NCERT Class 11 Chemistry Chapter 2 Structure Of Atom Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 2 Structure Of Atom Short Question And Answers 

Question 1. Which one of the following is associated with λ = A Broglie wave of longer wavelength a proton or an I electron moving with the same velocity?
Answer:

λ = \(\frac{h}{m v}\)

∴ \(\frac{\lambda_p}{\lambda_e}=\frac{m_e}{m_p}\)

⇒ \(m_p>m_e\)

∴ \(\lambda_e>\lambda_p\)

Question 2. Mention the difference in angular momentum of the electron belonging to 3p and 4p -subshell.
Answer:

In the case of p -p-orbitals, the value of the azimuthal quantum number Is 1. Hut the magnitude of angular momentum of an electron present in any subshell depends on the value of l. It is Independent ofthe value of the principal quantum number n.

Orbital angular momentum= \(\sqrt{l(l+1)} \times \frac{h}{2 \pi}\)

Thus, there is no difference in angular momentum of the electrons belonging to 3p and 4p -subshells.

Question 3. Are the differences in energy between successive energy levels of a hydrogen-like atom the same? Explain.
Answer:

No, the differences are not the same. The energy of an electron revolving in ‘ n ‘th orbit, En \(=-\frac{2 \pi^2 m z^2 e^4}{n^2 h^2}\)

Hence, the difference in energy between first (n = 1) and second (n = 2) shell

⇒  \(E_1-E_2=-\frac{2 \pi^2 m z^2 e^4}{h^2}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=-\frac{2 \pi^2 m z^2 e^4}{h^2} \times \frac{3}{4}\)

Similarly \(E_2-E_3=-\frac{2 \pi^2 m z^2 e^4}{h^2}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)

= \(-\frac{2 \pi^2 m z^2 e^4}{h^2} \times \frac{5}{36}\)

Obviously, E₁– E₂±E₂– E₃

Question 4. Energy by associated the expression, with the \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\) orbite V of. Show-atom \(E_{(n+I)}-E_n=\frac{13.6 \times 2}{n^3} \mathrm{eV},\)
Answer:

⇒ \(E_{(n+1)}-E_n=\left[-\frac{13.6}{(n+1)^2}-\left(-\frac{13.6}{n^2}\right)\right]\)

⇒  \(\left[\frac{13.6}{n^2}-\frac{13.6}{(n+1)^2}\right] \mathrm{eV}=\frac{13.6(2 n+1)}{n^2(n+1)^2}\)

If the value of n is very large, then (2n + 1)= 2n and

⇒ \((n+1) \approx n \quad\)

∴ \(E_{(n+1)}-E_n=\frac{13.6 \times 2 n}{n^2 \times n^2}=\frac{13.6 \times 2}{n^3} \mathrm{eV}\)

Question 5. de Broglie wavelength of the wave associated with a moving electron and a proton are equal. Show the velocity of the electron is greater than that of the proton.
Answer:

According to de-Broglie’s theory applicable to microscopic particles like electron-\(\lambda=\frac{h}{m v}\)[m =maSsofthe moving particle, v = velocity ofthe moving particle].

Now if the mass and velocity of the electron are me and ve and the mass and velocity of the proton are mp and vp respectively then according to the question

⇒  \(\frac{h}{m_e v_e}=\lambda=\frac{h}{m_p v_p}\)

∴ \(m_e v_e=m_p v_p \quad \text { or, } \frac{v_e}{v_p}=\frac{m_p}{m_e}\)

But, m_p > m_e so, v_e> v_p (proved)

Question 6. Calculate the accelerating potential that must be applied on a proton beam to give it an effective wavelength of 0.005 nm.
Answer:

λ = \(\frac{h}{m v}=\frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times v}\)

∴ v = 7.94 × 10⁴m.s^-1

If the accelerating potential is V volts, then energy acquired by the proton =eV. This becomes the kinetic energy ofthe proton.

Hence \(e V=\frac{1}{2} m v^2\)

e V = \(\frac{1}{2} m v^2\)

∴ v = 32.8v

Question 7. Arrange the following radiations in the order of their increasing frequencies
Answer:

1. The amber light of traffic signals,
2. FM radio waves
3. X-rays
4. Cosmic rays

Answer:

2 < 1 < 3 < 4

Question 8. In the case of a 15X-atom, five valence electrons are. If the spin quantum number of B and R is +1 then find the group(s) of electrons with three of the quantum numbers the same.
Answer:

The spin quantum number of ‘R’ is given as \(+\frac{1}{2}\) and hence that of ‘P’ and ‘Q’ will also be \(+\frac{1}{2}\).

Electrons P, Q, and R are in 3p -orbital, so their n and l values i.e., principal and azimuthal quantum numbers will also be the same.

Therefore, P, Q, and R form a group having three quantum numbers the same. Both A and B belong to 3s having the value of n = 3, l = 0 and m = 0. Hence they also have values of three quantum numbers the same.

Question 9. How are the following affected by the increase in intensity ofthe incident light?

1. Threshold frequency,
2. The kinetic energy of the emitted electrons,
3. Strength photoelectric current.

Answer: No effect

1. Remains the same
2. Increases.

Question 10. Give examples of the production of Photons from electrons and electrons from photons.
Answer:

When high-velocity electrons (cathode rays) strike the surface of hard metals like tungsten, platinum, etc., X – rays are produced.

When a light of suitable frequency or any other electromagnetic radiation strikes a metal surface, electrons are ejected from it.

Question 11. Mention die factors affecting the kinetic energy of the photoelectrons. Does the maximum kinetic energy depend on the intensity of light?
Answer:

1. The frequency of the incident radiation and
2. Work function ofthe metal.
3. Maximum kinetic energy does not depend on the intensity ofthe incident light.

Question 12. Why does the photoelectric work function differently for different metals?
Answer:

1. Electrons in a metal are delocalized and move freely throughout the crystal lattice of the metal.
2. Hence each electron has to do some work to overcome the force of attraction of the metal ions.
3. The amount of energy required to eject the electrons (known as work function) depends on the metal. Hence, different metals have different work functions.

Question 13. Explain The role of threshold frequency in photoelectric effect is in agreement with the particle nature of light and in disagreement with the wave nature of light’
Answer:

According to the wave theory of light, the photoelectric effect can occur by increasing the intensity of the incident light. However, according to particle theory, there is a minimum frequency (threshold frequency), for each metal below which, the photoelectric effect is not possible (no matter how high the intensity of light).

It has been experimentally proved that the photoelectric effect depends on the frequency of the incident light but not on its intensity. The threshold frequency of each metal is unique.

Hence photoelectric effect can be successfully explained with the help of the particle nature of light.

Question 14. Mention the property of electromagnetic radiation (wave nature or particle nature or both) that can best explain the following phenomena—

1. Photoelectric effect
2. Interference
3. Black body radiation
4. Diffraction
5. Einstein’sequation (e = hv)
6. Planck’s equation{e – me2).

Answer:

1. Particle nature
2. Wave nature
3. Particle nature
4. Wave nature
5. Both wave and particle nature
6. Particle nature

Question 15. Indicate spectral regions corresponding to Lyman, Balmer, Paschen & Brackettseries in the line spectrum of hydrogen.
Answer:

1. Lyman series →Ultraviolet
2. Balmer series → Visible
3. Paschen series → Infrared
4. Brackett series → Far infrared

Question 16. Give two examples of the particle nature of electromagnetic radiation.
Answer:

When light of a suitable frequency strikes a metal, photoelectrons are ejected from its surface. This phenomenon (of photoelectric effect) supports the particle nature of electromagnetic radiation.

