Download JEE MCQs for JEE Chemistry: Structure Of Atom
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Chapter-wise Objective Questions: Structure Of Atom
Access the complete set of multiple-choice questions for Structure Of Atom below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) has one angular node
(b) has no node
(c) has one radial node
(d) is dumb-bell shaped
Show Answer & Explanation
Answer: (c) has one radial node
Explanation:
1. Radial nodes \(= n-l-1 = 2-0-1 = 1\); angular nodes \(= 0\)
2. A 1s orbital has no node.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) 2
(b) 3
(c) 4
(d) 1
Show Answer & Explanation
Answer: (b) 3
Explanation:
1. Angular nodes \(= 2\); radial nodes \(= 4-2-1 = 1\)
2. Total \(= n-1 = 3\).
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) 5d
(b) 3d
(c) 4p
(d) 4f
Show Answer & Explanation
Answer: (a) 5d
Explanation:
1. Angular nodes = l = 2 means d
2. Radial nodes \(= n-2-1 = 2\) gives n = 5.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) decreases with n
(b) remains constant
(c) increases with n
(d) first increases and then decreases
Show Answer & Explanation
Answer: (c) increases with n
Explanation:
1. Higher shells are larger: the radial probability maximum moves out with increasing n.
2. Only option (c) agrees with this working.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) 3s has lower energy than 3p
(b) 3s, 3p and 3d orbitals have the same energy
(c) 3p has lower energy than 3d
(d) 3s and 3p are lower than 3d
Show Answer & Explanation
Answer: (b) 3s, 3p and 3d orbitals have the same energy
Explanation:
1. In a one-electron atom the energy depends only on n, so all orbitals of the same shell are degenerate.
2. Only option (b) agrees with this working.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) \(4\pi r^2\psi^2\,dr\)
(b) \(\pi r^2\psi^2\,dr\)
(c) \(2\pi r^2\psi^2\,dr\)
(d) \(\tfrac43\pi r^3\psi^2\,dr\)
Show Answer & Explanation
Answer: (a) \(4\pi r^2\psi^2\,dr\)
Explanation:
1. The shell volume is \(4\pi r^2dr\), and multiplying by the probability density \(\psi^2\) gives the probability.
2. Only option (a) agrees with this working.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) They can ionise gases
(b) They are deflected by electric and magnetic fields
(c) They produce fluorescence on ZnS
(d) They have wavelengths shorter than ultraviolet
Show Answer & Explanation
Answer: (b) They are deflected by electric and magnetic fields
Explanation:
1. X-rays are uncharged electromagnetic waves, so fields do not deflect them.
2. Only option (b) agrees with this working.
Teacher's Note:
1. Count nodes using radial nodes = n - l - 1 and angular nodes = l. Remember the geometry: s is spherical, p has one nodal plane, and dz2 has a ring in the xy plane.
(a) \([Xe]6s^2\)
(b) \([Xe]5d^2\)
(c) \([Kr]5s^25p^6\)
(d) \([Xe]4f^2\)
Show Answer & Explanation
Answer: (a) \([Xe]6s^2\)
Explanation:
1. 56 electrons: \([Xe]\) has 54, and the last two enter 6s (barium).
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.
(a) 2 and 4
(b) 3 and 1
(c) 0 and 2
(d) 4 and 2
Show Answer & Explanation
Answer: (d) 4 and 2
Explanation:
1. \(\sqrt{n(n+2)} = \sqrt{24}\) gives 4 unpaired electrons
2. Fe\(^{2+}\) (\(3d^6\)) has 4 unpaired electrons; Fe\(^{3+}\) has 5.
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.
(a) Zn\(^{2+}\)
(b) Fe\(^{2+}\)
(c) Ni\(^{3+}\)
(d) Cu\(^+\)
Show Answer & Explanation
Answer: (b) Fe\(^{2+}\)
Explanation:
1. Zn\(^{2+}\) is \(d^{10}\) (0), Fe\(^{2+}\) is \(d^6\) (4), Ni\(^{3+}\) is \(d^7\) (3), Cu\(^+\) is \(d^{10}\) (0).
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.
(a) 3
(b) 6
(c) 5
(d) 4
Show Answer & Explanation
Answer: (a) 3
Explanation:
1. The element has 23 electrons (vanadium)
2. M shell \(= 3s^23p^63d^3\), so three electrons have l = 2.
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.
(a) \(\tfrac12\cdot\tfrac{h}{2\pi}\)
(b) \(\dfrac{h}{2\pi}\)
(c) \(\sqrt2\,\dfrac{h}{2\pi}\)
(d) zero
Show Answer & Explanation
Answer: (d) zero
Explanation:
1. For s, l = 0, so \(\sqrt{l(l+1)}\,h/2\pi = 0\).
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.
