The number of protons, neutrons and electrons in ${}_{71}^{175}{\rm{Lu}}$ , respectively, are

${\rm{71,}}\,\,{\rm{104}}\,\,{\rm{and}}\,\,{\rm{71}}$

${\rm{104,}}\,\,{\rm{71}}\,\,{\rm{and}}\,\,{\rm{71}}$

${\rm{71,}}\,\,{\rm{71}}\,\,{\rm{and}}\,\,{\rm{104}}$

${\rm{175,}}\,\,{\rm{104}}\,\,{\rm{and}}\,\,{\rm{71}}$

STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 4

91 Easy

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Understanding +4 | -1

Assertion: Isobars have the same atomic mass number but different atomic numbers.

Reason: Isobars have different numbers of neutrons.

View answer and explanation

Correct option: A

Both Assertion and Reason are true and Reason is the correct explanation for Assertion

Option A (Both Assertion and Reason are true and Reason is the correct explanation for Assertion) is correct because isobars have the same mass number but different atomic numbers due to having different numbers of protons and neutrons, with the total number of nucleons (protons + neutrons) being the same.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is the correct explanation for Assertion. Isobars have the same mass number but different atomic numbers because they have different numbers of neutrons.

Step-by-Step Solution:

Step 1: Define isobars.\nIsobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z).

Step 2: Explain the relationship between mass number, atomic number, and number of neutrons.\nThe mass number (A) is the sum of the number of protons (Z) and the number of neutrons (N) in the nucleus of an atom: A = Z + N.\nSince isobars have the same mass number but different atomic numbers, they must have different numbers of neutrons to compensate for the difference in the number of protons.

Step 3: Evaluate the assertion and reason.\nThe assertion that isobars have the same atomic mass number but different atomic numbers is correct.\nThe reason that isobars have different numbers of neutrons is also correct and explains why they have the same mass number despite having different atomic numbers.

Step 4: Conclude the correct answer.\nBased on the evaluation, Option A is the correct answer.

Why other options are wrong:
- Option B: The reason is the correct explanation for the assertion.
- Option C: The reason is true.
- Option D: The assertion is true.

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

Memory Aids:
- Isobars have the same \"mass number,\" so think \"same mass.\"

Quick Tips:
- Remember the definitions of isotopes, isobars, and isotones.

92 Easy

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Remembering +4 | -1

The maximum number of electrons possible in a sublevel is equal to

View answer and explanation

Correct option: D

4l + 2

Option D (

4l + 2

) is correct because the number of orbitals in a sublevel is 2l+1, and each orbital can hold 2 electrons, so the maximum number of electrons is 2(2l+1) = 4l+2.

Solution approach

Correct Answer + Core Reason: $4l + 2$ . The maximum number of electrons in a sublevel is determined by the number of orbitals and the Pauli exclusion principle.

Step-by-Step Solution:

Step 1: Recall the relationship between the azimuthal quantum number ( $l$ ) and the number of orbitals in a sublevel.

Step 2: The number of orbitals in a sublevel is given by $2l + 1$ .

Step 3: Each orbital can hold a maximum of 2 electrons (Pauli exclusion principle).

Step 4: Therefore, the maximum number of electrons in a sublevel is $2(2l + 1) = 4l + 2$ .

Why other options are wrong:
- Option A: $2l + 1$ represents the number of orbitals in a sublevel, not the number of electrons.
- Option B: $2{n^2}$ represents the maximum number of electrons in a shell, not a sublevel.
- Option C: $2{l^2}$ is not a valid formula for the number of electrons in a sublevel.

Key Concepts:
- Quantum numbers
- Atomic orbitals
- Electron configuration
- Pauli exclusion principle

Memory Aids:
- Remember that the number of electrons in a sublevel is $4l + 2$ .

Quick Tips:
- Know the allowed values of $l$ for a given $n$ .

93 Easy

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Understanding +4 | -1

Assertion :- Bohr model is not suitable in case of multi-electron species.

Reason :- It does not tells about electron-electron interaction.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because Bohr's model, based on hydrogen-like species, fails to account for electron-electron interactions present in multi-electron systems, making it unsuitable for them.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. Bohr's model is limited to single-electron species due to its inability to handle electron-electron interactions.

Step-by-Step Solution:

Step 1: The assertion states that the Bohr model is not suitable for multi-electron species.

Step 2: The reason states that the Bohr model does not account for electron-electron interactions.