The phenomenon of black body radiation also supports the particle nature of electromagnetic radiation.

Question 17. Give the Rydberg formula for the calculation of the wave number of various spectral lines ofthe spectrum. What is the value Rydberg constant?
Answer:

Rydberg’s formula: \(\bar{v}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)

Where = 1, 2, 3, 4,…. etc;

n₂ =n₁+1. = n₁ + 2, = n₁ + 3, …………………… etc.

R = Rydberg’s constant = 109678cm^-1 ; v =wave number

Question 18. Indicate all the possible pathways (involving one or more steps)for the transition of an excited electron from the 4th orbit to the ground state.
Answer:

• n₄→ n₁
• n₄→ n₁→ n₁
• n₄→ n₃→ n₁ and
• n₄→ n₃ → n₂→ n₁

Question 19. What are the ground state and excited state of an electron?
Answer:

When the electrons in an atom are in their lowest (normal) energy state, they revolve in their respective orbits without losing energy. This state of the atom is called its ground state.

When energy is supplied to an atom by subjecting it to electric dischdt&eior high temperature, an electron in the atom may jump from its normal energy level (ground state) to some higher energy level, by absorbing a definite amount of energy. This state of the atom is called the excited state

Question 20. What do you understand by stationary states?
Answer:

According to Bohr’s theory of the hydrogen atom, electrons revolve around the nucleus in some fixed orbits, and during its motion, the electron does not lose energy. For this reason, these orbits are known as electronic orbits at stationary states.

When an electron stays in such an orbit, it does not remain stationary at all. Had it been so, the electron, being attracted by the nucleus would have fallen onto the nucleus. The electron always remains in motion to overcome the influence of nuclear attractive force

Question 21. Differentiate Itetween Rydbergformula & Balmerformula.
Answer:

The Rydberg formula is used to calculate the wave number of different series of lines of the spectrum of hydrogen or Hlike atoms. It is given by

⇒ \(\bar{v}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] Z^2(Z=1 \text { for hydrogen })\)

Where R = Rynx = 1, 2, 3, etc, n₂ = n₁ + 1 , n₁ + 2, n₁ + 3 , etc.

When n₁ = 2 in the Rydberg formula, it is called the Balmer formula.

Balmer formula is given by \(\bar{v}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right], \text { where } n=3,4,5 \cdots \text { etc. }\)

Question 22. Prove that, the velocity of an electron revolving in the first orbit is twice that revolving in the second orbit of the H-atom.
Answer:

The velocity of the electron in the nth orbit

v = \(\frac{2 \pi e^2}{n h}\)

∴  \(m v r=\frac{n h}{2 \pi} \text { and } r=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

The velocity of electron in second orbit \(v_2=\frac{2 \pi e^2}{2 \times h}\)

Therefore \(\frac{v_1}{v_2}=\frac{2 \pi e^2}{h} \times \frac{2 h}{2 \pi e^2}=2\)

therefore \(\frac{v_1}{v_2}=\frac{2 \pi e^2}{h} \times \frac{2 h}{2 \pi e^2}=2 \quad \text { or, } v_1=2 v_2\)

Question 23. Derive a relation between kinetic energy and de Broglie wavelength associated with a moving electron.
Answer:

We know, the kinetic energy (E) of the particle moving with velocity v, is given by, \(E=\frac{1}{2} m v^2\) or, 2E = mv₂

or, 2mE = m₂v₂

⇒ \(m v=\sqrt{2 m E}\)

Question 24. What happens to the de Broglie wavelength associated with a moving particle if its velocity is doubled?
Answer:

The de Broglie wavelength reduces to half its initial value

⇒ \(\text { [as } \left.\lambda=\frac{h}{m v} \text {, or } \lambda \propto \frac{1}{v}\right]\)

Question 25. A hard-struck cricket ball does not produce waves. Why?
Answer:

Due to the large size of the cricket ball, its mass is large and hence its wavelength is negligible.

Therefore \(\lambda \propto \frac{1}{m}\)

Question 26. Two particles P and Q are moving with the same velocity, but the de Broglie wavelength of P is thrice that of Q. What do you conclude?
Answer:

Since \(\lambda^{\circ} \frac{1}{m}, \lambda_P \propto \frac{1}{m_P} \text { and } \lambda_Q \propto \frac{1}{m_Q}\)

⇒\(\frac{\lambda_P}{\lambda_Q}=\frac{m_Q}{m_P}, \text { or } \frac{m_Q}{m_P}=\frac{\lambda_P}{\lambda_Q}=\frac{3}{1} \text { or, } m_Q=3 \times m_P\)

∴ Mass of Q is Thrice that of P.

Question 27. Compare the wavelengths of a molecule of each 02 and C02, travelling with the same velocity.
Answer:

Since \( \lambda \propto \frac{1}{m} ; \quad \lambda_{\mathrm{O}_2} \propto \frac{1}{32 \mathrm{u}} \text { and } \lambda_{\mathrm{CO}_2} \propto \frac{1}{44 \mathrm{u}}\)

[ V Molar mass of O₂ & CO₂ are 32u & 44u respectively]

Thus, the wavelengths of a molecule of each O₂ and CO₂ traveling with the same velocity is in the ratio 11:

Question 28. Is there any significance of Heisenberg’s uncertainty principle in our daily life?
Answer:

In our daily life, we deal with objects of ordinary size. So the uncertainties in their position and momentum are very small as compared to the size and momentum of the n- n-object respectively. So, such uncertainties may be neglected. Thus, the uncertainty principle has no significance in our daily life.

Question 29. Why does Bohr’s model contradict Heisenberg’s uncertainty principle?
Answer:

According to Bohr’s theory, negatively charged particles (electrons) inside an atom revolve around the nucleus in well-defined orbits having a fixed radius.

To balance the nuclear attractive force, electrons must move with a definite velocity. However, according to the uncertainty principle, it is impossible to determine simultaneously the exact position and the momentum (or velocity) of a microscopic particle like an electron. Thus, Bohr’s model contradicts Heisenberg’s uncertainty principle.

Question 30. Explain why the uncertainty principle is significant only for subatomic particles, but not for macroscopic objects.
Answer:

The position of a subatomic particle can be located accurately by illuminating it with some electromagnetic radiation.

The energy of the photon associated with such radiation is sufficient to disturb a subatomic particle so that there is uncertainty in the measurement of the position and momentum of the subatomic particles. However, this energy is insufficient to disturb a macroscopic object.

Question 31. Why is it not possible to overcome the uncertainty of Heisenberg’s principle using devices having high precision?
Answer:

Heisenberg’s uncertainty principle has no relation with the precisions of measuring devices.

AVe knows that the subatomic particles are very tiny and thus cannot be seen measured even under a powerful microscope; To velocity or to locate the position of the subatomic particles, they are illuminated (struck, protons) with suitable electromagnetic radiation.

Hence, the precision of measuring devices is not possible in overcoming Heisenberg’s uncertainty principle.

Question 32. How many radial nodes are present in

1. 3s -orbital and 
2. 2p -orbital?

Answer:

Radial nodes of 3s -orbital =n – l -1 = 3 – 0 – 1 = 2

Radial nodes of 2p -orbital = n- l -1 = 2 – 1 – 1 = 0

Question 33. How many radial nodes and planar nodes are present in 3p -orbital?
Answer:

No. of radial nodes = n-l-1

= 3-1-1

= 2-1

= 1

No. of planar nodes =l=1

Total no. of nodes = n- 1 = 3- 1 = 2

Question 34. What do you mean by the Acceptable values of e and Corresponding wave functions that are obtained by solving the Schrodinger wave equation for h-atom?
Answer:

No, atomic orbitals do not possess a sharp boundary. This is because the electron clouds are scattered to a large distance from the nucleus. The density of electron clouds decreases with increasing distance from the nucleus but theoretically, it never becomes zero

Question 35. In which direction the value of

1. ψ_2px
2. ψ_2py
3. ψ_{2pz ,} is the highest? Px

Answer:

• At the negative and positive direction of the x -x-axis.
• At the negative and positive direction of y -y-axis.
• At the negative and positive direction of the z-axis.

