STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 2

31 Easy

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Understanding +4 | -1

S-I: It is not possible to predict position and the velocity of an electron exactly and simultaneously.

S-II: Electron moving with high speed possesses both the particle nature and the wave nature.

View answer and explanation

Correct option: A

Both S-I and S-II are correct

The correct answer is A because S-I accurately describes Heisenberg's Uncertainty Principle, and S-II correctly states the wave-particle duality of electrons.

Solution approach

Correct Answer + Core Reason: A. Both S-I and S-II are correct. S-I describes Heisenberg's Uncertainty Principle, and S-II describes the wave-particle duality of electrons.

Step-by-Step Solution:

Step 1: Statement I: Heisenberg's Uncertainty Principle states that it is impossible to determine both the position and momentum (and therefore velocity) of a particle with perfect accuracy simultaneously. This statement is correct.

Step 2: Statement II: De Broglie's hypothesis and experimental evidence confirm that electrons exhibit both particle and wave nature. This statement is also correct.

Why other options are wrong:
- Option B: Both S-I and S-II are incorrect: Both statements are correct.
- Option C: S-I is correct and S-II is incorrect: S-II is also correct.
- Option D: S-I is incorrect and S-II is correct: S-I is also correct.

Key Concepts:
- Heisenberg's Uncertainty Principle
- Wave-Particle Duality

Memory Aids:
- Remember Heisenberg's Uncertainty Principle: You can't know both position and momentum perfectly.
- De Broglie's equation: $\lambda = \frac{h}{mv}$ shows the wave nature of particles.

Quick Tips:
- Understand the fundamental principles of quantum mechanics.

32 Easy

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Application +4 | -1

If Aufbau rule is not followed, 19th electron in

\operatorname{Sc}(Z=21)

will have

View answer and explanation

Correct option: C

n=3,l=2

Option C (

n=3,l=2

) is correct because if the Aufbau rule is not followed, the 19th electron would fill the 3d orbital, corresponding to n=3 and l=2.

Solution approach

Correct Answer + Core Reason: $n=3,l=2$ . If Aufbau rule is not followed, the 19th electron goes to the 3d orbital.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Sc (Z=21) following the Aufbau principle: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$ .

Step 2: Determine the electronic configuration of Sc (Z=21) if the Aufbau principle is not followed. Electrons fill the lower energy levels first within each shell before moving to the next shell. So, the order would be $1s, 2s, 2p, 3s, 3p, 3d, 4s...$

Step 3: Write the electronic configuration without following Aufbau rule: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$ .

Step 4: Identify the 19th electron. The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$ . The 19th electron is one of the 3d electrons.

Step 5: Determine the quantum numbers for the 3d electron. For a 3d electron, $n = 3$ and $l = 2$ (since $l = 0$ for s, $l = 1$ for p, and $l = 2$ for d).

Why other options are wrong:
- **Option A ( $n=3,l=0$ ): This corresponds to a 3s electron, but the 19th electron is a 3d electron when Aufbau rule is not followed.
- Option B ( $n=3,l=1$ ): This corresponds to a 3p electron, but the 19th electron is a 3d electron when Aufbau rule is not followed.
- Option D ( $n=4,l=0$ ): This corresponds to a 4s electron, which would be filled after the 3d electrons if Aufbau rule is followed.

Key Concepts:
- Aufbau Principle
- Electronic Configuration
- Quantum Numbers (n and l)

Memory Aids:
- Aufbau Principle: Electrons fill orbitals in order of increasing energy.
- Quantum Numbers: n = principal quantum number, l = azimuthal quantum number (0=s, 1=p, 2=d, 3=f).

Quick Tips:**
- Remember the order of filling orbitals according to the Aufbau principle.
- Know the relationship between the azimuthal quantum number (l) and the type of orbital (s, p, d, f).

33 Easy

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Remembering +4 | -1

The number of nucleons in the isotope of an atom

{}_z{X^m}

are

View answer and explanation

Correct option: A

m

Option A (m) is correct because the mass number (m) represents the total number of nucleons (protons and neutrons) in the nucleus of an atom.

Solution approach

Correct Answer + Core Reason: m. The mass number represents the total number of nucleons.

Step-by-Step Solution:

Step 1: Recall the definition of nucleons. Nucleons are the particles found in the nucleus of an atom, i.e., protons and neutrons.

Step 2: Understand the notation ${}_z{X^m}$ , where X is the element symbol, z is the atomic number (number of protons), and m is the mass number (number of protons + number of neutrons).

Step 3: Recognize that the mass number (m) represents the total number of nucleons in the atom.

Why other options are wrong:
- Option B (z): The atomic number (z) represents only the number of protons, not the total number of nucleons.
- Option C (m + z): This sum does not represent any meaningful quantity related to the nucleus.
- Option D (m - z): This difference represents the number of neutrons, not the total number of nucleons.

Key Concepts:
- Atomic Structure
- Nucleons (Protons and Neutrons)
- Atomic Number
- Mass Number

Memory Aids:
- Remember that the mass number is the total count of particles in the nucleus.

Quick Tips:
- Know the definitions of atomic number and mass number.

34 Easy

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Remembering +4 | -1

Electromagnetic radiation with maximum wavelength is

View answer and explanation

Correct option: B

Radio wave

Option B (Radio wave) is correct because radio waves have the maximum wavelength among the given options.

Solution approach

Correct Answer + Core Reason: Radio wave. Radio waves have the longest wavelength in the electromagnetic spectrum.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum.

Step 2: Arrange the given options in order of increasing wavelength: X-ray < UV < IR < Radio wave.

Step 3: Identify the radiation with the maximum wavelength.

Step 4: Radio waves have the maximum wavelength.