Step 3: The Bohr model is based on the assumption of a single electron orbiting a nucleus, neglecting the complex interactions between multiple electrons.

Step 4: Electron-electron interactions significantly affect the energy levels and behavior of electrons in atoms with more than one electron.

Step 5: Therefore, the reason correctly explains why the Bohr model is inadequate for multi-electron species.

Why other options are wrong:
- Option B: While both statements are true, the reason *is* the correct explanation.
- Option C: The assertion is true, so this option is incorrect.
- Option D: Both the assertion and reason are true, so this option is incorrect.

Key Concepts:
- Bohr Model Limitations
- Electron-Electron Interactions
- Multi-Electron Systems

Memory Aids:
- Bohr = Single Electron

Quick Tips:
- Remember that the Bohr model is a simplified model and has limitations, especially when dealing with complex atoms.

94 Easy

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Understanding +4 | -1

Assertion: Dalton's Atomic Theory states that atoms are indivisible and indestructible.
Reason: This statement is based on the observation that in chemical reactions, atoms are rearranged but not destroyed or created.

View answer and explanation

Correct option: A

Both Assertion and Reason are true and Reason is the correct explanation for Assertion

Option A is correct because Dalton's theory indeed states atoms are indivisible and indestructible, and this is based on observations of chemical reactions.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is the correct explanation for Assertion. Dalton's atomic theory states that atoms are indivisible and indestructible, and this is based on observations of chemical reactions.

Step-by-Step Solution:

Step 1: Recall Dalton's Atomic Theory. Dalton's Atomic Theory includes the postulates that atoms are indivisible and indestructible, and that in chemical reactions, atoms are rearranged but not created or destroyed.

Step 2: Evaluate the Assertion. The assertion that Dalton's Atomic Theory states that atoms are indivisible and indestructible is correct.

Step 3: Evaluate the Reason. The reason that this statement is based on the observation that in chemical reactions, atoms are rearranged but not destroyed or created is also correct.

Step 4: Determine if the Reason explains the Assertion. The reason directly explains why Dalton believed atoms were indivisible and indestructible, based on his observations of chemical reactions.

Why other options are wrong:
- Option B (Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion): The reason *is* the correct explanation for the assertion.
- Option C (Assertion is true, but Reason is false): Both the assertion and the reason are true.
- Option D (Assertion is false, but Reason is true): Both the assertion and the reason are true.

Key Concepts:
- Dalton's Atomic Theory
- Conservation of Atoms in Chemical Reactions

Memory Aids:
- Remember the key postulates of Dalton's Atomic Theory.

Quick Tips:
- Understand the historical context and experimental basis of scientific theories.

95 Easy

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Remembering +4 | -1

Maximum number of electrons in an orbit is given by:

View answer and explanation

Correct option: B

2{n^2}

Option B (

2{n^2}

) is correct because the maximum number of electrons that can occupy an orbit (energy level) is given by the formula

2n^2

, where n is the principal quantum number.

Solution approach

Correct Answer + Core Reason: $2{n^2}$ . The maximum number of electrons in an orbit is given by this formula.

Step-by-Step Solution:

Step 1: Recall the formula for the maximum number of electrons in an orbit.

Step 2: The formula is $2n^2$ , where n is the principal quantum number.

Step 3: This formula is derived from Pauli's exclusion principle and the number of orbitals in a given energy level.

Why other options are wrong:
- Option A: ${n^2}$ gives the number of orbitals in an energy level, not the maximum number of electrons.
- Option C: ${n^2}/2$ is not a valid formula for the number of electrons in an orbit.
- Option D: The correct answer is one of the given options.

Key Concepts:
- Electronic configuration
- Pauli's exclusion principle
- Quantum numbers

Memory Aids:
- $2n^2$ = Max electrons in an orbit

Quick Tips:
- Remember the relationship between principal quantum number and the number of electrons.

96 Easy

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Understanding +4 | -1

The quantum number values for the designation

3 d

are

View answer and explanation

Correct option: D

n=3, l=2

Option D (

n=3, l=2

) is correct because for a 3d orbital, the principal quantum number (n) is 3, and the azimuthal quantum number (l) is 2 (d corresponds to l=2).

Solution approach

Correct Answer + Core Reason: $n=3, l=2$ . For a 3d orbital, n=3 and l=2.

