RBSE Solutions Class 11 Chemistry Chapter 2 Atomic Structure

RBSE Class 11 Chemistry Chapter 2 Text Book Questions

RBSE Class 11 Chemistry Chapter 2 Multiple Choice Questions

 

Question 1. If atomic number of an element is 25 and its atomic mass is 55, then its nucleus contains:
(a) 25 protons and 30 neutrons
(b) 55 protons
(c) 30 protons and 25 neutrons
(d) 55 neutrons
Answer: (a) 25 protons and 30 neutrons
In simple words: The atomic number tells us the number of protons. The atomic mass is the sum of protons and neutrons. So, to find neutrons, we subtract the atomic number from the atomic mass.

🎯 Exam Tip: Remember that for a neutral atom, the number of electrons is equal to the number of protons.

 

Question 3. Energy of a photon is calculated by:
(a) \( E = h\nu \)
(b) \( h = E\nu \)
(c) \( h = E/\lambda \)
(d) \( E = n\lambda \)
Answer: (a) \( E = h\nu \)
In simple words: The energy of a tiny light particle, called a photon, is found by multiplying Planck's constant (h) by its frequency (\( \nu \)). This formula helps us understand how light carries energy.

🎯 Exam Tip: This formula, \( E = h\nu \), is fundamental in quantum mechanics and describes the energy of a single photon.

 

Question 5. In a sub shell of an atom, maximum number of electrons can be defined by which of the following?
(a) \( 4l + 2 \)
(b) \( 2l + 1 \)
(c) \( n^2 \)
(d) \( 2n^2 \)
Answer: (a) \( 4l + 2 \)
In simple words: To find the maximum number of electrons a subshell can hold, you use the azimuthal quantum number, \( l \). The formula \( 4l+2 \) gives you this maximum count. For example, if \( l=0 \) (s-subshell), it can hold 2 electrons. If \( l=1 \) (p-subshell), it can hold 6 electrons.

🎯 Exam Tip: Be careful not to confuse the formulas for total electrons in a shell (\( 2n^2 \)) with those for a subshell (\( 4l+2 \)) or the number of orbitals in a subshell (\( 2l+1 \)).

RBSE Class 11 Chemistry Chapter 2 Very Short Answer Type Questions

 

Question 6. Calculate the number of protons in 34 mg, \( \text{NH}_3 \) at N.T.P.
Answer: First, we find the molar mass of \( \text{NH}_3 \). Nitrogen (N) has 7 protons, and Hydrogen (H) has 1 proton. Since there are three hydrogen atoms, \( \text{NH}_3 \) has \( 7 + (3 \times 1) = 10 \) protons per molecule.
The molar mass of \( \text{NH}_3 \) is \( 14 \, \text{g/mol} \) (for N) \( + 3 \times 1 \, \text{g/mol} \) (for H) \( = 17 \, \text{g/mol} \).
Next, we calculate the number of moles of \( \text{NH}_3 \) in 34 mg. We convert 34 mg to grams: \( 34 \, \text{mg} = 0.034 \, \text{g} \).
Number of moles \( = \text{mass / molar mass} = 0.034 \, \text{g} / 17 \, \text{g/mol} = 0.002 \, \text{mol} \).
One mole of any substance contains Avogadro's number of molecules, which is \( 6.023 \times 10^{23} \).
So, \( 0.002 \, \text{mol} \) of \( \text{NH}_3 \) contains \( 0.002 \times 6.023 \times 10^{23} \) molecules.
\( \implies \) Number of \( \text{NH}_3 \) molecules \( = 1.2046 \times 10^{21} \).
Since each \( \text{NH}_3 \) molecule has 10 protons, the total number of protons is:
\( \implies 1.2046 \times 10^{21} \text{ molecules} \times 10 \text{ protons/molecule} = 1.2046 \times 10^{22} \text{ protons} \).
In simple words: We first figure out how many protons are in one \( \text{NH}_3 \) molecule (it's 10). Then, we find out how many \( \text{NH}_3 \) molecules are in 34 milligrams by using its weight and Avogadro's number. Finally, we multiply the total number of molecules by 10 to get the total number of protons.

🎯 Exam Tip: Pay close attention to unit conversions (mg to g) and correctly use Avogadro's number in your calculations.

 

Question 8. Out of 4f and 4d which orbital will experience high effective nuclear charge?
Answer: The effective nuclear charge is the net positive charge that an electron in a multi-electron atom feels. This charge depends on how close the orbital is to the nucleus. Orbitals that are closer to the nucleus experience a stronger attraction. The 4d orbital is closer to the nucleus compared to the 4f orbital. This is because d-orbitals penetrate more than f-orbitals for the same principal quantum number. Therefore, the 4d orbital will experience a higher effective nuclear charge than the 4f orbital.

🎯 Exam Tip: Remember that inner electrons shield outer electrons from the full nuclear charge. Orbitals with greater penetration (like s > p > d > f) experience less shielding and thus a higher effective nuclear charge.

 

Question 9. For \( n = 4 \) and \( m_s = -1/2 \) orbital, how many electrons are present?
Answer: For a principal quantum number \( n=4 \), there are \( n^2 \) orbitals in that shell.
\( \implies \) Number of orbitals \( = 4^2 = 16 \).
Each orbital can hold a maximum of two electrons, according to Pauli's exclusion principle. These two electrons have opposite spin quantum numbers: one with \( m_s = +1/2 \) and the other with \( m_s = -1/2 \). Since each of the 16 orbitals will have one electron with \( m_s = -1/2 \), there will be 16 electrons in total that have an \( m_s \) value of \( -1/2 \). This means half of the total electrons in the \( n=4 \) shell will have this spin value.
In simple words: If the main energy level is \( n=4 \), there are 16 different places (orbitals) where electrons can be. Each of these places can hold two electrons, one spinning up (\( +1/2 \)) and one spinning down (\( -1/2 \)). So, exactly half of the total electrons in the \( n=4 \) shell will have a spin of \( -1/2 \), which is 16 electrons.

🎯 Exam Tip: Always remember that each orbital can hold one electron with \( m_s = +1/2 \) and one electron with \( m_s = -1/2 \).

 

Question 10. Calculate the number of unpaired electrons in Fe and Cr.
Answer: To find the number of unpaired electrons, we write the electronic configuration for each element:
For Iron (Fe): The atomic number is \( Z = 26 \).
The electronic configuration is: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^6 \).
In the \( 3d \) subshell, there are 5 orbitals. According to Hund's rule, electrons will fill these orbitals singly first before pairing up.
The \( 3d^6 \) configuration means 5 orbitals are singly occupied and then one orbital gets a second electron. This leaves 4 orbitals with single (unpaired) electrons.
Thus, Iron (Fe) has 4 unpaired electrons.
For Chromium (Cr): The atomic number is \( Z = 24 \).
The electronic configuration is: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5 \). Chromium shows an exception to the Aufbau principle, where one electron from the 4s orbital moves to the 3d orbital to achieve a more stable half-filled 3d subshell.
In this configuration, the \( 4s \) orbital has 1 unpaired electron, and the \( 3d \) subshell has 5 orbitals, each with one unpaired electron.
Thus, Chromium (Cr) has \( 1 + 5 = 6 \) unpaired electrons.
In simple words: For Iron, the electrons in the \( 3d \) shell spread out as much as possible, leaving 4 electrons without a pair. For Chromium, it's special because one electron from the \( 4s \) shell moves to the \( 3d \) shell, making both the \( 4s \) and \( 3d \) shells half-full. This gives Chromium a total of 6 electrons that are all by themselves.

🎯 Exam Tip: Always remember Hund's rule for filling orbitals and be aware of exceptions to the Aufbau principle, especially for transition metals like Chromium and Copper, which achieve extra stability with half-filled or fully-filled d-orbitals.

 

Question 11. The energy of electrons in an orbital depends on which of the quantum number?
Answer: The energy of electrons in an orbital mainly depends on the principal quantum number (\( n \)) and the azimuthal (or angular momentum) quantum number (\( l \)). For single-electron systems (like hydrogen), the energy depends only on \( n \). However, for multi-electron atoms, it depends on both \( n \) and \( l \) (often expressed by the \( n+l \) rule), which determines the subshell's energy level. This means that an electron's energy is affected by its shell and subshell.
In simple words: How much energy an electron has in its orbital depends mostly on its main energy level (shell number, \( n \)) and also on the shape of its subshell (subshell number, \( l \)). For atoms with many electrons, both these numbers decide the electron's energy.