Why other options are wrong:
- Option A: UV has a shorter wavelength than radio waves.
- Option C: X-ray has the shortest wavelength among the given options.
- Option D: IR has a shorter wavelength than radio waves but longer than UV and X-rays.

Key Concepts:
- Electromagnetic spectrum
- Wavelength of electromagnetic radiation

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Microwave, IR, Visible, UV, X-ray, Gamma (from longest to shortest wavelength).

Quick Tips:
- Know the relative positions of different types of electromagnetic radiation in the spectrum.

35 Easy

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Remembering +4 | -1

The best metal to be used for photoemission is:

View answer and explanation

Correct option: C

Cesium

Option C (Cesium) is correct because Cesium has the lowest work function among the given options, making it the most efficient for photoemission.

Solution approach

Correct Answer + Core Reason: Cesium. Cesium has the lowest work function and is best for photoemission.

Step-by-Step Solution:
Step 1: Understand the question. The question asks for the best metal for photoemission.
Step 2: Recall the concept of work function. The work function is the minimum energy required to remove an electron from a solid.
Step 3: Relate work function to photoemission. A lower work function means less energy is needed for photoemission, making the metal more efficient.
Step 4: Consider the work functions of the given metals. Cesium has the lowest work function among the given options.
Step 5: Conclude the answer. Therefore, Cesium is the best metal for photoemission.

Why other options are wrong:
- Option A: Potassium has a higher work function than Cesium.
- Option B: Sodium has a higher work function than Cesium.
- Option D: Lithium has a higher work function than Cesium.

Key Concepts:
- Photoemission
- Work function
- Metals

Memory Aids:
- Remember that Cesium has a low work function and is good for photoemission.

Quick Tips:
- Consider the work function of metals when evaluating their suitability for photoemission.

36 Easy

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Analyzing +4 | -1

The electronic configuration

1{s^2},2{s^2}2{p^6},3{s^1}3{p^1}

correctly describes:

View answer and explanation

Correct option: C

Excited state of Mg

The correct answer is C (Excited state of Mg) because the ground state of Mg is

1s^22s^22p^63s^2

, and the given configuration has one electron promoted from 3s to 3p.

Solution approach

Correct Answer + Core Reason: Excited state of Mg. The given electronic configuration corresponds to an excited state of Magnesium.

Step-by-Step Solution:

Step 1: Determine the number of electrons in the given configuration: $1s^22s^22p^63s^13p^1$ has 2 + 2 + 6 + 1 + 1 = 12 electrons.

Step 2: Identify the element with 12 electrons: Magnesium (Mg) has an atomic number of 12.

Step 3: Write the ground state electronic configuration of Mg: $1s^22s^22p^63s^2$

Step 4: Compare the given configuration with the ground state configuration: The given configuration ( $1s^22s^22p^63s^13p^1$ ) has one electron promoted from the 3s orbital to the 3p orbital, indicating an excited state.

Why other options are wrong:
- Option A: Ground state of Na - Sodium (Na) has 11 electrons, so this is incorrect.
- Option B: Ground state of $\mathrm{Si}^+$ - Silicon (Si) has 14 electrons, and $\mathrm{Si}^+$ has 13 electrons, so this is incorrect.
- Option D: Excited state of $\mathrm{Al}^{3+}$ - Aluminum (Al) has 13 electrons, and $\mathrm{Al}^{3+}$ has 10 electrons, so this is incorrect.

Key Concepts:
- Electronic configuration
- Ground state
- Excited state
- Atomic number

Memory Aids:
- Remember that an excited state involves promoting electrons to higher energy levels.

Quick Tips:
- Determine the number of electrons first to identify the element.
- Compare the given configuration with the ground state configuration to determine if it's an excited state.

37 Easy

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Understanding +4 | -1

Wavelength associated with electron motion

View answer and explanation

Correct option: C

Decreases with increase of speed of

\mathrm{e}^{-}

(electron)

The correct answer is Decreases with increase of speed of

\mathrm{e}^{-}

(electron) because the wavelength associated with a moving electron is inversely proportional to its speed, as described by de Broglie's equation.

Solution approach

Correct Answer + Core Reason: Decreases with increase of speed of $\mathrm{e}^{-}$ (electron). De Broglie's equation shows an inverse relationship between wavelength and speed.

Step-by-Step Solution:
Step 1: Recall de Broglie's equation: $\lambda = \frac{{h}}{{mv}}$ , where $\lambda$ is the wavelength, $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity.
Step 2: Analyze the equation. Wavelength ( $\lambda$ ) is inversely proportional to the velocity ( $v$ ).
Step 3: Therefore, as the speed of the electron increases, the associated wavelength decreases.

Why other options are wrong:
- Option A: This is incorrect because wavelength and speed are inversely proportional.
- Option B: This is incorrect because the wavelength changes with the speed of the electron.
- Option D: This is incorrect because a moving electron has a non-zero wavelength.

Key Concepts:
- de Broglie's hypothesis
- Wave-particle duality
- Wavelength of matter waves

Memory Aids:
- Remember the inverse relationship: higher speed, shorter wavelength.

Quick Tips:
- de Broglie's equation is fundamental for understanding the wave nature of particles.

38 Easy

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Understanding +4 | -1

Combination of an α-particle with a nuclide results in the formation of a new nuclide which has:

View answer and explanation

Correct option: D

Higher atomic number

Option D (Higher atomic number) is correct because an alpha particle consists of 2 protons and 2 neutrons, so combining it with a nuclide increases the atomic number by 2.

Solution approach

Correct Answer + Core Reason: Higher atomic number. An alpha particle increases the atomic number by 2.

Step-by-Step Solution:

Step 1: Define an alpha particle.\nAn alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. It is represented as $^4_2He$ or $\alpha$ .

Step 2: Explain the effect of combining an alpha particle with a nuclide.\nWhen an alpha particle combines with a nuclide, the atomic number (number of protons) increases by 2, and the mass number (number of protons + neutrons) increases by 4.