Question 36. Why s -orbital does not possess directional properties?
Answer:

The angular part of the wave function of s-orbital does not depend on θ and ∅

As a result, a symmetrical distribution of electron density occurs with increasing distance from the nucleus. Thus, s -the orbital is spherically symmetrical and does not possess directional properties.

Question 37. Indicate the subshells present in the M -M-shell. How many orbitals are present in this shell?
Answer

: In the case of M-shell, the principal quantum number, n = 3. The values of azimuthal quantum no. Z are’ 0 ‘, ‘ 1’, and ‘2’.

This means that the M-shell contains three subshells, namely, ‘p’ and ‘ d’. For each value of‘ Z ’, the magnetic quantum number ‘ m ’ can have 2Z + 1 values. Therefore M-shell contains 2Z+ 1 orbital.

Question 38. Write the values of the azimuthal quantum number ‘l’ in the third energy level and(it) 3d -subshell of an atom.
Answer:

In the third energy level, the principal quantum number n = 3

Values of azimuthal quantum no., ‘ Z ’ are 0, 1, and 2.

For any d -subshell, 1 = 2.

Question 39. What is the maximum number of electrons that can be accommodated in the subshell with 1 = 3?
Answer:

For every value of’ Z ‘, ‘ m ’ can have 2Z + 1 values.

Since For Z = 3, m can have 2Z+ 1 values, i.e., 2 × 3 + 1 = 7 values. Therefore, the number of orbitals in the given shell = 7.

The maximum number of electrons that can be accommodated in these orbitals = 2×7 = 14 [v each orbital can accommodate a maximum of 2 electrons].

Question 40. What is the maximum number of electrons that can be accommodated in an orbital with m = +3?
Answer:

Each value of the magnetic quantum number ‘m’ indicates only one orbital and each orbital can accommodate a maximum of 2 electrons.

This means that the orbital Indicated by m = +3 can accommodate a maximum of 2 electrons.

Question 41. When Be is bombarded with a -particles, a new element viz carbon Is formed whereas, when gold is bombarded with a -particles, no new elements are formed. Explain.
Answer:

There are 79 protons in the nucleus of a gold (_7gAu) atom, while α -particles are helium nuclei with 2 unit positive charges.

The approaching α -particles are repelled strongly due to high positive charges of Au nuclei and thus suffer deflection.

On the other hand, there are only 4 protons in the nucleus of the beryllium (₄Be) atom are very weak compared to those between the gold nuclei and o – particles, due to the low positive charge of the Be nucleus. Thus, the fast-moving a -particles collide with Be nuclei and cause splitting

⇒  \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 n\)

Question 42. Why are atomic spectra not continuous?
Answer:

Each electron in an atom is associated with a definite energy corresponding to different energy levels. These electrons absorb energy from various external sources (like heat, light, etc.) and are promoted to higher energy levels. These excited electrons radiate different amounts of energy and return to the ground state.

Since the difference between any two energy levels is fixed, the atomic spectra obtained are discontinuous line spectra having fixed wavelengths. The spectrum so obtained consists of a few bright lines but does not contain all the possible spectral lines corresponding to a range of given wavelengths. Thus, atomic spectra are not continuous.

Question 43. With the help of Bohr’s theory, how will you determine the kinetic energy of hydrogen or hydrogen-like atoms?
Answer:

Let the no. of positive charges in the nucleus of a given atom or ion be Z.e (Z = atomic no., e = charge of a proton). According to Bohr’s theory, the electron present in that atom or ion revolves around the nucleus only in stationary orbits.

Let the radius ofthe stationary orbit be ‘r’ For the stability of the atom, the coulombic force must be equal to the centrifugal force of the electron moving with a velocity

⇒  \(\frac{Z e^2}{r^2}=\frac{m v^2}{r} \text { or } \frac{1}{2} m v^2=\frac{Z e^2}{2 r}\)

Question 44. What is the precessional motion of the orbit?
Answer:

According to Sommerfeld’s relativistic correction of the atomic model, an electron revolves in an elliptical orbit around the nucleus, which is located at the focus of the ellipse.

• This results in a continual change in the mass and velocity of the electron. The mass of the moving electron increases with its velocity.
• The velocity of this electron is maximum when closest to the focus of the ellipse (perihelion) and minimum when farthest from the focus (aphelion). Because of its increased mass at the perihelion, the electron experiences a stronger force of attraction from the nucleus.
• This compels the electron to deviate from its original orbit to a new and identical orbit, which lies in the same plane. The perihelion moves each time the electron completes a revolution.
• Thus the entire electron orbit moves about an axis passing through the nucleus. This phenomenon is known as Sommerfeld’s precession or precessional motion of the orbit.

Question 45. Name the noble gas and give its atomic number if the number of d -d-electrons present in this atom is equal to the difference in the no. of electrons present in the p and s- s-subshells
Answer:

The noble gas is krypton (Kr). Its atomic number = 36

Electronic configuration: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶

• Number of s -electrons = 2 + 2 + 2 + 2 = 8
• Number of p -electrons = 6 + 6 + 6 = 18
• Number of d -electrons = 10

∴ Number of p -electrons number of s -electrons =18-8 = 10= number of d -electrons

Question 46. There is a wavelength limit beyond which the spectrum of any given series of the H-atom becomes continuous. Why?
Answer:

The energy difference between the first and second orbits is maximum. With the increase in the value of the principal quantum number (n), the energy difference between two successive orbits decreases. Consequently, after a particular value of n, the energy levels become very closely spaced and as a result, they seem to be continuous.

Question 47. How will you prove that electrons are negatively charged particles with a definite mass?
Answer:

Under the influence of an electric field the cathode rays as well as the electron beam, are deflected towards the positive plate of the electric field. Cathode rays also neutralize the gold leaf ‘ electroscope, charged with positive electricity.

Thus it can be proved that electrons are negatively charged. A light paddle wheel placed in the path of cathode rays, begins to rotate, showing that cathode rays are made of matter particles.

Question 48. Calculate the number of particles present in 0.1 g electron.
Answer:

Number of electrons = \(\frac{\text { Total mass of electrons }}{\text { Mass of Celestron }}\)

= \(\frac{0.1 \mathrm{~g}}{9.11 \times 10^{-28} \mathrm{~g}}\)

=\(1.0977 \times 10^{26}\)

Question 49. The charge-to-mass ratio of an electron is 1836 times greater than that ofa proton. Establish a mathematical relation to compare their masses.
Answer:

Given \(\frac{e}{m_e}=1836 \times \frac{e}{m_p}\) However the charge on 1 electron Is the same as that of 1 proton

⇒ \(\frac{1}{m_e}=\frac{1836}{m_p}\)

or, m_p = 1836 x m_e

Question 50. Two discharge tubes containing H₂ and O₂ gas respectively are subjected to electrical discharge at low pressure. Will there be any difference like cathode rays and anode formed inside the tube?
Answer:

In both cases, cathode rays with identical properties are produced, because these rays are independent of the nature of the material of the cathode and the gas used in the discharge tube. However, in these two cases, anode rays with different properties are produced since these rays depend on the nature of the gas used in the discharge tube.

