NCERT MCQ CLASS-11 CHAPTER-2 | CHEMISTRY NCERT MCQ | | CHEMISTRY PART 1 | | STRUCTURE OF ATOM | EDUGROWN

In This Post we are  providing Chapter-2 Structure of Atom NCERT MCQ for Class 11 Chemistry which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON STRUCTURE OF ATOM

Question 1.
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

Answer: (d) n, α, p, e
Explanation:
(i) (e/m) for (i) neutron = (01) = 0
(ii) α− particle = (24) = 0.5
(iii) Proton = (11) = 1
(iv) electron = (11837) = 1837.

Question 2.
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 J mol^-1
(b) 6.56 × 10⁵ J mol^-1
(c) 7.56 × 10⁵ J mol^-1
(d) 9.84 × 10⁵ J mol^-1

Answer: (d) 9.84 × 10⁵ J mol^-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E₂=−(1.312 × 10⁶ × (1)²)/(2²)
= −3.28 × 10⁵ J mol^-1
E₁ = −1.312 × 10⁶ J mol^-1
ΔE = E₂ − E₁
=−3.28 × 10⁵−(−13.2 × 10⁶)
ΔE = 9.84×10⁵ J mol^-1

Question 3.
or a given principal level n = 4, the energy of its subshells is in the order
(a) s < p < d < f
(b) s > p > d > f
(c) s < p < f < d
(d) f < p < d < s

Answer: (a) s < p < d < f
Explanation:
Order of energy is:
s < p < d < f

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Question 4.
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) 325 nm
(d) 743 nm

Answer: (d) 743 nm
Explanation:
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 E₁ is Energy of first emitted photon emitted and E₂is Energy of second emitted photon.
Energy E and wavelength λ of a photon are related by the equation
E= (hc)/ (λ)….. (2)
Where Plancks constant = h, c is velocity of light.
Substituting the values from (2) in (1) we get
(hc/λ_T) = (hc)/ (λ₁) + (hc)/ (λ₂)
Or (1λT) = (1λ1) + (1λ2) …… (3)
Substituting given values in (3) we get
(1355) = (1680) + (1λ2)
Or (1)(λT) = (1355) − (1680)
⇒ (1λ2) = (680 − 355)/ (355 × 680)
⇒ λ₂ = 742.77nm

Question 5.
Which of the following statements in relation to the hydrogen atom is correct?
(a) 3s orbital is lower in energy than 3p orbital
(b) 3p orbital is lower in energy than 3d orbital
(c) 3s and 3p orbitals are of lower energy than 3d orbital
(d) 3s, 3p and 3d orbitals all have the same energy

Answer: (d) 3s, 3p and 3d orbitals all have the same energy
Explanation:
A hydrogen atom has 1st configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital.

Question 6.
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
Explanation:
The magnetic quantum number specifies orientation of orbitals.

Question 7.
The electronic configuration of silver atom in ground state is
(a) [Kr]3d¹⁰4s¹
(b) [Xe]4f¹⁴5d¹⁰6s¹
(c) [Kr]4d¹⁰5s¹
(d) [Kr]4d⁹5s²

Answer: (c) [Kr]4d¹⁰5s¹
Explanation:
The electronic configuration of Ag in ground state is [Kr]4d¹⁰5s¹

Question 8.
Which of the following element has least number of electrons in its M-shell?
(a) K
(b) Mn
(c) Ni
(d) Sc

Answer: (a) K
Explanation:
K = 19 = 1s²2s²2p⁶3s²3p⁶s¹
3s²3p⁶ = m-shell
= k has only 8 electrons in M shell

Question 9.
Which one of the following sets of ions represents a collection of isoelectronic species? (Atomic nos.: F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
(a) K⁺, Ca²⁺, Sc³⁺, Cl^–
(b) Na⁺, Ca²⁺ , Sc³⁺, F^–
(c) K⁺, Cl^–, Mg²⁺, Sc³⁺
(d) Na⁺, Mg²⁺, Al³⁺, Cl^–

Answer: (a) K⁺, Ca²⁺, Sc³⁺, Cl^–
Explanation:
Isoelectronic species are those which have same number of electrons.
K⁺ = 19 – 1 = 18; Ca²⁺ = 20 – 2 = 18; Sc³⁺ = 21 – 3 = 18; Cl^– = 17 + 1 = 18
Thus all these ions have 18 electrons in them.