Step 3: Analyze the given options.\nLess number of neutrons: The number of neutrons increases.\nEqual number of electrons: The number of electrons is not directly related to the combination with an alpha particle.\nLower mass number: The mass number increases.\nHigher atomic number: The atomic number increases.

Step 4: Conclude the correct answer.\nBased on the analysis, Option D is the correct answer.

Why other options are wrong:
- Option A: The number of neutrons increases, not decreases.
- Option B: The number of electrons changes to maintain charge balance, but it's not the primary effect.
- Option C: The mass number increases, not decreases.

Key Concepts:
- Alpha particle
- Atomic number
- Mass number
- Nuclear reactions

Memory Aids:
- Alpha particles are helium nuclei: 2 protons, 2 neutrons.

Quick Tips:
- Remember the composition of common nuclear particles (alpha, beta, gamma).

39 Easy

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Remembering +4 | -1

Compared to the mass of lightest nuclei, the mass of an electron is only (app.)

View answer and explanation

Correct option: C

\frac{1}{{1800}}

Option C (

\frac{1}{{1800}}

) is correct because the mass of an electron is approximately

\frac{1}{{1837}}

of the mass of the lightest nuclei (hydrogen).

Solution approach

Correct Answer + Core Reason: $\frac{1}{{1800}}$ . The mass of an electron is approximately 1/1800 of the mass of the lightest nuclei.

Step-by-Step Solution:
Step 1: Recall the relative masses of electrons and protons.
Step 2: The mass of an electron is approximately $\frac{1}{{1837}}$ of the mass of a proton (or hydrogen nucleus).
Step 3: Approximate $\frac{1}{{1837}}$ to $\frac{1}{{1800}}$ .

Why other options are wrong:
- Option A: $\frac{1}{{80}}$ is significantly larger than the actual ratio.
- Option B: $\frac{1}{{800}}$ is larger than the actual ratio.
- Option D: $\frac{1}{{2800}}$ is smaller than the actual ratio.

Key Concepts:
- Relative masses of subatomic particles
- Mass of electron
- Mass of proton

Memory Aids:
- Electron mass is very small compared to proton mass.

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

40 Easy

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Understanding +4 | -1

Which of the following is alway s a whole number

View answer and explanation

Correct option: D

Atomic number

Option D (Atomic number) is correct because the atomic number represents the number of protons in an atom's nucleus, which must be a whole number.

Solution approach

Correct Answer + Core Reason: Atomic number. Atomic number represents the number of protons, which must be a whole number.

Step-by-Step Solution:

Step 1: Define atomic weight.

Step 2: Atomic weight is the average mass of an atom of an element, which can be a decimal number due to the presence of isotopes.

Step 3: Define atomic radii.

Step 4: Atomic radii is the distance from the nucleus to the outermost electron, which can be a decimal number.

Step 5: Define equivalent weight.

Step 6: Equivalent weight is the molar mass divided by the n-factor, which can be a decimal number.

Step 7: Define atomic number.

Step 8: Atomic number is the number of protons in the nucleus of an atom, which must be a whole number.

Why other options are wrong:
- Option A: Atomic weight is incorrect because it can be a decimal number.
- Option B: Atomic radii is incorrect because it can be a decimal number.
- Option C: Equivalent weight is incorrect because it can be a decimal number.

Key Concepts:
- Atomic number
- Atomic weight
- Atomic radii
- Equivalent weight

Memory Aids:
- Atomic number = number of protons.

Quick Tips:
- Remember the definitions of atomic number, atomic weight, atomic radii, and equivalent weight.

41 Easy

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Analyzing +4 | -1

Assertion : A nuclear binding energy per nucleon is in the order

{ }_4^9 {Be}>_3^7 {Li}>_2^4 {He}

.

Reason : Binding energy per nuclear increases linearly with difference in number of neutrons and protons.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D is correct because both the assertion and the reason are false regarding nuclear binding energy and its relationship with neutron-proton difference.

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The binding energy per nucleon does not increase linearly with the difference in the number of neutrons and protons, and the given order of binding energies is incorrect.

Step-by-Step Solution:

Step 1: Analyze the assertion: The assertion states that the binding energy per nucleon is in the order ${ }_4^9 {Be}>_3^7 {Li}>_2^4 {He}$ . This is incorrect.

Step 2: Analyze the reason: The reason states that binding energy per nucleon increases linearly with the difference in the number of neutrons and protons. This is also incorrect. The binding energy per nucleon depends on several factors, and this statement is an oversimplification.

Step 3: Consider the actual binding energies: The binding energy per nucleon for ${ }_2^4 {He}$ is higher than that of ${ }_3^7 {Li}$ , which is higher than that of ${ }_4^9 {Be}$ .

Step 4: Conclusion: Since both the assertion and the reason are false, the correct answer is D.

Why other options are wrong:
- Option A: This is incorrect because both the assertion and the reason are false.
- Option B: This is incorrect because both the assertion and the reason are false.
- Option C: This is incorrect because both the assertion and the reason are false.

Key Concepts:
- Nuclear binding energy
- Binding energy per nucleon
- Nuclear stability

Memory Aids:
- Remember that ${ }_2^4 {He}$ is exceptionally stable.

Quick Tips:
- Always check the actual values or trends when dealing with binding energies.

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Remembering +4 | -1

The incorrect statement according to Dalton's atomic theory is

View answer and explanation

Correct option: D

Atoms can be created as well as destroyed in a chemical reaction

Option D (Atoms can be created as well as destroyed in a chemical reaction) is correct because Dalton's atomic theory states that atoms are neither created nor destroyed in a chemical reaction, only rearranged.