Question 51. Explain the generation ofthe positively charged particles in the discharge tube when hydrogen gas is used.
Answer:

Due to the high voltage in the discharge tube, H₂ and D₂ are dissociated into H and D atoms.

Due to the knockout of electrons from atoms or molecules present in the discharge tube by cathode rays, H⁺₂, D⁺₂, H⁺, and D⁺ ions are produced.

Similarly, HD+ ions are also produced by the knockout of electrons from a few HD molecules (produced by a combination of H and D atoms)

Question 52. How many protons will be needed to fill a spherical vessel of volume 10cm3? Also, calculate the mass of these protons.
Answer:

Volume of proton \(=\frac{4}{3} \pi r^3=\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-13}\right)^3 \mathrm{~cm}^3\)

⇒ \(\frac{\text { Volume of a sphere }}{\text { Volume of a proton }}\)

= \(\frac{10}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-13}\right)^3}\)

Mass of protons = No. of protons × Mass of a proton

= (1.382 × 10³⁹)× (1.6725× 10^-24)g =2.311 × 10¹⁵g.

Question 53. An element has an isotope with a mass number of 14. It contains 8 neutrons. Identify the element.
Answer:

Mass number = No. of protons+ No. of neutrons.

Number of protons =14 – 8 =6

The atomic number ofthe element = 6, which means, the element is ‘Carbon’.

Question 54. Why was it necessary to consider the existence of neutrons in the nucleus of an atom?
Answer:

The actual mass of an atom of an element, except hydrogen, Is greater than the sum of the masses of protons and electrons present in that atom. Hence, Rutherford in 1920, proposed the existence of an uncharged particle in an atom having unit mass. This particle was called the neutron.

Question 55. Identify the subshells denoted by the following:

n = 4,l = 2

n = 5, l = 3

 n = 6, l = 4

Answer:

1. 4d
2. 5f
3. 6g
4. 4s

Question 56. Which quantum number is to be mentioned to distinguish between the electrons present in the -K-shell? 
Answer:

For k-shell (n = 1), l = 0 and m = 0. This indicates that K-shell has only one orbital and this orbital can accommodate a maximum of 2 electrons having spin quantum no., ‘s’ with values +1/2 and -1/2.

So, to distinguish between the two electrons in the K-shell, it is important to indicate their spin quantum numbers.

Question 57. Write the values of n, l, and m for 3p -subshell.
Answer:

For 3p -orbitals, n = 3 , l = 1 and m = +1 , 0, -1 . Hence, 3p -subshell has 3 orbitals.

The values of and ‘ m ‘ for these orbitals are as follows:

1. n = 3, l = 1 , m = +1
2. n = 3,  l = l, m = 0
3. n = 3, l = 1 , m =-1

Question 58. Which of the following two orbitals is associated with a higher energy?

1. n = 3, l = 2, m = +1
2. n = 4, l = 0, m = 0

Answer: The algebraic sum of n and / determines the energy of a given subshell. The higher the value of (n + l), of an orbital, the higher its energy. Thus, the orbital with n = 3, l= 2 is associated with a higher energy.

Question 59. Is there any difference between the angular momentum of 3p and 4p -electrons?
Answer:

For any p -subshell, 1=1. The angular momentum of an electron depends on the values of all and is independent of the values of Angular momentum, \(L=\sqrt{l(l+1)} \frac{h}{2 \pi}.\).

Since 1=1 for both the p -subshells (3p and 4p), there is no difference in the angular momentum ofthe electrons occupying those subshells

Question 60. Mention the sequence in which the following orbitals are filled up by electrons: 3d and 4p.
Answer:

The energy of a given subshell is determined by the algebraic sum of’ and ‘Vue., n + 1.

11 the ‘n +l’ values of any two subshells are equal, then the electron enters the one with lower’ n ’. In the 3d -subshell, n = 3 , l = 2 n + l = 3 + 2 = 5 In the 4p -subshell, n = 4, 1=1

B +1 =4 +1 = 5 Since for the 3d -subshell, n = 3 which is lower than that of 4p where n = 4, the electron first enters the 3d -subshell.

Question 61. What is the maximum number of Ad -electrons with spin quantum number s =?
Answer:

For a 4d -subshell, n = 4, 1 = 2 and m = +2, +1,0, -1,-2. This implies that 5 orbitals in this sub-shell can accommodate a maximum of 10 electrons.

5 of these electrons have s = +| and the remaining 5 have -i . Hence maximum number of 4d -electrons with 2 i spin quantum no., s = -1/2 is 5.

Question 62. Is it possible for atoms with even atomic numbers to contain unpaired electrons?
Answer:

Atoms with even atomic numbers can have unpaired electrons. This is by Hund’s rule which states that the orbitals within the same subshell are at first filled up singly with electrons having parallel spin before pairing takes place.

For instance, in the case of a carbon atom (atomic number 6 and electronic configuration: ls²2s²2p_x¹2p_y¹2p_z⁰ ), there are two unpaired electrons

Question 63. Write the electronic configurations of Cu and Cr -atoms.
Answer:

Electronic configuration of ₂₉Cu

ls²2s²2p⁶3s²3p²3d¹⁰4s¹

Electronic configuration of ₂₄Cr :

ls²2s²2p²⁶3s²3p⁶3d⁵4s¹

Question 64. Write the electronic configurations of Fe²⁺ and Cu⁺ ions.
Answer:

Electronic configuration of Fe²⁺ (atomic number =26):

ls²2s²2p⁶3s²3p⁶3d²⁶

Electronic configuration of Cu⁺ (atomic number = 29 ):

1s²2s²2p⁶3s²3p⁶3d¹⁰

Question 65. Which of the following has a maximum number of unpaired electrons? (1) Mn²⁺Fe2+Cu2+ Cr
Answer:

The following are the electronic configurations of the given ions and atoms:

Mn2+: ls²2s²2p⁶3s²3p⁶3d⁵

Fe²⁺: ls²2s²2p⁶3s²3p⁶3d⁹

Cu²-: ls²2s²2p⁶3s²3p⁶3d⁹

Cr: ls²2s²2p⁶3s²3p⁶3d⁵4s¹

Question 66. Calculate the number of unpaired electrons in the N -atom.
Answer:

Electronic configuration of (atomic no. =7): ls²2s²2p³

According to Hund’s rule, the 3 electrons in the 2psubshell occupy the three p -p-orbitals (px, py, pz) singly.

Hence, the no. of unpaired electrons present in N = 3.

Question 67. How many electrons of the Ne -atom have clockwise spin?
Answer:

Electronic configuration of Ne (atomic number = 10)

Each of the pair of electrons present in each ofthe Is, 2s, 2px, 2py, and 2pz orbitals have a clockwise spin and the other, an anti-clockwise spin. Therefore no. of electrons of Ne-atom having clockwise spin = 5.

Question 68. Write the names and symbols of an atom, a cation, and an anion with the electronic configuration Is2.
Answer:

Atom: Helium (He);

Cation: Lithium-ion (Li⁺),

Anion: Hydride ion (H^– ).

Question 69. How many nodes are there in 3s -orbital?
Answer:

The node is the spherical shell (or region) inside the s -s-orbital where electron density is zero. In the case of 3s orbital, there are two such spherical shells where the electron density is zero.

So 3s -orbital has two radial nodes but no angular node (1 = 0). So the total no. of nodes is 2. [No. of nodal surfaces =n- 1, where n is the principal quantum number]

Question 70. How many nodal points are there in 3p -orbital?
Answer:

In a p -orbital the electron density, at the point where the two lobes meet is zero.

