Heisenberg’s uncertainty principle

STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 8

211 Easy

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Application +4 | -1

What will be the difference in mass number if the number of neutrons halved and the number of electrons double in

{{}_6{C^{12}}}

View answer and explanation

Correct option: A

3

Option A (3) is correct because halving the neutrons and doubling the electrons results in a mass number difference of 3.

Solution approach

Correct Answer + Core Reason: 3. The difference in mass number is calculated by considering the changes in the number of neutrons.

Step-by-Step Solution:

Step 1: Initial state: ${{}_6{C^{12}}}$ has 6 protons and 6 neutrons (mass number = 12).

Step 2: The number of neutrons is halved: $6 / 2 = 3$ neutrons.

Step 3: The number of electrons is doubled: This does not affect the mass number, as electrons have negligible mass.

Step 4: Final state: The new atom has 6 protons and 3 neutrons (mass number = 9).

Step 5: Calculate the difference in mass number: $12 - 9 = 3$ .

Step 6: Final Answer: 3

Why other options are wrong:
- Option B (6): This is incorrect. It doesn't result from the correct calculation of the change in mass number.
- Option C (Zero): This is incorrect. The number of neutrons changed, so the mass number also changed.
- Option D (9): This is incorrect. It doesn't result from the correct calculation of the change in mass number.

Key Concepts:
- Atomic structure
- Mass number
- Protons and neutrons

Memory Aids:
- Remember that the mass number is the sum of protons and neutrons.

Quick Tips:
- Pay attention to which particles affect the mass number.
- Double-check your calculations to avoid errors.

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According to Dalton, the smallest particle of matter which is indivisible, is called:

View answer and explanation

Correct option: B

Atom

Option B (Atom) is correct because, according to Dalton's atomic theory, the atom is the smallest indivisible particle of matter.

Solution approach

Correct Answer + Core Reason: Atom. Dalton's atomic theory defined atoms as indivisible.

Step-by-Step Solution:

Step 1: Recall Dalton's atomic theory: Dalton proposed that all matter is composed of indivisible and indestructible atoms.

Step 2: Identify the smallest particle according to Dalton: According to Dalton, the atom is the smallest particle of matter.

Why other options are wrong:
- Option A: Molecule is a group of atoms bonded together, not the smallest indivisible particle according to Dalton.
- Option B: Atom is the smallest indivisible particle according to Dalton.
- Option C: Compound is a substance made of atoms of different elements bonded together, not the smallest indivisible particle according to Dalton.
- Option D: Ion is an atom or molecule with an electrical charge, not the smallest indivisible particle according to Dalton.

Key Concepts:
- Dalton's atomic theory
- Atom
- Molecule
- Compound
- Ion

Memory Aids:
- Dalton = Atom (Think: DA)

Quick Tips:
- Remember the basic postulates of Dalton's atomic theory.

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What is the trend of energy of Bohr’s orbits?

View answer and explanation

Correct option: A

Energy of the orbit increases as we move away from the nucleus

Option A is correct because the energy of an electron in Bohr's orbit increases as the electron moves farther away from the nucleus (higher n value).

Solution approach

Correct Answer + Core Reason: Energy of the orbit increases as we move away from the nucleus. The energy levels in Bohr's model become less negative (increase) as the principal quantum number (n) increases.

Step-by-Step Solution:

Step 1: Recall Bohr's Model: In Bohr's model, electrons occupy specific energy levels or orbits around the nucleus.

Step 2: Energy Level Formula: The energy of an electron in the nth orbit is given by $E_n = -\frac{{13.6}}{{n^2}} eV$ , where n is the principal quantum number.

Step 3: Analyze the Formula: As n increases (electron moves away from the nucleus), the absolute value of $E_n$ decreases, meaning the energy becomes less negative and thus increases.

Why other options are wrong:
- Option B: Incorrect because energy increases, not decreases, as we move away from the nucleus.
- Option C: Incorrect because energy changes as we move away from the nucleus.
- Option D: Incorrect because the energy of Bohr's orbit can be calculated using the formula mentioned above.

Key Concepts:
- Bohr's Model
- Energy Levels
- Principal Quantum Number

Memory Aids:
- Higher orbit = Higher energy

Quick Tips:
- Remember the inverse square relationship between energy and the principal quantum number.

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What is the ration of mass of an electron to the mass of a proton?

View answer and explanation

Correct option: C

1:1837

Option C (

1:1837

) is correct because the mass of an electron is approximately 1/1837th the mass of a proton.

Solution approach

Correct Answer + Core Reason: $1:1837$ . This is the approximate ratio of electron mass to proton mass.

Step-by-Step Solution:
Step 1: Recall the masses of an electron and a proton.
Step 2: Mass of proton $\approx 1.6726 \times 10^{-27}$ kg\nMass of electron $\approx 9.109 \times 10^{-31}$ kg
Step 3: Calculate the ratio: $\frac{\text{Mass of electron}}{\text{Mass of proton}} = \frac{9.109 \times 10^{-31}}{1.6726 \times 10^{-27}} \approx \frac{1}{1837}$

Why other options are wrong:
- Option A: $1:2$ is incorrect; the proton is much heavier than the electron.
- Option B: $1:1$ is incorrect; the proton and electron have very different masses.
- Option D: $1:3$ is incorrect; the proton is much heavier than the electron.

Key Concepts:
- Atomic structure
- Mass of subatomic particles

Memory Aids:
- Proton is much heavier than electron.

Quick Tips:
- Remember the approximate mass ratio of proton to electron.

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Which of the following wave properties is proportional to
energy for electromagnetic radiation?

View answer and explanation

Correct option: B

Wave number

Option B (Wave number) is correct because energy is directly proportional to the wave number of electromagnetic radiation, as indicated by the formula E ∝ ν̄.

Solution approach

Correct Answer + Core Reason: Wave number. Energy is directly proportional to wave number.

Step-by-Step Solution:

Step 1: Recall the relationship between energy, frequency, wavelength, and wave number.

Step 2: The energy of electromagnetic radiation is given by $E = h\nu$ , where $h$ is Planck's constant and $\nu$ is frequency.