Solution approach

Correct Answer + Core Reason: Atoms can be created as well as destroyed in a chemical reaction. This statement contradicts Dalton's atomic theory.

Step-by-Step Solution:

Step 1: Recall Dalton's atomic theory.

Step 2: Dalton's atomic theory states:
- Matter is made up of indivisible particles called atoms.
- Atoms of a given element are identical.
- Atoms of different elements are different.
- Atoms are neither created nor destroyed in chemical reactions.

Step 3: The statement \"Atoms can be created as well as destroyed in a chemical reaction\" contradicts the fourth postulate of Dalton's atomic theory.

Why other options are wrong:
- Option A: Matter is made up of externally small, indivisible particles called atoms is a correct statement according to Dalton's atomic theory.
- Option B: Atoms of a given elements are identical in all respect is a correct statement according to Dalton's atomic theory.
- Option C: Atoms of different elements are different in all respects is a correct statement according to Dalton's atomic theory.

Key Concepts:
- Dalton's atomic theory
- Postulates of Dalton's atomic theory
- Conservation of mass

Memory Aids:
- Remember that Dalton's atomic theory states that atoms are \"indivisible\" and cannot be created or destroyed.

Quick Tips:
- Know the key postulates of Dalton's atomic theory.

43 Easy

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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 it is calculated using the formula

E = \frac{{hc}}{\lambda }

with correct unit conversions.

Solution approach

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

Step-by-Step Solution:

Step 1: Write down the formula for energy of a photon.
$E = \frac{{hc}}{\lambda }$

Step 2: Identify the given values.
$\lambda = 45 \;nm = 45 \times 10^{-9} \;m$ \nh = Planck's constant = $6.626 \times 10^{-34} \;Js$ \nc = speed of light = $3 \times 10^8 \;m/s$

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

Step 4: Calculate the energy.
$E = \frac{{19.878 \times 10^{-26}}}{{45 \times 10^{-9}}} \;J$
$E = 0.4417 \times 10^{-17} \;J$
$E = 4.417 \times 10^{-18} \;J \approx 4.42 \times 10^{-18} \;J$

Why other options are wrong:
- Option A: $6.67 \times {10^{15}}$ is incorrect due to a mistake in the calculation or incorrect units used.
- Option B: $6.67 \times {10^{11}}$ is incorrect due to a mistake in the calculation or incorrect units used.
- Option C: $4.42 \times {10^{ - 15}}$ is incorrect due to a mistake in the calculation, likely a unit conversion error.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda }$
- Planck's constant
- Speed of light
- Wavelength

Memory Aids:
- Remember the values of h and c.

Quick Tips:
- Ensure all units are in SI units before substituting into the formula.

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Remembering +4 | -1

Non-directional orbital is

View answer and explanation

Correct option: D

3s

Option D (3s) is correct because s-orbitals are spherically symmetrical and therefore non-directional.

Solution approach

Correct Answer + Core Reason: 3s orbital is non-directional because it is spherically symmetrical.

Step-by-Step Solution:

Step 1: Recall the shapes of different types of atomic orbitals.

Step 2: s-orbitals are spherical and have no directional preference.

Step 3: p-orbitals are dumbbell-shaped and are directional along the x, y, or z axes.

Step 4: d-orbitals have more complex shapes and are also directional.

Step 5: f-orbitals have even more complex shapes and are directional.

Why other options are wrong:
- Option A: 4p orbitals are directional.
- Option B: 4d orbitals are directional.
- Option C: 4f orbitals are directional.

Key Concepts:
- Atomic Orbitals
- Shapes of Atomic Orbitals (s, p, d, f)

Memory Aids:
- s: Spherical, non-directional.
- p, d, f: Directional.

Quick Tips:
- Visualize the shapes of the orbitals to determine their directionality.

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Remembering +4 | -1

Splitting of spectrum lines in magnetic field is

View answer and explanation

Correct option: C

Zeeman effect

Option C (Zeeman effect) is correct because the splitting of spectral lines in a magnetic field is known as the Zeeman effect.

Solution approach

Correct Answer + Core Reason: Zeeman effect. The splitting of spectral lines in a magnetic field is called the Zeeman effect.

Step-by-Step Solution:

Step 1: Recall the definitions of Stark, Raman, Zeeman, and Rutherford effects.

Step 2: The Zeeman effect is the splitting of spectral lines in the presence of a magnetic field.

Step 3: The Stark effect is the splitting of spectral lines in the presence of an electric field.

Step 4: The Raman effect involves the scattering of light by molecules with a change in energy.

Step 5: The Rutherford effect describes the scattering of alpha particles by atoms.

Why other options are wrong:
- Option A: Stark effect is the splitting of spectral lines in an electric field, not a magnetic field.
- Option B: Raman effect is the change in wavelength of light that occurs when a light beam is scattered by molecules.
- Option D: Rutherford effect describes the scattering of alpha particles by atoms.

Key Concepts:
- Zeeman effect
- Stark effect
- Raman effect
- Rutherford effect

Memory Aids:
- Zeeman = Magnetic field splitting

Quick Tips:
- Differentiate between the conditions for each effect.

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Understanding +4 | -1

When the value of azimuthal quantum number,

l=2

, value of '

n

' will be

View answer and explanation

Correct option: D

Any of these

Option D (Any of these) is correct because for a given azimuthal quantum number

l

, the principal quantum number

n

must be greater than or equal to

l+1

.

Solution approach

Correct Answer + Core Reason: Any of these. For $l=2$ , $n$ can be 3, 4, 5, or any higher integer.

Step-by-Step Solution:

Step 1: Define Azimuthal Quantum Number ( $l$ ): The azimuthal quantum number ( $l$ ) describes the shape of an atomic orbital and can have values from 0 to $n-1$ .

Step 2: Define Principal Quantum Number ( $n$ ): The principal quantum number ( $n$ ) describes the energy level of an electron and can have integer values 1, 2, 3, ...

