JEE Chemistry Structure Of Atom MCQs Set 10

Chemistry Objective Questions and Answers: Structure Of Atom

Review structured MCQ sets for JEE Chemistry Structure Of Atom. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Structure Of Atom MCQs with Answers

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Multiple Choice Questions

Question 1: The number of orbitals in the third shell is
(a) 3
(b) 9
(c) 5
(d) 18
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. Orbitals in a shell \(= n^2 = 9\) (one 3s, three 3p, five 3d).
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.

Question 2: The magnetic moment of a divalent ion of the element with Z = 24 in aqueous solution is
(a) 4.9 B.M.
(b) 2.45 B.M.
(c) 2.83 B.M.
(d) 1.73 B.M.
Show Answer & Explanation

Answer: (a) 4.9 B.M.

Explanation:
1. Cr\(^{2+}\) is \(3d^4\) with 4 unpaired electrons, so \(\sqrt{4\times6} = 4.9\) B.M.
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.

Question 3: Which pair of species is isoelectronic?
(a) Mn\(^{3+}\) and Fe\(^{2+}\)
(b) Cr\(^{3+}\) and Mn\(^{2+}\)
(c) Fe\(^{2+}\) and Co\(^{2+}\)
(d) Mn\(^{2+}\) and Fe\(^{3+}\)
Show Answer & Explanation

Answer: (d) Mn\(^{2+}\) and Fe\(^{3+}\)

Explanation:
1. Mn\(^{2+}\) has 23 electrons and Fe\(^{3+}\) has 23
2. The others differ: Mn\(^{3+}\) 22 and Fe\(^{2+}\) 24; Cr\(^{3+}\) 21 and Mn\(^{2+}\) 23; Fe\(^{2+}\) 24 and Co\(^{2+}\) 25.

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.

Question 4: The spin multiplicity of Ni\(^{2+}\), which has 2 unpaired electrons, is
(a) 3
(b) 1
(c) 2
(d) 4
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Multiplicity \(= 2S+1 = 2(1)+1 = 3\) for two unpaired electrons (S = 1).
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.

Question 5: The unpaired electrons in a scandium atom number
(a) 2
(b) 0
(c) 3
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Sc is \([Ar]3d^14s^2\), so there is 1 unpaired electron.
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.

Question 6: \(^{60}_{27}\)Co is radioactive because
(a) it has a high proton to neutron ratio
(b) its atomic number is high
(c) it has a high neutron to proton ratio
(d) its mass number is odd
Show Answer & Explanation

Answer: (c) it has a high neutron to proton ratio

Explanation:
1. n/p = 33/27 = 1.22, which is too high for Z = 27 (stable nuclei in this region have about 1.1 to 1.2 or lower), so it decays by beta emission.
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.

Question 7: For alpha, beta and gamma radiation of the same energy, the correct order of increasing penetrating power is
(a) \(\alpha < \beta < \gamma\)
(b) \(\gamma < \beta < \alpha\)
(c) \(\beta < \alpha < \gamma\)
(d) \(\alpha < \gamma < \beta\)
Show Answer & Explanation

Answer: (a) \(\alpha < \beta < \gamma\)

Explanation:
1. The heavy, doubly charged alpha particle is stopped by paper; beta particles go through thin metal; gamma rays are the most penetrating.
2. Only option (a) 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.

Question 8: Which nuclear emission gives an isotope of the parent element?
(a) alpha emission
(b) beta emission
(c) positron emission
(d) neutron emission
Show Answer & Explanation

Answer: (d) neutron emission

Explanation:
1. An isotope has the same Z. Only neutron emission leaves Z unchanged (and lowers A by 1); alpha, beta and positron emission change Z.
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.

Question 9: The number of alpha and beta particles lost when \(^{238}_{92}\)U changes to \(^{206}_{82}\)Pb is
(a) 6 alpha and 6 beta
(b) 6 alpha and 8 beta
(c) 8 alpha and 6 beta
(d) 4 alpha and 4 beta
Show Answer & Explanation

Answer: (c) 8 alpha and 6 beta

Explanation:
1. \(238-206 = 32\) gives 8 alpha particles; Z falls by 16 to 76, and 6 beta particles raise it to 82.
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.

