JEE Chemistry Structure of the Atom MCQs Set 04

Answer: (c) a nucleus

Explanation:
1. A few \(\alpha\)-particles were deflected through large angles and some bounced back.
2. This showed that the positive charge and most of the mass are packed into a tiny central region.
3. That region is the nucleus.

Teacher's Note:
1. The experiment used a thin gold foil and a beam of \(\alpha\)-particles.
2. Most particles passed straight through, so most of the atom is empty space.

Answer: (a) 27

Explanation:
1. Both have mass number 57.
2. Neutrons in B = mass number - atomic number = 57 - 30 = 27.

Teacher's Note:
1. Neutrons = A - Z.
2. Elements with the same mass number are called isobars.

Answer: (a) \(\ce{^{14}_{6}C}, \ce{^{15}_{7}N}, \ce{^{17}_{9}F}\)

Explanation:
1. Neutrons = A - Z. In set (a): C-14 has 8, N-15 has 8, F-17 has 8 neutrons.
2. In set (b): 6, 7 and 10 neutrons. In set (c): 8, 7 and 8 neutrons. In set (d): 8, 7 and 10 neutrons.
3. Only set (a) has equal neutron numbers.

Teacher's Note:
1. Atoms with the same number of neutrons are called isotones.
2. Work out neutrons for each atom separately.

Answer: (c) a, d, e

Explanation:
1. Electron counts: \(\ce{CN^-}\) = 6 + 7 + 1 = 14; \(\ce{OH^-}\) = 8 + 1 + 1 = 10; \(\ce{CH3^+}\) = 6 + 3 - 1 = 8.
2. \(\ce{N2}\) = 14 and \(\ce{CO}\) = 6 + 8 = 14.
3. So CN-, N2 and CO (a, d, e) have 14 electrons each.

Teacher's Note:
1. Isoelectronic species have the same number of electrons.
2. Add electrons for a negative charge and subtract for a positive charge.

Answer: (b) 14

Explanation:
1. \(\ce{X^{3-}}\) has 10 electrons, so the neutral atom has 7 electrons and Z = 7.
2. Neutrons per atom = 14 - 7 = 7.
3. An \(\ce{X2}\) molecule has two atoms, so 2 x 7 = 14 neutrons.

Teacher's Note:
1. A 3- charge means three extra electrons.
2. Count neutrons per atom first and then multiply by the number of atoms.

Answer: (b) Isotones \(\ce{^{30}_{14}Si}, \ce{^{31}_{15}P}, \ce{^{32}_{16}S}\)

Explanation:
1. Si-30, P-31 and S-32 have 16 neutrons each, so they are isotones.
2. Ca-40 and K-40 have the same mass number but different Z (isobars, not isotopes).
3. The oxygen atoms have the same Z (isotopes, not isobars).
4. \(\ce{N^{3-}}\) and \(\ce{O^{2-}}\) have 10 electrons but \(\ce{Cr^{3+}}\) has 21, so (d) is wrong.

Teacher's Note:
1. Isotopes: same Z. Isobars: same A. Isotones: same neutrons.
2. Check each definition by counting protons and neutrons.

Answer: (d) a, c, e only

Explanation:
1. Mass numbers: A = 236, B = 238, C = 238.
2. A and C have the same Z = 90, so they are isotopes (b is correct, a is wrong).
3. B and C have the same mass number 238, so they are isobars (d is correct, e is wrong).
4. A and B have different mass numbers, so (c) is wrong.
5. The wrong statements are a, c and e.

Teacher's Note:
1. Compute mass number = Z + n for each atom.
2. Same Z means isotopes, same mass number means isobars.

Answer: (a) (i)-a, (ii)-d, (iii)-b, (iv)-c, (v)-a

Explanation:
1. (i) and (v) have the same Z: isotopes.
2. (ii) both have mass number 3: isobars.
3. (iii) Ge-76 has 44 neutrons and As-77 has 44 neutrons: isotones.
4. (iv) U-235 has n - p = 143 - 92 = 51 and Th-231 has 141 - 90 = 51: isodiaphers.

Teacher's Note:
1. Isodiaphers have the same difference between neutrons and protons.
2. Learn all four relations: isotopes, isobars, isotones, isodiaphers.

Answer: (b) 150

Explanation:
1. On the usual scale oxygen is 16 and carbon is 12, so C/O = 12/16 = 3/4.
2. With oxygen = 200, carbon = \(200 \times \frac{12}{16} = 150\).

Teacher's Note:
1. Atomic masses are relative, so ratios stay the same on any scale.
2. Scale both values by the same factor.

Answer: (b) 85.5

Explanation:
1. Average = 0.75 x 85 + 0.25 x 87.
2. = 63.75 + 21.75 = 85.5.

Teacher's Note:
1. Average atomic mass is the abundance-weighted mean.
2. The answer must lie between the two isotopic masses and nearer the more abundant one.

Answer: (a) 2 : 1

Explanation:
1. Proton: charge e, mass m, so e/m.
2. \(\alpha\)-particle: charge 2e, mass 4m, so 2e/4m = e/2m.
3. Ratio = (e/m) : (e/2m) = 2 : 1.

Teacher's Note:
1. An alpha particle has 2 protons and 2 neutrons.
2. Specific charge = charge divided by mass.

Answer: (d) 10

Explanation:
1. Let the abundance of Ne-22 be x (fraction). \(20(1 - x) + 22x = 20.2\).
2. \(20 + 2x = 20.2\), so \(x = 0.1\).
3. The heavier isotope is 10%.

Teacher's Note:
1. Use the weighted-average formula and solve for the unknown fraction.
2. The average is close to 20, so the lighter isotope dominates.

