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🔥🌟 Grand Chemistry Interactive MCQs Quiz Test on Atomic Structure for Class FSC/MDCAT/ECAT | 🧪💡 | Learn & Practice!
1. Bohr's theory is not applicable to which of the following?
a) H
b) H⁺
c) He⁺
d) Li²⁺
✅ Correct Answer: (b)
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system). All the given species like H, He⁺ and Li²⁺ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H⁺ has no electron at all and hence cannot form spectrum.
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system). All the given species like H, He⁺ and Li²⁺ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H⁺ has no electron at all and hence cannot form spectrum.
2. Nitrogen has the electronic configuration 1s² 2s² 2pₓ¹ 2pᵧ¹ 2pz¹ and 1s² 2s² 2pₓ² 2pᵧ¹. This is determined by
a) Aufbau principle
b) Pauli’s rule
c) Hund's rule
d) n+l rule
✅ Correct Answer: (c)
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half-filled then pairing of electrons are allowed in them.
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half-filled then pairing of electrons are allowed in them.
3. Quantum number values for 3s orbital are
a) n = 0, l = 1
b) n=1, l=0
c) n=3, l=1
d) n=3, l=0
✅ Correct Answer: (d)
In the notation 3s (nl), 3 stands for n (principal quantum number) while s represent l (azimuthal quantum number). For 3s orbital, n =3 and l=0.
In the notation 3s (nl), 3 stands for n (principal quantum number) while s represent l (azimuthal quantum number). For 3s orbital, n =3 and l=0.
4. The radius of first orbit of hydrogen atom is
a) 529Å
b) 52.9 Å
c) 5.29 Å
d) 0.529 Å
✅ Correct Answer: (d)
The radius of first orbit of hydrogen atom is 0.529 × 10⁻¹⁰m or 0.529 × 10⁻⁸ cm or 0.529 Å or 0.0529 nm or 52.9 pm.
The radius of first orbit of hydrogen atom is 0.529 × 10⁻¹⁰m or 0.529 × 10⁻⁸ cm or 0.529 Å or 0.0529 nm or 52.9 pm.
5. Line spectrum is used as a tool for the identification of
a) Colors
b) Electrons
c) Elements
d) Molecules
✅ Correct Answer: (c)
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
6. The shape of orbital for which l = 0 is
a) Spherical
b) Dumbbell
c) Double dumbbell
d) Complicated
✅ Correct Answer: (a)
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g. l = 0, used for s-orbital, spherical in shape, l = 1, used for p-orbital, dumb-bell in shape l = 2, used for d-orbital, double dumb-bell in shape, l = 3, used for f-orbital, complicated shape
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g. l = 0, used for s-orbital, spherical in shape, l = 1, used for p-orbital, dumb-bell in shape l = 2, used for d-orbital, double dumb-bell in shape, l = 3, used for f-orbital, complicated shape
7. When 4d orbital is filled, the next electron enter into
a) 5s
b) 5p
c) 5d
d) 6s
✅ Correct Answer: (b)
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first. The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first. The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
8. Which of the following is not an iso electronic pair?
a) Na⁺ and Ne
b) Na⁺ and F⁻
c) Na and Ca
d) Na⁺ and Mg²⁺
✅ Correct Answer: (c)
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na⁺ and F⁻ are isoelectronic pair as both of them contains 10 electrons.
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na⁺ and F⁻ are isoelectronic pair as both of them contains 10 electrons.
9. Balmer series appears in the hydrogen spectrum if electron jumps from any appropriate higher energy orbit to
a) Second orbit
b) Third orbit
c) Fourth orbit
d) Fifth orbit
✅ Correct Answer: (a)
Balmer noted first that the Hydrogen spectrum consisted of some well-defined discrete lines in the visible region (i.e. having between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n₂ = 3, 4, 5, 6, 7 …. ∞) to 2nd orbit (n1 = 2).
Balmer noted first that the Hydrogen spectrum consisted of some well-defined discrete lines in the visible region (i.e. having between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n₂ = 3, 4, 5, 6, 7 …. ∞) to 2nd orbit (n1 = 2).
10. In 1935 A.D. James Chadwick was awarded Nobel Prize because
a) He discovered proton
b) He discovered neutron
c) He determined the radius of hydrogen atom
d) He gave the rules for electronic configuration
✅ Correct Answer: (b)
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
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