Question 10: The number of alpha and beta particles emitted when \(^{234}_{90}\)Th disintegrates to \(^{206}_{82}\)Pb is
(a) 7 alpha and 8 beta
(b) 8 alpha and 6 beta
(c) 6 alpha and 4 beta
(d) 7 alpha and 6 beta
Show Answer & Explanation

Answer: (d) 7 alpha and 6 beta

Explanation:
1. \(234-206 = 28\) gives 7 alpha particles; Z falls by 14 to 76, and 6 beta particles raise it to 82.
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.

Question 11: Isotones of \(^{76}_{32}\)Ge include
(a) \(^{77}_{32}\)Ge
(b) \(^{77}_{34}\)Se
(c) \(^{77}_{33}\)As
(d) \(^{79}_{34}\)Se
Show Answer & Explanation

Answer: (c) \(^{77}_{33}\)As

Explanation:
1. \(^{76}_{32}Ge\) has 44 neutrons
2. As-77 has \(77-33 = 44\); Ge-77 has 45; Se-77 has 43; Se-79 has 45.

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.

Question 12: Of alpha rays, beta rays, gamma rays and anode rays, the uncharged radiation is
(a) alpha rays
(b) gamma rays
(c) beta rays
(d) anode rays
Show Answer & Explanation

Answer: (b) gamma rays

Explanation:
1. Alpha rays are He\(^{2+}\) nuclei, beta rays are electrons, anode rays are positive ions; gamma rays are electromagnetic waves with no charge.
2. Only option (b) 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.

Question 13: Which of the following nuclear changes does NOT occur?
(a) \(^{113}_{48}Cd + n \to {}^{112}_{48}Cd + {}^{0}_{1}e\)
(b) \(^{24}_{12}Mg + {}^4_2He \to {}^{28}_{14}Si + n\)
(c) \(^{40}_{20}Ca + n \to {}^{40}_{19}K + {}^1_1H\)
(d) \(^{43}_{20}Ca + {}^4_2He \to {}^{46}_{21}Sc + {}^1_1H\)
Show Answer & Explanation

Answer: (a) \(^{113}_{48}Cd + n \to {}^{112}_{48}Cd + {}^{0}_{1}e\)

Explanation:
1. Check mass number and charge
2. First: \(113+1 = 114 \ne 112\), so it cannot occur
3. Others: \(24+4 = 28\), \(12+2 = 14\); \(40+1 = 40+1\) with \(20 = 19+1\); \(43+4 = 46+1\) with \(20+2 = 21+1\).

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.

Question 14: The element X formed in \(^{53}_{24}\)Cr + \(^{4}_{2}\)He \(\to\) X + \(^1_0\)n is
(a) \(^{55}_{25}\)Mn
(b) \(^{56}_{25}\)Mn
(c) \(^{56}_{26}\)Fe
(d) \(^{55}_{26}\)Fe
Show Answer & Explanation

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

Explanation:
1. Mass: \(53+4-1 = 56\)
2. Charge: \(24+2 = 26\) (iron)
3. X is \(^{56}_{26}Fe\).

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.

Question 15: \(^{37}_{18}\)Ar captures a K-electron into its nucleus. The product is
(a) \(^{38}_{18}\)Ar
(b) \(^{37}_{17}\)Cl
(c) \(^{36}_{18}\)Ar
(d) \(^{38}_{17}\)Cl
Show Answer & Explanation

Answer: (b) \(^{37}_{17}\)Cl

Explanation:
1. K-capture: \(p + e^- \to n\), so Z falls by 1 and A is unchanged: \(^{37}_{17}Cl\).
2. Only option (b) 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.

Question 16: Which of the following does NOT lead to the formation of isobars?
(a) emission of 1 alpha and 2 beta particles
(b) positron emission
(c) beta emission
(d) K-electron capture
Show Answer & Explanation

Answer: (a) emission of 1 alpha and 2 beta particles

Explanation:
1. Positron emission, beta emission and K capture change Z but not A, so they produce isobars
2. One alpha plus two beta changes A by 4 and leaves Z unchanged (an isotope).