Step 3: Frequency and wavelength are related by $c = \nu \lambda$ , where $c$ is the speed of light and $\lambda$ is the wavelength.

Step 4: Wave number is defined as the reciprocal of wavelength, $\bar{\nu} = \frac{1}{\lambda}$ .

Step 5: Therefore, $E = \frac{hc}{\lambda} = hc\bar{\nu}$ . Thus, energy is proportional to wave number.

Why other options are wrong:
- Option A: Velocity is constant for electromagnetic radiation in a given medium and is not proportional to energy.
- Option C: Wavelength is inversely proportional to energy ( $E = \frac{hc}{\lambda}$ ).
- Option D: Amplitude is related to the intensity of the radiation, not directly to its energy.

Key Concepts:
- Energy of Electromagnetic Radiation
- Frequency, Wavelength, and Wave Number
- Planck's Constant

Memory Aids:
- Remember the relationships: $E = h\nu$ , $c = \nu \lambda$ , and $\bar{\nu} = \frac{1}{\lambda}$ .

Quick Tips:
- Pay attention to the relationships between energy, frequency, wavelength, and wave number.

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Application +4 | -1

According to (n+l) rule after completing 'np' level the electron enters to:

View answer and explanation

Correct option: B

\left( {n + 1} \right)s

Option B ((n + 1)s) is correct because according to the (n+l) rule, after filling the np level, the next electron occupies the (n+1)s level due to its lower energy.

Solution approach

Correct Answer + Core Reason: (n + 1)s. The (n+l) rule dictates the filling of orbitals based on energy levels; (n+1)s has lower energy than nd or (n+1)p after np is filled.

Step-by-Step Solution:

Step 1: Understand the (n+l) rule. This rule states that electrons first fill the orbitals with the lowest (n+l) value. If two orbitals have the same (n+l) value, the orbital with the lower n value is filled first.

Step 2: Consider the given options after filling the 'np' level:
- (n - 1)d: n = n-1, l = 2, n+l = n+1
- (n + 1)s: n = n+1, l = 0, n+l = n+1
- nd: n = n, l = 2, n+l = n+2
- (n + 1)p: n = n+1, l = 1, n+l = n+2

Step 3: Compare the (n+l) values. After np, we need to compare (n-1)d and (n+1)s. However, the question implies that we are looking for the next orbital after 'np' in the same period or a subsequent period. Therefore, we should consider the relative energies of nd, (n+1)s, and (n+1)p.

Step 4: Apply the (n+l) rule. (n+1)s has an (n+l) value of (n+1), while nd and (n+1)p have (n+l) values of (n+2). Since (n+1)s has the lowest (n+l) value, it will be filled next.

Step 5: Conclusion: After filling the 'np' level, the electron enters the (n+1)s level.

Why other options are wrong:
- Option A: (n - 1)d: This orbital is filled before 'np' according to the (n+l) rule.
- Option C: nd: The 'nd' orbital has a higher (n+l) value (n+2) compared to (n+1)s (n+1), so it is filled after (n+1)s.
- Option D: (n + 1)p: The '(n+1)p' orbital also has a higher (n+l) value (n+2) compared to (n+1)s (n+1), so it is filled after (n+1)s.

Key Concepts:
- (n+l) rule
- Aufbau principle
- Electronic configuration

Memory Aids:
- Remember the order of filling orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

Quick Tips:
- Always remember the (n+l) rule and apply it systematically.
- When two orbitals have the same (n+l) value, the orbital with the lower 'n' value is filled first.

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Application +4 | -1

An electron has a velocity of

300 \mathrm{~m} / \mathrm{sec}

that is accurate to

0.01 \%

with what accuracy can be locate the position of this electron :

View answer and explanation

Correct option: D

0.0019

meter

Option D (

0.0019

meter) is correct because it is calculated using the Heisenberg uncertainty principle, relating the uncertainty in position and momentum.

Solution approach

Correct Answer + Core Reason: $0.0019$ meter. The uncertainty in position is calculated using the Heisenberg uncertainty principle.

Step-by-Step Solution:

Step 1: Identify the given information:\nVelocity (v) = 300 m/s\nAccuracy = 0.01%\nMass of electron (m) = $9.1 \times 10^{-31}$ kg\nPlanck's constant (h) = $6.626 \times 10^{-34}$ Js

Step 2: Calculate the uncertainty in velocity ( $\Delta v$ ):
$\Delta v = v \times accuracy = 300 \times (0.01/100) = 0.03$ m/s

Step 3: Apply the Heisenberg uncertainty principle: $\Delta x \times \Delta p \geq \frac{h}{4\pi}$ , where $\Delta p = m \times \Delta v$
$\Delta x \times m \times \Delta v \geq \frac{h}{4\pi}$

Step 4: Solve for the uncertainty in position ( $\Delta x$ ):
$\Delta x \geq \frac{h}{4\pi m \Delta v}$

Step 5: Substitute the values:
$\Delta x \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03}$

Step 6: Calculate $\Delta x$ :
$\Delta x \geq 0.001925$ m $\approx 0.0019$ m

Final Answer: $0.0019$ meter

Why other options are wrong:
- **Option A ( $0.019$ meter): This is incorrect because it likely involves an error in the calculation or an incorrect application of the Heisenberg uncertainty principle.
- Option B ( $0.190$ meter): This is incorrect because it likely involves an error in the calculation or an incorrect application of the Heisenberg uncertainty principle.
- Option C ( $0.029$ meter): This is incorrect because it likely involves an error in the calculation or an incorrect application of the Heisenberg uncertainty principle.

Key Concepts:
- Heisenberg uncertainty principle
- Uncertainty in position and momentum

Memory Aids:**
- Remember the Heisenberg uncertainty principle formula: $\Delta x \times \Delta p \geq \frac{h}{4\pi}$

Quick Tips:
- Ensure all units are consistent before performing calculations.
- Pay attention to the accuracy given in the problem.

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The equation,

\lambda = \frac{h}{{mv}}

was deduced by

View answer and explanation

Correct option: B

de-Broglie

Option B (de-Broglie) is correct because de-Broglie deduced the equation

\lambda = \frac{h}{{mv}}

relating wavelength to momentum.

