = 1240 eV nm / 300 nm = 4.13 eV.
2. The photoelectric effect occurs when the work function is below 4.13 eV, i.e. for Li, Na, K and Mg.

Teacher's Note:
1. E (eV) = 1240/λ (nm).
2. Compare the photon energy with each work function.

Question 17: Which transition in the hydrogen spectrum has the same wavelength as the Balmer transition n = 4 to n = 2 of the He+ spectrum?
(a) n = 4 to n = 3
(b) n = 3 to n = 2
(c) n = 4 to n = 2
(d) n = 2 to n = 1
Show Answer & Explanation

Answer: (d) n = 2 to n = 1

Explanation:
1. For He+, 1/λ = R × 4 (1/4 - 1/16) = 3R/4.
2. For hydrogen, 1/λ = R (1/1 - 1/4) = 3R/4, i.e. the transition n = 2 to n = 1.

Teacher's Note:
1. The Rydberg formula scales as Z2.
2. He+ (n1, n2) matches H (n1/2, n2/2).

Question 18: The number of photons emitted per second by a 60 W source of monochromatic light of wavelength 663 nm is (h = 6.63 × 10-34 J s):
(a) 4 × 10-20
(b) 1.54 × 1020
(c) 3 × 10-20
(d) 2 × 1020
Show Answer & Explanation

Answer: (d) 2 × 1020

Explanation:
1. Energy of one photon = hc/λ = 6.63 × 10-34 × 3 × 108 / 663 × 10-9 = 3 × 10-19 J.
2. Photons per second = 60 / 3 × 10-19 = 2 × 1020.

Teacher's Note:
1. Power = number of photons per second × energy per photon.
2. Convert nm to m.

Question 19: The work function of caesium is 1.9 eV. When it is irradiated with light of wavelength 500 nm, the velocity of the ejected photoelectron is about:
(a) 4.527 × 105 m/s
(b) 3.258 × 105 m/s
(c) 1.892 × 106 m/s
(d) 2.329 × 105 m/s
Show Answer & Explanation

Answer: (a) 4.527 × 105 m/s

Explanation:
1. Photon energy = 1240/500 = 2.48 eV, so KE = 2.48 - 1.9 = 0.58 eV = 9.29 × 10-20 J.
2. v = √(2KE/m) = √(2 × 9.29 × 10-20 / 9.11 × 10-31) = 4.5 × 105 m/s.

Teacher's Note:
1. KE = hν - work function.
2. Use 1 eV = 1.602 × 10-19 J.

Question 20: The ejection of photoelectrons from silver in a photoelectric experiment can be stopped by a voltage of 0.35 V when radiation of 256.7 nm is used. The work function of silver is:
(a) 4.83 eV
(b) 4.48 eV
(c) 4.45 eV
(d) 4.34 eV
Show Answer & Explanation

Answer: (b) 4.48 eV

Explanation:
1. Photon energy = 1240/256.7 = 4.83 eV and the stopping potential gives KEmax = 0.35 eV.
2. Work function = 4.83 - 0.35 = 4.48 eV.

Teacher's Note:
1. eV0 = hν - φ.
2. Stopping potential in volts equals KE in eV for a single charge.

Question 21: The quantum numbers of six electrons are: I (4, 2, -2, -1/2); II (3, 2, 1, +1/2); III (4, 1, 0, +1/2); IV (3, 2, -2, -1/2); V (3, 1, -1, +1/2); VI (4, 1, 0, +1/2). The correct increasing order of energy is:
(a) I < II = IV < III = VI < V
(b) V < II = VI < III = IV < I
(c) V < III = VI < II = IV < I
(d) V < II = IV < III = VI < I
Show Answer & Explanation

Answer: (d) V < II = IV < III = VI < I

Explanation:
1. The orbitals are I: 4d, II: 3d, III: 4p, IV: 3d, V: 3p and VI: 4p.
2. Energy follows (n + l): 3p (4) < 3d (5) = 4p (5) < 4d (6), and for equal (n + l) the lower n has lower energy, so 3d < 4p.

Teacher's Note:
1. Use the (n + l) rule.
2. Electrons in the same orbital have equal energy irrespective of spin.

Question 22: An electron beam is accelerated through a potential difference of 10,000 V. The de Broglie wavelength of the beam is:
(a) 0.123 Å
(b) 0.356 Å
(c) 0.186 Å
(d) 0.258 Å
Show Answer & Explanation

Answer: (a) 0.123 Å

Explanation:
1. For an electron, λ = 12.27/√V Å.
2. λ = 12.27/√10000 = 0.1227 Å.

Teacher's Note:
1. The formula λ = 12.27/√V Å is valid only for electrons.
2. Check by computing h/√(2meV).

Question 23: A particle A moving with a certain velocity has a de Broglie wavelength of 1 Å. Particle B has a mass 25% of that of A and a velocity 75% of A. The de Broglie wavelength of B is:
(a) 3 Å
(b) 5.33 Å
(c) 6.88 Å
(d) 0.48 Å
Show Answer & Explanation

Answer: (b) 5.33 Å

Explanation:
1. λ ∝ 1/(mv). For B, mv = 0.25 × 0.75 = 0.1875 times that of A.
2. λB = 1/0.1875 = 5.33 Å.

Teacher's Note:
1. Express the quantities as ratios.
2. Wavelength is inversely proportional to momentum.

Question 24: If the shortest wavelength in the Lyman series of hydrogen is x, the longest wavelength in the Balmer series of He+ is:
(a) 36x/5
(b) 5x/9
(c) x/4
(d) 9x/5
Show Answer & Explanation

Answer: (d) 9x/5

Explanation:
1. The shortest Lyman wavelength corresponds to n = ∞ to 1, so 1/x = R.
2. The longest Balmer wavelength of He+ is for n = 3 to 2: 1/λ = 4R (1/4 - 1/9) = 5R/9, so λ = 9x/5.

Teacher's Note:
1. The longest wavelength is for the transition with the smallest energy gap.
2. Z2 = 4 for He+.

Question 25: An ion of mass number 37 carries one unit of negative charge and contains 11.1% more neutrons than electrons. The symbol of the ion is:
(a) 3517Cl
(b) 3717Cl-
(c) 3719K+
(d) none of these
Show Answer & Explanation

Answer: (b) 3717Cl-

Explanation:
1. Let the electrons in the ion be x, so neutrons = 1.111x and the protons = x - 1.
2. 1.111x + x - 1 = 37 gives x = 18, so the protons = 17 and the ion is 37Cl-.

Teacher's Note:
1. Neutral atom electrons = protons; an anion has more electrons.
2. Mass number = protons + neutrons.

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