NCERT Solutions for Class 11 Chemistry Structure of Atom Part 2

NCERT Solutions for Class 11 Chemistry Structure of Atom Part 2 Class 11 Chemistry book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 11 Chemistry chapter wise NCERT solution for Class 11 Chemistry Some Basic Concepts of Chemistry Part 1 and all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

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Class 11 Chemistry Structure of Atom Part 2

NCERT Class 11 Chemistry Chapter wise Solutions

  • Chapter 1 – Some Basic Concepts of Chemistry
  • Chapter 2 – Structure of Atom
  • Chapter 3 – Classification of Elements and Periodicity in Properties
  • Chapter 4 – Chemical Bonding and Molecular Structure
  • Chapter 5 – States of Matter
  • Chapter 6 – Thermodynamics
  • Chapter 7 – Equilibrium
  • Chapter 8 – Redox Reactions
  • Chapter 9 – Hydrogen
  • Chapter 10 – The s-Block Elements
  • Chapter 11 – The p-Block Elements
  • Chapter 12 – Organic Chemistry – Some Basic Principles and Techniques
  • Chapter 13 – Hydrocarbons
  • Chapter 14 – Environmental Chemistry

NCERT Solutions for Class 11 Chemistry Structure of Atom Part 2

Class 11 Chemistry Structure of Atom Part 2

17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series, = 2. Thus, the expression of wave number is given by,

Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.

For to be minimum should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking = 3, we get:


18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is

Ans. Energy (E) of the Bohr orbit of an atom is given by,

Where,

Z = atomic number of the atom

Ground state energy =

Energy required to shift the electron from n= 1 to n= 5 is given as:

Wavelength of emitted light


19. The electron energy in hydrogen atom is given by Calculate the energy required to remove an electron completely from then= 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans. Given,

Energy required for ionization from n= 2 is given by,

Here, is the longest wavelength causing the transition.

= 3647


20. Calculate the wavelength of an electron moving with a velocity of

Ans. According to de Broglie’s equation,

Where,

= wavelength of moving particle

m = mass of particle

v = velocity of particle

h = Planck’s constant

Substituting the values in the expression of :

Hence, the wavelength of the electron moving with a velocity of


21. The mass of an electron is kg. If its K.E. is J, calculate its wavelength.

Ans. From de Broglie’s equation,

Given,

Kinetic energy (K.E) of the electron

Since, K.E.

Substituting the value in the expression of :

Hence, the wavelength of the electron is m.


Class 11 Chemistry Structure of Atom Part 2

22. Which of the following are isoelectronic species i.e., those having the same number of electrons?

Ans. Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na) = 11

Number of electrons in = 10

A positive charge denotes the loss of an electron.

Similarly,

Number of electrons in = 18

Number of electrons in = 10

Number of electrons in = 18

A negative charge denotes the gain of an electron by a species.

Number of electrons in sulphur (S) = 16

∴ Number of electrons in = 18

Number of electrons in argon (Ar) = 18

Hence, the following are isoelectronic species:

1) (10 electrons each)

2 (18 electrons each)


Class 11 Chemistry Structure of Atom Part 2

23. (i) Write the electronic configurations of the following ions:

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a ?

(iii) Which atoms are indicated by the following configurations?

Ans. (i) (a) ion

The electronic configuration of H atom is

A negative charge on the species indicates the gain of an electron by it.

Electronic configuration of

(b) ion

The electronic configuration of Na atom is

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of =

(c) ion

The electronic configuration of 0 atom is .

A dinegative charge on the species indicates that two electrons are gained by it.

Electronic configuration ofion =

(d) ion

The electronic configuration of F atom is .

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of F ion =

(ii)

Completing the electron configuration of the element as

Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11

(b)

Completing the electron configuration of the element as

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c)

Completing the electron configuration of the element as

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a

The electronic configuration of the element is [He]

Atomic number of the element = 3

Hence, the element with the electronic configuration [He]is lithium (Li).

(b

The electronic configuration of the element is [Ne]

Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] is phosphorus (P).

(c) [Ar

The electronic configuration of the element is [Ar]

Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] is scandium (Sc).


Class 11 Chemistry Structure of Atom Part 2

24. What is the lowest value of n that allows g orbitals to exist?

Ans. For g-orbitals, l = 4.

As for any value ‘n‘ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

For l = 4, minimum value of n = 5


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:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number = – 2, – 1, 0, 1, 2


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) For an atom to be neutral, the number of protons is equal to the number of electrons.

Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is


27. Give the number of electrons in the species ,and

Ans. : Number of electrons present in hydrogen molecule = 1 + 1 = 2

Number of electrons in = 2-1 = 1

H2: Number of electrons in H2 = 1 + 1 = 2

: Number of electrons present in oxygen molecule = 8 + 8 = 16

Number of electrons in = 16- 1 = 15


Class 11 Chemistry Structure of Atom Part 2

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

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

(iii) Which of the following orbitals are possible?

Ans. (i) n = 3 (Given)

For a given value of n, l can have values from 0 to (n1).

For n = 3

l = 0, 1, 2

For a given value of l,can have values.

For l = 0, m = 0

l = 1, m = – 1, 0, 1

l = 2, m = – 2, – 1, 0, 1, 2

For n = 3

l = 0, 1, 2

= 0

= – 1, 0, 1

= – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.

For a given value of l, can have (2l + 1) values i.e., 5 values.

For l = 2

m2 = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l = 1.

For a given value of n, l can have values from zero to (n – 1).

For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, l = 4.

For l = 4, the minimum value of n is 5.

Hence, 1p and 3f do not exist.


29. Using s, p, 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 (Given)

The orbital is 1s.

(b) For n = 3 and l = 1

The orbital is 3p.

(c) For n = 4 and l = 2

The orbital is 4d.

(d) For n = 4 and l = 3

The orbital is 4f.


Class 11 Chemistry Structure of Atom Part 2

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

Ans. (a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c) The given set of quantum numbers is not possible.

For a given value of n, ‘l‘ can have values from zero to (n – 1).

For n = 1, l = 0 and not 1.

(d) The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible.

For n = 3,

l = 0 to (3 – 1)

l = 0 to 2 i.e., 0, 1, 2

(f) The given set of quantum numbers is possible.


31. How many electrons in an atom may have the following quantum numbers?

(a)n = 4,

(b) n = 3, l = 0

Ans. (a) Total number of electrons in an atom for a value of n

For n = 4,

Total number of electrons

= 32

The given element has a fully filled orbital as

Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.


Class 11 Chemistry Structure of Atom Part 2

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.

Ans. Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

Where,

n = 1, 2, 3, …

According to de Broglie’s equation:

Substituting the value of ‘mv‘ from expression (2) in expression (1):

Since represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom Part 2

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