🎯 Exam Tip: For multi-electron atoms, higher values of \( n \) and \( l \) generally mean higher energy. Remember the \( n+l \) rule to compare the energies of different orbitals.

 

Question 13. Calculate the number of electrons in 1 gm atomic mass.
Answer: The mass of one electron is given as \( 9.10939 \times 10^{-31} \, \text{kg} \).
We need to find how many electrons are in 1 gram. First, convert 1 gram to kilograms:
\( 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \).
Number of electrons \( = (\text{total mass}) / (\text{mass of one electron}) \)
\( \implies \) Number of electrons \( = (1 \times 10^{-3} \, \text{kg}) / (9.10939 \times 10^{-31} \, \text{kg/electron}) \)
\( \implies \) Number of electrons \( = (1 / 9.10939) \times 10^{(-3 - (-31))} \)
\( \implies \) Number of electrons \( = 0.10978 \times 10^{28} \)
\( \implies \) Number of electrons \( = 1.0978 \times 10^{27} \).
Therefore, the number of electrons in 1 gm atomic mass is approximately \( 1.098 \times 10^{27} \) electrons. This calculation highlights the incredibly small mass of an individual electron.
In simple words: To find how many electrons make up 1 gram, we divide 1 gram (converted to kilograms) by the very tiny mass of just one electron. This shows that a huge number of electrons are needed to make up even a small mass like 1 gram.

🎯 Exam Tip: Always pay attention to units (grams vs. kilograms) and the correct handling of exponents in scientific notation.

 

Question 14. Determine the charges of 1 mole electrons
Answer: The charge on one electron is \( 1.6022 \times 10^{-19} \, \text{C} \) (Coulombs).
One mole of electrons contains Avogadro's number of electrons, which is \( 6.022 \times 10^{23} \).
To find the total charge of 1 mole of electrons, we multiply the charge of one electron by Avogadro's number:
\( \implies \) Total charge \( = (\text{charge of one electron}) \times (\text{Avogadro's number}) \)
\( \implies \) Total charge \( = (1.6022 \times 10^{-19} \, \text{C/electron}) \times (6.022 \times 10^{23} \, \text{electrons/mol}) \)
\( \implies \) Total charge \( = 9.648 \times 10^4 \, \text{C/mol} \).
The charge of 1 mole of electrons is approximately \( 9.65 \times 10^4 \, \text{Coulomb} \). This quantity is also known as 1 Faraday.
In simple words: We know the electrical charge of a single electron. To find the charge of one mole of electrons, we multiply the charge of one electron by the total number of electrons in one mole (Avogadro's number). This big number is also called one Faraday.

🎯 Exam Tip: Remember Avogadro's number and the elementary charge value. The product of these two constants is Faraday's constant, which is important in electrochemistry.

 

Question 15. The wavelength of yellow light emitted by a sodium lamp is 580 nm. Calculate its frequency.
Answer: We know that the relationship between wavelength (\( \lambda \)) and frequency (\( \nu \)) of light is given by the formula:
\( c = \lambda \times \nu \)
Where \( c \) is the velocity of light in vacuum, \( \lambda \) is the wavelength, and \( \nu \) is the frequency.
We can rearrange the formula to find frequency:
\( \nu = c / \lambda \)
Given values:
Velocity of light \( c = 3 \times 10^8 \, \text{m/s} \)
Wavelength \( \lambda = 580 \, \text{nm} \). First, convert nanometers to meters:
\( \lambda = 580 \times 10^{-9} \, \text{m} \)
Now, substitute the values into the formula:
\( \nu = (3 \times 10^8 \, \text{m/s}) / (580 \times 10^{-9} \, \text{m}) \)
\( \implies \nu = (3 / 580) \times 10^{(8 - (-9))} \)
\( \implies \nu = 0.005172 \times 10^{17} \)
\( \implies \nu = 5.172 \times 10^{14} \, \text{s}^{-1} \) or \( \text{Hz} \).
So, the frequency of the yellow light is approximately \( 5.17 \times 10^{14} \, \text{Hz} \).
In simple words: To find how fast the light waves are vibrating (frequency), we divide the speed of light by its wavelength. We first change the wavelength from nanometers to meters, and then use the simple formula to get the answer in Hertz.

🎯 Exam Tip: Always ensure consistent units in calculations; convert nanometers (nm) to meters (m) before using the speed of light in m/s.

 

Question 16. Calculate the number of protons and electrons in the following: \( ^{13}\text{C}_6 \), \( ^{16}\text{O}_8 \), \( ^{24}\text{Mg}_{12} \), \( ^{56}\text{Fe}_{26} \).
Answer: For a neutral atom, the atomic number (Z) gives both the number of protons and the number of electrons. The mass number (A) is the sum of protons and neutrons.
Here are the calculations for each given atom:
1. For \( ^{13}\text{C}_6 \) (Carbon):
Atomic number (Z) = 6
Mass number (A) = 13
Number of protons = Z = 6
Number of electrons = Z = 6 (since it's a neutral atom)
2. For \( ^{16}\text{O}_8 \) (Oxygen):
Atomic number (Z) = 8
Mass number (A) = 16
Number of protons = Z = 8
Number of electrons = Z = 8 (since it's a neutral atom)
3. For \( ^{24}\text{Mg}_{12} \) (Magnesium):
Atomic number (Z) = 12
Mass number (A) = 24
Number of protons = Z = 12
Number of electrons = Z = 12 (since it's a neutral atom)
4. For \( ^{56}\text{Fe}_{26} \) (Iron):
Atomic number (Z) = 26
Mass number (A) = 56
Number of protons = Z = 26
Number of electrons = Z = 26 (since it's a neutral atom)
In simple words: For any neutral atom, the number written at the bottom left (atomic number) tells you how many protons it has, and also how many electrons it has. The number at the top left is the total weight (mass number), which is protons plus neutrons.

🎯 Exam Tip: Always remember that the atomic number defines the element and, for neutral atoms, directly gives the number of protons and electrons.

 

Question 17. Identify the orbitals for the following quantum numbers:
\( n=1, l=0 \)
\( n=2, l=1 \)
\( n=4, l=3 \)
\( n=3, l=2 \)
Answer: The principal quantum number (\( n \)) indicates the main energy shell, and the azimuthal (or angular momentum) quantum number (\( l \)) indicates the subshell type (where \( l=0 \) for s, \( l=1 \) for p, \( l=2 \) for d, \( l=3 \) for f).
Let's identify the orbitals for each set of quantum numbers:
1. For \( n=1, l=0 \): This corresponds to the 1s orbital. This is the lowest energy orbital.
2. For \( n=2, l=1 \): This corresponds to the 2p orbital.
3. For \( n=4, l=3 \): This corresponds to the 4f orbital.
4. For \( n=3, l=2 \): This corresponds to the 3d orbital.
In simple words: The first number (\( n \)) tells you the main energy level, and the second number (\( l \)) tells you the type of orbital. When \( l=0 \), it's an 's' orbital; when \( l=1 \), it's a 'p' orbital; when \( l=2 \), it's a 'd' orbital; and when \( l=3 \), it's an 'f' orbital. Putting these together gives the orbital name.

🎯 Exam Tip: Memorize the correspondence between the \( l \) values and orbital designations (s, p, d, f) as this is fundamental for writing electronic configurations and understanding atomic structure.

 

Question 19. Who discovered protons and neutrons?
Answer: Protons were discovered by Eugen Goldstein. While Rutherford is credited with the discovery of the nucleus and naming the proton, Goldstein's experiments with canal rays in 1886 first indicated the existence of positively charged particles. Neutrons were discovered by James Chadwick in 1932. He proposed that the neutral particles in the nucleus, which account for the missing mass, were neutrons.
In simple words: Eugen Goldstein found protons, which are positively charged particles in an atom. James Chadwick found neutrons, which are neutral particles inside the atom's center.

🎯 Exam Tip: It's important to remember the key scientists associated with the discovery of atomic sub-particles: J.J. Thomson (electron), Eugen Goldstein (proton), and James Chadwick (neutron).

 

Question 20. What are the fundamental particles?
Answer: The fundamental particles that make up an atom are electrons, protons, and neutrons. These particles are the basic building blocks of all matter. Electrons carry a negative charge, protons carry a positive charge, and neutrons have no charge. Protons and neutrons reside in the nucleus, while electrons orbit the nucleus.
In simple words: The three most basic parts that make up every atom are electrons (which are negative), protons (which are positive), and neutrons (which have no charge).