Step-by-Step Solution:

Step 1: Define the quantum numbers.
- Principal quantum number (n): Determines the energy level of the electron (n = 1, 2, 3, ...).
- Azimuthal quantum number (l): Determines the shape of the orbital (l = 0, 1, 2, ..., n-1). l = 0 corresponds to s orbital, l = 1 corresponds to p orbital, l = 2 corresponds to d orbital, l = 3 corresponds to f orbital.

Step 2: Identify the values of n and l for the 3d orbital.
- For 3d, n = 3 (from the number 3).
- For 3d, l = 2 (since d corresponds to l = 2).

Why other options are wrong:
- **Option A ( $n=2, m=-3$ ): n=3 for 3d orbital. Also, m is the magnetic quantum number, not the azimuthal quantum number.
- Option B ( $n=3, l=3$ ): For n=3, the possible values of l are 0, 1, and 2. l=3 is not possible.
- Option C ( $n=4, l=1$ ): n=3 for 3d orbital.

Key Concepts:
- Quantum numbers
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Atomic orbitals

Memory Aids:
- n = energy level
- l = shape (s=0, p=1, d=2, f=3)

Quick Tips:**
- Remember the relationship between the azimuthal quantum number (l) and the shape of the orbital.

97 Easy

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Understanding +4 | -1

S-II: Electrons may be considered as particles and waves.

S-II: An electron in an atom may be described as occupying at atomic orbital or by a wave function

\psi

, which is a solution to the schrodinger wave equation.

View answer and explanation

Correct option: A

Both S-I and S-II are correct

Option A (Both S-I and S-II are correct) is correct because both statements accurately describe the dual nature of electrons and their behavior within atoms.

Solution approach

Correct Answer + Core Reason: Both S-I and S-II are correct. Both statements are fundamental principles in quantum mechanics.

Step-by-Step Solution:

Step 1: Analyze statement S-I: Electrons may be considered as particles and waves. This statement reflects the wave-particle duality of electrons, a cornerstone of quantum mechanics.

Step 2: Analyze statement S-II: An electron in an atom may be described as occupying at atomic orbital or by a wave function $\psi$ , which is a solution to the Schrodinger wave equation. This statement accurately describes how electrons are treated in quantum mechanics, using atomic orbitals and wave functions derived from the Schrodinger equation.

Step 3: Conclude that both statements are correct.

Why other options are wrong:
- Option B: Incorrect because both statements are correct.
- Option C: Incorrect because S-II is also correct.
- Option D: Incorrect because S-I is also correct.

Key Concepts:
- Wave-Particle Duality
- Quantum Mechanics
- Atomic Orbitals
- Schrodinger Wave Equation

Memory Aids:
- Electrons act as both waves and particles.
- Wave functions describe electron behavior in atoms.

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

98 Easy

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Analyzing +4 | -1

Which one is wrong statement?

View answer and explanation

Correct option: C

The magnetic quantum number of an electron present in 2nd orbit can be ‘ –2’

The statement that the magnetic quantum number of an electron present in the 2nd orbit can be '-2' is wrong because for n=2, l can be 0 or 1, and ml can range from -l to +l, so it can only be -1, 0, or +1.

Solution approach

Correct Answer + Core Reason: The magnetic quantum number of an electron present in 2nd orbit can be ‘ –2’. The magnetic quantum number (ml) ranges from -l to +l, and for n=2, the maximum value of l is 1, so ml cannot be -2.

Step-by-Step Solution:
Step 1: Understand the quantum numbers: Principal quantum number (n), azimuthal quantum number (l), and magnetic quantum number (ml).
Step 2: Relate n and l: For a given n, l can have values from 0 to n-1.
Step 3: Relate l and ml: For a given l, ml can have values from -l to +l, including 0.
Step 4: Apply to the problem: For n=2, l can be 0 or 1. If l=0, ml=0. If l=1, ml can be -1, 0, or +1.
Step 5: Check the options: The magnetic quantum number cannot be -2 for an electron in the 2nd orbit.

Why other options are wrong:
- Option A: Schrodinger wave equation cannot give information about spin quantum number: This is a correct statement. The Schrodinger equation does not directly provide information about the spin quantum number.
- Option B: The orbital angular momentum of a s – electron in 3rd orbit zero: This is a correct statement. For an s-electron (l=0), the orbital angular momentum is zero.
- Option D: In H – atom, the order of energies of orbitals is: 3s = 3p = 3d: This is a correct statement. In a hydrogen atom, orbitals with the same principal quantum number have the same energy.

