Important Questions for CBSE Class 11 Chemistry Chapter 2 - Structure of Atom

1 Mark Questions

1. Name the sub – atomic particles of an atom.

Ans: The subatomic particles of an atom are:

• Proton

• Neutron

• Electron
  
  

2. Name the scientist who first formulated the atomic structure.

Ans: The atomic structure was formulated by John Dalton, a British teacher in 1808 .  He first proposed a firm scientific basis named Dalton’s atomic theory.

3. What is the e/m ratio of an electron?

Ans: According to Thomson’s experiment, e/m ratio for an electron is $1.76\times {{10}^{3}}\text{c}{{\text{g}}^{-1}}$ . It means that every gram of an electron carries charge of $1.76\times {{10}^{3}}C$

4. What is the charge (e) of an electron?

Ans: From Millikan’s experiment, the charge of an electron (e) is $-1.602\times {{10}^{-19}}\text{C}$ . This is universal, and is the same for each and every electron.

5. (i) What is the mass of a proton?

(ii) What is the charge of a proton?

Ans: (i) The mass of a proton is $1.676\times {{10}^{-27}}\text{kg}$

(ii) The charge of a proton is $+1.602\times {{10}^{-19}}\text{C}$.

6. (i) What is the mass of a neutron?

(ii) What is the charge of a neutron?

Ans: (i) The mass of a neutron is $1.676\times {{10}^{-24}}\text{g}$ .

(ii) Neutrons are electrically neutral i.e. it has a charge of 0.

7. Name the scientist who first gave the atomic model.

Ans: J.J. Thomson, in 1898 first proposed the atomic model called raising-pudding model. The negative charges(raisins) were spread around in the plum pudding.

8. What is an isotope?

Ans: Atoms of the same elements which have the same atomic number but different mass number are called isotopes. Chemical properties of isotopes are almost similar.

e.g. ${{^{1}}_{1}}H{{.}^{2}}_{1}H\text{ and}{{\text{ }}^{3}}_{1}H$ ,${{^{35}}_{17}}Cl{{,}^{37}}_{17}Cl{{,}^{12}}_{6}C{{,}^{13}}_{6}C{{,}^{14}}_{6}C$

9. What are isobars?

Ans: Atoms of different elements which have the same mass number but different atomic numbers are called isobars. Isobars differ in chemical property but have same physical properties

e.g. ${{^{14}}_{6}}C{{,}^{14}}_{7}N$ , ${{^{40}}_{18}}Ar{{,}^{40}}_{19}K{{,}^{40}}_{20}Ca$

10. What are isotones?

Ans: Atoms of different elements which contain the same number of neutrons. Isotones have no similarity, when it comes to chemical properties.

e.g. ${{^{14}}_{6}}C{{,}^{15}}_{7}N{{,}^{16}}_{8}O$

11. What is an atomic number?

Ans: Atomic number is the measure of the number of protons which is present in the nucleus of an atom. For example, ${}_{1}^{2}H$ has an atomic number of 1.

12. What is a mass number?

Ans: Mass number of an element is the sum of the number of protons and neutrons present in the nucleus of an atom. For example, ${}_{1}^{2}H$ has a mass number of 2, as it has 1 proton and 1 electron.

13. Give the drawbacks of J.J. Thomson’s experiment.

Ans: The drawbacks are:

(i) It failed to explain the origin of the spectral lines of hydrogen and other atoms. 

(ii) It failed to explain scattering of $\alpha -$ particles in Rutherford’s scattering experiment.

14. Why Rutherford’s model could not explain the stability of an atom?

Ans: According to the electromagnetic theory of Maxwell, when charged particles are accelerated then they should emit electromagnetic radiation. Therefore, an electron in an orbit will continue to emit radiation for infinite time ; the orbit will then continue to shrink which is not the case in an atom.

15. Define photoelectric effect.

Ans: The phenomenon in which the surface of alkali metals like potassium and calcium emit electrons when a beam of light with high frequency is projected on them is called the photoelectric effect.

16. How does the intensity of light affect photoelectrons?

Ans: The number of electrons ejected and kinetic energy associated with them is directly proportional to the intensity of light projected towards the metal.

17. What is the threshold frequency?

Ans: The minimum frequency below which electrons are not ejected is called threshold frequency (${{v}_{0}}$ ). Threshold frequency is different for different metals.