Solution approach

Correct Answer + Core Reason: de-Broglie. The equation $\lambda = \frac{h}{mv}$ represents the de Broglie wavelength, demonstrating the wave-particle duality of matter.

Step-by-Step Solution:

Step 1: Recall the concept of wave-particle duality.

Step 2: Remember that de Broglie proposed that all matter has wave-like properties.

Step 3: Recognize the equation $\lambda = \frac{h}{mv}$ as the de Broglie wavelength equation.

Why other options are wrong:
- Option A: Newton is known for classical mechanics and gravity.
- Option C: Planck is known for quantum theory and the quantization of energy.
- Option D: Heisenberg is known for the uncertainty principle.

Key Concepts:
- Wave-particle duality
- de Broglie wavelength
- Quantum mechanics

Memory Aids:
- de Broglie = wavelength.

Quick Tips:
- Know the key contributions of different scientists to physics and chemistry.

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Analyzing +4 | -1

Statement (S): Humphrey series discovered in

\mathrm{H}

atomic spectra has lowest energy radiations.

Explanation (E): The series belongs to

n_1=6.

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

Option C is correct because the Humphrey series corresponds to transitions to n1=6, and transitions to higher energy levels result in lower energy radiations.

Solution approach

Correct Answer + Core Reason: Both S and E are correct and E is correct explanation of S. Humphrey series has $n_1 = 6$ , and transitions to higher energy levels result in lower energy radiations.

Step-by-Step Solution:
Step 1: Recall the different series in the hydrogen atomic spectrum: Lyman, Balmer, Paschen, Brackett, Pfund, and Humphrey.
Step 2: Understand that the Humphrey series corresponds to transitions where the electron falls to the $n_1 = 6$ energy level.
Step 3: Recognize that the energy of the emitted radiation is related to the difference in energy levels between the initial and final states: $\Delta E = E_{n_2} - E_{n_1}$ , where $n_2 > n_1$ .
Step 4: Since the Humphrey series has the largest value of $n_1$ (6) among the common series, the energy difference $\Delta E$ for transitions to this level will be the smallest, resulting in the lowest energy radiations.
Step 5: Therefore, both the statement (S) and the explanation (E) are correct, and the explanation correctly explains the statement.

Why other options are wrong:
- Option A: S is correct but E is wrong: E is correct.
- Option B: S is wrong but E is correct: S is correct.
- Option D: Both S and E are correct but E is not correct explanation of S: E is the correct explanation of S.

Key Concepts:
- Hydrogen atomic spectrum
- Energy levels
- Spectral series (Lyman, Balmer, Paschen, Brackett, Pfund, Humphrey)

Memory Aids:
- Remember the order of the spectral series and their corresponding $n_1$ values.

Quick Tips:
- The series with the highest $n_1$ value will have the lowest energy radiations.

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Analyzing +4 | -1

S-I:

\mathrm{Fe}^{+3}

ion is less stable than

\mathrm{Fe}^{+2}.

S-II :

\mathrm{Fe}^{+3}

ion has less number of unpaired electrons than

\mathrm{Fe}^{+2}.

View answer and explanation

Correct option: B

Both S-I and S-II are incorrect

The correct answer is B (Both S-I and S-II are incorrect) because Fe+3 is more stable than Fe+2 due to its half-filled d-orbital configuration, and it has more unpaired electrons.

Solution approach

Correct Answer + Core Reason: Both S-I and S-II are incorrect. Fe+3 is more stable and has more unpaired electrons.

Step-by-Step Solution:

Step 1: Write the electronic configurations of $\mathrm{Fe}^{3+}$ and $\mathrm{Fe}^{2+}$ .

Step 2: $\mathrm{Fe}^{3+} = 3d^5$ . This is a half-filled configuration, which is more stable.

Step 3: $\mathrm{Fe}^{2+} = 3d^6$ .

Step 4: Determine the number of unpaired electrons in each ion.

Step 5: $\mathrm{Fe}^{3+}$ has 5 unpaired electrons.

Step 6: $\mathrm{Fe}^{2+}$ has 4 unpaired electrons.

Step 7: Analyze the given statements.

Step 8: S-I: $\mathrm{Fe}^{+3}$ ion is less stable than $\mathrm{Fe}^{+2}$ . This is incorrect.

Step 9: S-II : $\mathrm{Fe}^{+3}$ ion has less number of unpaired electrons than $\mathrm{Fe}^{+2}$ . This is incorrect.

Step 10: Final Answer: Both S-I and S-II are incorrect

Why other options are wrong:
- Option A: Both statements are incorrect, so this option is wrong.
- Option C: S-I is incorrect, but S-II is also incorrect, so this option is wrong.
- Option D: S-I is incorrect, and S-II is also incorrect, so this option is wrong.

Key Concepts:
- Electronic configuration
- Stability of half-filled and fully-filled orbitals
- Hund's rule

Memory Aids:
- Half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Write the electronic configuration to determine the number of unpaired electrons and the stability of the ion.

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Analyzing +4 | -1

S-I: Line spectrum of

\mathrm{Li}^{+2}

and

\mathrm{He}^{+}

are identical.

S-II: Isoelectronic species produce identical spectrum.

View answer and explanation

Correct option: A

Both S-I and S-II are correct

Option A is correct because both statements are correct. Both Li+2 and He+ have one electron and thus identical spectra. Isoelectronic species with the same number of electrons produce identical spectra.

Solution approach

Correct Answer + Core Reason: Both S-I and S-II are correct.

Step-by-Step Solution:

Step 1: Statement I: Li+2 and He+ are isoelectronic species, both having only one electron.

Step 2: Species with the same number of electrons will have similar electronic transitions and hence similar line spectra.

Step 3: Statement II: Isoelectronic species have the same electronic configuration and therefore produce identical spectra.

Step 4: Both statements are correct and logically consistent.

Why other options are wrong:
- Option B: Both statements are correct.
- Option C: Both statements are correct.
- Option D: Both statements are correct.

Key Concepts:
- Isoelectronic species
- Line spectra
- Electronic transitions

Memory Aids:
- Isoelectronic: "iso" means same, "electronic" refers to electrons.

Quick Tips:
- Remember that species with the same number of electrons will exhibit similar spectral properties.

