Atomic Structure Important Questions And Answers
The practice set of Atomic Structure with Important Questions And Answers helps students of class 11 and also for students studing for various competitive exams.Students are advised to practice and understand all the questions accordingily.
1. Which of the following atoms and ions are isoelectronic i.e. have the same number of electrons with the neon atom.
a) F−
b) Oxygen atom
c) Mg
d) N−
ANS: F− have the same number of electrons with the neon atom.
2. Atoms consist of protons, neutrons, and electrons. If the mass of neutrons and electrons were made half and two times respectively to their actual masses, then the atomic mass of 6C12
a)Will remain approximately the same
b)Will become approximately two times
c)Will remain approximately half
d)Will be reduced by 25%
ANS: No change by doubling the mass of electrons however by reducing the mass of neutron to half total atomic mass becomes 6 + 3 instead of 6 + 6. Thus reduced by 25%.
3.The increasing order (lowest first) for the values of e/m (charge/mass) for:
a) e, p, n, α
b) n, p, e, α
c) n, p, α, e
d ) n, α, p, e
ANS:
\(\frac{e}{m}\) for
(i) neutron = \(\frac{0}{1}\) = 0
(ii) α —particle=\(\frac{2}{4}\) = 0.5
(iii) Proton = \(\frac{1}{1}\) = 1
(iv) electron = \(\frac{11}{1837}\) = 1837
4. The ratio of the energy of a photon of 2000AA wavelength radiation to that of 4000AA radiation is.
a) \(\frac{1}{4}\)
b) 4
c) \(\frac{1}{2}\)
d) 2
ANS:
E= hv = \(\frac{hc}{λ}\) = E ∝ \(\frac{1}{λ}\)
= \(\frac{E_1}{E_2}\) = \(\frac{λ_2}{λ_1}\) = \(\frac{4000}{2000}\) = 2
5. In a Bohr’s model of an atom when an electron jumps from n=1 to n=3, how much energy will be emitted or absorbed.
a) 2.15×10−11erg
b) 0.1911×10−10erg
c) 2.389×10−12erg
d) 0.239×10−10erg
ANS:
According to Bohr’s model ΔE= E1 − E3
2.179×10−11 — \(\frac{{2.179}\times{10^{-11}}}{9}\)
\(\frac{8}{9}\)\(\times{2.179}\)\(\times{10^{-11}}\) = 0.1911×10−10erg
from n=1 to n=3 hence energy is absorbed.
6.Rydberg gave the formula as:
a) \(\bar{v} = \frac{1}{\bar{v}}\) = \(R[\frac{1}{{1^n}^2} – \frac{1}{{2^n}^2}]\)
b) \(\bar{v} = \frac{1}{\bar{v}}\) = \(R[\frac{1}{{2^n}^2} – \frac{1}{{1^n}^2}]\)
c) \(\bar{v} = \frac{1}{\bar{v}}\) = \(R[\frac{1}{{1_n}^2} + \frac{1}{{2^n}^2}]\)
d) \(\bar{v} = \frac{1}{\bar{λ}}\) = \(R[\frac{1}{{1^n}_2} – \frac{1}{{2^n}^2}]\)
7.The energy of an electron in the first Bohr orbit of Hatom is −13.6eV. The possible energy value(s) of the excited state (s) for electrons in Bohr orbits to hydrogen is(are):
a) −3.4eV
b) −4.2eV
c) −6.8eV
d) +6.8eV
ANS:
Values of energy in the excited state =\(\frac{-13.6}{n^2}\)eV = \(\frac{-13.6}{4}\) = −3.4eV n = 2,3,4
8.Which of the following electron transition in a hydrogen atom will require the largest amount of energy.
a) From n=1 to n=2
b) From n=2 to n=3
c) From n=∞ to n=1
d) From n=3 to n=5
ANS:
E ∝ \(\frac{1}{n{_2}{^2}}\) – \(\frac{1}{n{_1}{^2}}\)
9.In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen.