18. Name the scientist who demonstrated the photoelectric effect experiment.  

Ans: In 1887, H. Hertz demonstrated photo electric effect. He observed the photoelectric effect, while working on radio waves.

19. What did Einstein explain about the photoelectric effect?

Ans: Einstein was able to explain the photoelectric effect using Planck’s quantum theory of electromagnetic radiation in 1905. Energy in each quantum of light is equal to a constant multiplied with the speed of light.

20. Calculate energy if 2mole of photons of radiation whose frequency is $5\times {{10}^{14}}\text{Hz}$. 1 Mark

Ans: Energy (E) of one photon $E=hv$

Where $h=6.626\times {{10}^{-34}}\text{Js}$ 

$v=5\times {{10}^{14}}{{s}^{-1}}$ 

$\therefore E=(6.626\times {{10}^{-34}}\times 5\times {{10}^{14}})$ 

$=3.313\times {{10}^{-19}}J$ 

Energy of 2 mole of photon $=(3.313\times {{10}^{-19}}J)\times (2\times 6.022\times {{10}^{23}}mo{{l}^{-1}})$ 

$=3990.2klmo{{l}^{-1}}$

21. States Heisenberg’s Uncertainty Principle.

Ans: It states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron. Mathematically, $\Delta x\Delta p\ge \frac{h}{4\pi }$.

22. How would the velocity be affected if the position is known?

Ans: If the position of the electron is known with high degree of accuracy ($\Delta x$ is small), then, according to the equation, the velocity of the electron will be uncertain $(\Delta ({{V}_{x}})$ is large).

23. We don’t see a car moving as a wave on the road why?

Ans: According to de Broglie’s relation, $\lambda ={}^{h}/{}_{mv}\text{ i}\text{.e}\text{. }\lambda \alpha \frac{1}{m}$ the mass of the car is very large and its wavelength $(\lambda )$ or wave character is negligible. Therefore, we do not see a car moving like a wave.

24. Give the de-Broglie’s relation.

Ans: Every  particle in motion is associated with a wavelength and other wave characteristics. The wavelength $(\lambda )$ of a particle in motion is equal to the Planck’s constant (h) divided by the momentum (p) of the particle. 

i.e. $\lambda =\frac{h}{p}=\frac{1}{mv}$ 

Where $m$ is the mass, $v$ is the velocity of the particles.

25. Calculate the uncertainty in the velocity of a wagon of mass $4000\text{ kg}$ whose position is known accurately of $\pm 10\text{m}$.

Ans: $\Delta v\ge \frac{h}{4\pi m\Delta x}$ 

$=\frac{6.6\times {{10}^{-34}}\text{kg}{{\text{m}}^{2}}{{s}^{-1}}}{4\times \frac{22}{7}\times 4\times {{10}^{3}}\text{kg}\times (\pm 10\text{m)}}$ 

$=1.3\times {{10}^{-39}}m{{s}^{-1}}$ 

$\therefore $ The uncertainty in the velocity of the wagon is $=1.3\times {{10}^{-39}}m{{s}^{-1}}$.

26. What is the physical significance of ${{\Psi }^{2}}$ up?

Ans: ${{\Psi }^{2}}$ represents the probability of finding an electron. It is the probability of finding a particle specified by a particular wave function.

27. Which orbital is non-directional?

Ans: S- orbital is spherically symmetrical i.e. it is non-directional.  It has a spherical shape, like a hollow ball.

28. What is the meaning of quantization of energy?

Ans: Quantization of energy means the energy is distributed and transmitted in the form of packets. These packets are called photons.

29. Why is the energy of 1s electron lower than 2s electron?

Ans: 1s electron being close to the nucleus experiences more force of attraction than 2s-electron which is away from the nucleus. Force of attraction is inversely proportional to the square of distance between the particles.

30. What is a nodal surface or nodes?

Ans: The region where the probability of finding an electron is zero i.e. ${{\text{Y}}^{2}}=0$ . This means that the value of the wave function is also zero.

31. How many spherical nodal surfaces are there in 4s – subshell?

Ans: In ns orbital, the number of spherical nodal surfaces are $(n-1)$, hence is $4s(4-1)=3$ nodal surfaces are present.