Question 10.
In the ground state, an element has 13 electrons in its M-shell. The element is_____.
(a) Copper
(b) Chromium
(c) Nickel
(d) Iron

Answer: (b) Chromium
Explanation:
M shell means it is third shell ⇒ n = 3
Number of electrons in M shell = 13
⇒ 3s²3p⁶3d⁵
The electronic configuration is: (1s²) (2s² 2p⁶) (3s² 3p⁶ 3d⁵) (4s¹)
The element is chromium is Cr.

Question 11.
The electrons of the same orbitals can be distinguished by
(a) Principal quantum number
(b) Azimuthal quantum number
(c) Spin quantum number
(d) Magnetic quantum number

Answer: (c) Spin quantum number
Explanation:
Electrons occupying the same orbital are distinguished by Spin quantum number.
For spin Quantum number it has two values +1/2 or -1/2,
Hence the value of n, l , m are same for the two electrons occupying in the same orbitals, but only the is different, which is
Therefore, Spin quantum number explains the direction through which the electron spins in an orbital. so obviously there are only 2 possible directions. Which is either clockwise or anticlockwise.
So the electron which are available in the same orbitals, must have opposite spins. Hence spin quantum number distinguished b/w the two electrons.

Question 12.
Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively:
(a) 12 and 4
(b) 12 and 5
(c) 16 and 4
(d) 16 and 5Answer

Answer: (b) 12 and 5
Explanation:
₂₄Cr → 1s² 2s²2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
As we know for p, l = 1 and d, l = 2
For l = 1, total number of electrons = 12 [2p⁶ and 3p⁶]
For l = 2, total number of electrons = 5 [3d⁵]

Question 13.
A body of mass 10 mg is moving with a velocity of 100 ms^-1. The wavelength of de-Broglie wave associated with it would be (Note: h = 6.63 × 10^-34 Js)
(a) 6.63 × 10^-37 m
(b) 6.63 × 10^-31 m
(c) 6.63 × 10^-34 m
(d) 6.63 × 10^-35 m

Answer: (b) 6.63 × 10^-31 m
Explanation:
m = 10 mg
= 10 × 10^-6 kg
v = 100 ms^-1
λ = (h)(mv)
= (6.63×10^-34)/ (10 × 10^-6 × 100)
= 6.63 × 10^-31 m

Question 14.
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 J mol^-1
(b) 6.56 × 10⁵ J mol^-1
(c) 7.56 × 10⁵ J mol^-1
(d) 9.84 × 10⁵ J mol^-1

Answer: (d) 9.84 × 10⁵ J mol^-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E₂ = −(1.312 × 10⁶ × (1)²)/(2²)
= −3.28 × 10⁵ J mol^-1
E₁ = −1.312 × 10⁶ J mol^-1
ΔE = E₂ − E₁
= −3.28 × 10⁵−(−13.2 × 10⁶)
ΔE = 9.84 × 10⁵ J mol^-1

Question 15.
In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atom
(a) 3.4 eV
(b) 6.8 eV
(c) -13.6 eV
(d) +13.6 eV

Answer: (a) 3.4 eV
Explanation:
For hydrogen atom,
The kinetic energy is equal to the negative of the total energy.
And the potential energy is equal to the twice of the total energy.
The first excited state energy of orbital = -3.4 eV
and The kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV
Therefore, the kinetic energy of same orbit of hydrogen atom is 3.4 eV.

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