a)3 ® 2
b) 5 ® 2
c) 4 ® 1
d) 2 ® 5
10.The frequency of one of the lines in Paschen series of hydrogen atom is 2.340×1011Hz. The quantum number n2 which produces this transition is:
a) 6
b) 5
c) 4
d) 3
ANS:
\(\bar{v} = \frac{1}{λ}\) = \(R{_h}[\frac{1}{n{_1}{^2}}\) – \(\frac{1}{n{_2}{^2}}\)]
\(\frac{1}{λ}\) = \(R{_h}[\frac{1}{3^2}\) – \(\frac{1}{n{_2}{^2}}\) = n2 = 3 for Paschen
11. Which of the following has maximum number of unpaired electron (atomic number of Fe 26).
a) Fe
b)Fe (II)
c)Fe (III)
d)Fe (IV)
12.The third line in Balmer series corresponds to an electronic transition between which Bohr?s orbits in hydrogen:
a) 5 ® 3
b)5 ® 2
c)4 ® 3
d)4 ® 2
ANS:
In Balmer series of hydrogen atomic spectrum which electronic transition causes third line O→L, n2 = 5, n1 =2.
13.The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by 1.25×10−13×A1/3cm Radius of atom is one AA. If the mass number is 64, then the fraction of the atomic volume that is occupied by the nucleus is.
a) 1.0×10−3
b) 5.0×10−5
c) 2.5×10−2
d) 1.25×10−13
ANS:
Radius of nucleus =1.25×10−13×A1/3cm = 1.25×10−13×641/3= 5×10−13cm
\(\frac{Volume \,of \,nucleus}{Volume \,of \,atom}\) = \(\frac{\frac{4}{3}\pi(5\times 10^{-13})^3}{\frac{4}{3}\pi(10^{-8})^3}\)
=1.25×10−13.
14.In a Bohr’s model of atom when an electron jumps from n=1 to n=3, how much energy will be emitted or absorbed.
a) 2.15×10−11eV
b) 0.1911×10−10erg
c) 2.389×10−12eV
d) 0.239×10−10eV
ANS:
According to Bohr?s model ΔE = E1 − E3
= 2.179\(\times{10^{-11}}\) -\(\frac{{2.179}\times{10^{-11}}}{9}\)
\(\frac{8}{9}\)\(\times{2.179}\)\(\times{10^{-11}}\)
=1.91×10−11=0.191×10−10erg
Since electron is going from n=1 to n=3 hence energy is absorbed.
15. Maximum number of electrons in a subshell with l = 3 and n = 4 is:
a) 10
b) 12
c) 14
d) 16
ANS:
n = main energy level
l = subshell
If n = 4 and l = 3, the subshell is 4f. In f subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum no. of electrons in 4f subshell = 7 x 2 = 14.
16.The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:
a) 5, 0, 0, + \(\frac{1}{2}\)
b )5, 1, 0, + \(\frac{1}{2}\)
c) 5, 1, 1, + \(\frac{1}{2}\)
d) 6, 0, 0, + \(\frac{1}{2}\)
ANS:
37Rb = 36[Kr]5s1
Its valance electron is 5s1
So, n = 5 , l = 0 m = 0 s = + \(\frac{1}{2}\)
17.A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at:
a) 518 nm
b) 1035 nm
c) 352 nm
d) 743 nm
ANS:
From Law of Conservation of energy, energy of absorbed photon must be equal to combined energy of two emitted photons.
ET=E1+E2… ..(1)
where E1is Energy of first emitted photon emitted and Eis Energy of the second emitted photon.
Energy E and wavelength λof a photon are related by the equation
E = \(\frac{hc}{λ}\)….(2)
where h is Planck’s constant, c is the velocity of light.