Answer: (c) isodiaphers

Explanation:
1. C-13 has 7 neutrons and 6 protons, so n - p = 1.
2. O-17 has 9 neutrons and 8 protons, so n - p = 1.
3. Equal n - p means they are isodiaphers.

Teacher's Note:
1. Isotopes need the same Z, isobars the same A, isotones the same n.
2. Here only n - p matches.

Answer: (c) \(\ce{H+}\)

Explanation:
1. Approximate e/m in units of e/u: \(\ce{D+}\) = 1/2, \(\ce{He+}\) = 1/4, \(\ce{H+}\) = 1/1, \(\ce{He^{2+}}\) = 2/4 = 1/2.
2. The largest value is for \(\ce{H+}\).

Teacher's Note:
1. Charge divided by mass is largest for the lightest singly charged ion.
2. The proton is the lightest nucleus.

Answer: (a) increase by 37.5%

Explanation:
1. Originally \(\ce{^{16}O}\) has 8 protons and 8 neutrons, mass about 8 + 8 = 16 units.
2. New mass = 8 x 2 + 8 x 0.75 = 22 units.
3. Increase = (22 - 16)/16 = 37.5%.

Teacher's Note:
1. Electron mass is negligible, so only protons and neutrons count.
2. Change in percentage is the change divided by the original value.

Answer: (c) (ii) and (iv)

Explanation:
1. Ge-76 has 76 - 32 = 44 neutrons.
2. (i) 45, (ii) 44, (iii) 43, (iv) 44 neutrons.
3. Isotones have the same neutron number, so (ii) and (iv).

Teacher's Note:
1. Isotones: same neutrons, different protons.
2. Calculate neutrons for every option.

Answer: (c) reduced by 25%

Explanation:
1. Original mass is about 7 (protons) + 7 (neutrons) = 14 units.
2. New mass = 3.5 (protons) + 7 (neutrons) + negligible electron mass = 10.5.
3. Reduction = 3.5/14 = 25%.

Teacher's Note:
1. The electrons contribute almost nothing even when doubled.
2. Only the nuclear particles matter for the mass.

Answer: (d) \(3.75 \times 10^{-8}\)

Explanation:
1. \(\lambda = \frac{c}{\nu}\).
2. \(\lambda = \frac{3 \times 10^{8}}{8 \times 10^{15}} = 3.75 \times 10^{-8}\) m.

Teacher's Note:
1. Wavelength is speed divided by frequency.
2. Planck's constant is not needed here.

Answer: (c) \(6.64 \times 10^{-8}\) Å

Explanation:
1. \(\lambda = \frac{hc}{\Delta E} = \frac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-8}} = 6.64 \times 10^{-18}\) m.
2. Since 1 Å = \(10^{-10}\) m, \(\lambda = 6.64 \times 10^{-8}\) Å.

Teacher's Note:
1. Convert metres to angstrom by dividing by \(10^{-10}\).
2. A very large energy gives a very short wavelength.

Answer: (d) x

Explanation:
1. Angular momentum \(= \frac{nh}{2\pi}\), which depends only on n.
2. The first excited state of \(\ce{Li^{2+}}\) has n = 2, the same as the second orbit of hydrogen.
3. So the angular momentum is x.

Teacher's Note:
1. Bohr angular momentum does not depend on the nuclear charge.
2. First excited state means n = 2.

Answer: (a) \(\frac{3h}{\pi}\)

Explanation:
1. The shells K, L, M, N, O, P correspond to n = 1 to 6, so P has n = 6.
2. Angular momentum \(= \frac{nh}{2\pi} = \frac{6h}{2\pi} = \frac{3h}{\pi}\).

Teacher's Note:
1. Remember the shell letters in order: K, L, M, N, O, P.
2. Use Bohr's quantisation rule.

Answer: (c) inversely proportional to the mass of the electron

Explanation:
1. Bohr's rule: \(mvr = \frac{nh}{2\pi}\).
2. So \(vr = \frac{nh}{2\pi m}\).
3. Hence vr is inversely proportional to the mass m.

Teacher's Note:
1. Start from the angular momentum quantisation condition.
2. Rearrange for the required product.

Answer: (a) 3

Explanation:
1. \(r = 0.529\,\frac{n^2}{Z}\) Å with Z = 1.
2. \(n^2 = \frac{4.761}{0.529} = 9\).
3. So n = 3.

Teacher's Note:
1. The Bohr radius is 0.529 Å for hydrogen.
2. Take the square root to get n.

Answer: (a) 0.159 nm

Explanation:
1. \(r_n = 0.053\,\frac{n^2}{Z}\) nm.
2. For \(\ce{Li^{2+}}\), Z = 3 and n = 3.
3. \(r = 0.053 \times \frac{9}{3} = 0.159\) nm.

Teacher's Note:
1. Radius is proportional to \(n^2/Z\).
2. Here the n^2 and Z factors partly cancel.

Answer: (c) 1 : 4 : 9

Explanation:
1. \(r \propto n^2\).
2. For n = 1, 2, 3 the ratio is 1 : 4 : 9.

Teacher's Note:
1. Bohr radii grow as the square of n.
2. The spacing between orbits increases outwards.

Answer: (d) both (b) and (c)

Explanation:
1. \(r \propto n^2\), so \(\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = 4\) means \(\frac{n_1}{n_2} = 2\).
2. L and K (n = 2 and 1) give 4 : 1, and N and L (n = 4 and 2) also give 16 : 4 = 4 : 1.
3. Both (b) and (c) are correct, while K and L would give 1 : 4.

Teacher's Note:
1. Any pair of shells with n in the ratio 2 : 1 works.
2. Check the order in which the ratio is stated.