🎯 Exam Tip: Understand the charge and location of each fundamental particle within an atom, as this is crucial for explaining atomic structure and chemical behavior.

RBSE Class 11 Chemistry Chapter 2 Short Answer Type Questions

 

Question 21. Calculate radius of fifth Bohr orbital of hydrogen
Answer: The radius of Bohr's \( n^{\text{th}} \) orbit for a hydrogen atom is given by the formula:
\( r_n = (0.0529 \, \text{nm}) n^2 \)
Here, \( r_n \) is the radius of the \( n^{\text{th}} \) orbit, and \( n \) is the principal quantum number.
We need to calculate the radius for the fifth Bohr orbital, so \( n = 5 \).
Substitute \( n=5 \) into the formula:
\( r_5 = (0.0529 \, \text{nm}) \times (5)^2 \)
\( r_5 = (0.0529 \, \text{nm}) \times 25 \)
\( r_5 = 1.3225 \, \text{nm} \)
Therefore, the radius of the fifth Bohr orbital of hydrogen is \( 1.3225 \, \text{nm} \). The radius of an orbit increases significantly as the principal quantum number increases.
In simple words: To find the size of the fifth orbit in a hydrogen atom, we use a special formula. We multiply a small number (0.0529 nm) by the square of the orbit number (which is 5). This calculation gives us the radius, showing that higher orbits are much larger.

🎯 Exam Tip: Memorize Bohr's radius formula for hydrogen-like species. Remember that the radius increases quadratically with \( n \), so higher orbits are much larger.

 

Question 23. What will be the angular momentum of an electron is 5th energy state, according to Bohr's model?
Answer: According to Bohr's model, the angular momentum of an electron in a specific orbit is quantized. It can only take on certain discrete values. The formula for the angular momentum (L) of an electron in the \( n^{\text{th}} \) orbit is:
\( L = n \frac{h}{2\pi} \)
Where \( n \) is the principal quantum number (the energy state or orbit number) and \( h \) is Planck's constant.
For the 5th energy state, \( n = 5 \).
Substitute \( n=5 \) into the formula:
\( L = 5 \frac{h}{2\pi} \)
This can also be written as \( L = 2.5 \frac{h}{\pi} \).
So, the angular momentum of an electron in the 5th energy state is \( 5h/2\pi \) or \( 2.5h/\pi \). Each orbit has a unique, fixed value for angular momentum.
In simple words: Bohr's model says that electrons in specific orbits have a fixed amount of "spinning" energy, called angular momentum. For the fifth energy level, you just put 5 into the formula \( n \frac{h}{2\pi} \). This gives you \( 5h/2\pi \).

🎯 Exam Tip: Understand that angular momentum in Bohr's model is quantized, meaning it only exists in integer multiples of \( h/2\pi \).

 

Question 24. De broglie wavelength of an object having mass 'x' and moving with velocity 100 ms\(^{-1}\) is 6.62 × 10\(^{-35}\) m. Find the value of x
Answer: According to de Broglie's hypothesis, the wavelength (\( \lambda \)) of a particle is related to its momentum by the formula:
\( \lambda = \frac{h}{mv} \)
Where \( h \) is Planck's constant, \( m \) is the mass of the object, and \( v \) is its velocity.
We are given:
Wavelength \( \lambda = 6.62 \times 10^{-35} \, \text{m} \)
Velocity \( v = 100 \, \text{m/s} \)
Planck's constant \( h = 6.6 \times 10^{-34} \, \text{Js} \)
We need to find the mass \( x \) (i.e., \( m \)). Rearrange the formula to solve for \( m \):
\( m = \frac{h}{\lambda v} \)
Now, substitute the given values:
\( m = \frac{6.6 \times 10^{-34} \, \text{Js}}{(6.62 \times 10^{-35} \, \text{m}) \times (100 \, \text{m/s})} \)
\( m = \frac{6.6 \times 10^{-34}}{6.62 \times 10^{-35} \times 10^2} \)
\( m = \frac{6.6 \times 10^{-34}}{6.62 \times 10^{-33}} \)
\( m = (6.6 / 6.62) \times 10^{(-34 - (-33))} \)
\( m \approx 0.9969 \times 10^{-1} \)
\( m \approx 0.09969 \, \text{kg} \).
Therefore, the value of \( x \) (mass) is approximately \( 0.0997 \, \text{kg} \). This shows that objects with everyday speeds but very small masses can still have a measurable de Broglie wavelength.
In simple words: We are given the de Broglie wavelength of an object, its speed, and Planck's constant. We use the de Broglie formula, \( \lambda = h / mv \), and rearrange it to find the mass. We put in the numbers and calculate the mass, which is a small amount in kilograms.

🎯 Exam Tip: Ensure that all units are consistent (e.g., meters, kilograms, seconds) before performing calculations using de Broglie's equation.

 

Question 25. If uncertainty in momentum of an electron is 1 × 10\(^{-5}\) kg ms\(^{-1}\), then what will be the uncertainity in position?
Answer: According to Heisenberg's Uncertainty Principle, it is impossible to know both the exact position and exact momentum of a particle simultaneously. The relationship is given by:
\( \Delta x \times \Delta p \ge \frac{h}{4\pi} \)
Where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( h \) is Planck's constant.
We need to find the uncertainty in position (\( \Delta x \)). So, we can write the minimum uncertainty as:
\( \Delta x = \frac{h}{4\pi \Delta p} \)
Given values:
Uncertainty in momentum \( \Delta p = 1 \times 10^{-5} \, \text{kg m/s} \)
Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
\( \pi \approx 3.14 \)
Substitute the values into the formula:
\( \Delta x = \frac{6.626 \times 10^{-34} \, \text{Js}}{4 \times 3.14 \times (1 \times 10^{-5} \, \text{kg m/s})} \)
\( \Delta x = \frac{6.626 \times 10^{-34}}{12.56 \times 10^{-5}} \)
\( \Delta x = (6.626 / 12.56) \times 10^{(-34 - (-5))} \)
\( \Delta x = 0.5275 \times 10^{-29} \)
\( \Delta x = 5.275 \times 10^{-30} \, \text{m} \)
Therefore, the uncertainty in position is approximately \( 5.28 \times 10^{-30} \, \text{m} \). This extremely small uncertainty value highlights the quantum nature of particles like electrons.
In simple words: Heisenberg's rule says we can't know an electron's exact position and exact speed at the same time. If we know the uncertainty in its push (momentum), we can find the minimum uncertainty in its location using a specific formula. The answer shows that if we know the momentum fairly well, the position becomes very uncertain, but still a very tiny number.

🎯 Exam Tip: Remember the Heisenberg Uncertainty Principle equation. Pay attention to the constants \( h \) and \( \pi \) and ensure correct handling of scientific notation.

 

Question 27. Calculate the value of all four quantum numbers for valence electron of rubidium, (Z = 37)
Answer: Rubidium (Rb) has an atomic number \( Z = 37 \). To find the quantum numbers for its valence electron, we first write its electronic configuration:
\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 5s^1 \)
The valence electron is the outermost electron, which in this case is in the \( 5s^1 \) orbital.
Now, let's determine the four quantum numbers for this electron:
1. Principal quantum number (\( n \)): This indicates the main energy shell. For \( 5s^1 \), \( n = 5 \).
2. Azimuthal (or angular momentum) quantum number (\( l \)): This indicates the subshell type. For an s-orbital, \( l = 0 \).
3. Magnetic quantum number (\( m_l \)): This specifies the orientation of the orbital in space. For \( l = 0 \) (s-orbital), there is only one possible value for \( m_l \), which is \( 0 \).
4. Spin quantum number (\( m_s \)): This describes the intrinsic angular momentum (spin) of the electron. Since there is only one electron in the \( 5s \) orbital, it can be either spin-up or spin-down. By convention, we often assign the first electron in an orbital as spin-up.
\( m_s = +1/2 \) (or \( -1/2 \), as it's the only electron).
Therefore, the four quantum numbers for the valence electron of Rubidium are \( n=5, l=0, m_l=0, m_s=+1/2 \).
In simple words: Rubidium has 37 electrons. We fill them into orbitals until we find the very last electron, which is in the \( 5s \) orbital. For this electron, the main energy level (\( n \)) is 5, the orbital type (\( l \)) is 0 (for 's'), its position in space (\( m_l \)) is 0, and its spin (\( m_s \)) is \( +1/2 \). These four numbers describe exactly where this electron is.