\(\frac{hc}{λ_T}\) = \(\frac{hc}{λ_1}\) + \(\frac{hc}{λ_2}\)
\(\frac{1}{λ_T}\) = \(\frac{1}{λ_1}\) + \(\frac{1}{λ_2}\) ….(3)
Substituting given values in (3) we get:
\(\frac{1}{355}\) = \(\frac{1}{680}\) + \(\frac{1}{λ_2}\)
\(\frac{1}{λ_2}\) = \(\frac{1}{355}\) – \(\frac{1}{680}\)
λ2 = 742.77nm
18.Ionisation energy of He+ is 19.6*10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is
a) −4.41×10−17Jatom−1
b) −4.91×10−17Jatom−1
c) −36.41×10−17Jatom−1
d) −4.431×10−17Jatom−1
ANS:
Ionization energy of He+= 19.6×10−18J/atom−1
IE = E1(forH) \(\times{Z^2}\)
E1 \(\times{4}\) = 19.6 \(\times{10_{18}}\)
E1(forLi2+) = E1forH \(\times{9}\)
−4.41×10−17Jatom−1
19. Mg2+ is isoelectronic with:
a) Ca+2
b) Na+
c) Zn+2
d) Cu+2
ANS:
Its configuration is like that of Ne. Thus it is isoelectronic with any element having 10e- or we can say 8e-in its valence shell. (2s2 2p6)
Thus Mg2+ is isoelectronic with Na+
20.Bond order of 1.5 is shown by:
a) O22-
b) O2
c) O2+
d) O2–
20.Atom of an element has Z electrons and its atomic mass is 2Z + 3. The number of neutrons in its nucleus will be:
a) 2Z b) Z + 3
c) Z + 2 d) Z
ANS:
here , no of protons = no of electrons.
atomic mass = Proton + Neutron
2Z + 3 = Z + Neutron
21.The orbital angular momentum of an electron is 2s orbital is:
a) + \(\frac{1}{2}\) . \(\frac{h}{2π}\)
b) Zero
c) \(\frac{h}{2π}\)
d) + \(\frac{1}{20}\)
ANS:
orbital angular momentum =
\(\sqrt{{l}\times{(l+1)}}\) \(\frac{h}{2π}\)
For S orbital, l=0
Orbital angular momentum = 0
22.Which of the following has more unpaired d-electrons?
a) Fe+2
b) Zn+
c)Ni+3
d) Cu+
ANS:
The electronic configurations for various species are:
Zn+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10
Fe+2: 1s2 2s2 2p6 3s2 3p6 3d6
Ni+3 : 1s2 2s2 2p6 3s2 3p6 3d7
Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10
Thus maximum number of unpaired electrons are present in Fe+2
23. The atomic orbitals are progressively filled in order of increasing energy . This principle is called:
a) Hunds rule
b) Aufbau principle
c) Exclusion principle
d) de – Broglie rule
24.What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10ms–1 ?
a) 6.626 × 10–34 m (J = kg m2s-2)
b) 6.626 × 10–30 m (J = kg m2s-2)
c) 6.626 × 10–30 m (J = kg m2 s-2)
d) 6.626 × 10–33 m (J = kg m2s-2)
ANS:
According to de Brogile equation,
λ = \(\frac{h}{mv}\)
6.626 × 10–34 m (J = kg m2s-2)
25.The ionisation energy of H-atom (in the ground state) is x kJ. The energy required for an electron to jump from 2nd to 3rd orbit will be:
a) \(\frac{x}{6}\)
b) 5x
c) 7.2x
d) \(\frac{5x}{36}\)
ANS:
The ionisation energy in ground state is x kJ,
so the energy of 1st orbit (E1 ) = –xkJ
Energy of 2nd orbit (E2 ) = – \(\frac{x}{2^2}\) = – \(\frac{x}{4}\) kJ
Energy of 3rd orbit (E3 ) = – \(\frac{x}{3^2}\) = – \(\frac{x}{9}\) kJ
Energy required to jump from 2nd to 3rd orbit:
is E3 – E2 = – \(\frac{x}{9}\) – – \(\frac{x}{4}\) kJ
\(\frac{x}{4}\) – \(\frac{x}{9}\) = \(\frac{5x}{36}\)
26.Number of planar or angular nodes is:
a) n+1
b) n-1
c) l
d) l + 1
27.Number of nodes in any orbital can be calculated by:
a) n – l -1
b) n-1
c) l + 1
d) l
28.The most abundant isotope of Oxygen is:
a) O-18
b) O-17
c) O-16
d) O-15
29.O-16 is used in:-
a) production of radioactive
b) food labeling
c) a tracer in the study of cerebral oxygen utilization.
