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Structure of Atom Class 11 MCQs Questions with Answers
Which of the following pair of ions have same paramagnetic moment?
(a) Cu+2, Ti+3
(b) Mn+2, Cu+2
(c) Ti+4, Cu+2
(d) Ti+3, Ni+2
Answer: (a) Cu+2, Ti+3
29Cu = [Ar] 3d10 4s1
Cu2+ = [Ar] 3d9 (n = 1)
22Ti = [Ar] 3d² 4s²
Ti3+ = [Ar] 3d² (n = 1)
Both of these ions have one unpaired electron, hence these have same paramagnetic moment.
The charge to mass ratio of α – particles is approximately …… the charge to mass ratio of protons
(c) Four times
(d) Six times
Answer: (b) Half
Let charge of proton be +e, then charge of alpha particle will be + 2e. Similarly let mass of proton be m, then mass of alpha particle will be 4m.
Now, specific charge = (charge) / (mass of the substance.)
For proton, specific charge = (e/m)
For alpha particle, specific charge = (2e)/ (4m)
Therefore there ratio is: (e/m) × (4m)/ (2e) = 2 : 1
The frequency of a wave of light is 12 × 1014s-1. The wave number associated with this light
(a) 5 × 10-7m
(b) 4 × 10-8cm-1
(c) 2 × 10-7m-1
(d) 4 × 104cm-1
Answer: (d) 4 × 104 cm-1
Frequency ν = 12 × 1014s-1 and
velocity of light c = 3 × 1010 cms-1.
We know that the wave number ν– = (v/c)
= (12 × 1014)/(3 × 1010)
= 4 × 104 cm-1
In a multi – electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic acid and electric fields? (a) n = 1, l = 0, m = 0 (b) n = 2, l = 0, m = 0 (c) n = 2, l = 1, m = 1 (d) n = 3, l = 2, m = 1 (e) n = 3, l = 2, m = 0
(a) (a) and (b)
(b) (b) and (c)
(c) (c) and (d)
(d) (d) and (e)
Answer: (d) (d) and (e)
In the absence of magnetic and electric fields, the orbitals defined by magnetic quantum number are degenerate (of same energy). So, energy of the orbital, in the absence of magnetic and electric fields depends on the (n + l) value.
Higher the (n + l) value, larger the energy of the orbital.
Orbitals with same n and n + 1 values are degenerate and have same energy.
In the given combinations, d and e have same n and n + l value and so, have same energy in the absence of magnetic and electric fields.
The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state)
Answer: (a) Li2+
(1/λ) = Z²RH [(1/n1²) – (1/n2²)], n1 = 1, n2 = 2
Therefore, (1/ λ) = Z²RH [(1/1) – (1/4)]
= (3/4) Z²RH
λ ∝ (1/Z)²
Therefore, Li2+ will produce shortest wave length.
In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is
(a) 1.51 eV
(b) 3.4 eV
(c) 6.04 eV
(d) 13.6 eV
Answer: (a) 1.51 eV
En = (13.6)/(n²) eV
or E = (13.6)/(9)eV
= 1.51 eV
The credit of discovering neutron goes to
Answer: (d) Chadwick
The essential nature of the atomic nucleus was established with the discovery of the neutron by James Chadwick in 1932 and the determination that it was a new elementary particle, distinct from the proton.
The maximum number of electrons that can be accommodated in fifth energy level is
Answer: (c) 50
The maximum number of electrons that can occupy a given energy level n is given by
max. no. of electrons = 2n²
So the number of orbitals that are present in an energy level n is given by
of orbitals = n²
Also a given orbital can hold a maximum of 2 electrons, which is why the maximum number of electrons that can be added to a given energy level n is twice the number of orbitals present on said energy level.
According to given,
n = 5, it refers to the fifth energy level, holds
of orbitals = 5²
of orbitals = 25
This means that the maximum number of electrons that can be added to the fifth energy level is
max no. of electrons = 2 × 25
= 50 electrons
According to Aufbaus principle, which of the three 4d, 5p and 5s will be filled with electrons first
(d) 4d and 5s will be filled simultaneously
Answer: (c) 5s
According to the Aufbaus principle, electron will be first enters in those orbital which have least energy. So decreasing order of energy is 5p > 4d > 5s.
