NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

“Structure of Atom” is the second chapter in the NCERT Class 11 Chemistry textbook. The topics covered under this chapter include subatomic particles, Thomson’s atomic model, Rutherford’s atomic model, Bohr’s model and the quantum mechanical model of the atom.  The types of questions asked in the NCERT exercise section for this chapter include:

• Basic calculations regarding subatomic particles (protons, electrons and neutrons).
• Numericals based on the relationship between wavelength and frequency.
• Numericals based on calculating the energy associated with electromagnetic radiation.
• Electron transitions to different shells.
• Writing electron configurations.
• Questions related to quantum numbers and their combinations (for electrons).

Students can note that these NCERT Solutions have been prepared and solved by our experienced subject experts, as per the latest CBSE Syllabus 2023-24 and its guidelines. The NCERT Solutions for Class 11 Chemistry provided on this page (for Chapter 2) provide detailed explanations of the steps to be followed while solving the numerical value questions that are frequently asked in the examinations. The subtopics covered in the chapter are listed below.

Subtopics of NCERT Solutions for Class 11 Chemistry Chapter 2 Structure Of Atom

1. Sub-atomic Particles
   • Discovery Of Electron
   • Charge To Mass Ratio Of Electron
   • Charge On The Electron
   • Discovery Of Protons And Neutrons
2. Atomic Models
   • Thomson Model Of Atom
   • Rutherford’s Nuclear Model Of Atom
   • Atomic Number And Mass Number
   • Isobars And Isotopes
   • Drawbacks Of Rutherford Model
3. Developments Leading To The Bohr’s Model Of Atom
   • Wave Nature Of Electromagnetic Radiation
   • Particle Nature Of Electromagnetic Radiation: Planck’s Quantum Theory
   • Evidence For The Quantized* Electronic Energy Levels: Atomic Spectra
4. Bohr’s Model For Hydrogen Atom
   • Explanation Of Line Spectrum Of Hydrogen
   • Limitations Of Bohr’s Model
5. Towards Quantum Mechanical Model Of The Atom
   • Dual Behaviour Of Matter
   • Heisenberg’s Uncertainty Principle
6. Quantum Mechanical Model Of Atom
   • Orbitals And Quantum Numbers
   • Shapes Of Atomic Orbitals
   • Energies Of Orbitals
   • Filling Of Orbitals In Atom
   • Electronic Configuration Of Atoms
   • Stability Of Completely Filled And Half Filled Subshells.

Students can make use of the NCERT Solutions for Class 11 exclusively to comprehend key topics and concepts, along with precise answers to exercise questions in the NCERT Class 11 Chemistry Textbook.

Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 2

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Q.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Ans:

(i)  Mass of 1 electron = 9.108 x 10-28 g

Hence, 

1 g = 1/(9.108 x 10-28) = 1.098 x 1027 electrons

(ii) Mass of one mole of electron = 9.108 x 10-28 x 6.022 x 1023

We get,

= 5.48 x 10-4 g

Charge on one mole of electron = 1.6 x 10-19 C x 6.022 x 1023

We get,

= 9.63 x 104 C

 

Q.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 × 10–27 kg).
(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?

Ans:

(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen)

Therefore, 1 mole of methane contains = 10 x 6.022 x 1023 = 6.022 x 1024 electrons.

(ii) (a)1 g atom of 14C = 14 g 

Since 14 g = 14000 mg have 6.022 x 1023 x 8 neutrons

Therefore, 7 mg will have neutrons = (6.022 x 1023 x 8)/14000 x 7 = 2.4088 x 1022

(b) mass of 1 neutron = 1.675 x 10-27 kg

Hence, mass of 2.4088 x 1021 neutrons = 2.4088 x 1021 x 1.67 x 10-27 = 4.0347 x 10-6 kg

(iii) Step I. Calculation of total number of NH3 molecules

Gram molecular mass of ammonia (NH3) = 17 g = 17 x 103 mg

17 x 103  mg of NH3 have molecules = 6.022 x 1023

34 mg of NH3 have molecules = 

Therefore, the kinetic energy of the emission = 0.97 eV.

(iii)

The velocity of the photoelectron (v) can be determined using the following expression:

Hence, the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is 15

Q.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.