🎯 Exam Tip: To correctly assign quantum numbers, always start by writing the full electronic configuration to identify the valence electron's orbital.

 

Question 28. Which property is depicted by three quantum number n, l and m?
Answer: The three quantum numbers \( n \), \( l \), and \( m_l \) describe various properties of an electron in an atom:
* **Principal Quantum Number (\( n \)):** This number describes the main energy shell or energy level of an electron. It indicates the probable distance of the electron from the nucleus. Higher \( n \) values mean higher energy and larger orbitals.
* **Azimuthal (or Angular Momentum) Quantum Number (\( l \)):** Also known as the orbital quantum number, this describes the subshell within an energy shell (e.g., s, p, d, f) and determines the shape of the orbital. It also gives the magnitude of the orbital angular momentum.
* **Magnetic Quantum Number (\( m_l \)):** This number describes the orientation of the orbital in space. For a given \( l \), there are \( 2l+1 \) possible \( m_l \) values, which correspond to the different spatial arrangements of orbitals (e.g., \( p_x, p_y, p_z \)).
These three numbers together define a specific orbital and the energy, shape, and spatial orientation of an electron's wave function within an atom.
In simple words: The first number (\( n \)) tells you the electron's main energy level and how far it is from the center. The second number (\( l \)) tells you the shape of the electron's path (like a ball or a dumbbell). The third number (\( m_l \)) tells you which way that shape is pointed in space. Together, they describe where an electron is likely to be found.

🎯 Exam Tip: Clearly differentiate the role of each quantum number. \( n \) (size/energy), \( l \) (shape), \( m_l \) (orientation), and \( m_s \) (spin).

 

Question 29. Which elements contain six unpaired electrons according to Hund's Rules?
Answer: According to Hund's rule of maximum multiplicity, electrons fill degenerate orbitals (orbitals of the same energy) singly with parallel spins before any orbital is doubly occupied. This maximizes the total spin and leads to a more stable configuration.
We are looking for an element with six unpaired electrons. Let's examine the electronic configuration of Chromium (Cr), which has an atomic number \( Z = 24 \).
The electronic configuration of Chromium is \( [Ar] \, 3d^5 \, 4s^1 \). This is an exception to the Aufbau principle, as a half-filled d-subshell and a half-filled s-subshell provide extra stability.
In this configuration:
* The \( 4s \) orbital has 1 electron (which is unpaired).
* The \( 3d \) subshell has 5 orbitals, each containing 1 electron (all 5 are unpaired).
So, in total, Chromium has \( 1 + 5 = 6 \) unpaired electrons. This stable configuration helps explain many of chromium's chemical properties.
In simple words: Hund's rule says electrons like to spread out in separate slots before pairing up. Chromium is a special atom that ends up with six electrons all by themselves (unpaired). This happens because its outer \( 4s \) and \( 3d \) shells prefer to be half-full for extra stability.

🎯 Exam Tip: Remember Hund's rule for filling degenerate orbitals. Also, be aware of the special electronic configurations of elements like Chromium and Copper that achieve extra stability through half-filled or fully-filled d-orbitals.

 

Question 30. What will be the orbital angular momentum for d-electron?
Answer: The orbital angular momentum (L) of an electron is determined by the azimuthal quantum number (\( l \)). The formula for orbital angular momentum is:
\( L = \sqrt{l(l+1)} \frac{h}{2\pi} \)
For a d-electron, the azimuthal quantum number \( l = 2 \).
Now, substitute \( l=2 \) into the formula:
\( L = \sqrt{2(2+1)} \frac{h}{2\pi} \)
\( L = \sqrt{2(3)} \frac{h}{2\pi} \)
\( L = \sqrt{6} \frac{h}{2\pi} \)
Therefore, the orbital angular momentum for a d-electron is \( \sqrt{6} \frac{h}{2\pi} \). This value is characteristic for any electron in a d-orbital, regardless of its principal quantum number.
In simple words: The "spinning" energy of an electron as it moves around the nucleus (called orbital angular momentum) depends on what type of orbital it's in. For a 'd' electron, we use a specific formula and replace 'l' with 2. This gives us the value \( \sqrt{6} \frac{h}{2\pi} \).

🎯 Exam Tip: Distinguish between orbital angular momentum (related to \( l \)) and spin angular momentum (related to \( s \)). Memorize the formula \( L = \sqrt{l(l+1)} \frac{h}{2\pi} \).

 

Question 31. Calculate number of electrons, protons and neutrons in element \( ^{89}\text{Y}_{39} \).
Answer: For the element \( ^{89}\text{Y}_{39} \) (Yttrium):
The atomic number (Z) is the subscript, which is 39.
The mass number (A) is the superscript, which is 89.
Now, we can calculate the number of protons, electrons, and neutrons:
1. **Number of Protons:** The atomic number (Z) directly gives the number of protons.
Number of protons \( = Z = 39 \).
2. **Number of Electrons:** For a neutral atom, the number of electrons is equal to the number of protons.
Number of electrons \( = Z = 39 \).
3. **Number of Neutrons:** The number of neutrons can be found by subtracting the atomic number from the mass number.
Number of neutrons \( = A - Z = 89 - 39 = 50 \).
So, element \( ^{89}\text{Y}_{39} \) has 39 protons, 39 electrons, and 50 neutrons.
In simple words: For the element Yttrium, the bottom number (39) tells us it has 39 protons and 39 electrons. The top number (89) is the total weight. If we subtract the protons (39) from the total weight (89), we find that it has 50 neutrons.

🎯 Exam Tip: Clearly identify the atomic number (Z) and mass number (A) from the isotopic notation. Remember Z = protons = electrons (for neutral atoms) and A - Z = neutrons.

 

Question 32. What will be the ratio between areas covered by first and second orbits?
Answer: According to Bohr's model, the radius of an orbit (\( r_n \)) is proportional to the square of the principal quantum number (\( n \)).
\( r_n \propto n^2 \)
The area covered by a circular orbit (A) is given by \( A = \pi r^2 \).
Since \( r_n \propto n^2 \), then \( r_n = k n^2 \) for some constant \( k \).
So, the area \( A_n = \pi (k n^2)^2 = \pi k^2 n^4 \).
This means the area of an orbit is proportional to \( n^4 \).
\( A_n \propto n^4 \)
For the first orbit, \( n_1 = 1 \), so \( A_1 \propto (1)^4 = 1 \).
For the second orbit, \( n_2 = 2 \), so \( A_2 \propto (2)^4 = 16 \).
The ratio of the area of the first orbit to the second orbit is:
\( \frac{A_1}{A_2} = \frac{n_1^4}{n_2^4} = \frac{1^4}{2^4} = \frac{1}{16} \)
Therefore, the ratio between the areas covered by the first and second orbits is \( 1:16 \).
In simple words: The size of an electron's orbit grows much faster than its number. The area of an orbit is proportional to the orbit number raised to the power of four. So, the second orbit's area will be \( 2 \times 2 \times 2 \times 2 = 16 \) times larger than the first orbit's area.

🎯 Exam Tip: Remember the proportional relationships in Bohr's model: radius \( \propto n^2 \) and energy \( \propto 1/n^2 \). Extend this to area by \( A = \pi r^2 \).

 

Question 33. What is the reason of presence of three unpaired electrons in N-atom?
Answer: The presence of three unpaired electrons in a Nitrogen (N) atom is explained by Hund's rule of maximum multiplicity.
Nitrogen has an atomic number \( Z = 7 \). Its electronic configuration is:
\( 1s^2 \, 2s^2 \, 2p^3 \)
The outermost shell is the 2nd shell, which contains the \( 2s \) and \( 2p \) subshells.
* The \( 2s \) subshell has 1 orbital and contains 2 electrons, which are paired.
* The \( 2p \) subshell has 3 degenerate orbitals (i.e., orbitals of equal energy): \( 2p_x, 2p_y, 2p_z \).
According to Hund's rule, electrons will occupy each of these degenerate \( 2p \) orbitals singly with parallel spins before any pairing occurs. Since there are three electrons in the \( 2p \) subshell (\( 2p^3 \)), each of these three electrons will go into a different \( 2p \) orbital.
So, the configuration for the \( 2p^3 \) subshell is \( 2p_x^1 \, 2p_y^1 \, 2p_z^1 \). This results in three unpaired electrons. This configuration is very stable due to the half-filled p-subshell.
In simple words: Nitrogen has 7 electrons. The last three electrons go into the \( 2p \) shell. Because of Hund's rule, these three electrons spread out into three different \( p \) "slots" instead of crowding into one. This makes them all single (unpaired) and gives the atom extra stability.