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Statement (S): Specific charge of

\alpha

-particles is twice to that of proton.

Explanation (E): Specific charge is given by

e / m.

View answer and explanation

Correct option: B

S is wrong but E is correct

Option B (S is wrong but E is correct) is correct because the specific charge of an alpha particle is half that of a proton, and the explanation correctly defines specific charge.

Solution approach

Correct Answer + Core Reason: S is wrong but E is correct. Specific charge of alpha particle is half that of proton.

Step-by-Step Solution:
Step 1: Define specific charge: Specific charge = charge/mass = e/m
Step 2: Calculate specific charge of proton: e/m
Step 3: Calculate specific charge of alpha particle: (2e)/(4m) = e/(2m)
Step 4: Compare specific charges: Specific charge of alpha particle is half that of proton.

Why other options are wrong:
- Option A: S is wrong.
- Option C: S is wrong.
- Option D: S is wrong.

Key Concepts:
- Specific charge
- Charge and mass of proton and alpha particle

Memory Aids:
- Alpha particle: 2 protons + 2 neutrons

Quick Tips:
- Remember the charge and mass of proton and alpha particle.

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Application +4 | -1

Which of the following particles has more electrons than neutrons

View answer and explanation

Correct option: C

{O}^{-2}

Option C (

{O}^{-2}

) is correct because it has 10 electrons and 8 neutrons, thus having more electrons than neutrons.

Solution approach

Correct Answer + Core Reason: ${O}^{-2}$ . Oxygen has more electrons than neutrons.

Step-by-Step Solution:\nStep 1: Analyze the given options:\nStep 2: Option A: Carbon (C): ${ }_6^{12} \mathrm{C}$ has 6 protons, 6 electrons, and 6 neutrons.\nStep 3: Option B: Fluoride ion ( $F^{-}$ ): ${ }_9^{19} \mathrm{F}^{-}$ has 9 protons, 10 electrons, and 10 neutrons.\nStep 4: Option C: Oxide ion ( ${O}^{-2}$ ): ${ }_8^{16} \mathrm{O}^{-2}$ has 8 protons, 10 electrons, and 8 neutrons.\nStep 5: Option D: Aluminum ion ( ${Al}^{+3}$ ): ${ }_{13}^{27} \mathrm{Al}^{+3}$ has 13 protons, 10 electrons, and 14 neutrons.\nStep 6: Compare the number of electrons and neutrons for each option.\nStep 7: Only ${O}^{-2}$ has more electrons (10) than neutrons (8).

Why other options are wrong:
- Option A: Carbon has equal number of electrons and neutrons.
- Option B: Fluoride has equal number of electrons and neutrons.
- Option D: Aluminum has more neutrons than electrons.

Key Concepts:
- Atomic structure
- Number of protons, neutrons, and electrons
- Ions

Memory Aids:
- Number of protons = atomic number
- Number of neutrons = mass number - atomic number
- Number of electrons = number of protons - charge

Quick Tips:
- Remember to account for the charge when calculating the number of electrons in an ion.

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The H-spectrum show

View answer and explanation

Correct option: D

Presence of quantised energy level

Option D (Presence of quantised energy level) is correct because the hydrogen spectrum provides evidence for the existence of discrete, quantized energy levels within the hydrogen atom.

Solution approach

Correct Answer + Core Reason: Presence of quantised energy level. The discrete lines in the hydrogen spectrum indicate quantized energy levels.

Step-by-Step Solution:

Step 1: Understand the origin of the hydrogen spectrum.\nThe hydrogen spectrum is an emission spectrum produced when excited hydrogen atoms release energy in the form of photons as electrons transition between energy levels.

Step 2: Relate the spectral lines to energy levels.\nEach spectral line corresponds to a specific wavelength (and thus energy) of light emitted during an electron transition. The fact that the spectrum consists of discrete lines, rather than a continuous range of wavelengths, implies that the energy levels within the hydrogen atom are quantized (i.e., can only take on specific, discrete values).

Step 3: Evaluate the other options.
- Heisenberg's uncertainty principle describes the limits on the precision with which certain pairs of physical properties (e.g., position and momentum) can be known simultaneously. It doesn't directly explain the discrete nature of the hydrogen spectrum.
- Diffraction is the bending of waves around obstacles or through apertures. While diffraction can be observed with light, it doesn't explain the specific wavelengths observed in the hydrogen spectrum.
- Polarization is a property of transverse waves (like light) that describes the direction of the oscillation. It doesn't explain the discrete nature of the hydrogen spectrum.

Step 4: Conclude that the hydrogen spectrum shows the presence of quantized energy levels.

Why other options are wrong:
- Option A: Heisenberg's uncertainty principle does not directly explain the discrete lines in the H-spectrum.
- Option B: Diffraction is a wave phenomenon but does not explain the quantized energy levels.
- Option C: Polarization is a property of transverse waves and is not directly related to the H-spectrum's discrete lines.

Key Concepts:
- Atomic spectra (emission and absorption spectra)
- Quantization of energy levels in atoms
- Bohr model of the hydrogen atom
- Heisenberg's Uncertainty Principle
- Diffraction
- Polarization

Memory Aids:
- Remember that discrete spectral lines imply discrete energy levels.

Quick Tips:
- Understand the relationship between electron transitions and the emission/absorption of photons.
- Know the basic principles behind atomic spectra.

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In case of

d_{x^2-y^2}

orbital:

View answer and explanation

Correct option: C

Probability of finding the electron is maximum along
x and y-axes

Option C (Probability of finding the electron is maximum along x and y-axes) is correct because the

d_{x^2-y^2}

orbital has lobes along the x and y axes, indicating a higher probability of finding electrons along those axes.

Solution approach

Correct Answer + Core Reason: Probability of finding the electron is maximum along x and y-axes. The $d_{x^2-y^2}$ orbital has lobes along the x and y axes.

Step-by-Step Solution:

Step 1: Recall the shape of the $d_{x^2-y^2}$ orbital: It has four lobes, with two lobes lying along the x-axis and two lobes lying along the y-axis.

Step 2: Relate the shape to electron probability: The lobes represent regions of high electron density, indicating a higher probability of finding electrons in those regions.