Step 3: Relate $n$ and $l$ : For a given value of $n$ , $l$ can have values from 0 to $n-1$ . Conversely, for a given value of $l$ , $n$ must be greater than or equal to $l+1$ .

Step 4: Apply to the given problem: If $l=2$ , then $n \geq 3$ . Therefore, $n$ can be 3, 4, 5, or any higher integer.

Why other options are wrong:
- Option A: While 3 is a possible value for n, it's not the only one.
- Option B: While 4 is a possible value for n, it's not the only one.
- Option C: While 5 is a possible value for n, it's not the only one.

Key Concepts:
- Quantum Numbers
- Principal Quantum Number ( $n$ )
- Azimuthal Quantum Number ( $l$ )
- Atomic Orbitals

Memory Aids:
- $l$ ranges from 0 to $n-1$ .
- $n$ must be at least $l+1$ .

Quick Tips:
- Remember the relationship between $n$ and $l$ when determining possible quantum number combinations.

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Application +4 | -1

An excited hydrogen atom emits a photon of wavelength

\lambda

in returning to the ground state. If

R

is the Rydberg constant, then the quantum number

n

of the excited state is

View answer and explanation

Correct option: C

\sqrt{\frac{\lambda R}{\lambda R-1}}

Option C (

\sqrt{\frac{\lambda R}{\lambda R-1}}

) is correct because it is the quantum number derived from the Rydberg formula for the given transition.

Solution approach

Correct Answer + Core Reason: $\sqrt{\frac{\lambda R}{\lambda R-1}}$ . The quantum number of the excited state is derived from the Rydberg formula.

Step-by-Step Solution:

Step 1: The hydrogen atom returns to the ground state (n=1) and emits a photon of wavelength $\lambda$ .

Step 2: Use the Rydberg formula: $\frac{1}{\lambda} = R \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ , where $R$ is the Rydberg constant, $Z$ is the atomic number (1 for hydrogen), $n_1$ is the final energy level (1), and $n_2$ is the initial energy level (n).

Step 3: Substitute the values: $\frac{1}{\lambda} = R \times 1^2 \times (\frac{1}{1^2} - \frac{1}{n^2}) \Rightarrow \frac{1}{\lambda} = R (1 - \frac{1}{n^2})$ .

Step 4: Solve for n: $\frac{1}{\lambda R} = 1 - \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R} \Rightarrow n^2 = \frac{\lambda R}{\lambda R - 1} \Rightarrow n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$ .

Step 5: Final Answer: The quantum number n of the excited state is $\sqrt{\frac{\lambda R}{\lambda R - 1}}$ .

Why other options are wrong:
- **Option A ( $\sqrt{\lambda R}$ ): This is incorrect due to an incorrect rearrangement of the Rydberg formula.
- Option B ( $\sqrt{\lambda R-1}$ ): This is incorrect due to an incorrect rearrangement of the Rydberg formula.
- Option D ( $\sqrt{\lambda R(\lambda R-1)}$ ): This is incorrect due to an incorrect rearrangement of the Rydberg formula.

Key Concepts:
- Rydberg formula
- Hydrogen spectrum
- Quantum numbers

Memory Aids:**
- Remember the Rydberg formula: $\frac{1}{\lambda} = R (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ .

Quick Tips:
- Identify the initial and final energy levels correctly.
- Rearrange the Rydberg formula carefully to solve for the unknown.

48 Easy

48 of 239

Understanding +4 | -1

Which of the following ion is expected to be coloured?

View answer and explanation

Correct option: D

\mathrm{V}^{2+}(Z=23)

Option D is correct because

V^{2+}

has unpaired d-electrons, allowing for d-d transitions and thus exhibiting color.

Solution approach

Correct Answer + Core Reason: $\mathrm{V}^{2+}(Z=23)$ . Ions with unpaired d-electrons are expected to be colored due to d-d transitions.

Step-by-Step Solution:
Step 1: Write the electronic configuration of each ion:
- $\mathrm{Zn}^{2+}$ : $[Ar] 3d^{10}$ (no unpaired electrons)
- $\mathrm{Ca}^{2+}$ : $[Ar]$ (no unpaired electrons)
- $\mathrm{Sn}^{2+}$ : $[Kr] 4d^{10} 5s^2$ (inert pair effect, but no unpaired d electrons)
- $\mathrm{V}^{2+}$ : $[Ar] 3d^3$ (3 unpaired electrons)
Step 2: Ions with unpaired d-electrons can undergo d-d transitions, absorbing light and appearing colored.
Step 3: $\mathrm{V}^{2+}$ has 3 unpaired d-electrons, so it is expected to be colored.

Why other options are wrong:
- Option A: $\mathrm{Zn}^{2+}$ has a completely filled d-orbital and therefore is colorless.
- Option B: $\mathrm{Ca}^{2+}$ has no d-electrons and therefore is colorless.
- Option C: $\mathrm{Sn}^{2+}$ has a completely filled d-orbital and therefore is colorless.

Key Concepts:
- d-d transitions
- Electronic configuration
- Color of transition metal ions

Memory Aids:
- Unpaired d-electrons = colored ion.

Quick Tips:
- Check for unpaired d-electrons to determine if an ion is colored.

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One electron volt is:

View answer and explanation

Correct option: B

1.6 \times {10^{ - 12}}\mathrm{erg}

Option B (

1.6 \times {10^{ - 12}}\mathrm{erg}

) is correct because 1 eV is equal to

1.602 \times 10^{-19}

J, which is equivalent to

1.602 \times 10^{-12}

erg.

Solution approach

Correct Answer + Core Reason: $1.6 \times {10^{ - 12}}\mathrm{erg}$ . This is the correct conversion of electron volt to erg.