🎯 Exam Tip: Hund's rule is key to understanding the number of unpaired electrons. Always fill degenerate orbitals singly first before pairing them up.

 

Question 34. Determine the number of radial nodes of 3s and 2p orbitals.
Answer: The number of radial nodes in an orbital can be determined using the formula:
Number of radial nodes \( = n - l - 1 \)
Where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.
1. For the **3s orbital**:
Here, \( n = 3 \) (from '3s') and \( l = 0 \) (for an 's' orbital).
Number of radial nodes \( = 3 - 0 - 1 = 2 \).
So, the 3s orbital has 2 radial nodes.
2. For the **2p orbital**:
Here, \( n = 2 \) (from '2p') and \( l = 1 \) (for a 'p' orbital).
Number of radial nodes \( = 2 - 1 - 1 = 0 \).
So, the 2p orbital has 0 radial nodes.
Radial nodes are spherical regions where the probability of finding an electron is zero. The presence of these nodes helps characterize the shape and complexity of orbitals.
In simple words: Radial nodes are like empty layers within an electron cloud where you won't find the electron. To find them, we subtract 1 from the main energy level (\( n \)) and then subtract the orbital type number (\( l \)). For a 3s orbital, it has 2 nodes. For a 2p orbital, it has no radial nodes.

🎯 Exam Tip: Remember the formula for radial nodes. Also, be aware of angular nodes, which are given by \( l \), and total nodes, which are \( n-1 \).

 

Question 35. If \( (n + l) = 6 \), then what will be the possible number of the sub-shells?
Answer: We are given that \( (n+l) = 6 \). We need to find the possible combinations of \( n \) and \( l \) that satisfy this condition, keeping in mind that \( n \) must be a positive integer (\( 1, 2, 3, \ldots \)) and \( l \) can range from \( 0 \) to \( n-1 \).
Let's list the possible combinations:
1. If \( n = 6 \): Then \( l = 6 - n = 6 - 6 = 0 \).
This corresponds to the \( 6s \) subshell. (Here, \( l=0 \) is less than \( n-1=5 \), so it's valid).
2. If \( n = 5 \): Then \( l = 6 - n = 6 - 5 = 1 \).
This corresponds to the \( 5p \) subshell. (Here, \( l=1 \) is less than \( n-1=4 \), so it's valid).
3. If \( n = 4 \): Then \( l = 6 - n = 6 - 4 = 2 \).
This corresponds to the \( 4d \) subshell. (Here, \( l=2 \) is less than \( n-1=3 \), so it's valid).
4. If \( n = 3 \): Then \( l = 6 - n = 6 - 3 = 3 \).
This corresponds to the \( 3f \) subshell. (Here, \( l=3 \) is equal to \( n-1=2 \). *Correction*: \( l \) must be *less than* \( n \). Here, \( l=3 \) is *equal to* \( n \). This combination is **not possible** because \( l \) cannot be equal to or greater than \( n \). It must be \( l \le n-1 \)).
Therefore, the possible subshells are \( 6s, 5p, \) and \( 4d \). There are 3 such possible subshells.
In simple words: We are looking for combinations of main energy level (\( n \)) and orbital type (\( l \)) that add up to 6. We list all options, like \( 6s, 5p, 4d \). We must remember that the orbital type number (\( l \)) can never be as big as or bigger than the main energy level number (\( n \)). Based on this, there are three possible subshells.

🎯 Exam Tip: Always verify that the calculated \( l \) value is less than \( n \) (\( l \le n-1 \)). This is a common point of error when determining valid quantum number combinations.

RBSE Class 11 Chemistry Chapter 2 Long Answer Type Questions

 

Question 36. Write the postulates of Bohr's Model. What are the demerits of the Bohr's Model?
Answer: Niels Bohr (1913) developed an atomic model that successfully explained the spectrum of the hydrogen atom and overcame some limitations of Rutherford's model. His model was based on Planck's quantum theory and radiation principles.
**Postulates of Bohr's Model:**
1. **Stationary Orbits:** Electrons revolve around the nucleus in specific circular paths called stationary orbits or shells. These orbits have fixed energy levels and are denoted by integers (\( n=1, 2, 3, \ldots \)) or letters (K, L, M, N, \ldots). As long as an electron stays in a particular orbit, it does not lose or gain energy.
2. **Quantized Angular Momentum:** An electron can only revolve in those orbits for which its angular momentum is an integral multiple of \( h/2\pi \).
\( L = m_e v r = n \frac{h}{2\pi} \), where \( n = 1, 2, 3, \ldots \)
3. **Energy Absorption/Emission:** Energy changes occur only when an electron jumps from one stationary orbit to another.
* When an electron absorbs energy, it jumps to a higher energy orbit (excitation).
* When an electron falls from a higher energy orbit to a lower energy orbit, it emits energy in the form of electromagnetic radiation (a photon).
4. **Frequency Condition:** The energy of the emitted or absorbed radiation is equal to the difference in energy between the two orbits. This is known as Bohr's frequency rule:
\( \Delta E = E_2 - E_1 = h\nu \)
\( \implies \nu = \frac{E_2 - E_1}{h} \)
**Demerits of Bohr's Model:**
1. **Limited to Hydrogen:** Bohr's model could only explain the spectra of hydrogen and hydrogen-like ions (e.g., \( \text{He}^+, \text{Li}^{2+} \)) which have only one electron. It failed to explain the spectra of multi-electron atoms.
2. **Fine Structure:** It could not explain the fine structure of spectral lines (i.e., when observed with high-resolution spectroscopes, single spectral lines appear as a group of closely spaced lines). This fine structure is explained by the Zeeman effect (splitting of spectral lines in a magnetic field) and the Stark effect (splitting in an electric field), which Bohr's model could not account for.
3. **Arbitrary Quantization:** The postulate of quantized angular momentum (\( L = n \frac{h}{2\pi} \)) was proposed arbitrarily without theoretical justification.
4. **No Wave Nature:** Bohr's model treated electrons as particles moving in well-defined orbits. It did not consider the wave-particle duality of matter, which was later proposed by de Broglie.
5. **Uncertainty Principle:** It violated Heisenberg's Uncertainty Principle, which states that both the position and momentum of an electron cannot be known simultaneously with absolute certainty.
Despite its limitations, Bohr's model was a crucial step in the development of quantum mechanics, introducing the concept of quantized energy levels.
*In simple words: Bohr's model said electrons orbit the atom's center in fixed paths with fixed energy, like planets around the sun. They only change energy when jumping between these paths. But this model only worked for simple atoms like hydrogen and couldn't explain why some light lines split into even finer lines when looked at closely or why electrons also act like waves. Also, it wrongly assumed we could know an electron's exact position and speed at the same time.*

🎯 Exam Tip: When listing postulates, focus on quantization of energy and angular momentum. For demerits, emphasize its failure for multi-electron atoms and its inability to explain fine spectral structures or wave-particle duality.

 

Question 37. Explain Planck's Quantum theory.
Answer: Max Planck (1900) proposed the Quantum Theory of Radiation to explain the emission and absorption of energy. This theory marked a significant departure from classical physics.
**Main Postulates of Planck's Quantum Theory:**
1. **Quantized Energy:** Energy is not emitted or absorbed continuously but in discrete packets or bundles. These small bundles of energy are called 'quanta'. In the case of light, these quanta are known as 'photons'. This idea was revolutionary because classical physics assumed continuous energy exchange.
2. **Proportionality to Frequency:** The energy of each quantum (or photon) is directly proportional to the frequency (\( \nu \)) of the radiation.
\( E \propto \nu \)
\( \implies E = h\nu \)
Where \( E \) is the energy of a quantum, \( \nu \) is the frequency, and \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)). This constant links the energy of a photon to its frequency.
3. **Integral Multiples:** Any body can emit or absorb energy only in whole number multiples of a quantum. This means the total energy (E) emitted or absorbed is an integral multiple of \( h\nu \).
\( E = nh\nu \)
Where \( n \) is a positive integer (\( 1, 2, 3, \ldots \)).
Planck's quantum theory was groundbreaking and laid the foundation for quantum mechanics. It successfully explained phenomena like black-body radiation and the photoelectric effect, which classical physics could not.
In simple words: Planck's theory states that energy is not smooth and continuous but comes in tiny, fixed packets called 'quanta' (or 'photons' for light). The amount of energy in each packet depends on its vibration speed (frequency) multiplied by Planck's constant. Also, energy can only be exchanged in whole numbers of these packets, never in bits and pieces.