This point is called the nodal point of the p-orbital. So each of the three 3p -orbitals (viz., px, py, and pz ) has only one nodal point.

Question 71. Indicate principal and azimuthal quantum numbers for the subshells:

1. 4s
2. 5d
3. 2p
4. 6

Answer:

1. n = 4, 1 = 0
2. n = 5,1 =2
3. n = 2, 1=1
4. n = 6, 1 = 3

Question 72. An element (symbol M) has 26 protons in the nucleus. Write the electronic configuration of M²⁺ and M³⁺.
Answer:

₂₆M: ls²2s²2p⁶3s²3p⁶3d⁶4s²

₂₆M²⁺: ls²2s²2p⁶3s²3p⁶3d⁶

₂₆M³⁺ : ls²2s²2p⁶3s²3p⁶3d⁵

Question 73. There are 8 electrons in the 3d -subshell of an atom. Among these, what will be the maximum number of electrons with similar spin? What is the number of odd electrons?
Answer:

1. Electronic configuration of 3d -subshell
2. Maximum number of electrons with the same spin = 5.
3. Number of odd electrons in that atom = 2.

Question 74. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

1. 2s and
2. 3s,
3. 4d and
4. 4f,
5. 3d and
6. 3p

Answer:

1. 2s
2. 4d
3. 3p

Question 75. Indicate the number of impaired electrons in:

1. P
2. Si
3. Cr
4. Fe and
5. kr

Answer:

1. ₁₅P = ls²2s²2p⁶3s²3p_x¹3p_y¹3p_z¹ ; number of unpaired electrons = 3.
2. ₁₄Si = ls²2s²2p⁶3s²3p_x¹3p_y¹ number of unpaired electrons = 2.
3. ₂₄Cr = ls²2s²2p⁶3s²3p⁶3d⁵4s¹; number of paired electrons = 6 (5 in d-subshell &1 in s-subshell).
4. ₂₆Fe = ls²2s²2p⁶3s²3p⁶3d⁶4s²; number of unpaired electrons in d-subshell = 4.
5. ₃₆Kr = ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶; number of unpaired electrons = 0.

Question 76. How many subshells are associated with n =4? How many electrons will be present in the subshells having ms value of \(-\frac{1}{2}\) for n = 4?
Answer:

For n = 4, l = 0, 1, 2, 3. Thus, the energy level with n = 4 contains four subshell (4s, 4p, 4d and 4f).

For n = 4, the number of orbitals (n)2 = (4)2 = 16.

Each orbital will have only one electron with ms = Hence, for n = 4, 16 electrons will be present in the subshells with the value of ms \(=-\frac{1}{2}.\)

Question 77. Write the complete symbol for the atom with the h = 6.626 × 10^-34J.s, c = 3.0× 10⁸m-s^-1 ] given atomic number (Z) and atomic mass (A)

1. Z = 17, A = 35
2. Z= 92, A=233
3. Z = 4, A = 9

Answer:

⇒ \({ }_{17}^{35} \mathrm{Cl}\)

⇒  \({ }_{92}^{233} \mathrm{U}\)

⇒  \({ }_4^9 \mathrm{Be}\)

Question 78. Yellow light emitted from a sodium lamp has a wavelength (A) of 580 nm. Calculate the frequency (v) and wave number (v) of the yellow light.
Answer:

Wavelength, A =580nm =580 × 10^-9m =5.80 ×10^-7m.

Frequency of yellow light.

v = \(\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{5.80 \times 10^{-7} \mathrm{~m}}=5.17 \times 10^{14} \mathrm{~s}^{-1}\)

Speed of light, c = 3 × 10⁸ m.s^-1  and wave number,

⇒ \(\bar{v}=\frac{1}{\lambda}=\frac{1}{5.80 \times 10^{-7} \mathrm{~m}}\)

= 1.72 × 10⁸ m.s^-1

Question 79. Find the energy of each of the photons which corresponds to light of frequency 3 × × 10¹⁵ Hz. Have a wavelength of 0.50A
Answer:

⇒ \(=h v=\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^{15} \mathrm{~s}^{-1}\right)\)

=1.988× 10^-18J

E =  \(h v=\frac{h c}{\lambda}\)

= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} \times 3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{0.50 \times 10^{-10} \mathrm{~m}}\)

= 3.978 × 10^-15J

Question 80. Calculate the wavelength, frequency, and wavenumber of a light wave whose period is 2.0 × 10^-10s.
Answer:

⇒ \(\text { Frequency }(v)=\frac{1}{\text { period }}=\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}=5 \times 10^9 \mathrm{~s}^{-1}\)

⇒ \(\text { Wavelength }(\lambda)=\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{5 \times 10^9 \mathrm{~s}^{-1}}=6 \times 10^{-2} \mathrm{~m}\)

And wavelength \((\bar{v})=\frac{1}{\lambda}=\frac{1}{6 \times 10^{-2} \mathrm{~m}}=16.66 \mathrm{~m}^{-1}\)

Question 81. Electromagnetic radiation of wavelength 242 is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJ.mol^-1
Answer:

Ionisation energy ofsodium (E) = Nhv \(=N h \frac{c}{\lambda}\)

E = \(\frac{\left(6.022 \times 10^{23}\right) \times\left(6.626 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{242 \times 10^{-9}}\)

= 4.946  ×  10⁵ J.mol^-1

= 494.6 kJ-mol^-1

Question 82. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A. Calculate threshold frequency (vQ) and work function (W0) of the meta
Answer:

Threshold frequency \(v_0=\frac{c}{\lambda_0}=\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\)

= 4.41 ×  10¹⁴ s^-1

And the work function of the metal ( wQ) = hv0

= (6.626 × 10³⁴ J.s) ×  (4.41 × 10^-14.s^-1) =2.92 ×10^-19J

Question 83. Which of the following are isoelectronic species i.e., those having the same number of electrons? Na⁺,K⁺,Mg²⁺, Ca²⁺, S^2-,Ar
Answer:

The number of electrons present in

Na⁺=(11 – 1) =10

K⁺=( 19 – 1) =18

Mg²⁺ =(12 – 2) = 10

Ca²⁺ =(20 – 2)=18

S^2- =(1 6+ 2) =18

Ar = 18.

Elance, Na⁺, and Mg²⁺ are isoelectronic because they contain 10 electrons each. K⁺, Ca²⁺, S^2-, and Ar are isoelectronic because they contain 18 electrons each.

Question 84.

1. What is the orbital angular momentum of a p -electron in \(\frac{h}{2 \pi}\) unit?
2. The atomic numbers of two elements X and Y are 15 and 27 respectively. Write down the electronic configuration of X^3- and Y³⁺ ions.
   

Answer:

1. The orbital angular momentum in \(\frac{h}{2 \pi}\) unit is given by \(\sqrt{l(l+1)}\) where l = azimuthal quantum number.

For p -orbital , l = 1.

∴ The orbital angular momentum of p -orbital in \(\frac{h}{2 \pi}\) unit

= \( \sqrt{l(l+1)}=\sqrt{1(1+1)}=\sqrt{2} .\)

2. ₁₅X^3- : ls²2s²2p⁶3s²3p⁶

₂₇Y³⁺ : ls²2s²2p⁶3s²3p⁶3d⁶

Question 85.