Step-by-Step Solution:

Step 1: Define electron volt (eV). One electron volt is the amount of energy gained or lost by a single electron when it traverses an electric potential difference of one volt.

Step 2: Convert eV to Joules (J). 1 eV = $1.602 \times 10^{-19}$ J

Step 3: Convert Joules (J) to ergs. 1 J = $10^7$ ergs

Step 4: Multiply the value in Joules by the conversion factor to get the value in ergs: $1.602 \times 10^{-19} \text{ J} \times 10^7 \text{ ergs/J} = 1.602 \times 10^{-12} \text{ ergs}$

Step 5: Final Answer: $1.6 \times {10^{ - 12}}\mathrm{erg}$

Why other options are wrong:
- Option A: $1.6 \times {10^{ - 19}}\;\mathrm{erg}$ : This is the value of eV in Joules, not ergs.
- Option C: $1.6 \times {10^{ - 8}}\mathrm{\;erg}$ : This is an incorrect value.
- Option D: $1.6 \times {10^8}\;\mathrm{erg}$ : This is an incorrect value.

Key Concepts:
- Electron volt (eV)
- Conversion between energy units (Joules and ergs)

Memory Aids:
- Remember the conversion factor: 1 eV = $1.602 \times 10^{-19}$ J and 1 J = $10^7$ ergs

Quick Tips:
- Always pay attention to the units and use the correct conversion factors.

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Transition from

n = 4,5,6

to n=3 in hydrogen spectrum gives:

View answer and explanation

Correct option: B

Paschen series

Option B (Paschen series) is correct because the Paschen series corresponds to transitions where electrons fall to the n=3 energy level.

Solution approach

Correct Answer + Core Reason: Paschen series corresponds to transitions to the n=3 energy level.

Step-by-Step Solution:

Step 1: Recall the different series in the hydrogen spectrum.

Step 2: Lyman series: transitions to n=1\nBalmer series: transitions to n=2\nPaschen series: transitions to n=3\nBrackett series: transitions to n=4\nPfund series: transitions to n=5

Step 3: Since the transition is from n = 4, 5, 6 to n = 3, it belongs to the Paschen series.

Why other options are wrong:
- Option A: Lyman series corresponds to transitions to n=1.
- Option C: Balmer series corresponds to transitions to n=2.
- Option D: Pfund series corresponds to transitions to n=5.

Key Concepts:
- Hydrogen spectrum
- Lyman series
- Balmer series
- Paschen series
- Brackett series
- Pfund series

Memory Aids:
- Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5)

Quick Tips:
- Remember the energy levels corresponding to each series in the hydrogen spectrum.

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Orbital angular momentum depends on

View answer and explanation

Correct option: A

l

Option A (

l

) is correct because the orbital angular momentum depends only on the azimuthal quantum number (

l

).

Solution approach

Correct Answer + Core Reason: Orbital angular momentum depends on $l$ .

Step-by-Step Solution:

Step 1: Recall the formula for orbital angular momentum.

Step 2: Understand that the orbital angular momentum is given by $\sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }}$ , where $l$ is the azimuthal quantum number.

Step 3: Observe that the formula depends only on $l$ .

Why other options are wrong:
- Option B: $n$ and $l$ is incorrect because orbital angular momentum does not depend on the principal quantum number ( $n$ ).
- Option C: $n$ and $m$ is incorrect because orbital angular momentum does not depend on the principal ( $n$ ) or magnetic ( $m$ ) quantum numbers.
- Option D: $m$ and $s$ is incorrect because orbital angular momentum does not depend on the magnetic ( $m$ ) or spin ( $s$ ) quantum numbers.

Key Concepts:
- Orbital angular momentum
- Azimuthal quantum number ( $l$ )
- Quantum numbers

Memory Aids:
- Remember the formula for orbital angular momentum: $\sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }}$

Quick Tips:
- Know the relationship between quantum numbers and atomic properties.

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Rutherford’s experiment on the scattering of α-particles showed for the first time that the atom has

View answer and explanation

Correct option: C

Nucleus

Option C (Nucleus) is correct because Rutherford's alpha-particle scattering experiment demonstrated the existence of a small, dense, positively charged nucleus within the atom.

Solution approach

Correct Answer + Core Reason: Nucleus. Rutherford's experiment revealed the presence of a nucleus in the atom.

Step-by-Step Solution:

Step 1: Recall Rutherford's alpha-particle scattering experiment.

Step 2: Understand the experimental setup and observations. Alpha particles were directed at a thin gold foil.

Step 3: Remember the conclusions drawn from the experiment. Most alpha particles passed through undeflected, some were deflected at small angles, and a few were deflected at large angles.

Step 4: Relate the observations to the structure of the atom. The large deflections indicated the presence of a small, dense, positively charged nucleus.

Why other options are wrong:
- Option A: Electrons were discovered earlier by J.J. Thomson.
- Option B: Protons are found in the nucleus, but Rutherford's experiment primarily demonstrated the existence of the nucleus itself.
- Option D: Neutrons were discovered later by Chadwick.

Key Concepts:
- Rutherford's alpha-particle scattering experiment
- Atomic structure
- Nucleus
- Discovery of the nucleus

Memory Aids:
- Remember that Rutherford's experiment led to the discovery of the nucleus.

Quick Tips:
- Know the key experiments that led to the discovery of subatomic particles.

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Deuterium nucleus contains:

View answer and explanation

Correct option: B

1 proton, 1 neutron

Option B (1 proton, 1 neutron) is correct because a deuterium nucleus is an isotope of hydrogen with one proton and one neutron.

Solution approach

Correct Answer + Core Reason: 1 proton, 1 neutron. Deuterium is an isotope of hydrogen with one proton and one neutron.

Step-by-Step Solution:

Step 1: Define deuterium.

