## Ncert Solutions For Class 11 Chemistry Chapter 2

Adda247 provides NCERT solutions for class 11 chemistry chapter 2. The NCERT Solutions provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers.

These NCERT Solutions Class 11 chemistry are presented in a very simple language so that you can understand the basic of chemistry with ease. These NCERT Solutions class 11 Chemistry cover chapters 1 to 14 with all important questions and answers explained in a detailed way.

Students can download the Class 11 Chemistry NCERT Solutions, which they want to study with the comfort of their house.

The solutions available are in depth and simplest way. Thus will help the students beyond examination marks. This will help them develop a core understanding of the subject. Because this subject demands to understand rather than just memorizing solutions of Class 11 Chemistry. Here below we are providing you with the overview of all the chemistry Class 11 Chapters that are there in the NCERT textbook.

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**Benefits of Solutions of NCERT class 11 Chemistry:**

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## NCERT Solutions for Class 11 Chemistry Chapter 2: Structure of Atom

Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).

An atom is a complex arrangement of negatively charged electrons arranged in defined shells about a positively charged nucleus. This nucleus contains most of the atom’s mass and is composed of protons and neutrons (except for common hydrogen which has only one proton).

Atoms are extremely important structures that make up all of the materials on earth. Atoms are in our bodies and they bond together to form molecules, which make up matter. What is Matter? Matter is any substance that takes up space, meaning it has mass and volume.

Basic Atomic Structure. The idea that everything is made of atoms was pioneered by John Dalton (1766-1844) in a book he published in 1808. He is sometimes called the “father” of atomic theory, but judging from this photo on the right “grandfather” might be a better term.

NCERT Solutions for Class 11 Chemistry Chapter 2 are available for you to download online in PDF format. The Class 11 Chemistry NCERT Solutions Chapter 2 are prepared by the experts of Adda247 who are working in this field for decades now. Our subject matter expert will get the right answers for all questions that are asked in the examinations relating to Class 11 Chemistry Chapter 2. You can also download the Structure of Atom Class 11 questions and answers PDF, and go through them.

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**Chapter 2 Chemistry Class 11 covers the below topics:**

**Discovery of Subatomic Particles****Discovery of Electron.****Charge to Mass Ratio of Electron****Charge on the Electron****Discovery of Protons and Neutrons****Atomic models****Thomson Model of Atom****Rutherford’s Nuclear Model of Atom****Atomic Number and Mass Number****Isobars and Isotopes****Drawbacks of Rutherford Model****Developments Leading to the Bohr’s Model of Atom****Wave Nature of Electromagnetic Radiation****Particle Nature of Electromagnetic****Photoelectric Effect****Dual Behaviour of Electromagnetic Radiation****Evidence for the quantized Electronic Energy Levels: Atomic spectra****Emission and Absorption Spectra****Line Spectrum of Hydrogen****The Spectral Lines for Atomic Hydrogen****Bohr’s Model for Hydrogen Atom****Explanation of Line Spectrum of Hydrogen****Limitations of Bohr’s Model****Towards Quantum Mechanical Model of the Atom****Dual Behaviour of Matter****Heisenberg’s Uncertainty Principle****Significance of the Uncertainty Principle****Reasons for the Failure of the Bohr Model****Quantum Mechanical Model of Atom****Hydrogen Atom and the Schrodinger Equation****Orbitals and Quantum Numbers****Shapes of Atomic Orbitals****Energies of Orbitals****Filling of Orbitals in Atom****Aufbau Principle****Pauli Exclusion Principle****Hund’s Rule of Maximum Multiplicity****Electronic Configuration of Atoms****Stability of Completely Filled and Half- Filled Sunshells**

** **

### Key features of NCERT solutions class 11 Chemistry chapter 2

- The NCERT solution provide clear and precise answer.
- The columns are used wherever necessary.

## Important questions of NCERT Solutions of Class 11 Chemistry Chapter 2

**Question: 1 i) Calculate the number of electrons that will together weight**** gram. **

**ii) Calculate the mass and charge of one mole of electrons.**

Answer:

- Mass of electron = 9.10939 x 10
^{-31}kg

Number of electrons that weigh = 9.10939 x 10^{-31}kg = 1

Number of electrons that will weigh 1 g = (1 x 10^{-3} kg)

1 (1 x 10^{-3}kg) / 9.10939 x 10^{-31} kg

= 0.1098 x 10^{-3 + 31}

= 0.1098 x 10^{28}

= 1.098 x 10^{27}

- Mass of electron = 9.10939 x 10
^{-31 }kg

Mass of one mole of electron = (6.022 x 10^{23})(9.10939 x 10^{-31})

= 5.48 x 10^{-7}kg

Charge on one electron = 1.6022 x 10^{-19}columb

Charge on one mole of electron = (1.6022 x 10^{-19}C)(6.022 x 10^{23})

