## 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 ^{14}C. (Assume that mass of a neutron = 1.675 × 10^{-27}kg).**

**(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH**

_{3}at STP.**Will the answer change if the temperature and pressure are changed ?**

**Answer:**

**(i)**One mole of methane (CH

_{4}) has molecules = 6.022 × 10

^{23}

No. of electrons present in one molecule of CH

_{4}= 6 + 4 = 10

No. of electrons present in 6.022 × 10

^{23}molecules of CH

_{4}= 6.022 × 10

^{23}× 10

= 6.022 × 10

^{24}electrons

**(ii)Step I.** Calculation of total number of carbon atoms

Gram atomic mass of carbon (C-14) = 14 g = 14 × 10^{3} mg

14 × 10^{3} mg of carbon (C-14) have atoms = 6.022 × 10^{23}

**Step II.** Calculation of total number and mass of protons

No. of protons present in one molecule of NH_{3} = 7 + 3 = 10 .

No. of protons present in 12.044 × 10^{20} molecules of NH_{3} = 12.044 × 10^{20} × 10

= 1.2044 × 10^{22} protons

Mass of one proton = 1.67 × 10^{-27} kg

Mass of 1.2044 × 10^{22} protons = (1.67 × 10^{-27} kg) × 1.2044 × 10^{22}

= 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 = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{1}.

**Question 27.****Give the number of electrons in the species : H _{2}^{+}, H_{2} and 0_{2}^{+}.**

**Answer:**

H

_{2}

^{+}= one ; H

_{2}= two ; 0

_{2}

^{+}= 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 = 2n

^{2}= 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]^{10}3s^{2}3p^{1} ; 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]^{10}3s^{2}3p^{3} ; No. of unpaired electrons = 3**(b)** Si (z=14) : [Ne]^{10}3s^{2}3p^{2} ; No. of unpaired electrons = 2**(c)** Cr (z=24): [Ar]^{18}4s^{1}3d^{5} ; No. of unpaired electrons = 6**(d)** Fe (z=26): [Ar]^{18}4s^{2}3d^{6} ; No. of unpaired electrons = 4**(e)** Kr (z=36) : [Ar]^{18}4s^{2}3d^{10}4p^{6} ; 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 = n^{2} = 4^{2} = 16.

Each orbital can have an electron with m_{s} = – 1/2 -‘. Total no. of electrons with m, = – 1/2 is 16.