Step 2: Deuterium is an isotope of hydrogen with a mass number of 2.

Step 3: Recall the composition of a nucleus.

Step 4: A nucleus contains protons and neutrons.

Step 5: Hydrogen has 1 proton and 0 neutrons. Deuterium has 1 proton and 1 neutron.

Why other options are wrong:
- Option A: 1 proton, 1 electron is incorrect because electrons are not in the nucleus.
- Option C: 2 protons, 1 electron is incorrect because deuterium has only 1 proton and electrons are not in the nucleus.
- Option D: 1 proton, 2 electrons is incorrect because deuterium has only 1 proton and electrons are not in the nucleus.

Key Concepts:
- Isotopes
- Atomic structure
- Nucleus composition

Memory Aids:
- Deuterium has 1 proton and 1 neutron.

Quick Tips:
- Remember the definitions of isotopes and their nuclear composition.

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Splitting of spectral line under the influence of magnetic field is called ____ while in electric field is called _____ resp.

View answer and explanation

Correct option: B

Zeernan effect, stark effect

Option B (Zeernan effect, stark effect) is correct because the splitting of spectral lines in a magnetic field is called the Zeeman effect, while the splitting in an electric field is called the Stark effect.

Solution approach

Correct Answer + Core Reason: Zeernan effect, stark effect. These are the definitions of spectral line splitting in magnetic and electric fields, respectively.

Step-by-Step Solution:

Step 1: Recall the definition of the Zeeman effect.

Step 2: The Zeeman effect is the splitting of spectral lines when a static magnetic field is applied.

Step 3: Recall the definition of the Stark effect.

Step 4: The Stark effect is the splitting of spectral lines when a static electric field is applied.

Why other options are wrong:
- Option A: Stark effect, Zeeman effect: This reverses the correct order.
- Option C: Compton effect, stark effect: The Compton effect is related to the scattering of photons by charged particles.
- Option D: Zeeman effect, photoelectric effect: The photoelectric effect is the emission of electrons when light hits a material.

Key Concepts:
- Zeeman effect
- Stark effect
- Spectroscopy

Memory Aids:
- Zeeman effect: Think \"Zeeman = Magnetic\" (both start with Z and M are close in the alphabet).
- Stark effect: Think \"Stark = Electric\" (both start with S and E).

Quick Tips:
- Memorize the definitions of the Zeeman and Stark effects.

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Application +4 | -1

Which has the maximum magnetic moment

View answer and explanation

Correct option: A

M{n^{ + 2}}

Option A (

M{n^{ + 2}}

) is correct because it has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

Solution approach

Correct Answer + Core Reason: $M{n^{ + 2}}$ . It has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the given ions:
- $Mn^{2+}$ : [Ar] $3d^5$
- $Fe^{2+}$ : [Ar] $3d^6$
- $Ti^{2+}$ : [Ar] $3d^2$
- $Cr^{2+}$ : [Ar] $3d^4$

Step 2: Determine the number of unpaired electrons in each ion:
- $Mn^{2+}$ : 5 unpaired electrons
- $Fe^{2+}$ : 4 unpaired electrons
- $Ti^{2+}$ : 2 unpaired electrons
- $Cr^{2+}$ : 4 unpaired electrons

Step 3: Recall the formula for magnetic moment: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Step 4: Since the magnetic moment is directly proportional to the number of unpaired electrons, the ion with the maximum number of unpaired electrons will have the maximum magnetic moment.

Step 5: $Mn^{2+}$ has the maximum number of unpaired electrons (5).

Step 6: Final Answer: $M{n^{ + 2}}$

Why other options are wrong:
- **Option B ( $F{e^{ + 2}}$ ):** This is incorrect because $Fe^{2+}$ has 4 unpaired electrons, which is less than $Mn^{2+}$ .
- **Option C ( $T{i^{ + 2}}$ ):** This is incorrect because $Ti^{2+}$ has only 2 unpaired electrons.
- **Option D ( $C{r^{ + 2}}$ ):** This is incorrect because $Cr^{2+}$ has 4 unpaired electrons, which is less than $Mn^{2+}$ .

Key Concepts:
- Electronic Configuration
- Hund's Rule
- Magnetic Moment
- Unpaired Electrons

Memory Aids:
- Remember that magnetic moment increases with the number of unpaired electrons.

Quick Tips:
- Write the electronic configuration correctly to determine the number of unpaired electrons.
- Use Hund's rule to fill the orbitals.

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Penetration power of proton is:

View answer and explanation

Correct option: B

Less than electron

Option B (Less than electron) is correct because protons are much more massive than electrons, and therefore have lower penetration power.

Solution approach

Correct Answer + Core Reason: Less than electron. Protons are more massive than electrons.

Step-by-Step Solution:

Step 1: Define penetration power: Penetration power refers to the ability of a particle to pass through matter.

Step 2: Consider the factors affecting penetration power: Penetration power depends on the particle's mass, charge, and energy. Higher mass and charge generally decrease penetration power.

Step 3: Compare proton and electron: A proton has a much larger mass than an electron (approximately 1836 times greater). Both have the same magnitude of charge, but opposite signs.

Step 4: Conclude about penetration power: Due to its significantly larger mass, a proton has less penetration power than an electron.

Why other options are wrong:
- Option A: Penetration power of a proton is not greater than an electron.
- Option C: Penetration power of a proton is not greater than a neutron. Neutrons have no charge and can penetrate further than protons.
- Option D: The correct answer is B.

Key Concepts:
- Penetration power
- Mass and charge of particles
- Properties of protons and electrons

Memory Aids:
- Heavier particles penetrate less.

Quick Tips:
- Remember that mass is a primary factor affecting penetration power.

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The total number of nodes are given by

View answer and explanation

Correct option: C

(n-1)

Option C (

(n-1)

) is correct because the total number of nodes in an atomic orbital is given by (n-1), where n is the principal quantum number.