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.

Question 17: The radiation which is NOT deflected in an electric field is
(a) beta rays
(b) gamma rays
(c) alpha rays
(d) anode rays
Show Answer & Explanation

Answer: (b) gamma rays

Explanation:
1. Gamma rays are electromagnetic radiation with no charge, so they pass straight through an electric field.
2. Only option (b) 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.

Question 18: The decay of \(^{13}_7\)N gives
(a) \(^{13}_6\)C + positron
(b) \(^{13}_8\)O + beta particle
(c) \(^{9}_5\)Be + alpha particle
(d) \(^{13}_{8}\)O + positron
Show Answer & Explanation

Answer: (a) \(^{13}_6\)C + positron

Explanation:
1. \(^{13}_7N\) has n/p = 6/7, below 1, so it emits a positron and goes to \(^{13}_6C\).
2. Only option (a) 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.

Question 19: If the binding energy of the 2nd excited state of a hydrogen-like sample is 24 eV, its ionisation energy is
(a) 54.4 eV
(b) 24 eV
(c) 122.4 eV
(d) 216 eV
Show Answer & Explanation

Answer: (d) 216 eV

Explanation:
1. The 2nd excited state is n = 3
2. \(13.6Z^2/9 = 24\), so \(13.6Z^2 = 216\) eV, which is the ionisation energy.

Teacher's Note:
1. Use the scaling laws for hydrogen-like species first and substitute numbers last; this avoids long arithmetic and exposes the intended ratio.

Question 20: The ionisation energy of H is \(21.79\times10^{-19}\) J. The binding energy of the second excited state of Li\(^{2+}\) is
(a) \(9\times21.79\times10^{-19}\) J
(b) \(21.79\times10^{-19}\) J
(c) \(\tfrac13\times21.79\times10^{-19}\) J
(d) \(\tfrac19\times21.79\times10^{-19}\) J
Show Answer & Explanation

Answer: (b) \(21.79\times10^{-19}\) J

Explanation:
1. For Li\(^{2+}\) (Z = 3), n = 3: \(E\propto Z^2/n^2 = 9/9 = 1\), so it equals that of the H ground state.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Use the scaling laws for hydrogen-like species first and substitute numbers last; this avoids long arithmetic and exposes the intended ratio.

Question 21: The maximum number of emission lines of atomic hydrogen visible to the naked eye when electrons fall among levels n = 2 to n = 6 is
(a) 4
(b) 6
(c) 5
(d) 15
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Only the Balmer series (ending at n = 2) is visible: 6-2, 5-2, 4-2 and 3-2 give 4 lines.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Use the scaling laws for hydrogen-like species first and substitute numbers last; this avoids long arithmetic and exposes the intended ratio.

Question 22: What is the total spin of Fe\(^{3+}\) (Z = 26)?
(a) \(\pm1\)
(b) \(\pm2\)
(c) \(\pm3\)
(d) \(\pm5/2\)
Show Answer & Explanation

Answer: (d) \(\pm5/2\)

Explanation:
1. Fe\(^{3+}\) is \(3d^5\) with 5 unpaired electrons
2. Total spin \(= 5\times\tfrac12 = \pm5/2\).

Teacher's Note:
1. Use the scaling laws for hydrogen-like species first and substitute numbers last; this avoids long arithmetic and exposes the intended ratio.

Question 23: The total spin and spin magnetic moment of Cu\(^+\) are
(a) 0 and 0
(b) 1/2 and 1.73 B.M.
(c) 1 and 2.83 B.M.
(d) 5/2 and 5.92 B.M.
Show Answer & Explanation

Answer: (a) 0 and 0

Explanation:
1. Cu\(^+\) is \(3d^{10}\), fully paired, so total spin and magnetic moment are both zero.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Use the scaling laws for hydrogen-like species first and substitute numbers last; this avoids long arithmetic and exposes the intended ratio.

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