1. The electronic configuration of an atom is \([Z](n-2) f^{14}(n-1) d^1 n s^2\). What is the minimum position of the atom in the periodic table and correspondingly what is the atomic number of Z?
2. Find the number of impaired electrons in the atom of the element having atomic number 16
3. Which of the following ions does not obey Bohr’s atomic theory? He²⁺+,Li²⁺,B³⁺,Be³⁺
   

Answer:

Electronic configuration of the given atom:

1. (n-2)f^1-14-14(n-1)d^0-1 ns². So, it can be stated that the given atom belongs f-block. Hence, the element is of group IIIB and its electronic configuration is identical to ₇₁Lu of the lanthanoids and Lu of the actinoids. The lowest position available to the atom ofthe element is the 6th period and group-IIlB(3). Thus, it belongs to the lanthanoids and has atomic number 71.

2. The electronic configuration of an atom ofthe element with atomic number 16 is 1s²2s²2p⁶3s²3p_x²3p_y¹ 3p_z¹. Thus, number of unpaired electrons is two.

3.  Be³⁺does not obey Bohr’s atomic theory because it is 2- an electron system

Question 86. Using s, p, and d notations, describe the orbital with the following quantum numbers.

• n- 1,
• 1=0;
• n = 3;
• 1 = 1
• n = 4;
• 1 = 2;
• n = 4;
• l = 3

Answer:

• 1s
• 2s
• 4d
• 4f

Question 87. Explain, giving reasons, which of the following sets of quantum numbers are not possible

• n = 0, l = 0, m₁ = 0, m_s = \(+\frac{1}{2}\)
• n = 1, l = 0, m₁ = 0, m_s = \(-\frac{1}{2}\)
• n = 1, l =1, m₁ = 0, m_s = \(+\frac{1}{2}\)
• n = 2, l = 1, m₁ = 0, m_s = \(-\frac{1}{2}\)
• n = 3, l = 3, m₁ = -3, m_s = \(+\frac{1}{2}\)
• n = 3, l = 1, m₁ = 0, m_s = \(+\frac{1}{2}\)

Answer: This is not possible because n cannot be zero. and are not possible because the value of n cannot be equal to 1.

Question 88. How many electrons in an atom may have the following quantum numbers?

1. n = 4,
2. m_s = \(+\frac{1}{2}\)
3. n = 3, l = 0

Answer: For n = 4, the total number of electron

= 2n² = 2 × 4² = 32 . Among these 32 electrons, half,16 electrons will have s or m_s = \(+\frac{1}{2}\) and the other 16 electrons will have ms = \(-\frac{1}{2}\)

n = 3 , l = 0 means 3s -subshell. The maximum number of electrons in this subshell is two.

Question 89. 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.
Answer:

According to Bohr’s postulate, angular momentum \(m v r=\frac{n h}{2 \pi}\) [n = principle quantum number = 1, 2, 3, 4 …………..

Again, according to the de Broglie equation, for a revolving electron, wavelength \((\lambda)=\frac{h}{m v}\)

Substituting the value of A from (2) in (1) we have, 2pi r = n lambda (n = 1,2, 3, 4….)

Thus, the circumference of the Bohr orbit is an integral multiple ofthe de Broglie wavelength ofthe revolving electron.

Question 90. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than electrons, find the symbol of the ion
Answer:

As the ion contains one unit of negative charge, the ion has one electron more than the number of protons. Total number of electrons and neutrons =37 + 1 = 38. Let the number of electrons in the ion = x.

Hence, number of neutron \(=x+\frac{11.1}{100} \times x=1.111 x\)

Again, x+ 1. 111 x = 38 or, x = 18

Thus, the number of electrons present in the ion = 18.

Thus, the number of protons present in the ion =18-1 = 17

So, the element’s atomic number is 17 i.e., the atom is chlorine. its symbol ,\({ }_{17}^{37} \mathrm{Cl}^{-}\)

Question 91. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
Answer:

The ion contains three units of positive charge. Thus, the number of electrons in the ion is three less than that of the number of protons. Number of protons + number of neutrons mass number =56. let, the number of electrons in the ion be x.

Number of neutrons present in the ion \(=x+\frac{30.4}{100} \times x=1.304 x\)

Again, the total number of electrons and neutrons = 53. x + 1.304x = 53 or, 2.304x = 53 Thus, number of electrons present in the ion = 23 and the number of protons =  23 + 3 = 26 .

The element with atomic number 26 is Fe and their symbol will be \({ }_{26}^{56} \mathrm{Fe}^{3+}\).

Question 92. Nitrogen laser produces radiation at a ‘wavelength of 337.1 nm. If the die number of photons emitted is 5.6 × 10²⁴, calculate the power of this laser.
Answer:

⇒ \(E=N h v=N h \frac{c}{\lambda}\)

= \(\frac{\left(5.6 \times 10^{24}\right) \times\left(6.626 \times 10^{-34}\right) \times\left(3.0 \times 10^8\right)}{\left(337.1 \times 10^{-9}\right)}\)

= 33 × 10⁶J

Question 93. The dual behavior of matter proposed by de Broglie led to = 1.52 × 10^-38, the discovery of an electron microscope often used for the highly magnified images of biological molecules and another type of material. If the velocity of the electron in this microscope is 1.6 × 10⁶m.s^-1, calculate the de Broglie wavelength associated with this electron.
Answer:

Velocity of an electron (y) = 1.6 ×10⁶m.s^-1 and mass of an electron (m) = 9.108 × 10^-31 kg

∴ de Broglie wavelength \((\lambda)=\frac{h}{m v}\)

= \(\frac{6.626 \times 10^{-34}}{9.108 \times 10^{-31} \times 1.6 \times 10^6}\)

= \(4.55 \times 10^{-10} \mathrm{~m}=455 \mathrm{pm}\)

Question 94. Similar to electron diffraction, a 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 the neutron.
Answer:

Mass of neutron (m) = 1.675 × 10²⁷kg

According to de Broglie equation, wavelength \((\lambda)=\frac{h}{m v}\)

∴ Velocity ofa neutron \(v=\frac{h}{m \lambda}\)

Or, \(v=\frac{6.626 \times 10^{-34}}{1.675 \times 10^{-27} \times 800 \times 10^{-12}}\)

or, v = 494m.s^-1

Question 95. The bromine atom possesses 35 electrons. It contains 6 electrons in a 2p orbital. 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electrons experiences the lowest effective nuclear charge?
Answer:

The value of n for the 4p electrons is highest and hence they are the furthest from the nucleus and thus experience the lowest effective nuclear charge.

In a given orbit, for the same type of subshells, the higher the value of n, the lower the value of effective nuclear charge.

Question 96. 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. What difference would be observed from the above results if the thin foil of light atoms like aluminum etc. is used?
Answer:

The nucleus of heavy atoms contains a large amount of chlorine. positive charge. Thus, the a -particles that move towards the nucleus are deflected back due to strong repulsion by the nucleus. Those particles which pass through the region closer to the nucleus are deflected in different directions. On the other hand, the nucleus of light atoms contains a small amount of positive charge. Hence, a negligible number of particles are deflected back or are deflected by small angles.

Question 97. Symbol \({ }_{35}^{79} \mathrm{Br}\) and \({ }^{79} \mathrm{Br}\) symbol \({ }_{35}^{79} \mathrm{Br}\) and \({ }^{35} \mathrm{Br}\) are not acceptable. Answer briefly.
Answer:

⇒ \({ }_{35}^{79} \mathrm{Br}\) is not acceptable because mass number should be written as superscript and atomic number as subscript. ₃₅Br is not acceptable because the atomic number of an element is fixed but the mass number is not fixed. It depends on the isotopes. Thus, an indication of mass number is essential.

Question 98. Why is the line spectrum of an element known as the fingerprint of its atoms?
Answer:

The line spectrum of any element consists of several lines having different wavelengths. It is observed that each element has its characteristic spectrum, different from those of all other elements.

