JEE Chemistry Structure Of Atom MCQs Set 13

Answer: (c) x/4

Explanation:
1. \(\bar\nu\propto Z^2\); \(Z^2 = 16\) for Be\(^{3+}\) and 4 for He\(^+\), so the ratio is 1/4.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Answer: (a) \(3h/2\pi\)

Explanation:
1. The third Balmer line is n = 5 to n = 2
2. \(\Delta L = (5-2)h/2\pi = 3h/2\pi\), independent of Z.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Answer: (d) 9x/5

Explanation:
1. Lyman limit: \(1/x = R\)
2. First Balmer line of He\(^+\): \(\bar\nu = 4R(1/4-1/9) = 5R/9\), so \(\lambda = 9/5R = 9x/5\).

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Answer: (a) 3

Explanation:
1. 12.1 eV takes the electron from n = 1 to n = 3 (\(13.6(1-1/9) = 12.09\) eV)
2. From n = 3 the possible lines are 3-2, 3-1 and 2-1, which makes 3.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Answer: (c) n = 6 to n = 4

Explanation:
1. H: \(R(1/4-1/9) = 5R/36\)
2. He\(^+\): \(4R(1/16-1/36) = 4R\times5/144 = 5R/36\).

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Answer: (d) 1

Explanation:
1. The three quantum numbers n, l and m define one orbital (4d with m = +1)
2. Two electrons can occupy it.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (c) 2 electrons

Explanation:
1. Each orbital holds a maximum of two electrons; a d subshell (five orbitals) holds 10.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (a) spin quantum number

Explanation:
1. Both are in 1s (n = 1, l = 0, m = 0), so only s differs.
2. Only option (a) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (b) s orbital

Explanation:
1. For an s orbital, m = 0 only, so m = -1 is impossible.
2. Only option (b) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (a) d

Explanation:
1. Angular nodes \(= l\); two nodes mean l = 2 (d).
2. Only option (a) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (c) 5

Explanation:
1. There are five d orbitals and each can contain one electron with s = +1/2.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (d) \(4l+2\)

Explanation:
1. A subshell has \(2l+1\) orbitals, each with two electrons: \(2(2l+1) = 4l+2\).
2. Only option (d) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (b) Cu\(^{2+}\), Fe\(^{2+}\), Ni\(^{2+}\)

Explanation:
1. Cu\(^{2+}\) \(3d^9\), Fe\(^{2+}\) \(3d^6\) and Ni\(^{2+}\) \(3d^8\) each have 1s, 2s and 3s filled (6 s electrons)
2. C has 4 and Zn has 8; S\(^{2-}\) has 6 and Zn 8.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (a) \(1s^22s^22p^63s^3\)

Explanation:
1. An s subshell has one orbital, so it holds at most two electrons; \(3s^3\) breaks the Pauli principle
2. The others violate no rule on electron count (some are only excited states).

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (b) \(\sqrt6\,\dfrac{h}{2\pi}\)

Explanation:
1. For l = 2, \(L = \sqrt{2\times3}\,h/2\pi = \sqrt6\,h/2\pi\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (d) \(2d^5\)

Explanation:
1. For n = 2 the allowed subshells are 2s and 2p, so there is no 2d subshell.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (c) yz plane

Explanation:
1. The p\(_x\) orbital has a nodal plane (the yz plane) so the probability is zero there
2. It is non-zero in the xy plane and along x.

Teacher's Note:
1. For electrons in an atom, check the allowed ranges: l from 0 to n - 1, m from -l to +l and s = +1/2 or -1/2. For ions remove electrons from the outermost shell (4s) before 3d. Magnetic moment = sqrt(n(n+2)) B.M.

Answer: (d) two beta particles

Explanation:
1. Alpha emission gives \(^{234}_{90}\)X. To return to Z = 92, two beta particles (each +1) are needed.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (b) 41/62

Explanation:
1. Mass number \(= 238-32 = 206\); atomic number \(= 92-16+6 = 82\)
2. Neutrons \(= 124\)
3. Ratio \(= 82/124 = 41/62\).

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (c) \(^{59}\)Cu

Explanation:
1. Stable copper isotopes are 63 and 65
2. \(^{59}\)Cu has the lowest neutron to proton ratio, so it is the most neutron-deficient and emits positrons.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (b) beta emission, giving \(^{24}_{12}\)Mg

Explanation:
1. \(^{24}Na\) has n/p = 13/11, higher than the stable isotope's 12/11
2. It lowers n/p by converting a neutron to a proton: beta emission.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (d) \(^{56}_{26}\)Fe

Explanation:
1. Binding energy per nucleon rises to a maximum of about 8.78 MeV at \(^{56}_{26}Fe\), which is therefore the most stable nucleus.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (a) 6 alpha and 4 beta particles

Explanation:
1. \(232-208 = 24\) means 6 alpha particles, lowering Z by 12 to 78
2. Four beta particles raise Z to 82.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.

Answer: (c) 8 alpha and 5 beta

Explanation:
1. \(241-209 = 32\) means 8 alpha particles; Z falls by 16 to 78, and 5 beta particles raise it to 83.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Balance mass number and atomic number on both sides of a nuclear equation. High n/p gives beta decay, low n/p gives positron emission or K-capture, and heavy nuclei emit alpha particles.