🎯 Exam Tip: The core idea of Planck's theory is the quantization of energy. Remember the key formula \( E = h\nu \) and the definition of 'quanta' or 'photons'.

 

Question 38. What is Rutherford's Atomic Model? What are the reasons of failure?
Answer: Ernest Rutherford, through his gold foil experiment (1911), proposed a new model for the atom known as the Rutherford's Nuclear Model or Planetary Model of Atom.
**Main Features of Rutherford's Atomic Model:**
1. **Nucleus at the Center:** An atom consists of a tiny, dense, positively charged nucleus at its center. This nucleus contains almost all the mass of the atom.
2. **Empty Space:** Most of the space within an atom is empty. This was inferred from the fact that most alpha particles passed straight through the gold foil without deflection.
3. **Electrons Orbit the Nucleus:** Electrons, which are negatively charged, revolve around the nucleus in circular paths, much like planets orbit the sun.
4. **Electrostatic Attraction:** The strong electrostatic force of attraction between the positively charged nucleus and the negatively charged electrons holds the electrons in their orbits.
5. **Atom is Electrically Neutral:** The total positive charge on the nucleus is balanced by the total negative charge of the electrons, making the atom electrically neutral.
**Reasons for Failure of Rutherford's Model:**
1. **Stability of the Atom:** According to classical electromagnetic theory, a moving charged particle (like an electron) in a circular orbit should continuously emit radiation. If an electron continuously loses energy, its orbit should continuously shrink, and it should eventually spiral into the nucleus. This would make the atom unstable, but atoms are known to be stable. Rutherford's model could not explain this stability.
2. **Line Spectrum:** If electrons continuously emit radiation, the spectrum of an atom should be continuous (like a rainbow), covering all wavelengths. However, atoms are known to produce discrete line spectra (specific wavelengths of light). Rutherford's model could not explain the origin or nature of these line spectra.
Rutherford's model was a significant improvement over previous models, introducing the concept of a nucleus, but its inability to explain atomic stability and discrete spectra led to further developments in atomic theory.
In simple words: Rutherford's model said an atom has a tiny, heavy center (nucleus) with positive charge, and light electrons circle it like planets. Most of the atom is empty space. But this model couldn't explain why atoms are stable and don't collapse (because electrons should lose energy and fall into the nucleus). It also couldn't explain why atoms emit light in specific colors, not a continuous rainbow.

🎯 Exam Tip: Focus on the two main failures: inability to explain atomic stability (electron spiraling into the nucleus) and the inability to explain discrete line spectra.

 

Question 39. Write short notes on:
(a) Dual nature of Particle de-Broglie Hypothesis
(b) Heisenberg's Uncertainty Principle
(c) Quantum number and Orbitals
(d) Shapes and Orbitals
Answer:
**(a) Dual Nature of Particle de-Broglie Hypothesis:**
In 1924, Louis de Broglie proposed that, similar to light, matter also exhibits dual behavior, meaning it can behave as both a particle and a wave. This is known as the wave-particle duality. For any moving particle, a wave is associated with it, and its wavelength (\( \lambda \)) is inversely proportional to its momentum (\( p \)).
The de Broglie wavelength is given by the equation:
\( \lambda = \frac{h}{p} = \frac{h}{mv} \)
Where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity. This wave nature is significant only for microscopic particles like electrons, protons, and atoms, because their masses are so small that their wavelengths are measurable. For macroscopic objects (like a baseball), their mass is so large that their de Broglie wavelength is too tiny to be observed, making their wave nature negligible. This hypothesis was later experimentally confirmed by Davisson and Germer.
**(b) Heisenberg's Uncertainty Principle:**
This principle, formulated by Werner Heisenberg in 1927, states that it is fundamentally impossible to simultaneously know with perfect accuracy both the position and the momentum (mass times velocity) of a microscopic particle, such as an electron. The more precisely one property is measured, the less precisely the other can be known. Mathematically, it is expressed as:
\( \Delta x \times \Delta p \ge \frac{h}{4\pi} \)
Where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( h \) is Planck's constant. This principle arises from the wave nature of matter and the interaction between the measuring instrument and the particle. It is a fundamental concept in quantum mechanics and explains why electrons cannot have well-defined orbits like planets.
**(c) Quantum Number and Orbitals:**
Quantum numbers are a set of four numbers that completely describe the state of an electron in an atom. They provide information about the electron's energy, location, the type of orbital it occupies, and its orientation in space.
The four quantum numbers are:
1. **Principal Quantum Number (\( n \)):** Defines the electron shell or energy level and the size of the orbital. Values: \( 1, 2, 3, \ldots \).
2. **Azimuthal (Angular Momentum) Quantum Number (\( l \)):** Defines the subshell (s, p, d, f) and the shape of the orbital. Values: \( 0, 1, 2, \ldots, (n-1) \).
3. **Magnetic Quantum Number (\( m_l \)):** Defines the orientation of the orbital in space. Values: \( -l, \ldots, 0, \ldots, +l \).
4. **Spin Quantum Number (\( m_s \)):** Describes the intrinsic angular momentum or spin of the electron. Values: \( +1/2 \) or \( -1/2 \).
An atomic orbital is a mathematical function that describes the wave-like behavior of an electron in an atom. It defines a region around the nucleus where there is a high probability of finding an electron. Each orbital can hold a maximum of two electrons with opposite spins, as per the Pauli exclusion principle.
**(d) Shapes and Orbitals:**
The shapes of atomic orbitals are primarily determined by the azimuthal quantum number (\( l \)). The different values of \( l \) correspond to different types of subshells (s, p, d, f) with distinct spatial forms.
* **s-orbitals (\( l=0 \)):** These are spherically symmetric, meaning they are shaped like a hollow ball around the nucleus. The probability of finding an electron is the same in all directions at a given distance from the nucleus. As \( n \) increases (e.g., from 1s to 2s to 3s), the s-orbital becomes larger and contains radial nodes.
* **p-orbitals (\( l=1 \)):** There are three p-orbitals (\( p_x, p_y, p_z \)) for each principal quantum number \( n \ge 2 \). Each p-orbital has a dumbbell shape, consisting of two lobes on opposite sides of the nucleus, with a nodal plane passing through the nucleus. The \( p_x, p_y, p_z \) orbitals are oriented along the x, y, and z axes, respectively.
* **d-orbitals (\( l=2 \)):** There are five d-orbitals (\( d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, d_{z^2} \)) for each \( n \ge 3 \). Four of these ( \( d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2} \)) have a cloverleaf or double-dumbbell shape, while the \( d_{z^2} \) orbital has a unique dumbbell shape with a donut-shaped ring around the middle.
* **f-orbitals (\( l=3 \)):** There are seven f-orbitals for \( n \ge 4 \). These have even more complex shapes with multiple lobes, making them difficult to visualize.
The specific shapes of these orbitals arise from the mathematical solutions of the Schrödinger wave equation and determine how electrons will interact with other atoms during chemical bonding.
In simple words: De Broglie said that tiny particles, like electrons, act like both little balls and waves at the same time. Heisenberg's rule says you can't perfectly know both where an electron is and how fast it's moving at the same time. Quantum numbers are like an address system (four numbers) that tells you everything about an electron in an atom. And finally, these numbers also describe the specific shapes of the electron's cloud-like home (orbital), which can be like a ball (s-orbital) or a dumbbell (p-orbital) or more complex shapes (d- and f-orbitals).

🎯 Exam Tip: For the de Broglie hypothesis, highlight the equation and its relevance for microscopic particles. For Heisenberg, state the principle and its mathematical form. For quantum numbers, list each and its significance. For orbital shapes, describe the s and p orbitals, and generally mention d and f are more complex.