The spectra of any two elements can never be identical. Hence, the line spectrum of an element is known as the fingerprint of its atoms

Question 99. Does atomic orbitals possess a sharp boundary? Explain.
Answer:

No, atomic orbitals do not possess a sharp boundary. This is because the electron clouds are scattered to a large distance from the nucleus.

The density of electron clouds decreases with increasing distance from the nucleus but theoretically, it never becomes zero

Question 100. Why do we consider each stationary state as an energy level with a definite value?
Answer:

Electrons in a particular orbit do not lose or gain energy. In other words, the energy of an electron in a particular orbit remains constant. Hence, these orbits or stationary states are known as energy levels having definite values.

Question 101. There are nine electrons in the 5f-orbital of an atom of an element Mention the maximum number of electrons that have the same spin and number of impaired electrons.
Answer:

Element Mention the maximum number of electrons that have the same spin and number of impaired electrons-

Thus, the maximum number of electrons with the same spin will be 7, and several unpaired electrons will be 5.

Question 102. Explain whether 3f-orbital is present in P-atom. State the rule
Answer:

Electronic configuration: 1s²2s²2p⁶3s²3p¹_x3p_y¹3_z¹

Pliosphorous atom does not contain 3f- subshell.

For 3f-subshell, n = 3 . Hence, the maximum value of f = (n- l) = 2 i.e., d -subshell.

Question 103.

1. What are the quantum numbers by which an electron In an atom can be designated?
2. What Is the maximum number of quantum numbers that may be the same or two electrons of an atom?

Answer:

1. Principal quantum number (n), azimuthal quantum number (Z), magnetic quantum number (m), and spin quantum number (s) are required to designate an electron in an atom.
2. The maximum number of quantum numbers that may be the same for two electrons of an atom is 3.

NCERT Solutions For Class 11 Chemistry Chapter 2 Structure Of Atom Warm-Up Exercise Question And Answers

Question 1.  If the energy of the first Bohr’s orbit is -13.58 eV of a hydrogen atom, calculate the energy of the third Bohr’s orbit of that atom.
Answer:

According to Bohr’s theory

⇒ \(E_n=-13.58 \times \frac{Z^2}{n^2} \mathrm{eV}=-13.58 \times \frac{1^2}{3^2} \mathrm{eV}=-1.5089 \mathrm{eV}\)

Question 2. Which quantum numbers specify the size and the shape of electronic orbital?
Answer:

The size of an electronic orbital is determined by the principal quantum number (n) and the azimuthal quantum number (f) determines the shape of an electronic orbital.

Question 3. Write down the values of the quantum numbers of the electron in the outermost shell of sodium.
Answer:

The electron present in the outermost shell of sodium is identified as 3s. Its principal quantum number n = 3, azimuthal quantum number 1 = 0, magnetic quantum number, m = 0 and spin quantum number, s = +1/2.

Question 4. Which is the lowest principal energy level that permits the existence of off-subshell?
Answer:

For f-subshell, the value of the azimuthal quantum number Z is 3. So the lowest principal energy level that permits the existence of an f -subshell is the fourth shell (i.e, N -N-shell)

Question 5. The unpaired electrons in A1 and Si are present in the 3p orbital. Which electrons will experience a more effective nuclear charge from the nucleus?
Answer:

The nuclear charge of silicon (+14) is greater than that of aluminum (+13). Hence the impaired 3p -electron of silicon will experience a more effective nuclear charge.

Question 6. Mention the most important application of the de Broglie concept.
Answer:

The de-Broglie concept is utilized in the construction of an electron microscope used for the measurement of the size of very small objects.

Question 7. What is the physical significance of Ψ and Ψ²?
Answer:

The wave function has no physical significance, while Ψ² gives the probability density i.e., the probability of finding the electron at any point around the nucleus.

Question 8. Write Schrodinger’s wave equation in the briefest possible form.
Answer:

The briefest form of Schrodinger’s wave equation is, \(\widehat{\mathrm{H}} \psi=E \psi\), where H is known as the Hamiltonian operator.

Question 9. What do you mean by ‘doughnut’?
Answer:

The two lobes of d -the orbital are distributed along the z-axis and a sphere is situated with the nucleus at its center. This sphere is called a ‘doughnut’

Question 10. How many angular nodes are present in dÿ Identify them.
Answer:

Two angular nodes are present (they pass through the origin and lie at an angle of 45° with the xz and zipline and themselves lie perpendicular to each other)

Question 11. Why do p-orbitals possess directional properties?
Answer:

The angular part of the wave function of p -p-orbital depends on the value of 6 and <p. Thus, p -p-orbitals possess directional properties.

Question 12. Why is the de Broglie wave termed a matter wave?
Answer:

Since the de Broglie wave is associated with fast-moving tiny material particles, it is also known as matter wave. The wavelength of such waves depends on the mass and velocity of the particles.

Question 13. Write the mathematical expression for Heisenberg’s uncertainty principle-
Answer:

⇒ \(\Delta x \cdot \Delta p \frac{h}{4 \pi},\) where Ax and Ap are uncertainties in the determination of exact position and momentum respectively.

Question 14. How many photons are emitted in the transition of the electron from the first to the first energy level of the H-atom?
Answer:

There is only one electron in the H-atom. Hence, during the transition of electrons from the fourth to the first energy level, only one photon is emitted.

Question 15. How is the radius of an electronic orbit related to the principal quantum number?
Answer:

The relation between the radius (r) ofthe electronic orbit and the principal quantum number (n) is given by, Therefore, the radius of the orbit is directly proportional to the square ofthe principal quantum number.

Question 16. How would you obtain the line spectrum of hydrogen?
Answer:

When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing electric discharge is resolved in a spectroscope, the spectrum obtained is the line spectrum of hydrogen.

Question 17. Explain why Rutherford did not mention the presence of neutrons in the proposed nuclear model of the atom.
Answer:

In the year 1911, when Rutherford proposed the nuclear model of the atom, the existence of neutrons was still not known (In fact, neutrons were discovered in 1932). Hence, Rutherford did not mention the presence of neutrons.

Question 18. From this experiment, it was concluded that the entire mass and positive ‘charge is present at the center of an atom.
Answer:

From Rutherford’s -particle scattering experiment, it was concluded that the entire mass and positive charge are present at the center of an atom.

Question 19. What is the nuclear model of the atom?
Answer:

The atomic model which describes the rotation of electrons in different orbitals around the positively charged nucleus is called the nuclear model of the atom.

Question 20. Identify the isotopes and isobars from the following list of atoms with a given number of protons and neutrons.
Answer:

A and B have the same number of protons but different numbers of neutrons. Hence, A and B are isotopes. C and D have different numbers of protons, but the sum of the protons and neutrons, in both cases, are the same. Hence, C and D are isobars.

Question 21. Find the total number of electrons present in l mol methane.
Answer:

1 methane (CH₄) molecules 1 C-atom + 4 H- atoms Number of electrons in CH₄ molecule =1  6 + 4 × 1

= 10. Therefore, the total number of electrons in l mol of CH4 = 6.022 × 10²³ × 10 = 6.022 × 10²⁴

Question 22. What are electromagnetic radiations? What is their velocity in a vacuum?
Answer:

Electromagnetic waves with wavelength ranging between 0.003 and 0.3m are known as microwaves. As these waves collectively travel in the same direction over a long distance, they are used in radars.

Question 23. State the principle of the formation of electromagnetic radiation.
Answer:

Electromagnetic waves with wavelengths ranging between 0.003 and 0.3m are known as microwaves. As these waves collectively travel in the same direction over a long distance, they are used in radars.