Step 3: Conclude the probability distribution: The probability of finding the electron is maximum along the x and y-axes.

Why other options are wrong:
- Option A: Probability of finding the electron along x-axis is zero: Incorrect, as the lobes lie along the x-axis.
- Option B: Probability of finding the electron along y-axis is zero: Incorrect, as the lobes lie along the y-axis.
- Option D: Probability of finding the electron is zero in xy plane: Incorrect, as the lobes lie in the xy plane.

Key Concepts:
- d-orbitals
- $d_{x^2-y^2}$ orbital shape
- Electron probability distribution

Memory Aids:
- $d_{x^2-y^2}$ lies along x and y axes.

Quick Tips:
- Visualize the shapes of different orbitals to understand electron probability distributions.

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The electronic energy levels of the hydrogen atom in the Bohr^' s theory are called:

View answer and explanation

Correct option: B

Orbits

Option B (Orbits) is correct because Bohr's theory postulates that electrons revolve around the nucleus in specific, quantized orbits.

Solution approach

Correct Answer + Core Reason: Orbits. Bohr's theory describes electrons orbiting the nucleus in specific energy levels called orbits.

Step-by-Step Solution:

Step 1: Recall Bohr's theory: Bohr's model of the atom postulates that electrons revolve around the nucleus in specific, quantized energy levels.

Step 2: Define the terms:
- Orbitals: Regions of space where an electron is likely to be found (quantum mechanical model).
- Orbits: Fixed paths around the nucleus (Bohr's model).
- Rydberg levels: Energy levels related to the Rydberg formula for spectral lines.
- Ground states: The lowest energy state of an atom.

Step 3: Match the term to Bohr's theory: In Bohr's theory, electrons occupy specific orbits.

Why other options are wrong:
- Option A: Orbitals are a concept from the quantum mechanical model of the atom, not Bohr's theory.
- Option C: Rydberg levels are related to the Rydberg formula for calculating spectral lines, but the energy levels themselves are called orbits in Bohr's theory.
- Option D: Ground state refers to the lowest energy level, but the energy levels in Bohr's theory are generally called orbits.

Key Concepts:
- Bohr's Atomic Model
- Orbits
- Orbitals
- Energy Levels

Memory Aids:
- Remember that Bohr's model introduced the concept of quantized orbits.

Quick Tips:
- Differentiate between Bohr's model and the quantum mechanical model of the atom.

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Application +4 | -1

If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by

View answer and explanation

Correct option: A

\lambda = \frac{{36}}{{5R}}

Option A (

\lambda = \frac{{36}}{{5R}}

) is correct because it is derived from the Rydberg formula for the wavelength of emitted radiation when an electron transitions between energy levels.

Solution approach

Correct Answer + Core Reason: $\lambda = \frac{{36}}{{5R}}$ . This is derived from the Rydberg formula for the wavelength of emitted radiation.

Step-by-Step Solution:

Step 1: Identify the given information: The electron jumps from the third orbit ( $n_2 = 3$ ) to the second orbit ( $n_1 = 2$ ).

Step 2: Recall the Rydberg formula: $\frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}}} \right]$ , where R is the Rydberg constant and Z is the atomic number.

Step 3: Apply the Rydberg formula for hydrogen (Z=1): $\frac{1}{\lambda } = R{(1)^2}\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right]$

Step 4: Simplify the expression: $\frac{1}{\lambda } = R\left[ {\frac{1}{4} - \frac{1}{9}} \right] = R\left[ {\frac{{9 - 4}}{{36}}} \right] = R\left[ {\frac{5}{{36}}} \right]$

Step 5: Solve for $\lambda$ : $\lambda = \frac{{36}}{{5R}}$

Why other options are wrong:
- Option B: This option is incorrect because it is the inverse of the correct expression for $\lambda$ .
- Option C: This option is incorrect because it does not correctly apply the Rydberg formula with the given values of $n_1$ and $n_2$ .
- Option D: This option is incorrect because it does not correctly apply the Rydberg formula with the given values of $n_1$ and $n_2$ .

Key Concepts:
- Rydberg formula
- Atomic spectra
- Electronic transitions

Memory Aids:
- Remember the Rydberg formula: $\frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}}} \right]$

Quick Tips:
- Always remember to square the principal quantum numbers ( $n_1$ and $n_2$ ) in the Rydberg formula.
- For hydrogen, Z = 1.

228 Easy

228 of 239

Understanding +4 | -1

Which is not an atomic orbital?

View answer and explanation

Correct option: A

2d

Option A (2d) is correct because the 2d orbital does not exist according to quantum mechanical principles.

Solution approach

Correct Answer + Core Reason: 2d. The 2d orbital does not exist because the principal quantum number (n) must be greater than the azimuthal quantum number (l).

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the azimuthal quantum number (l).
- n = 1, 2, 3, ... (energy level)
- l = 0, 1, 2, ..., (n-1) (shape of the orbital: s, p, d, f, ...)

Step 2: For n = 2, the possible values of l are 0 and 1, corresponding to 2s and 2p orbitals, respectively.

Step 3: The value l = 2 corresponds to a d orbital. However, for n = 2, l cannot be 2.

Step 4: Therefore, the 2d orbital does not exist.

Why other options are wrong:
- Option B: 5p exists because for n = 5, l can be 1 (p orbital).
- Option C: 3p exists because for n = 3, l can be 1 (p orbital).
- Option D: 4d exists because for n = 4, l can be 2 (d orbital).

Key Concepts:
- Quantum numbers (n and l)
- Atomic orbitals (s, p, d, f)

Memory Aids:
- l ranges from 0 to n-1.

Quick Tips:
- Remember the allowed values of quantum numbers.

229 Easy

229 of 239

Application +4 | -1

The energy in joule corresponding to light of wavelength 45 nm is

View answer and explanation

Correct option: D

4.42 \times {10^{ - 18}}

Option D (

4.42 \times {10^{ - 18}}

) is correct because the energy is calculated using the Planck's constant, speed of light, and the wavelength of the light.

Solution approach

Correct Answer + Core Reason: $4.42 \times {10^{ - 18}}$ . The energy is calculated using the formula $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given information: wavelength $\lambda = 45 \text{{ nm}} = 45 \times 10^{-9} \text{{ m}}$ .