d) All of these
30. Isotope of nitrogen N-13 used in:-
a) food labeling
b) PET imaging
c) both (b) and (d)
d) myocardial perfusion
31.The study of spectra is called:-
a) Spectrography
b) Analogy
c) Spectroscopy
d) None of these
32. Which shape of D- orbital is called as doughnut shape or baby soother type shape:-
a) dxy
b) dx2-y2
c) dyz
d) dz2
ANS:
The dz2 orbital consists of two lobes along the z-axis with a ring of high electron density in the XY plane. Hence the shape is not a closed one it can be divided into 3 parts ie.-2 Lobes and 1 rings.
33.No two electrons can have all the four quantum numbers to be the same or, if two electrons have to placed in an energy state they should be placed with opposite spies.
a) Pauli’s exclusion principle
b) Hund’s rule of maximum multiplicity
c) Aufbau’s principle
d) None of these
34. Distinguish between Isotopes, Isobars , Isotones & Isodiaphers:-
ANS:
| Isotopes | Isobars | Isotones | Isodiaphers |
| Isotopes are atom have same atomic number but different mass number or atomic mass. | Isobars are atoms of different elements which have the same mass number but a different atomic number. | Isotone is the atom of a different element which contains a same number of a neutron with a different mass number and atomic number. | Isodiaphers are atoms, having a different atomic number and mass number but have the same difference between the number of neutron and number of proton-proton number |
| hydrogen has 3 isotopes as protium deuterium and tritium having same atomic number 1 with different mass number 1,2,3 | argon and calcium have the same mass number 40 but different atomic number 18,20. | silicon and phosphorous have same number of neutron 16 with different mass number 30,31 and atomic number 14,15. | in thorium, proton number is 90 and neutron number is 144 144-90 = 54 in uranium, proton number is 92 and neutron number is 146 146 – 92 = 54 |
35. The most abundant isotope of hydrogen on the planet Earth is:
a) Protium
b) Deuterium
c) Tritium
d) All of these.
36. How does the Binding Energy per nucleon vary with the increase in the number of nucleons?
a) Increases continuously with mass number
b) Decrease continuously with mass number
c) First decreases and then increases with increase in mass number
d) First increases and then decreases with increase in mass number
37.If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus, yet 1s7 is not observed because it violates:
a) Heisenberg uncertainty principle
b) Hund’s rule
c) Pauli exclusion principle
d) Bohr postulate of stationary orbit
38.Difference between Orbit and Orbital:
ANS:
| Orbit | Orbital |
| An is simply present in a body with a certain mass | An orbital exists for an electron as well as an atom. |
| An orbit is the simple planar representation of an electron | An orbital refers to the dimensional motion of an electron around the nucleus in a three dimensional motion. |
39..Difference between Shell and Subshell:
ANS:
| Shell | Subshell |
| Orbits / Shells are the circular paths around the nucleus of an atom along which the electrons traverse. | Each shell consists of one or more sub-shells. |
| (According to Bohr Model). The shells are represented by the principal quantum number, n. | There are four sub-shells – s, p, d, f. |
40.Difference between Emission Spectra and Absorption Spectra:
ANS:
| Emission Spectra | Absorption Spectra |
| Produced when atoms release energy | Produced when atoms absorb energy |
| Comprise of coloured lines in the spectrum | Comprise of dark lines or gaps in the spectrum |
| Is in helpful in figuring out the composition of a certain matter | Can be used to figure out the ability of certain objects to retian heat and its absorption level |
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