NCERT Solutions | Class 11 Chemistry Chapter 2

NCERT Solutions | Class 11 Chemistry Chapter 2 | Structure of The Atom 

CBSE Solutions | Chemistry Class 11

Check the below NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of The Atom Pdf free download. NCERT Solutions Class 11 Chemistry  were prepared based on the latest exam pattern. We have Provided Structure of The Atom Class 11 Chemistry NCERT Solutions to help students understand the concept very well.

NCERT | Class 11 Chemistry

NCERT Solutions Class 11 Chemistry

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	11th
Subject:	Chemistry
Chapter:	2
Chapters Name:	Structure of The Atom
Medium:	English

Structure of The Atom | Class 11 Chemistry | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Chemistry Chapter 2 Structure of The Atom to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Exercises

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 1.

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

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, 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 x 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?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 3.

How many neutrons and protons are there in the following nuclei?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 4.

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A).
(i) Z= 17, 4 = 35.
(ii) Z= 92, A = 233.
(iii) Z = 4, 4 = 9.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 5.

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (υ) and wavenumber \(\overline { v } \) of the yellow light.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 6.

Find energy of each of the photons which
(i) corresponds to light of frequency 3 x 10¹⁵Hz.
(ii) have wavelength of 0.50 A.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 7.

Calculate the wavelength, frequency and wavenumber of a light wave whose time period is 2.0 x 10^-10 s.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 8.

What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 9.

A photon of wavelength 4 x 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
(iii) the velocity of the photoelectron (1 eV = 1.6020 x 10^-19 J).

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 10.

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 11.

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 12.

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 \(\AA \). Calculate threshold frequency \({ v }_{ 0 } \)and work function \({ W }_{ 0 } \) of the metal.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 13.

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Solution.

According to Rydberg equation,

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 14.

How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionisation enthalpy of H atom (energy required to remove the electron from n = 1

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 15.

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

Solution.

\(=\frac { n(n-1) }{ 2 } =\quad \frac { 6(6-1) }{ 2 } =\frac { 30 }{ 2 } =15lines\)

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 16.

(i) The energy associated with the first orbit in the hydrogen atom is -2.18 x 1O^-18 J atom^-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 17.

Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Solution.

According to Rydberg equation,

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 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 returnstothegroundstate?Theground state electron energy is -2.18 x 10^-11 ergs.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 19.

The electron energy in hydrogen atom is given by E_n = (-2.18 x 10^-18)/n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 20.

Calculate the wavelength of an electron moving with a velocity of 2.05 x 10⁷ m s^-1.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 21.

The mass of an electron is 9.1 x 10^-31 kg. If its K.E. is 3.0 x 10^-25 J, calculate its wavelength.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 22.

Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar

Solution.

Species Number of electrons
Na⁺ 11 – 1 = 10
K⁺ 19 – 1 = 18
Mg²⁺ 12 – 2 = 10
Ca²⁺ 20 – 2 = 18
S^2-, 16 + 2 = 18
Ar = 18
Thus, Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, and Ar have the same number of electrons.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 23.

(i) Write the electronic configurations of the following ions:
(a) H^–
(b) Na⁺
(c) O^2-
(d) F ^–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s^-1
(b) 2p³ and
(c) 3p⁵?

(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s^-1
(b) [Ne] 3s²3p³
(c) [Ar] 4s²3d¹.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 24.

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

Solution.

For g-orbital, l = 4. For a value of n, possible values of l are 0 to n – 1. Thus,
l = 4 = n-1 ⇒ n = 5
The lowest value of n that allows y-orbitals to exist is 5.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 25.

An electron is in one of the 3 d orbitals. Give the possible values of n, I and m, for this electron.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 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.

Solution.

Number of electrons = 29, number of neutrons = 35
(i) We know that Z = c = p
∵ c = 29 ∴ p = 29
(ii) Electronic configuration :
1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 27.

Give the number of electrons in the species \({ H }_{ 2 }^{ + }\), H₂ and \({ O}_{ 2 }^{ + }\).

Solution.

Species Number of electro:
\({ H }_{ 2 }^{ + }\) 1(1 + 1 – 1 = 1)
H₂ 2 (1 + 1 =2)
\({ O}_{ 2 }^{ + }\) 15(8 + 8 – 1 = 15)

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 28.