Ans. The expression for the energy associated with nth orbit in hydrogen atom is

En = (-2.18 x 10-18)/n2 J/atom

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

Q.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

Ans: For the 3d orbital:

Possible values of the Principal quantum number (n) = 3

Possible values of the Azimuthal quantum number (l) = 2

Possible values of the Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2

 

Q.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.

Ans:

(i)

In a neutral atom, number of protons = number of electrons.

∴ Number of protons present in the atoms of the element = 29

(ii)

The electronic configuration of this element (atomic number 29) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 . The element is copper (Cu).

 

Q.27.Give the number of electrons in the species , H2+, H2 and O2+

Ans: No. electrons present in H2 = 1 + 1 = 2.

∴ Number of electrons in H2+ = 2 – 1 = 1

Number of electrons in H2 = 1 + 1 = 2

Number of electrons in O2 = 8 + 8 = 16.

∴ Number of electrons in O2+= 16 – 1 = 15

 

Q.28. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?

(II)List the quantum numbers (ml and l) of electrons for 3d orbital.

(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

Ans.

(I)

The possible values of ‘l’ range from 0 to (n – 1). Thus, for n = 3, the possible values of l are 0, 1, and 2.

The total number of possible values for ml = (2l + 1). Its values range from -l to l.

For n = 3 and l = 0, 1, 2:

m0 = 0

m1 = – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2

(II)

For 3d orbitals, n = 3 and  l = 2. For l = 2 , possible values of m2 = –2, –1, 0, 1, 2

(III)

It is possible for the 2s and 2p orbitals to exist. The 1p and 3f cannot exist.

For the 1p orbital, n=1 and l=1, which is not possible since the value of l must always be lower than that of n.

Similarly, for the 3f orbital, n =3 and l = 3, which is not possible.

Q.29. Using s, p and d notations, describe the orbital with the following quantum numbers.

(a)n = 1, l = 0;

(b)n = 3; l =1

(c) n = 4; l = 2;

(d) n = 4; l =3.

Ans:

(a)n = 1, l = 0 implies a 1s orbital.

(b)n = 3 and l = 1 implies a 3p orbital.

(c)n = 4 and l = 2 implies a 4d orbital.

(d)n = 4 and l = 3 implies a 4f orbital.

Q.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

a) n = 0, l = 0, ml= 0, ms =+

Q.35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across the length of the scale of length 20 cm long.

Ans. 1 m = 100 cm

1 cm = 10–2 m

Length of the scale = 20 cm =

Q 36. atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans. Length of the arrangement = 2.4 cm

Number of carbon atoms present = 2 × 108

The diameter of the carbon atom =

Q.37. The diameter of the zinc atom is 2.6Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Ans. (a) Given 

Diameter of zinc atom = 2.6 Å

= 2.6 x 10-10 m

= 260 x 10-12 m

= 260 pm

Radius of zinc atom in pm = (260 pm)/2 = 130 pm

(b) Given

Length of the arrangement = 1.6 cm

= 1.6 x 10-2 m

Diameter of a zinc atom = 2.6 Å = 2.6 x 10-10 m

∴ Number of zinc atom present in the arrangement

= (1.6 x 10-2 m)/(2.6 x 10-10 m)

= 0.6153 x 108 m

= 6.153 x 107 m

 

Q.38. A certain particle carries

Q.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum, etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like Aluminium etc. is used, what difference would be observed from the above results?

Ans.

Rutherford used alpha rays scattering to reveal the structure of the atom in 1911. The nucleus of heavy atoms is big and carries a lot of positive charge. As a result, when certain alpha particles hit the nucleus, they are easily deflected back. A number of alpha particles are also deflected at tiny angles due to the nucleus’s substantial positive charge. If light atoms are used, their nuclei will be light and their nuclei will have a modest positive charge. As a result, the number of particles deflected back and those deflected at an angle will be insignificant.

 

Q.41. Symbols

Q.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans.

Given, Mass number of the element = 81, which implies that (number of protons + number of neutrons) = 81

Let the number of protons in the element be x.

Therefore, number of neutrons in the element = x + 31.7% of x

= x + 0.317 x

= 1.317 x

Q.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans.

Let the number of electrons in the negatively charged ion be x.