Solution approach

Correct Answer + Core Reason: $(n-1)$ . The total number of nodes is given by $n-1$ .

Step-by-Step Solution:

Step 1: Define nodes: Nodes are regions in space where the probability of finding an electron is zero.

Step 2: Identify types of nodes: There are two types of nodes: radial nodes and angular nodes.

Step 3: State the formula for total number of nodes: Total number of nodes = n - 1, where n is the principal quantum number.

Step 4: State the formula for number of angular nodes: Number of angular nodes = l, where l is the azimuthal quantum number.

Step 5: State the formula for number of radial nodes: Number of radial nodes = n - l - 1.

Step 6: Relate the formulas: Total number of nodes = Number of radial nodes + Number of angular nodes = (n - l - 1) + l = n - 1.

Why other options are wrong:
- Option A: $(n+1)$ is incorrect; it does not represent the total number of nodes.
- Option B: $(n-l-1)$ represents the number of radial nodes, not the total number of nodes.
- Option D: $(n-l+1)$ is incorrect; it does not represent the total number of nodes.

Key Concepts:
- Nodes
- Radial nodes
- Angular nodes
- Principal quantum number (n)
- Azimuthal quantum number (l)

Memory Aids:
- Total nodes = n - 1

Quick Tips:
- Remember the formulas for total, radial, and angular nodes.

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The quantum number which is responsible for the size of electron cloud is

View answer and explanation

Correct option: C

Principal

Option C (Principal) is correct because the principal quantum number (n) determines the energy level and hence the size of the electron cloud.

Solution approach

Correct Answer + Core Reason: The principal quantum number determines the size of the electron cloud.

Step-by-Step Solution:

Step 1: Define the quantum numbers.

Step 2: The principal quantum number ( $n$ ) determines the energy level and the size of the electron cloud.

Step 3: The azimuthal quantum number ( $l$ ) determines the shape of the electron cloud.

Step 4: The magnetic quantum number ( $m_l$ ) determines the orientation of the electron cloud in space.

Step 5: The spin quantum number ( $m_s$ ) describes the intrinsic angular momentum of the electron.

Why other options are wrong:
- Option A: Spin quantum number describes the spin of the electron.
- Option B: Azimuthal quantum number describes the shape of the electron cloud.
- Option D: Magnetic quantum number describes the orientation of the electron cloud.

Key Concepts:
- Quantum numbers
- Atomic structure

Memory Aids:
- n = size, l = shape, ml = orientation, ms = spin

Quick Tips:
- Know the significance of each quantum number.

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Probability of finding an electron at the nodal surface is:

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because the probability of finding an electron at a nodal surface is zero, as it's a region where the electron density is zero.

Solution approach

Correct Answer + Core Reason: zero. Electron density is zero at a node.

Step-by-Step Solution:

Step 1: Define Nodal Surface.

Step 2: A nodal surface is a region where the probability of finding an electron is zero.

Step 3: Therefore, the probability at the nodal surface is zero.

Why other options are wrong:
- Option A: unity means 100% probability, which is incorrect at a node.
- Option B: low is incorrect because the probability is exactly zero.
- Option C: high is incorrect because the probability is zero.

Key Concepts:
- Atomic Orbitals
- Nodal Surfaces

Memory Aids:
- Node = zero probability.

Quick Tips:
- Understand the shapes of atomic orbitals and the location of nodes.

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Application +4 | -1

The number of orbitals having

(n+l) < 5

is

View answer and explanation

Correct option: D

10

Option D (10) is correct because it accurately counts the number of orbitals satisfying the condition (n+l) < 5 based on the quantum numbers n and l.

Solution approach

Correct Answer + Core Reason: 10. The number of orbitals is determined by considering all possible combinations of n and l such that (n+l) < 5.

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the azimuthal quantum number (l): $n = 1, 2, 3, ...$ and $l = 0, 1, 2, ..., (n-1)$ .

Step 2: Consider the case when $n = 1$ . Then $l = 0$ . So, $n + l = 1 < 5$ . The number of orbitals is 1 (1s).

Step 3: Consider the case when $n = 2$ . Then $l = 0$ and $l = 1$ . For $l = 0$ , $n + l = 2 < 5$ . The number of orbitals is 1 (2s). For $l = 1$ , $n + l = 3 < 5$ . The number of orbitals is 3 (2p).

Step 4: Consider the case when $n = 3$ . Then $l = 0, 1, 2$ . For $l = 0$ , $n + l = 3 < 5$ . The number of orbitals is 1 (3s). For $l = 1$ , $n + l = 4 < 5$ . The number of orbitals is 3 (3p). For $l = 2$ , $n + l = 5$ , which does not satisfy the condition.

Step 5: Consider the case when $n = 4$ . Then $l = 0, 1, 2, 3$ . For $l = 0$ , $n + l = 4 < 5$ . The number of orbitals is 1 (4s). For $l = 1$ , $n + l = 5$ , which does not satisfy the condition.

Step 6: Sum the number of orbitals: $1 (1s) + 1 (2s) + 3 (2p) + 1 (3s) + 3 (3p) + 1 (4s) = 1 + 1 + 3 + 1 + 3 + 1 = 10$ .

Why other options are wrong:
- Option A (9): This is an incorrect count of the orbitals satisfying the condition.
- Option B (8): This is an incorrect count of the orbitals satisfying the condition.
- Option C (4): This is an incorrect count of the orbitals satisfying the condition.

Key Concepts:
- Quantum numbers (n and l)
- Atomic orbitals

Memory Aids:
- Remember the possible values of l for a given n: $l = 0, 1, 2, ..., (n-1)$ .

Quick Tips:
- Systematically consider each value of n and its corresponding l values to avoid missing any orbitals.

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