Step 2: Recall the values of Planck's constant $h = 6.626 \times 10^{-34} \text{{ J s}}$ and the speed of light $c = 3 \times 10^8 \text{{ m/s}}$ .

Step 3: Use the formula $E = \frac{{hc}}{\lambda}$ .

Step 4: Substitute the values: $E = \frac{{(6.626 \times 10^{-34} \text{{ J s}})(3 \times 10^8 \text{{ m/s}})}}{{45 \times 10^{-9} \text{{ m}}}}$ .

Step 5: Calculate the energy: $E = \frac{{19.878 \times 10^{-26}}}{{45 \times 10^{-9}}} = 0.4417 \times 10^{-17} \text{{ J}} = 4.417 \times 10^{-18} \text{{ J}} \approx 4.42 \times 10^{-18} \text{{ J}}$ .

Why other options are wrong:
- **Option A ( $6.67 \times {10^{15}}$ ): This is incorrect because the calculation is wrong, and the exponent is incorrect.
- Option B ( $6.67 \times {10^{11}}$ ): This is incorrect because the calculation is wrong, and the exponent is incorrect.
- Option C ( $4.42 \times {10^{ - 15}}$ ): This is incorrect because the exponent is incorrect. The calculation is off by a factor of 1000.

Key Concepts:
- Planck's Quantum Theory
- Energy of a Photon

Memory Aids:**
- Remember the formula $E = \frac{{hc}}{\lambda}$ .

Quick Tips:
- Always use consistent units (SI units).
- Remember the values of Planck's constant and the speed of light.

230 Easy

230 of 239

Application +4 | -1

A metallic ion

{M^{2 + }}

ion has a electronic configuration of 2, 8, 14 and the ionic weight is 56 amu. The number of neutrons in its nucleus are

View answer and explanation

Correct option: A

30

Option A (30) is correct because the number of neutrons is the difference between the mass number and the atomic number.

Solution approach

Correct Answer + Core Reason: 30. Number of neutrons = Mass number - Atomic number.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the neutral atom M. Since ${M^{2+}}$ has 2, 8, 14 configuration, M has 2, 8, 16 configuration. Therefore, the atomic number (Z) is 26.

Step 2: Identify the mass number (A) as 56.

Step 3: Calculate the number of neutrons: Number of neutrons = A - Z = 56 - 26 = 30.

Why other options are wrong:
- Option B (32): This is incorrect because it involves an incorrect calculation.
- Option C (34): This is incorrect because it involves an incorrect calculation.
- Option D (42): This is incorrect because it involves an incorrect calculation.

Key Concepts:
- Atomic number
- Mass number
- Number of neutrons
- Electronic configuration

Memory Aids:
- Remember that the number of neutrons is the difference between the mass number and the atomic number.

Quick Tips:
- Ensure you correctly determine the atomic number from the electronic configuration.

231 Easy

231 of 239

Remembering +4 | -1

Which among the following radiations has maximum frequency?

View answer and explanation

Correct option: C

\gamma

-rays

Option C (

\gamma

-rays) is correct because gamma rays have the highest frequency among the given options in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: $\gamma$ -rays. Gamma rays have the highest frequency in the electromagnetic spectrum among the given options.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum.

Step 2: The electromagnetic spectrum consists of radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Step 3: The frequency increases in the order: Radio waves < Microwaves < X-rays < Gamma rays.

Step 4: Therefore, gamma rays have the highest frequency.

Why other options are wrong:
- Option A: X-rays have a lower frequency than gamma rays.
- Option B: Radio waves have the lowest frequency in the electromagnetic spectrum.
- Option D: Microwaves have a lower frequency than X-rays and gamma rays.

Key Concepts:
- Electromagnetic spectrum
- Frequency of electromagnetic radiation

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Micro, Infra, Visible, UV, X-ray, Gamma.

Quick Tips:
- Know the order of the electromagnetic spectrum by frequency.

232 Easy

232 of 239

Remembering +4 | -1

How many neutrons are present in tritium nucleus?

View answer and explanation

Correct option: A

2

Option A (2) is correct because tritium, an isotope of hydrogen, has 1 proton and 2 neutrons in its nucleus (mass number 3, atomic number 1).

Solution approach

Correct Answer + Core Reason: 2. Tritium has 1 proton and 2 neutrons.

Step-by-Step Solution:

Step 1: Define tritium. Tritium is an isotope of hydrogen with the symbol $^3H$ or $T$ .

Step 2: Determine the number of protons. Hydrogen has an atomic number of 1, so tritium has 1 proton.

Step 3: Determine the number of neutrons. The mass number of tritium is 3. The number of neutrons is the mass number minus the number of protons: $3 - 1 = 2$ .

Step 4: Final Answer: Tritium has 2 neutrons.

Why other options are wrong:
- Option B (3): This would be the total number of nucleons (protons + neutrons) in tritium, not just neutrons.
- Option C (1): This is the number of protons in tritium, not neutrons.
- Option D (0): Tritium must have neutrons to have a mass number greater than 1.

Key Concepts:
- Isotopes
- Atomic number
- Mass number
- Neutrons

Memory Aids:
- Mass number = Protons + Neutrons

Quick Tips:
- Remember that isotopes of an element have the same number of protons but different numbers of neutrons.

233 Easy

233 of 239

Understanding +4 | -1

Which one of the following set of ions represents a collection of isoelectronic species

View answer and explanation

Correct option: B

C{N^ - },\,C_2^{ - 2},\,{N_2}

Option B (

CN^-, C_2^{-2}, N_2

) is correct because all three species have the same number of electrons (14).

Solution approach

Correct Answer + Core Reason: $C{N^ - },\,C_2^{ - 2},\,{N_2}$ . All species have 14 electrons.

Step-by-Step Solution:

Step 1: Define isoelectronic species: Species with the same number of electrons.