2 Marks Questions

1. What is the mass (m) of an electron?

Ans: mass of an electron $(m)=\frac{e}{({}^{e}/{}_{m})}$ 

$=\frac{1.602\times {{10}^{-19}}C}{1.76\times {{10}^{3}}C{{g}^{-1}}}$ 

$=9.10\times {{10}^{-28}}g$ 

$=9.1\times {{10}^{-31}}kg$ 

So, the mass of an electron is $=9.1\times {{10}^{-31}}kg$or ${}^{1}/{}_{1837}$  of the mass of a hydrogen atom.

2. Which experiment let to the discovery of electrons and how?

Ans: The beam discharge tube experiment performed by J.J. Thomson led to the invention of negativity charged particles called electron.

A beam tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a pump and therefore the pressure inside the tube is reduced to 0.01mm . When a fairly high voltage of 10,000 volts is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the invention electrons.

3. Give the main properties of the canal ray experiment.

Ans: The canal ray experiment led to the discovery of:

(i) The canal rays travel in a straight line. 

(ii) They have the ability to penetrate through small openings. 

(iii) They are positively charged as they get deflected from electric and magnetic fields.

4. Find out atomic number, mass number, number of electrons and neutrons in an element $\frac{40}{20}\times $ ?

Ans: The mass no. of $\times $ is $40$ 

The atomic no. of $\times $ is $20$ 

No. of proton is $=Z-A=40-20=20$ 

No. of electron is $(\text{A)}=20$ 

No. of proton is $(\text{A)}=20$

5. Give the main features of Thomson’s Model for an atom.

Ans: The features are:

• An atom hosts a sphere, which is positively charged, and the electrons are present in it and spread all over.

• There is balancing of charges, total positive charge is equal to the total negative charge.
  
  

6. What did Rutherford conclude from the observations of the $\alpha -ray$ scattering experiment?

Ans: Rutherford proposed the nuclear model of an atom as

(i) The positive charge was concentrated in a small space at the centre, which is the nucleus. 

(ii) The nucleus is surrounded by electrons that move around it in orbits with a very high speed.

(iii) Most of the space inside the atom is empty as most of the rays pass undeflected.

7. What is the relation between kinetic energy and frequency of the photoelectrons?

Ans: Kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation. As the kinetic energy is increased, the light incident on the metal has more energy, and for that, the time period of the electrons ejected reduces, which increases the frequency.

8. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, $\text{n}=4$ to $\text{n}=2$ of $\text{H}{{\text{e}}^{+}}$ spectrum?

Ans: For the Balmer transition, $\text{n}=4$ to $\text{n}=2$ in a $\text{H}{{\text{e}}^{+}}$ ion, we can write.

$\frac{1}{\lambda }={{\text{Z}}^{2}}{{\text{R}}_{H}}\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right]$ 

$={{\text{Z}}^{2}}{{\text{R}}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]$ 

$=\frac{3}{4}{{\text{R}}_{H}}$  ……(i)

For a hydrogen atom

$\frac{1}{\lambda }={{\text{R}}_{H}}\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right]$ ……(ii)

Equating equation (ii) and (i), we get

$\frac{1}{{{\text{n}}_{1}}^{2}}-\frac{1}{{{\text{n}}_{2}}^{2}}=\frac{3}{4}$ 

This equation gives ${{\text{n}}_{1}}=1$ and $\text{n=2}$ . Thus the transition $\text{n=2}$ to $\text{n=1}$ in hydrogen atom will have same wavelength as transition, $\text{n}=4$to $\text{n=2}$ in $\text{H}{{\text{e}}^{+}}$.

9. Spectral lines are regarded as the fingerprints of the elements. Why?   

Ans: Spectral lines are regarded as the fingerprints of the elements because identification of the elements can be done from these lines. Just like fingerprints, the spectral lines of no two elements resemble each other.  Spectral lines are only observed when electrons jump from one energy level to another.

10. Why cannot the motion of an electron around the nucleus be determined accurately?
Ans: We cannot determine the motion of an electron around the nucleus accurately, due to Heisenberg's uncertainty principle. The mass of an electron is fixed, but it is almost impossible to predict the velocity of the electron, for which its position cannot be determined easily, at a specific time.

11. Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length $1\times {{10}^{-10}}$.

Ans: According to the Uncertainty Principle.