 

Question 39. Write short notes on:
(a) Dual nature of Particle de-Broglie Hypothesis
(b) Heisenberg's Uncertainty Principle
(c) Quantum number and Orbitals
(d) Shapes and Orbitals
Answer:
(a) **Dual Nature of Particle de-Broglie Hypothesis:** In 1924, Louis de Broglie suggested that matter can act as both a particle and a wave. This means it has both momentum and wavelength. According to de Broglie, the wavelength \( (\lambda) \) of a particle with mass 'm' moving at velocity 'v' is given by the equation: \( \lambda = \frac{h}{mv} \). Here, \( h \) is Planck's constant. This equation is useful for very small, microscopic particles. We can also relate energy with frequency \( E = h\nu \) or \( E = \frac{hc}{\lambda} \) from Planck's quantum theory and \( E = mc^2 \) from Einstein's mass-energy equivalence. Combining these ideas, we can derive the de Broglie equation \( \lambda = \frac{h}{mc} \) or \( \lambda = \frac{h}{mv} \).
(b) **Heisenberg's Uncertainty Principle:** Werner Heisenberg proposed this principle in 1927. It states that it is impossible to precisely measure both the position and momentum of a microscopic particle at the same time. If you measure one with very high accuracy, the other quantity becomes less certain. This principle shows that there are fundamental limits to how much we can know about tiny particles simultaneously.
(c) **Quantum number and Orbitals:** Quantum numbers are a set of four values that give complete information about all electrons in an atom. They describe an electron's location, energy, the type of orbital it occupies, and its shape and orientation. The four quantum numbers are described below:

🎯 Exam Tip: Remember to use simple analogies for complex concepts like dual nature and uncertainty, as these help in better understanding and recall.

 

Answer: (Continued from (c))

(d) **Shapes and Orbitals:** An orbital is a mathematical function that describes the wave-like behavior of an electron. It is also called an atomic orbital or electron orbital. There are four main types of orbitals: s, p, d, and f.
**s-orbitals:** An s-orbital is spherical in shape, meaning it looks like a hollow ball around the nucleus. It has no direction preference.
**p-orbitals:** P-orbitals have a dumbbell shape. They are oriented along the x, y, and z axes, so there are three p-orbitals: \( p_x, p_y, \) and \( p_z \).
**d-orbitals:** D-orbitals have more complex shapes, typically double dumbbell shapes, except for \( d_{z^2} \). There are five d-orbitals: \( d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, \) and \( d_{z^2} \).
In simple words: Quantum numbers are like special codes that tell us everything about an electron in an atom - where it is, how much energy it has, and what shape its space looks like. Orbitals are the actual shapes these electrons are likely to be found in, like spherical for 's' or dumbbell for 'p'.

🎯 Exam Tip: When describing quantum numbers, focus on what each one *determines* (e.g., n for energy level, l for shape) rather than just listing its possible values.

 

Numerical Problems

 

Question 40. The wavelength of yellow light emitted by sodium lamp is 580 nm. Calculate its frequency and wave number \( (\overline{\nu}) \).
Answer: We know that the relationship between wavelength \( (\lambda) \) and frequency \( (\nu) \) is given by: \( c = \nu \lambda \).
So, frequency \( \nu = \frac{c}{\lambda} \).
Given:
Velocity of light in vacuum \( c = 3 \times 10^8 \text{ m/s} \)
Wavelength of yellow light \( \lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m} \)
Substituting the values, we get:
\( \nu = \frac{3 \times 10^8 \text{ m/s}}{580 \times 10^{-9} \text{ m}} \)
\( \nu = 5.17 \times 10^{14} \text{ s}^{-1} \)
Now, the wave number \( (\overline{\nu}) \) of yellow light is given by: \( \overline{\nu} = \frac{1}{\lambda} \)
Substituting the value of \( \lambda \):
\( \overline{\nu} = \frac{1}{580 \times 10^{-9} \text{ m}} \)
\( \overline{\nu} = 1.72 \times 10^6 \text{ m}^{-1} \). Wave number tells us how many waves fit into a certain length.
In simple words: To find how many waves pass a point per second (frequency), we divide the speed of light by the wavelength. To find the wave number, which is how many waves fit into one meter, we simply take one divided by the wavelength.

🎯 Exam Tip: Always convert all units to SI units (meters, seconds) before starting calculations to avoid errors.

 

Question 41. Calculate the energy of photon of light having frequency \( 3 \times 10^{15} \) Hz. If the wavelength of the light is \( (0.50 \text{ Å}) \), then calculate its energy.
Answer:
(i) **Energy (E) of a photon given frequency:**
The energy of a photon is given by Planck's equation: \( E = h\nu \)
Where:
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Frequency \( \nu = 3 \times 10^{15} \text{ Hz (Given)} \)
Substituting the values:
\( E = (6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^{15} \text{ s}^{-1}) \)
\( E = 1.988 \times 10^{-18} \text{ J} \). This is the energy carried by a single photon.
(ii) **Energy (E) of a photon given wavelength:**
The energy of a photon having wavelength \( (\lambda) \) is given by the expression: \( E = \frac{hc}{\lambda} \)
Where:
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Velocity of light in vacuum \( c = 3 \times 10^8 \text{ m/s} \)
Wavelength \( \lambda = 0.50 \text{ Å} = 0.50 \times 10^{-10} \text{ m} \)
Substituting the values:
\( E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{0.50 \times 10^{-10} \text{ m}} \)
\( E = 3.98 \times 10^{-15} \text{ J} \)
In simple words: A photon's energy can be calculated using its frequency or wavelength. If you have the frequency, multiply by Planck's constant. If you have the wavelength, multiply Planck's constant by the speed of light, then divide by the wavelength.

🎯 Exam Tip: Be careful with units for wavelength; Angstroms (\( \text{Å} \)) and nanometers (\( \text{nm} \)) need to be converted to meters for calculations.

 

Question 42. The work function of a metal is 2.13 eV. A photon having wavelength of \( 4 \times 10^{-7} \text{m} \) strikes the metal then calculate (i) energy of the incident photon, (ii) kinetic energy of emitted electron, and (iii) the velocity of a photoelectron.
Answer:
Given:
Work function of metal \( W = 2.13 \text{ eV} \)
Wavelength of photon \( \lambda = 4 \times 10^{-7} \text{ m} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Velocity of light in vacuum \( c = 3 \times 10^8 \text{ m/s} \)
Mass of electron \( m_e = 9.10939 \times 10^{-31} \text{ kg} \)
Conversion: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \)

(i) **Energy of the incident photon (E):**
Using the formula: \( E = \frac{hc}{\lambda} \)
\( E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{4 \times 10^{-7} \text{ m}} \)
\( E = 4.9695 \times 10^{-19} \text{ J} \)
Hence, the energy of the photon is approximately \( 4.97 \times 10^{-19} \text{ J} \). This energy is used to release the electron and give it kinetic energy.

(ii) **Kinetic energy of the emission (KE):**
The kinetic energy of the emitted electron is given by the photoelectric equation: \( KE = E - W \)
First, convert the work function to Joules:
\( W = 2.13 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 3.41226 \times 10^{-19} \text{ J} \)
Now, calculate kinetic energy in Joules:
\( KE = (4.9695 \times 10^{-19} \text{ J}) - (3.41226 \times 10^{-19} \text{ J}) \)
\( KE = 1.55724 \times 10^{-19} \text{ J} \)
Or in eV:
\( E = \frac{4.9695 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 3.102 \text{ eV} \)
\( KE = 3.102 \text{ eV} - 2.13 \text{ eV} = 0.972 \text{ eV} \)
Hence, the kinetic energy of emission is \( 1.56 \times 10^{-19} \text{ J} \) or \( 0.972 \text{ eV} \).

(iii) **The velocity of a photoelectron (v):**
The kinetic energy is also given by: \( KE = \frac{1}{2} mv^2 \)
So, \( v = \sqrt{\frac{2 \times KE}{m_e}} \)
Using \( KE = 1.55724 \times 10^{-19} \text{ J} \) and \( m_e = 9.10939 \times 10^{-31} \text{ kg} \):
\( v = \sqrt{\frac{2 \times 1.55724 \times 10^{-19} \text{ J}}{9.10939 \times 10^{-31} \text{ kg}}} \)
\( v = \sqrt{\frac{3.11448 \times 10^{-19}}{9.10939 \times 10^{-31}}} \)
\( v = \sqrt{0.34189 \times 10^{12}} \)
\( v = \sqrt{3.4189 \times 10^{11}} \)
\( v = 5.847 \times 10^5 \text{ m/s} \)
Hence, the velocity of the photoelectron is approximately \( 5.85 \times 10^5 \text{ m/s} \).
In simple words: When light hits a metal, some of its energy helps electrons escape (work function), and the rest becomes the electron's movement energy (kinetic energy). From this kinetic energy, we can then calculate how fast the electron is moving.