A hydrogen atom in its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by (Given Planck constant h = 6.6 × 10-34 Jsec)
(a) 1.05 × 10-34 Jsec
(b) 3.16 × 10-34 Jsec
(c) 2.11 × 10-34 Jsec
(d) 4.22 × 10-34 Jsec
Answer: (a) 1.05 × 10-34 Jsec
Electron after absorbing 10.2 eV energy goes to its first excited state (n = 2) from ground state (n = 1).
Therefore, Increase in momentum = (h)/(2π)
= (6.6 × 10-34)/ (6.28)
= 1.05 × 10-34 Js.
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
(a) 8.51 × 105 Jmol-1
(b) 6.56 × 105 Jmol-1
(c) 7.56 × 105 Jmol-1
(d) 9.84 × 105 Jmol-1/sup>
Answer: (d) 9.84 × 105J mol-1
Energy required when an electron makes transition from n = 1 to n = 2
E2 = −(1.312 × 106 × (1)²)/(2²)
= −3.28 × 105J mol-1
E1 = −1.312 × 106J mol-1
ΔE = E2 − E1
= −3.28 × 105 – (−13.2 × 106)
ΔE = 9.84 × 105J mol-1
For principal quantum number n = 4, the total number of orbitals having l = 3 is
Answer: (b) 7
For n = 4 and l = 3, the orbital is 4f.
Number of values of m= no. of orbitals= (2l + 1) = 7.
Maximum number of electrons in a subshell with l = 3 and n = 4 is
Answer: (c) 14
n = 4, so 4th shell and l = 3 so it is f subshell.
Thus n = 4, l = 3 indicates 4f orbitals.
In f subshell there are 7 orbitals and each orbital can accommodate a maximum of 2 electrons. So, maximum no. of electrons in 4f subshell = 7 × 2 = 14.
Which hydrogen-like species will have same radius as that of Bohr orbit of hydrogen atom?
(a) n = 2, Li2+
(b) n = 2, Be3+
(c) n = 2, He+
(d) n = 3, Li2+
Answer: (b) n = 2, Be3+
r = (r0) × (n²/Z) Å
r0 = radius of 1st Bohrs orbit of hydrogen atom = 0.529 Å
For r = r0–
(n²/Z) = 1
n² = Z (1)
Because for n = 2 , Be+3
Z = 4, which satisfies equation (1).
Hence, n = 2,Be+3 will have the same radius of 1 st Bohrs orbit of a hydrogen atom.
The magnetic quantum number specifies
(a) Size of orbitals
(b) Shape of orbitals
(c) Orientation of orbitals
(d) Nuclear Stability
Answer: (c) Orientation of orbitals
The magnetic quantum number specifies orientation of orbitals.
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-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
Answer: (b) 5 → 2
Since the line in hydrogen spectrum lies within visible region that is at red end, therefore it is corresponds to the Balmer series.
The line at the red end suggests that
The first line of Balmer series is n = 3 to n =2
The second line of Balmer series is n = 4 to n = 2
The third line of Balmer series is n = 5 to n = 2
In chromium atom, in ground state, the number of occupied orbitals is
Answer: (b) 15
The configuration of
24Cr is 1s², 2s², 2p6, 3s², 3p6, 3d5, 4s1
Therefore, total s-orbitals = 4
total p-orbitals = 6
total d-orbitals = 5 and thus
Thus, total orbitals = 4 + 6 + 5 = 15.
A sub-shell with n = 6 , l = 2 can accommodate a maximum of
(a) 12 electrons
(b) 36 electrons
(c) 10 electrons
(d) 72 electrons
Answer: (c) 10 electrons
n = 6, ℓ = 2 means 6d → will have 5 orbitals.
Therefore max 10 electrons can be accommodated as each orbital can have maximum of 2 electrons.
Which of the following sets of quantum numbers represents the highest energy of an atom?
(a) n = 3, l = 0, m = 0, s = +1/2
(b) n = 3, l = 1, m = 1, s = +1/2
(c) n = 3, l = 2, m = 1, s = +1/2
(d) n = 4, l = 0, m = 0, s = +1/2
Answer: (c) n = 3, l = 2, m = 1, s = + 1/2
n = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d.
The value of Plancks constant is 6.63 × 10-34Js. The velocity of light is 3.0 × 108ms-1. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of 8 × 1015s-1
(a) 3 × 107
(b) 2 × 10-25
(c) 5 × 10-18
(d) 4 × 101
Answer: (d) 4 × 101
λ = (c/v) = (3 × 108)/(8 × 1015)
= 3.75 × 10-8
= 3.75 × 10-8 × 109 nm
= 4×101 nm.
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