NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10^-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?
Answer:
(i) One mole of methane (CH₄) has molecules = 6.022 × 10²³
No. of electrons present in one molecule of CH₄ = 6 + 4 = 10
No. of electrons present in 6.022 × 10²³ molecules of CH₄ = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ electrons

(ii)Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 10³ mg

14 × 10³ mg of carbon (C-14) have atoms = 6.022 × 10²³

Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH₃ = 7 + 3 = 10 .
No. of protons present in 12.044 × 10²⁰ molecules of NH₃ = 12.044 × 10²⁰ × 10
= 1.2044 × 10²² protons
Mass of one proton = 1.67 × 10^-27 kg
Mass of 1.2044 × 10²² protons = (1.67 × 10^-27 kg) × 1.2044 × 10²²
= 2.01 × 10^-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

Question 25.
An electron is in one of the 3d orbitals. Give the possible values of n, l and nil for the electron.
Answer:
For electron in 3d orbital, n = 3, l = 2, mi = -2, -1, 0, +1, +2.

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences lowest effective nuclear charge ?
Answer:
4p electron experiences lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect. It is farthest from the nucleus.

Question 64.
Among the following pairs of orbitals, which orbital will experience more effective nuclear charge (i) 2s and 3s (ii) 4d and 4f (iii) 3d and 3p ?
Answer:
Please note that greater the penetration of the electron present in a particular orbital towards the nucleus, more will be the magnitude of the effective nuclear charge. Based upon this,
(i) 2s electron will experience more effective nuclear charge.
(ii) 4d electron will experience more effective nuclear charge.
(iii) 3p electron will experience more effective nuclear charge.

Question 65.
The unpaired electrons in A1 and Si are present in the 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer:
Configuration of the two elements are :
A1 (Z = 13) : [Ne]¹⁰3s²3p¹ ; Si (Z = 14) : [Ne] 103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element Si is more than that of A1.

Question 66.
Indicate the number of unpaired electrons in :
(a) P (b) Si (c) Cr (d) Fe and (e) Kr.
Answer:
(a) P (z=15) : [Ne]¹⁰3s²3p³ ; No. of unpaired electrons = 3
(b) Si (z=14) : [Ne]¹⁰3s²3p² ; No. of unpaired electrons = 2
(c) Cr (z=24): [Ar]¹⁸4s¹3d⁵ ; No. of unpaired electrons = 6
(d) Fe (z=26): [Ar]¹⁸4s²3d⁶ ; No. of unpaired electrons = 4
(e) Kr (z=36) : [Ar]¹⁸4s²3d¹⁰4p⁶ ; No. of unpaired electrons = Nil.

Question 67.
(a) How many sub-shells are associated with n = 4 ?
(b) How many electrons will be present in the sub-shells having ms value of -1/2 for n = 4 ?
Answer:
(a) For n = 4 ; No. of sub-shells = (l = 0, l = 1, l = 2, l = 3) = 4.
(b) Total number of orbitals which can be present = n² = 4² = 16.
Each orbital can have an electron with m_s = – 1/2 -‘. Total no. of electrons with m, = – 1/2 is 16.