🎯 Exam Tip: In photoelectric effect problems, ensure consistent units (Joules or eV) throughout the calculation, especially when subtracting energies.

 

Question 43. A bulb of 25 W emits yellow colour monochromatic light of wavelength 0.57 m. Calculate number of photon emitted per second.
Answer:
Given:
Power of bulb \( P = 25 \text{ W} = 25 \text{ Js}^{-1} \) (Energy emitted per second)
Wavelength of light \( \lambda = 0.57 \text{ m} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Velocity of light \( c = 3 \times 10^8 \text{ m/s} \)

First, calculate the energy of one photon (E):
\( E = \frac{hc}{\lambda} \)
\( E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{0.57 \text{ m}} \)
\( E = 34.8736 \times 10^{-20} \text{ J} \)
\( E \approx 3.487 \times 10^{-19} \text{ J} \). Each photon carries this small amount of energy.

Now, calculate the number of photons emitted per second:
Number of photons per second \( = \frac{\text{Total energy emitted per second}}{\text{Energy of one photon}} \)
Number of photons per second \( = \frac{25 \text{ J}}{3.48736 \times 10^{-19} \text{ J/photon}} \)
Number of photons per second \( = 7.169 \times 10^{19} \text{ photons/second} \)
In simple words: A bulb's power tells us how much energy it gives out each second. If we know the energy of one tiny light particle (photon), we can find out how many of these photons the bulb sends out every second by dividing the total energy by the energy of one photon.

🎯 Exam Tip: Always relate power (Watts) to energy per unit time (Joules/second) when calculating photon emission rates.

 

Question 44. When radiation of \( 6800 \text{ Å} \) wavelength strikes in metal surface, the electron with zero velocity emitted. Calculate the threshold frequency and work function \( (W_0) \) of the metal.
Answer:
Given:
Wavelength of radiation \( \lambda_0 = 6800 \text{ Å} = 6800 \times 10^{-10} \text{ m} = 6.8 \times 10^{-7} \text{ m} \)
Since electrons are emitted with zero velocity, this wavelength is the threshold wavelength.
Velocity of light \( c = 3 \times 10^8 \text{ m/s} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)

First, calculate the threshold frequency \( (\nu_0) \):
\( \nu_0 = \frac{c}{\lambda_0} \)
\( \nu_0 = \frac{3 \times 10^8 \text{ m/s}}{6.8 \times 10^{-7} \text{ m}} \)
\( \nu_0 = 4.41 \times 10^{14} \text{ s}^{-1} \). This is the minimum frequency of light needed to eject an electron.

Now, calculate the work function \( (W_0) \) of the metal:
\( W_0 = h\nu_0 \)
\( W_0 = (6.626 \times 10^{-34} \text{ Js}) \times (4.41 \times 10^{14} \text{ s}^{-1}) \)
\( W_0 = 2.922 \times 10^{-19} \text{ J} \)
In simple words: When light just barely makes electrons leave a metal surface without moving, the wavelength of that light is called the threshold wavelength. From this, we can find the minimum frequency (threshold frequency) and the energy needed to free an electron (work function).

🎯 Exam Tip: Remember that "zero velocity emitted" implies the incident light's energy equals the work function, and its wavelength is the threshold wavelength.

 

Question 45. If an electron of hydrogen atom jumps from \( n=4 \) to \( n=2 \) energy state, which wavelength of light will be emitted?
Answer:
The transition from \( n_i = 4 \) (initial energy level) to \( n_f = 2 \) (final energy level) for a hydrogen atom will emit light in the Balmer series.
The energy involved in this transition is given by the Rydberg formula for energy change:
\( \Delta E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \) where \( R_H \) is the Rydberg constant for energy, \( 2.18 \times 10^{-18} \text{ J} \).
So, \( \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \)
\( \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{4} - \frac{1}{16} \right) \)
\( \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{4 - 1}{16} \right) \)
\( \Delta E = 2.18 \times 10^{-18} \text{ J} \times \left( \frac{3}{16} \right) \)
\( \Delta E = 0.40875 \times 10^{-18} \text{ J} \)
\( \Delta E = 4.0875 \times 10^{-19} \text{ J} \). This is the specific amount of energy released when the electron moves from a higher to a lower energy level.

Now, we find the wavelength \( (\lambda) \) of the emitted light using the relation \( \Delta E = \frac{hc}{\lambda} \):
\( \lambda = \frac{hc}{\Delta E} \)
Where:
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Velocity of light \( c = 3 \times 10^8 \text{ m/s} \)
\( \lambda = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{4.0875 \times 10^{-19} \text{ J}} \)
\( \lambda = \frac{1.9878 \times 10^{-25}}{4.0875 \times 10^{-19}} \text{ m} \)
\( \lambda = 4.863 \times 10^{-7} \text{ m} \)
Converting to nanometers: \( \lambda = 4.863 \times 10^{-7} \times 10^9 \text{ nm} = 486.3 \text{ nm} \)
Hence, the wavelength of light emitted is approximately \( 486 \text{ nm} \).
In simple words: When an electron in an atom drops from a higher energy level to a lower one, it releases energy as light. We can calculate the exact amount of this energy, and then use it to find the color (wavelength) of the light that comes out.

🎯 Exam Tip: For electron transitions in hydrogen-like atoms, remember to use the correct Rydberg formula and ensure the initial and final energy levels are properly identified.

 

Question 46. If the value of energy of 1st orbit of hydrogen atom is \( -2.18 \times 10^{18} \text{ J atm}^{-1} \), then what will be energy associated with 5th orbit?
Answer:
The energy of an electron in the \( n^{th} \) orbit of a hydrogen atom is given by the formula:
\( E_n = \frac{-2.18 \times 10^{-18} \times Z^2}{n^2} \text{ J/atom} \)
Given for 1st orbit (\( n=1 \)): \( E_1 = -2.18 \times 10^{-18} \text{ J/atom} \) (Note: The source text has "J atm-1", but the standard unit is J/atom).
For a hydrogen atom, the atomic number \( Z = 1 \).
We need to find the energy associated with the 5th orbit (\( n=5 \)):
\( E_5 = \frac{-2.18 \times 10^{-18} \times (1)^2}{(5)^2} \)
\( E_5 = \frac{-2.18 \times 10^{-18}}{25} \)
\( E_5 = -0.0872 \times 10^{-18} \text{ J/atom} \)
\( E_5 = -8.72 \times 10^{-20} \text{ J/atom} \). This calculation shows how the electron's energy becomes less negative (higher) as it moves to orbits further from the nucleus.
In simple words: The energy of an electron in a hydrogen atom depends on which energy level (orbit) it is in. We can use a simple formula to calculate the energy for any orbit if we know the energy of the first orbit.

🎯 Exam Tip: Pay close attention to units and ensure the Rydberg constant's value matches the required units for the final answer (Joules or eV).

 

Question 47. Calculate the wavelength of electron if it is moving with a velocity of \( 2.05 \times 10^7 \text{ ms}^{-1} \).
Answer:
According to de Broglie's equation, the wavelength \( (\lambda) \) of a moving particle is given by:
\( \lambda = \frac{h}{mv} \)
Where:
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Mass of particle (electron) \( m = 9.109 \times 10^{-31} \text{ Kg} \)
Velocity of particle (electron) \( v = 2.05 \times 10^7 \text{ ms}^{-1} \)
Substituting the values into the equation:
\( \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{(9.109 \times 10^{-31} \text{ Kg}) \times (2.05 \times 10^7 \text{ ms}^{-1})} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{1.867345 \times 10^{-23}} \)
\( \lambda = 3.548 \times 10^{-11} \text{ m} \)
Hence, the wavelength of the electron is approximately \( 3.55 \times 10^{-11} \text{ m} \). This wavelength is very small, which is typical for microscopic particles moving at high speeds.
In simple words: The de Broglie equation tells us that every moving particle, even an electron, has a wave-like nature. We can find this wavelength by dividing Planck's constant by the electron's mass and its speed.

🎯 Exam Tip: Always use the mass of the electron (\( 9.109 \times 10^{-31} \text{ kg} \)) for calculations involving electron wavelength.