(i) An atomic orbital has n = 3. What are the possible values of l and m₁?
(ii) List the quantum numbers (m, and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1 p, 2s, 2p and 3f

Solution.

(i) When n = 3, l = 0, 1, 2
When l = 0, m₁ = 0. When l = 1, m₁ = -1, 0, +1. When l = 2, m₁ = -2, -1, 0, +1, +2.
(ii) For 3d-orbital l = 2, m₁ = -2, -1, 0, +1, +2
(iii) For a particular value of n, the allowed values of l are 0 to n – 1 only. Hence, 2s and 2p are the only possible orbitals.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 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.

Solution.

(a) 1s
(b) 3p
(c) 4d
(d) 4f

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 30.

Explain, giving reasons, which of the following sets of quantum numbers are not possible,
(a) n = 0, l = 0, m₁ = 0, m_s = + 1/2
(b) n = 1, l = 0, m₁ = 0, m_s = -1/2
(c) n= 1, l= 1,m₁ = 0, m_s = + 1/2
(d) n = 2, l= 1, m₁ = 0, m_s = -1/2
(e) n = 3, l- 3, m₁ = -3, m_s = +1/2
(f) n = 3, l= 1, m₁ = 0, m_s = + 1/2

Solution.

(a) Not possible because n ≠ 0
(c) Not possible because when n = 1, l = 0
(e) Not possible because when n = 3, l = 0, 1, 2, and not equal to 3.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 31.

How many electrons in an atom may have the following quantum number?
(a) n = 4,m_s = -1/2
(b) n = 3, l=0

Solution.

(a) Total number of electrons in n = 4 is 2n² = 2(4)² = 32
But half of these electrons have m_s = -1/2
∴ Number of electrons = 16
(b) Number of electrons = 2 [ ∵ 3s subshell]

NCERT Solutions for Class 11 Chemistry Chapter 2, 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.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 33.

What transition in the hydrogen spectrumwould have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 34.

Calculate the energy required for the process

The ionization energy for the H atom in the ground state is 2.18 x 10^-18 J atom^-1.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 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 length of scale of length 20 cm long.

Solution.

A carbon atom covers length = diameter of atom = 0.15 nm
= 0.15 x 10^-9 x 10² cm = 0.15 x 10^-7 cm
∴ Number of carbon atoms which can be placed on 20 cm length
= \(\frac { 20 }{ 0.15\times { 10 }^{ -7 } } =1.33\times { 10 }^{ 9 }\)

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 36.

2 x 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Solution.

An atom will cover length equal to its diameter.
Diameter of carbon atom = \(\frac { 2.4 }{ 2\times { 10 }^{ 8 } } =1.2\times { 10 }^{ -8 }cm \)
Radius of carbon atom = \(\frac { 1.2\times { 10 }^{ -8 } }{ 2 } \)
= 0.6 x 10^-8 cm = 0.6 x 10^-8 x 10^-2 m
= 0.06 x 10^-8 m = 0.06 nm

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 37.

The diameter of zinc atom is 2.6 A. 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.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 38.

A certain particle carries 2.5 x 10^-16C of static electric charge. Calculate the number of electrons present in it.

Solution.

Charge of one electron = 1.602 x 10^-19 C
Number of electrons present = \(\frac { 2.5\times { 10 }^{ -16 } }{ 1.602\times { 10 }^{ -19 } } \)
= 1.56 x 10³

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 39.

In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is – 1.282 x 10^-18 C, what will be the number of electrons present on it?

Solution.

Charge of one electron = -1.602 x 10^-19 C
Number ot electrons = \(\frac { -1.282\times { 10 }^{ -18 } }{ -1.602\times { 10 }^{ -19 } } \approx 8.\)

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 40.

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

Solution.

Lesser number of a-particles will be deflected because nucleus of lighter atoms have smaller positive charge on their nuclei.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 41.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 42.

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

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 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.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 44.

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol of this ion.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 45.

Arrange the following type of radiations in increasing order of frequency :
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space and
(e) X-rays.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 46.

Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴, calculate the power of this laser.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 47.

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate
(a) the frequency of emission,
(b) distance travelled by this radiation in 30 s,
(c) energy of quantum and
(d) number of quanta present if it produces 2 J of energy.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 48.

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 49.

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 50.

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 51.

The work function for caesium atom is 1.9 eV. Calculate
(a) the threshold frequency and
(b) the threshold wavelength of the radiation.
(c) If the caesium elements is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 52.

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
λ(nm) 500 450 400
v x 10⁵ (m s^-1) 2.55 4.35 5.20

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 53.

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 54.

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ ms^-1, calculate the energy with which it is bound to the nucleus

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 55.

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be
represented as \(\upsilon =3.29\times { 10 }^{ 15 }(HZ)\left[ \frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { n }^{ 2 } } \right] \) Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 56.

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 57.

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 106 ms”1, calculate de Broglie wavelength associated with this electron.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 58.

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with neutron.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 59.

If the velocity of the electron in Bohr’s first orbit is 2.19 x 10⁶ ms^-1, calculate the de Broglie wavelength associated with it.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 60.

The velocity associated with a proton moving in a potential differenceof 1000Vis4.37 x 10⁵ ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 61.

If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac { h }{ 4\Pi r\times 0.05nm } \) ,is there any problem in defining this value.

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 62.

Thequantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination (s) has l have the same energy lists :
1. n = 4, l = 2, m₁ = -2, m_s = -1/2
2. n = 3, l = 2, m₁= 1, m_s =+1/2
3. n = 4, l= 1, m₁ = 0, m_s = +1/2
4. n = 3, l = 2,m₁ = -2, m_s = -M2
5. n = 3, l = 1, m₁=-1, m_s =+1/2
6. n = 4, l = 1, m₁ = 0, m_ss =+1/2

Solution.

1. 4d(n + l = 4 + 2 = 6)
2. 3d (n + l = 3 + 2 = 5)
3. 4p (n + l = 4 + 1 = 5)
4. 3d (n + l = 3 + 2 = 5)
5. 3p (n + l = 3 + 1 = 4)
6. 4p (n + l = 4 + 1 = 5)
Greater the value of n + l, higher will be the energy of orbital. If two orbitals have same n + l value then the orbital having higher n value will possess higher energy.
Therefore, the required order is : 5<2 = 4<6 = 3<1

NCERT Solutions for Class 11 Chemistry Chapter 2, 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 electron experiences the lowest effective nuclear charge?

Solution.

The electron in 4 p orbital experiences lowest effective nuclear charge because it is farthest from the nucleus.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 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.

Solution.

Orbital closer to the nucleus will experience larger effective nuclear charge.
(i) 2s
(ii) 4 d
(iii) 3 p

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 65.

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

Solution.

Si has nuclear charge, Z = 14 while Al has Z = 13. Thus, the electrons of 3p-orbital of Si will experience more effective nuclear charge from the nucleus.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 66.

Indicate the number of unpaired electrons in :
(a) P,
(b) Si,
(c) Cr,
(d) Fe and
(e) Kr

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 2, Question 67.

(a) How many sub-shells are associated with n = 4?
(b) How many electrons will be present in the sub-shells having m5 value of -1/2 for n = 4?

Solution.

NCERT Class 11 Chemistry

Class 11 Chemistry Chapters | Chemistry Class 11 Chapter 2

Chapterwise NCERT Solutions for Class 11 Chemistry

NCERT Solutions For Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
NCERT Solutions For Class 11 Chemistry Chapter 2 Structure of The Atom
NCERT Solutions For Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
NCERT Solutions For Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
NCERT Solutions For Class 11 Chemistry Chapter 5 States of Matter
NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics
NCERT Solutions For Class 11 Chemistry Chapter 7 Equilibrium
NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions
NCERT Solutions For Class 11 Chemistry Chapter 9 Hydrogen
NCERT Solutions For Class 11 Chemistry Chapter 10 The sBlock Elements
NCERT Solutions For Class 11 Chemistry Chapter 11 The pBlock Elements
NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques
NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons
NCERT Solutions For Class 11 Chemistry Chapter 14 Environmental Chemistry

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