Question 24. What are microwaves? Why are they used in radars?
Answer:

Electromagnetic waves with wavelength ranging between 0.003-0.3m is known as microwaves. As these waves collectively travel in the same direction over a long distance, they are used in radars.

Question 25. Arrange the various types of radiations constituting the electromagnetic spectrum, in the decreasing order of their frequencies.
Answer:

The various radiations in the electromagnetic spectrum in decreasing order of their frequencies are as follows:

Cosmic rays > γ-rays> X-rays> UV-rays > visible rays > microwaves > radio waves.

Question 26. What is black body radiation Out of red and blue light, which one is associated with photons possessing higher energy?
Answer:

Blue light has a higher frequency (v) than red light. The energy of each photon = hv. Consequently, the photons associated with blue light have higher energy.

Question 27. Energy associated with X-rays is higher than that of visible light— explain.
Answer:

The energy of electromagnetic radiation refers to the energy of its photons (hv).

Since, ν_x-ray > ν_{visible light} -hence, hν_x-ray > hν_{visible light}  Thus, the energy of X-rays is higher than that of visible light.

Question 28. State the Pauli exclusion principle. Write the electronic configurations of ₂₄Cr³⁺ and ₂₇CO³⁺
Answer:

₂₄Cr³⁺: ls²2s²2p⁶3s²3p⁶3d³

₂₇CO³⁺: ls²2s²2p⁶3s²3p⁶3d⁶

Question 29. What is the maximum number of emission lines when the excited electron of the H atom in n = 6 drops to the ground state?
Answer:

The number of lines produced in a spectrum when an electron returns from the nth energy state to the ground state

= ∑(n₂-n₁) =∑(6- 1)

= ∑(5) = 5 + 4 + 3 + 2+1

= 15.

Question 30. A certain particle carries 2.5 = 1.602 × 10^-19 of static electric charge. Find the no. electrons present in It.
Answer:

Charge of electron = 1.602 × 10^-19 C (excluding -ye sign

∴ No. electrons contained in \(=\frac{2.5 \times 10^{-16} \mathrm{C}}{1.602 \times 10^{-19} \mathrm{C}}=1560\)

Question 31. Calculate the mass of 1 mol electron.
Answer:

Mass = Avogrado number x Mass of one electron

= (6.022 × 10²³) x (9.11 × 10^-28)

= 0.5486 mg

Question 32. A discharge tube containing H2 gas at low pressure is subjected to high voltage. Will there be er Ission of protons from the anode?
Answer:

In a discharge tube, anode rays are not emitted from the anode. Therefore protons are not emitted from the anode. However, they are produced from H2gas, by the knockout of the electrons by high-speed cathode rays.

Question 33. Write the nuclear reaction for the emission of neutrons. Indicate the e/m value of the neutron.
Answer:

When beryllium foil is bombarded with or -particles, it undergoes a nuclear reaction which primarily leads to the emission of chargeless particles called the neutron.

⇒ \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 n\)

Question 34. Mention the symbol, charge, and names of the discoverers of positron, -meson, and neutrino.
Answer:

When beryllium foil is bombarded with or -particles, it undergoes a nuclear reaction which primarily leads to the emission of chargeless particles called neutrons.

Question 35. Write the symbols of two anions isoelectronic with K⁺ 
Answer:

K⁺ion: ls²2s²2p⁶3s²3p⁶.

Two anions that are isoelectronic with K⁺ -ion are S^2- and Cl^–

Question 36.

1. What is the stationary energy level of an electron? 
2. Write the electronic configurations of CO²⁺ and As³⁺ ions. (Atomic number of Co is 27 and As is 33).

Answer:

According to Bohr’s theory, the energy level in which an electron emits no energy is known as the stationary energy level of that electron.

₂₇CO²⁺ : ls²2s²2p⁶3s²3p⁶3d⁷

₃₃AS³⁺ : ls²2s²2p⁶3s²3p⁶3d¹⁰4s

Question 37. State the Heisenberg’s uncertainty principle. Calculate the de Broglie wavelength associated with an electron moving with a velocity of 1.0 × 10⁷ m/s. (Mass of an electron: 9.1 x 10-31kg
Answer:

Wavelength \(\lambda=\frac{h}{m v}\) [h = Planck’s constant, m = mass of an electron, v = velocity of an electron]

λ = \(\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.0 \times 10^7}=7.281 \times 10^{-11} \mathrm{~m}\)

Question 38. Write down the electronic configuration of 14Si and 25Mn stating the underlying principle. Which of the following orbitals is not possible? 1 p, 2d, 3s, 3f
Answer:

First Part: The underlying principle is ‘The Aufbau principle! For electronic configuration see article no. 2.10.3.

Second Part: Ip, 2d, and 3f orbitals are not possible.

Question 39. When an electron jumps down from the 5th Bohr orbit to the 3rd Bohr orbit in the H atom, how many numbers of spectral lines will be formed?
Answer:

When an electron jumps down from the 5th Bohr orbit to the 3rd Bohr orbit in an H atom it can jump directly or it can jump to the 4th Bohr orbit first and from it, jump to the 3rd Bohr orbit. Thus, we get 3 spectral lines for these 3 transitions.

Question 40. Write the values of the four quantum numbers of the electron(s) in the outermost shell of the Cr-atom.
Answer:

According to the electronic configuration of b 24Cr(ls²2s²2p⁶3s²3p⁶3d⁵4s¹), the outermost shell configuration of the Cr-atom is: 4s¹. therefore Quantum numbers for 4s¹: n = 4, l = 0, m = 0 , s= \(+\frac{1}{2}\)

Question 41. Which is most paramagnetic among Cu²⁺, and Fe²⁺, and why?
Answer:

From the electronic configurations it is observed that the number of unpaired electrons in Cu²⁺, Fe²⁺, and Cr³⁺, ions are 1, 4, and 3 respectively. So Fe²⁺, ion, containing the highest number of unpaired electrons, will be most paramagnetic.

Question 42. Why splitting of spectral lines is observed when the source producing the atomic spectrum is placed in a magnetic field?
Answer:

In the presence of a magnetic field, the orbitals present in a sub-shell (which were originally degenerate) take up different orientations and hence their degeneracy is lost.

Electronic configuration: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹.

Question 43. Write two differences between orbit and orbital. Two sets of four quantum numbers of an electron are (n = 4, 1 = 3, m = 3, s = -) and n = 3, 1 = 2, 2 m = -2, s = 0). Which one of these sets is not correct and why?
Answer:

The second set is incorrect because the value of s can be either \(\frac{1}{2} \text { or }-\frac{1}{2}\) but can never be zero.

Question 44. Write the electronic configuration of Cu⁺ and Cr²⁺ ions (Atomic numbers of Cu and Cr are 29 and 24 respectively).
Answer:

Cu⁺: ls²2s²2p⁶3s²3p⁶3d¹⁰

Cr²⁺: 1s²2s²2p⁶3s²3p⁶3d⁴

Question 45. A cation M³⁺ has 23 electrons. What is the atomic number of M?
Answer:

Number of electrons present in the neutral M atom = 23 + 3 = 26. So, die number of protons in the nucleus = 26. Hence, the atomic number of M is 26.

Question 46. What is the unit of Planck’s constant in S.I.? What other physical quantity has the same unit?
Answer:

The SI unit of Planck’s constant ‘h’ =kg-m².s^–1

The SI unit of angular momentum (mvr) is also kg-m².s^–1

Question 47. Do atomic orbitals have sharp boundaries?
Answer:

No, atomic orbitals do not have sharp boundaries because the probability of finding an electron even at a large distance is never zero, although it may be very small.