Step 2: Calculate the number of electrons in each species:
- $CN^-$ : 6 (C) + 7 (N) + 1 (charge) = 14 electrons
- $C_2^{-2}$ : 2 * 6 (C) + 2 (charge) = 14 electrons
- $N_2$ : 2 * 7 (N) = 14 electrons

Why other options are wrong:
- **Option A ( $SO_4^{2 - },\,PO_4^{3 - },BO_3^{3 - }$ ):** These species do not have the same number of electrons. $SO_4^{2-}$ has 50, $PO_4^{3-}$ has 50, and $BO_3^{3-}$ has 32 electrons.
- **Option C ( ${O_2},{F_2}{N_2}$ ):** These species do not have the same number of electrons. $O_2$ has 16, $F_2$ has 18, and $N_2$ has 14 electrons.
- **Option D ( $CO_3^{2 - },NO_3^ - ,ClO_4^ -$ ):** These species do not have the same number of electrons. $CO_3^{2-}$ has 32, $NO_3^-$ has 32, and $ClO_4^-$ has 42 electrons.

Key Concepts:
- Isoelectronic species
- Counting electrons in ions and molecules

Memory Aids:
- Iso- means "same", so isoelectronic means "same electrons".

Quick Tips:
- Carefully count the electrons, including the charge on the ions.

234 Easy

234 of 239

Application +4 | -1

The de Broglie's wavelength of electron present in first Bohr orbit of

\mathrm{H}

atom is

View answer and explanation

Correct option: B

2 \pi \times 0.529 \, \mathop {\text{A}}\limits^ \circ

The correct answer is B;

2 \pi \times 0.529 \, \mathop{{\text{A}}}\limits^ \circ

. The de Broglie wavelength is equal to the circumference of the first Bohr orbit.

Solution approach

Correct Answer + Core Reason: $2 \pi \times 0.529 \, \mathop{{\text{A}}}\limits^ \circ$ . The de Broglie wavelength is found by relating it to the circumference of the first Bohr orbit.

Step-by-Step Solution:

Step 1: Recall Bohr's quantization condition: $mvr = \frac{{nh}}{{2\pi}}$ , where $n$ is the principal quantum number, $m$ is the mass of the electron, $v$ is its velocity, and $r$ is the radius of the orbit.

Step 2: For the first Bohr orbit, $n = 1$ , so $mvr = \frac{{h}}{{2\pi}}$ .

Step 3: Rearrange the equation to get $2\pi r = \frac{{h}}{{mv}}$ .

Step 4: Recognize that the de Broglie wavelength is given by $\lambda = \frac{{h}}{{mv}}$ .

Step 5: Substitute the de Broglie wavelength into the equation from Step 3: $2\pi r = \lambda$ .

Step 6: The radius of the first Bohr orbit is given as $r = 0.529 \, \mathop{{\text{A}}}\limits^ \circ$ .

Step 7: Substitute the value of $r$ into the equation: $\lambda = 2\pi \times 0.529 \, \mathop{{\text{A}}}\limits^ \circ$ .

Step 8: Final Answer: $2 \pi \times 0.529 \, \mathop{{\text{A}}}\limits^ \circ$

Why other options are wrong:
- Option A: This is the radius of the first Bohr orbit, not the de Broglie wavelength.
- Option C: This is the radius divided by $2\pi$ , which is not the de Broglie wavelength.
- Option D: This is twice the diameter of the first Bohr orbit, which is not the de Broglie wavelength.

Key Concepts:
- Bohr's quantization condition
- de Broglie wavelength

Memory Aids:
- de Broglie wavelength: $\lambda = \frac{{h}}{{mv}}$
- Bohr's quantization condition: $mvr = n\frac{{h}}{{2\pi}}$

Quick Tips:
- Remember the relationship between the de Broglie wavelength and the momentum of a particle.
- Know the value of the radius of the first Bohr orbit.

235 Easy

235 of 239

Application +4 | -1

If a species has 16 protons, 18 electrons and 16 neutrons. Find the species and its charge

View answer and explanation

Correct option: D

\mathrm{S}^{2-}

Option D (

\mathrm{S}^{2-}

) is correct because the species has 16 protons (Sulfur), and 18 electrons, indicating a 2- charge.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}^{2-}$ . The species is identified by the number of protons, and the charge is determined by the difference between the number of electrons and protons.

Step-by-Step Solution:

Step 1: Determine the element: The number of protons (16) identifies the element as Sulfur (S).

Step 2: Calculate the charge: The number of electrons (18) is 2 more than the number of protons (16), indicating a 2- charge.

Step 3: Write the species with its charge: $\mathrm{S}^{2-}$ .

Why other options are wrong:
- Option A: $\mathrm{S}^{1-}$ : This is incorrect because the species has 18 electrons, which is 2 more than the number of protons, indicating a 2- charge, not 1-.
- Option B: $\mathrm{Si}^{2+}$ : This is incorrect because the species has 16 protons, which corresponds to Sulfur (S), not Silicon (Si). Also, the charge is negative, not positive.
- Option C: $\mathrm{P}^{3-}$ : This is incorrect because the species has 16 protons, which corresponds to Sulfur (S), not Phosphorus (P).

Key Concepts:
- Atomic number
- Number of protons, neutrons, and electrons
- Ion formation

Memory Aids:
- Number of protons = Atomic number
- Charge = Number of protons - Number of electrons

Quick Tips:
- Use the number of protons to identify the element.

236 Easy

236 of 239

Application +4 | -1

The time period of revolution of electron in

\mathrm{H}

-atom is directly proportional to

View answer and explanation

Correct option: C

r^{3 / 2}

The correct answer is C (

r^{3 / 2}

) because the time period of revolution is proportional to n³ and the radius is proportional to n², therefore T is proportional to r^(3/2).

Solution approach

Correct Answer + Core Reason: $r^{3 / 2}$ . The time period of revolution is directly proportional to $r^{3/2}$ .

Step-by-Step Solution:

Step 1: Recall the relationship between the time period ( $T$ ) and the principal quantum number ( $n$ ): $T \propto n^3$ .

Step 2: Recall the relationship between the radius ( $r$ ) and the principal quantum number ( $n$ ): $r \propto n^2$ .

Step 3: Express $n$ in terms of $r$ : $n \propto \sqrt{r}$ .

Step 4: Substitute the expression for $n$ in terms of $r$ into the equation for $T$ : $T \propto (\sqrt{r})^3 = r^{3/2}$ .