$\Delta x.\Delta p=\frac{h}{4\pi }$ 

or $\Delta p=\frac{h}{4\pi \Delta x}$ 

or $\Delta p=\frac{6.626\times {{10}^{34}}\text{kg}{{\text{m}}^{2}}{{s}^{-1}}}{4\times 3.143\times {{10}^{10}}\text{m)}}$ 

$=5.27\times {{10}^{-26}}\text{kgm}{{\text{s}}^{-1}}$

12. Give the mathematical expression of uncertainty principle.

Ans: Mathematically, it can be given as

$\Delta x\times \Delta {{p}_{x}}\ge {}^{h}/{}_{4\pi }$

or, $\Delta x\times \Delta (m{{v}_{z}})\ge {}^{h}/{}_{4\pi }$

or, $\Delta x\times \Delta {{V}_{x}}\ge {}^{h}/{}_{4\pi m}$ 

Where $\Delta x$ is the uncertainty in position and $\Delta {{p}_{x}}(\Delta {{v}_{x}})$ is the uncertainty in momentum (or velocity) of the particle.

13. Which quantum number determines.

(i) energy of electron

(ii) Orientation of orbitals. 

Ans: 

(i) Principal quantum number (n), and 

(ii) Magnetic quantum number (m).

14. Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy.

1. $\text{n}=4,\text{I}=0,\text{m}=0,\text{s}=+1/2$ 

2. $\text{n}=3,\text{I}=1,\text{m}=1,\text{s}=-1/2$ 

3. $\text{n}=3,\text{I}=2,\text{m}=0,\text{s}-+1/2$ 

Ans: (i) Represents 4s orbital

(ii) Represents 3p orbital

(iii) Represents 3d orbital

(iv) Represents 3s orbital

The decreasing order of energy $3\text{d4s3p3s}$ 

$\text{n}=3,\text{I}=0,\text{m}=0,\text{s}=-1/2$

3 Marks Questions

1. What designations are given to the orbitals having

(i) $\text{n}=2,\text{I}=1$  (ii) $\text{n}=2,\text{I}=0$  (iii) $\text{n}=4,\text{I}=3$ 

(iv) $\text{n}=4,\text{I}=2$ (v) $\text{n}=1,\text{I}=1$ ?

Ans: (i) Here $\text{n}=2$ and $\text{I}=1$ 

Since $\text{I}=1$ it means a p=orbital, hence the given orbital is designated as 2p.

(ii) Here, $n=2$ and $\text{I}=0$ 

Since $\text{I}=0$ means s-orbital, hence the given orbital is 2s.

(iii) Here, $\text{n}=4$ and $\text{I}=3$ 

Since, $\text{I}=3$ represents f-orbital, hence the given orbital is a 4f orbital.

(iv) Here, $\text{n}=4$ and $\text{I}=2$ 

Since, $\text{I}=2$ represents a d-orbital, hence the given orbital is a 4d-orbital. 

(v) $\text{n}=4$ and $\text{I}=1$ 

Since, $\text{I}=1$ means it is a p-orbital, hence the given orbital can be designated as -4p orbital.

2. Write the electronic configuration of (i) $\text{M}{{\text{n}}^{4+}}$ ,(ii)$\text{F}{{\text{e}}^{3+}}$ ,(iii)$\text{C}{{\text{r}}^{2+}}$ and $\text{Z}{{\text{n}}^{2+}}$ . Mention the number of unpaired electrons in each case. 3 Marks

Ans.(i) $\text{Mn(z}=25),\text{ M}{{\text{n}}^{4+}}(\text{z}=21)$ 

The electronic configuration of $\text{M}{{\text{n}}^{4+}}$ is given by

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}3{{\text{d}}^{3}}$ 

As the outermost shell $3\text{d}$ has 3 electrons, thus the number of unpaired electrons is 3.

(ii) $\text{Fe (z}=26),\text{ F}{{\text{e}}^{3+}}(\text{z}=23)$  

The electronic configuration of $\text{F}{{\text{e}}^{3+}}$ is given by

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}3{{\text{d}}^{5}}$ 

The number of unpaired electrons is $5$ .

(iii) $\text{Cr(z}=24),\text{ C}{{\text{r}}^{2+}}(\text{z}=22)$ 

The electronic configuration of $\text{C}{{\text{r}}^{2+}}$ is

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}\text{3}{{\text{d}}^{4}}$ 

The number of unpaired electrons is $4$ .

(iv) $\text{Zn(z}=30),\text{Z}{{\text{n}}^{2+}}(\text{z}=28)$ 

The electronic configuration of $\text{Z}{{\text{n}}^{2+}}$ is

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}3{{\text{d}}^{10}}$ 

The number of unpaired electrons is $0$.

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