Then, number of neutrons present = x + 11.1% of x = x + 0.111 x = 1.111 x

Number of electrons present in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

Therefore, number of protons present in the neutral atom = x – 1

Given, mass number of the ion = 37

Mass number = number of protons + number of neutrons

(x – 1) + 1.111x = 37

2.111x = 38

x = 18

∴ Number of protons = Atomic number = x – 1 = 18 – 1 = 17

Therefore, the symbol of the ion is

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy (E):

Q.62: The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

Ans.

Electrons  1, 2, 3, 4, 5, and 6 reside in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals (respectively). Ranking these orbitals in the increasing order of energies: (3p) < (3d) < (4p)  < (4d).

 

Q.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 electron experiences the lowest effective nuclear charge?

Ans.

The nuclear charge that is experienced by electrons (which are present in atoms containing multiple electrons) depends on the distance between its orbital and the nucleus of the atom. The greater the distance, the lower the effective nuclear charge. Among p-orbitals, 4p orbitals are the farthest from the nucleus of the bromine atom with (+35) charge. Hence, the electrons that reside in the 4p orbital are the ones to experience the lowest effective nuclear charge. These electrons are also shielded by electrons that are present in the 2p and 3p orbitals along with the s-orbitals.

 

Q.64. Among the following pairs of orbitals, which orbital will experience the larger effective nuclear charge?

(i) 2s and 3s,

(ii) 4d and 4f,

(iii) 3d and 3p

Ans.

The nuclear charge can be defined as the net positive charge that acts on an electron in the orbital of an atom that has more than 1 electrons. It is inversely proportional to the distance between the orbital and the nucleus.

(i)Electrons that reside in the 2s orbital are closer to the nucleus than those residing in the 3s orbital and will, therefore, experience greater nuclear charge.

(ii)4d orbital is closer to the nucleus than 4f orbital and will, therefore, experience greater nuclear charge.

(iii)3p will experience greater nuclear charge (since it is closer to the nucleus than the 3f orbital).

 

Q.65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Ans.

The nuclear charge can be defined as the net positive charge that acts on an electron in the orbital of an atom that has more than 1 electrons.  The greater the atomic number, the greater the nuclear charge. Silicon holds 14 protons while aluminium holds only 13. Therefore, the nuclear charge of silicon is greater than that of aluminium, implying that the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

 

Q.66. Indicate the number of unpaired electrons in:

(a)P

(b)Si

(c)Cr

(d)Fe

(e)Kr

Ans.

(a)Phosphorus (P):

The atomic number of phosphorus is 15

Electronic configuration of Phosphorus:

1s 2 2s 2 2p 6 3s 2 3p 3

This can be represented as follows:

From the diagram, it can be observed that phosphorus has three unpaired electrons.

(b) Silicon (Si):

The atomic number of Silicon is 14

Electronic configuration of Silicon:

1s 2 2s 2 2p 6 3s 2 3p 2

This can be represented as follows:

From the diagram, it can be observed that silicon has two unpaired electrons.

(c) Chromium (Cr):

The atomic number of Cr is 24

Electronic configuration of Chromium:

1s 2 2s2 2p 6 3s 2 3p 6 4s 1 3d 5

This can be represented as follows:

From the diagram, it can be observed that chromium has six unpaired electrons.

(d) Iron (Fe):

The atomic number of iron is 26

Electronic configuration of Fe:

1s 2 2s2 2p 6 3s 2 3p 6 4s 2 3d 6

This can be represented as follows:

From the diagram, it can be observed that iron has four unpaired electrons.

(e) Krypton (Kr):

The atomic number of Krypton is 36

Its electronic configuration is:

1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d10 4p 6

This can be represented as follows:

From the diagram, it can be observed that krypton has no unpaired electrons.

 

Q.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?

Ans.

(a)n = 4 (Given)

For a given value of ‘n’, the values of ‘l’ range from 0 to (n – 1).

Here, the possible values of l are 0, 1, 2, and 3

Therefore, 4 subshells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the nth shell = n2

For n = 4

Therefore, number of orbitals when n = 4 is 16

Each orbital has 1 electron with ms value of -1/2.

Therefore, total number of electrons present in the subshell having ms value of (