= 9.65 x 10^{4}c

**Question: 2 i) Calculate the total number of electrons present in one mole of **

** Methane. **

**ii) Find the total number and the total mass of neutrons in 7 mg of 14C.**

** iii) Find the total number and 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) Molecule of CH4 (methane) contains electron = 10

Therefore 1 mole (6.022 x 10^{23} atoms) contains electron = 6.022 x 10^{24}

(ii) a) 1g atom of 14C = 14g = 6.022 x 10^{23} atoms = 6.022 x 10^{24} x 8 neutrons

Thus 14g or 14000 mg have 6.022 x 10^{24} x 8 neutrons

Therefore 7 mg will have neutrons = 6.022 x 10^{24} x 8 / 14000 x 7 = 2.4088 x 10^{22}

- b) Mass of 1 neutron = 1.675 x 10
^{-27}kg

Therefore mass of 2.4088 x 10^{21} neutrons = 2.4088 x 10^{21} x 1.67 x 10^{-27} = 4.0347 x 10^{-6} kg

(iii) a) 1 mol of NH3 = 17g NH3 = 6.022 x 10^{23} molecules of NH3 = (6.022×10^{23})(7 + 3) proton = 6.022 x 10^{24} protons

Therefore 34 mg i.e 0.034 g NH3 = 6.022 x 10^{24} x 0.034/1 = 1.2044 x 10^{22} protons

- b) mass of 1 proton = 1.6726 x 10
^{-27}kg

Therefore mass of 1.2044 x 10^{22} protons = (1.6726 x 10^{-27})(1.2044 x 10^{22}) kg = 2.0145 x 10^{-5} kg

No, the answer will not change with change in temperature & pressure

**Question: 3 How many neutrons and protons are there in the following nuclei? **

^{13}_{6}**C, _{8}^{16}O, ^{24}_{12}Mg, _{26}^{56}Fe, ^{88}_{38}Sr.**

Answer:

_{6}^{13}C:

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) – (Atomic number) = 13 – 6 = 7

^{16}_{8} O:

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass) – (Atomic number) = 16 – 8 = 8

^{24}_{12} MG:

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass) – (Atomic number) = 24 – 12 = 12

_{26}^{56} Fe:

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) – (Atomic number) = 56 – 26 = 30

^{88}_{38} Sr:

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass) – (Atomic number) = 88 – 38 = 50

** **

**Question: 4 Write the complete symbol for the atom with the given atomic number **

**(Z) and atomic mass (A) **

**Z = 17 , A = 35****Z = 92 , A = 233**

**Z = 4 , A = 9**

Answer:

- The element with atomic number(Z) 17 & mass number (A) 35 is chlorine =
^{35}_{17}Cl - The element with atomic number(Z) 92 & mass number (A)233 is uranium =
^{233}_{92}U - The element with atomic number(Z) 4 & mass number (A) 9 is berellium =
_{4}^{9}Be

**Question: 5 Yellow light emitted from a sodium lamp has a wavelength of 580 nm. Calculate the frequency (v) and wavenumber (v-) of the yellow light. **

Answer:

From the expression, = c/ v

We get,

V = c / ….(i)

Where,

V = frequency of yellow light

C = velocity of the light in vacuum = 3 x 10^{8}m / s

= wavelength of yellow light = 580 nm = 580 x 10^{-9}m substituting the values of in expression (i)

V = 3 x 10^{8} / 580 x ^{10-9 } = 5.17 x 10^{14} / S

Thus, frequency of yellow light emitted from the sodium lamp

= 5.17 x 10^{14} / s

Wave number of yellow light = 1 /

= 1/ 580 x 10^{-9} = 1.72 x 10^{8} / m

**Question: 6 Find the energy of each of the photons which**

**Correspond to light of frequency 3 × 10 ^ 15 Hz****Have wavelength of 0.50 A°.**

Answer:

- Energy (E) of a photon is given by the expression,

E = hv

Where, h = Planck’s constant = 6.626 × 10^{-34} Js

ν = frequency of light = 3 × 10^{15} Hz

Substituting the values in the given expression of E:

E = (6.626 × 10^{-34}) (3 × 10^{15}) E = 1.988 × 10^{-18} J

- Energy (E) of a photon having wavelength (λ) is given by the expression,

E = hc/λ

h = Planck’s constant = 6.626 × 10^{-34} Js

c = velocity of light in vacuum = 3 × 10^{8} m/s

Substituting the values in the given expression of E:

E=(6.626 x 10^{-34})(3 x 10^{8})/0.50 x 10^{-10} = 3.976 x 10^{-15} J E = 3.98 x 10^{-15} J

** **

**Question:7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10^-10 s. **

Answer:

Frequency (v) of light = 1 / period

= 1 / 2.0 x 10^{-10} = 5.0 x 10^{9} / s

Wavelength of light = c / v

Where,

C = velocity of light in vacuum

Substituting the value in the given equation of wavelength,

= 3 x 10^{8} / 5.0 x 10^{9} = 6.0 x 10^{-2} m

Wave number of light = 1 / 6.0 x 10^{-2} = 1.66 x 10^{1} / m = 16.66 m

**Question:8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy? **

Answer:

Energy of a photon = hv

Energy on n photon = nhv

N = E_{n} / hc

Where,

= wavelength of light = 4000 x 10^{-12} m

c = velocity of light in vacuum

h = Plank’s constant

Substituting the values in expression of n:

n = 1 x (4000 x 10^{-12}) / (6.626 x 10^{-34}) (3 x 10^{8}) = 2.012 x 10^{16}

Hence, the number of photons with a wavelength of 4000 and energy of 1J are 2.012 x 10^{16}

** **

**Question :9 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 ? **

Answer:

According to Formula :

Wave number=ν=R[1/n₁² -1/n₂²]

where

R=109678 cm⁻¹n1=2n2=4

ν=109678[1/2² – 1/4²]

=109678[(4-1)/16]

=109678×3/16

As we know that wave number=1/Wavelength

⇒ ν=1/λ

λ=1/ν

= 1/[109678×3/16]

= 16/109678×3

= 486×10⁻⁷ cm

= 486x 10⁻⁹m

= 486 nm

∴wavelength of light emitted is 486 nm

** **

**Question :10 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Campare your answer with the ionization enthalpy of H atom. **

Answer:

The expression of energy is given by,

Where,

Z = atomic number of the atom

n = principal quantum number

For ionization from n1 = 5 to ,

Therefore ΔE = E2- E1 = – 21.8 X10-19 (1/n22-1/n12)

= 21.8 X10-19 (1/n22-1/n12)

= 21.8 X10-19 (1/52-1/∞)

= 8.72 x 10-20 J

For ionization from 1st orbit, n1= 1,

Therefore ΔE’ = 21.8×10-19 (1/12-1/∞)

= 21.8×10-19 J

Now ΔE’/ ΔE = 21.8×10-19 / 8.72×10-20 = 25

Thus the energy required to remove electrons from 1st orbit is 25 times than the required to electron from 5th orbit.

## FAQs on NCERT Solutions Chemistry Class 11 Chapter 2

**What are the important questions of Structures of Atoms?**

Ans. The topics covered are

- Subatomic Particles

- Discovery of Electron
- Charge To Mass Ratio of Electron
- Charge On the Electron
- Discovery Of Protons And Neutrons

- Atomic Models

- Thomson Model of Atom
- Rutherford’s Nuclear and Mass Number
- Isobars And Isotopes
- Drawbacks of Rutherford Model

- Developments Leading to the Bohr’s Model of Atom

- Wave Nature of Electromagnetic Radiation
- Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory
- The Quantized Electronic Energy Levels: Atomic Spectra

- Bohr’s Model For Hydrogen Atom

- Explanation of Line Spectrum Of Hydrogen
- Limitations Of Bohr’s Model

- Towards Quantum Mechanical Model of the Atom

- Dual Behaviour of Matter
- Heisenberg’s Uncertainty Principles

- 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.

**What is discussed in the chapter?**

Ans. This chapter introduces the concepts of atoms, electrons, protons, and neutrons. Also, the students will understand the concepts of isotopes, atomic number, and isobars. These concepts are important for them to understand, as these will build a strong foundation of Chemistry. This chapter will also introduce the students to advanced theories like Rutherford’s model, Thomson’s model and Bohr’s model, their uses and their limitations.

Also, we will discuss the concepts of subshells, shells, de Broglie’s relationship, dual nature of light and matter, Heisenberg uncertainty principle, shapes of s, the concept of orbitals, p and d orbitals, quantum numbers etc. The chapter will also teach them the usage of principles like Hund’s rule, the Aufbau principle and Pauli’s exclusion principle.

**What is the photoelectric effect in simple words?**

Ans. The photoelectric effect is a phenomenon in which electrons are ejected from the surface of a metal when light is incident on it. These ejected electrons are called photoelectrons. It is important to note that the emission of photoelectrons and the kinetic energy of the ejected photoelectrons is dependent on the frequency of the light that is incident on the metal’s surface. The process through which photoelectrons are ejected from the surface of the metal due to the action of light is commonly referred to as photoemission.

The photoelectric effect occurs because the electrons at the surface of the metal tend to absorb energy from the incident light and use it to overcome the attractive forces that bind them to the metallic nuclei. An illustration detailing the emission of photoelectrons as a result of the photoelectric effect is provided below.

**How to prepare for Chemistry with Adda247?**

Ans. Preparing in a step by step and structured manner will help in understanding the concepts in Chemistry easier. The most scoring and difficult concept in this subject is the structure of the Atom. The weightage of this topic is also high. You may think of Qualitative analysis towards the end, as it needs very less time.

Our NCERT Solutions gives you an in-depth knowledge of the conceptual topics. NCERT Solutions have been drafted as per the latest CBSE Class 11 Science Syllabus. Students will feel that the solutions are in a simple language and can understand the difficult topics easily.