Final Answer: $r^{3 / 2}$

Why other options are wrong:
- Option A (r): This is incorrect because the time period is not directly proportional to the radius.
- **Option B ( $r^{1 / 2}$ ): This is incorrect because the time period is not directly proportional to the square root of the radius.
- Option D ( $r^2$ ): This is incorrect because the time period is not directly proportional to the square of the radius.

Key Concepts:
- Bohr's model of the atom
- Time period of revolution of an electron
- Radius of an electron orbit
- Principal quantum number

Memory Aids:**
- Remember the relationships: $T \propto n^3$ and $r \propto n^2$ .

Quick Tips:
- Understand the relationships between the time period, radius, and principal quantum number in Bohr's model.
- Use the proportionality relationships to derive the relationship between $T$ and $r$ .

237 Easy

237 of 239

Understanding +4 | -1

For sets of values of quantum numbers

(\mathrm{n}, 1, \mathrm{~m}

and

\mathrm{s})

are given below. Which of these does not provide a permissible solution of the wave equation?

View answer and explanation

Correct option: B

3,3,1,-\frac{1}{2}

Option B (

3,3,1,-\frac{1}{2}

) is correct because the value of 'l' must be less than 'n'. In this case, n=3 and l=3, which is not permissible.

Solution approach

Correct Answer + Core Reason: $3,3,1,-\frac{1}{2}$ . The value of l must be less than n.

Step-by-Step Solution:

Step 1: Recall the rules for quantum numbers:
- n (principal quantum number): 1, 2, 3, ...
- l (azimuthal quantum number): 0, 1, 2, ..., n-1
- m (magnetic quantum number): -l, -l+1, ..., 0, ..., l-1, l
- s (spin quantum number): +1/2, -1/2

Step 2: Analyze each option:
- A. $3,2,-2, \frac{1}{2}$ : Possible (n=3, l=2, m=-2, s=1/2)
- B. $3,3,1,-\frac{1}{2}$ : Not possible (n=3, l=3 - l must be less than n)
- C. $3,2,1, \frac{1}{2}$ : Possible (n=3, l=2, m=1, s=1/2)
- D. $3,1,1,-\frac{1}{2}$ : Possible (n=3, l=1, m=1, s=-1/2)

Step 3: In option B, n=3 and l=3, which violates the rule that l must be less than n.

Why other options are wrong:
- **Option A ( $3,2,-2, \frac{1}{2}$ ): This is a valid set of quantum numbers.
- Option C ( $3,2,1, \frac{1}{2}$ ): This is a valid set of quantum numbers.
- Option D ( $3,1,1,-\frac{1}{2}$ ): This is a valid set of quantum numbers.

Key Concepts:
- Quantum numbers
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Magnetic quantum number (m)
- Spin quantum number (s)

Memory Aids:
- Remember that l < n.

Quick Tips:**
- Always check the relationship between n and l.

238 Easy

238 of 239

Understanding +4 | -1

State the number of protons, neutrons and electrons in

\mathrm{C}^{14}

.

View answer and explanation

Correct option: A

6, 8, 2006

Option A (6, 8, 6) is correct because Carbon-14 (

C^{14}

) has 6 protons (atomic number), 8 neutrons (mass number - atomic number), and 6 electrons (equal to the number of protons in a neutral atom).

Solution approach

Correct Answer + Core Reason: 6, 8, 6. Carbon-14 has 6 protons, 8 neutrons, and 6 electrons.

Step-by-Step Solution:

Step 1: Recall the definitions of atomic number, mass number, number of protons, number of neutrons, and number of electrons.

Step 2: Identify the atomic number of carbon (C): 6.

Step 3: Determine the number of protons: equal to the atomic number, so 6.

Step 4: Identify the mass number of carbon-14 ( $C^{14}$ ): 14.

Step 5: Calculate the number of neutrons: mass number - atomic number = 14 - 6 = 8.

Step 6: Determine the number of electrons: equal to the number of protons in a neutral atom, so 6.

Why other options are wrong:
- Option B: 6, 6, 9: Incorrect number of neutrons and electrons.
- Option C: 6, 7, 8: Incorrect number of neutrons and electrons.
- Option D: 7, 6, 9: Incorrect number of protons, neutrons, and electrons.

Key Concepts:
- Atomic number
- Mass number
- Protons
- Neutrons
- Electrons
- Isotopes

Memory Aids:
- Remember that the atomic number defines the element and equals the number of protons.

Quick Tips:
- Use the atomic number and mass number to quickly determine the number of protons, neutrons, and electrons in an atom or ion.

239 Easy

239 of 239

Application +4 | -1

According to Bohr’s theory, the angular momentum of an electron in 5th orbit is

View answer and explanation

Correct option: B

2.5\,h/\pi

Option B (

2.5\,h/\pi

) is correct because, according to Bohr's theory, the angular momentum of an electron in the nth orbit is given by

n*h/2\pi

, and for the 5th orbit, it's

5h/2\pi = 2.5h/\pi

.

Solution approach

Correct Answer + Core Reason: $2.5\,h/\pi$ . Angular momentum is calculated using Bohr's quantization condition.

Step-by-Step Solution:

Step 1: Recall Bohr's quantization condition for angular momentum.
$L = \frac{{nh}}{{2\pi}}$ where n is the orbit number.

Step 2: Substitute n = 5 for the 5th orbit.
$L = \frac{{5h}}{{2\pi}}$

Step 3: Simplify the expression.
$L = 2.5 \frac{{h}}{{\pi}}$

Why other options are wrong:
- Option A: $10\,h/\pi$ - This is incorrect because it seems to be multiplying n by 2 instead of dividing by 2 $\pi$ .
- Option C: $25\,h/\pi$ - This is incorrect, possibly obtained by squaring the orbit number.
- Option D: $h/\pi$ - This is incorrect, likely representing the angular momentum for n=2 orbit multiplied by 2.

Key Concepts:
- Bohr's Atomic Model
- Quantization of Angular Momentum

Memory Aids:
- Remember the formula: $L = \frac{{nh}}{{2\pi}}$

Quick Tips:
- Always use the correct formula for angular momentum in Bohr's model.
- Ensure you are using the correct value of 'n' for the given orbit.

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