(a) zero
(b) \(\dfrac{h}{2\pi}\)
(c) \(\sqrt2\,\dfrac{h}{2\pi}\)
(d) \(\sqrt6\,\dfrac{h}{2\pi}\)
Show Answer & Explanation
Answer: (c) \(\sqrt2\,\dfrac{h}{2\pi}\)
Explanation:
1. Radial nodes \(= n-l-1 = 1\) gives l = 1
2. \(L = \sqrt{1\times2}\,h/2\pi = \sqrt2\,h/2\pi\).
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.
(a) n = 5, l = 1, m = 0, s = +1/2
(b) n = 6, l = 0, m = 0, s = +1/2
(c) n = 5, l = 1, m = 1, s = +1/2
(d) n = 5, l = 0, m = 0, s = +1/2
Show Answer & Explanation
Answer: (d) n = 5, l = 0, m = 0, s = +1/2
Explanation:
1. Rb is \([Kr]5s^1\), so n = 5, l = 0, m = 0.
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.
(a) 1, 1, 4, 2
(b) 0, 1, \(\sqrt6\), \(\sqrt3\)
(c) 0, 0, \(\sqrt6\), \(\sqrt2\)
(d) 0, 0, \(\sqrt{20}\), \(\sqrt6\)
Show Answer & Explanation
Answer: (c) 0, 0, \(\sqrt6\), \(\sqrt2\)
Explanation:
1. \(L = \sqrt{l(l+1)}\): s gives 0; d (l = 2) gives \(\sqrt6\); p (l = 1) gives \(\sqrt2\).
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.
(a) \(\sum_{l=1}^{n}2(2l+1)\)
(b) \(\sum_{l=0}^{n-1}2(2l+1)\)
(c) \(\sum_{l=0}^{n}2(2l+1)\)
(d) \(\sum_{l=1}^{n-1}2(2l+1)\)
Show Answer & Explanation
Answer: (b) \(\sum_{l=0}^{n-1}2(2l+1)\)
Explanation:
1. For a given n, l runs from 0 to n - 1 and each subshell holds \(2(2l+1)\) electrons, which sums to \(2n^2\).
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.
(a) clockwise and anticlockwise rotation
(b) anticlockwise and clockwise rotation
(c) magnetic fields in the same direction
(d) two quantum mechanical spin states which have no classical analogue
Show Answer & Explanation
Answer: (d) two quantum mechanical spin states which have no classical analogue
Explanation:
1. Electron spin has no classical equivalent; clockwise or anticlockwise spinning is only a pictorial aid.
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.
(a) n = 4, l = 3, m = -3, s = +1/2
(b) n = 3, l = 0, m = 0, s = -1/2
(c) n = 2, l = 0, m = 1, s = +1/2
(d) n = 5, l = 2, m = 1, s = -1/2
Show Answer & Explanation
Answer: (c) n = 2, l = 0, m = 1, s = +1/2
Explanation:
1. For l = 0, m must be 0, so m = 1 is not allowed.
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.
(a) shape of the orbital
(b) size and energy of the shell
(c) orientation of the orbital
(d) spin of the electron
Show Answer & Explanation
Answer: (b) size and energy of the shell
Explanation:
1. The principal quantum number fixes the size and the main energy of the orbital
2. The azimuthal quantum number gives the shape.
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.
(a) Ti
(b) V
(c) Mn
(d) Al
Show Answer & Explanation
Answer: (c) Mn
Explanation:
1. Mn \(3d^54s^2\): 5; Ti \(3d^24s^2\): 2; V \(3d^34s^2\): 3; Al \(3p^1\): 1.
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.
(a) Cr\(^{3+}\)
(b) Fe\(^{3+}\)
(c) N\(^{3+}\)
(d) Co\(^{3+}\)
Show Answer & Explanation
Answer: (b) Fe\(^{3+}\)
Explanation:
1. Fe\(^{3+}\) (\(d^5\)): 5 unpaired
2. Cr\(^{3+}\) (\(d^3\)): 3
3. N\(^{3+}\) (\(1s^22s^2\)): 0
4. Co\(^{3+}\) (\(d^6\)): 4.
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.
(a) (d) < (b) < (c) < (a)
(b) (c) < (d) < (b) < (a)
(c) (b) < (d) < (a) < (c)
(d) (a) < (c) < (b) < (d)
Show Answer & Explanation
Answer: (a) (d) < (b) < (c) < (a)
Explanation:
1. By the (n + l) rule: (d) 3p = 4 (n = 3), (b) 4s = 4 (n = 4), (c) 3d = 5 (n = 3), (a) 4p = 5 (n = 4)
2. Equal (n + l): lower n first.
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.
(a) refers to a 2d electron
(b) denotes a 2p electron
(c) is not allowed
(d) describes one of the 7f orbitals
Show Answer & Explanation
Answer: (c) is not allowed
Explanation:
1. For n = 2, l can only be 0 or 1, so l = 2 is impossible.
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.
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Structure Of Atom Objective Questions & Solutions for JEE Chemistry
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FAQs
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