Key Concepts:
- Quantum numbers (n, l, ml, s)
- Schrodinger wave equation
- Atomic orbitals
- Energy levels in hydrogen atom

Memory Aids:
- Remember the relationships between n, l, and ml.

Quick Tips:
- Always remember the allowed values for each quantum number.

99 Easy

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Understanding +4 | -1

The ratio of speed of γ-rays and X-rays is:

View answer and explanation

Correct option: A

1

Option A (1) is correct because both γ-rays and X-rays are electromagnetic waves and travel at the speed of light in a vacuum, so their speeds are equal.

Solution approach

Correct Answer + Core Reason: 1. Both gamma rays and X-rays are electromagnetic waves and travel at the speed of light.

Step-by-Step Solution:

Step 1: Recall that gamma rays and X-rays are both part of the electromagnetic spectrum.

Step 2: Understand that all electromagnetic waves travel at the speed of light (c) in a vacuum, which is approximately $3 \times 10^8$ m/s.

Step 3: Calculate the ratio of their speeds: $\frac{{Speed \, of \, \gamma - rays}}{{Speed \, of \, X - rays}} = \frac{c}{c} = 1$ .

Why other options are wrong:
- Option B: < 1 implies that X-rays travel faster than gamma rays, which is incorrect.
- Option C: > 1 implies that gamma rays travel faster than X-rays, which is incorrect.
- Option D: None of these is incorrect because the ratio is 1.

Key Concepts:
- Electromagnetic spectrum
- Speed of light
- Gamma rays
- X-rays

Memory Aids:
- All electromagnetic waves travel at the same speed in a vacuum.

Quick Tips:
- Remember that the speed of light is constant for all electromagnetic waves in a vacuum.

100 Easy

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Remembering +4 | -1

Stark effect refers to the

View answer and explanation

Correct option: A

Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field

Option A (Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field) is correct because the Stark effect is defined as the splitting of spectral lines in the presence of an external electric field.

Solution approach

Correct Answer + Core Reason: Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field. This is the definition of the Stark effect.

Step-by-Step Solution:

Step 1: Define Stark effect: The Stark effect is the splitting of spectral lines into several components in the presence of an external electric field.

Step 2: Relate the definition to the options.

Why other options are wrong:
- Option B: This describes the Tyndall effect, not the Stark effect.
- Option C: This describes the Zeeman effect, which involves a magnetic field, not an electric field.
- Option D: This describes the photoelectric effect, not the Stark effect.

Key Concepts:
- Stark effect
- Zeeman effect
- Photoelectric effect
- Tyndall effect

Memory Aids:
- Stark = Electric field, Zeeman = Magnetic field

Quick Tips:
- Differentiate between the Stark and Zeeman effects based on the field involved.

101 Easy

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Understanding +4 | -1

Which of the following is never true for cathode rays?

View answer and explanation

Correct option: B

They are electromagnetic waves.

Option B (They are electromagnetic waves) is correct because cathode rays are streams of electrons, not electromagnetic waves.

Solution approach

Correct Answer + Core Reason: They are electromagnetic waves. Cathode rays are streams of electrons.

Step-by-Step Solution:

Step 1: Define cathode rays: Cathode rays are streams of electrons.

Step 2: Recall properties of cathode rays: Cathode rays possess kinetic energy, produce heat, and exert mechanical pressure.

Step 3: Compare with electromagnetic waves: Cathode rays are not electromagnetic waves.

Step 4: Conclude that cathode rays are not electromagnetic waves.

Why other options are wrong:
- Option A: Cathode rays do possess kinetic energy.
- Option C: Cathode rays do produce heat when they strike a surface.
- Option D: Cathode rays do produce mechanical pressure when they strike a surface.

Key Concepts:
- Cathode rays
- Electromagnetic waves

Memory Aids:
- Cathode rays: Electrons

Quick Tips:
- Remember that cathode rays are particles (electrons), not waves.

102 Easy

102 of 239

Application +4 | -1

If radius of second orbit of

\mathrm{He}^{+}

is

\mathrm{x} \mathrm{nm}

then the radius of the first orbit of

\mathrm{Li}^{2+}

will be

View answer and explanation

Correct option: C

\frac{x}{6} \mathrm{~nm}

Option C (

\frac{x}{6} \mathrm{~nm}

) is correct because the radius of an orbit is proportional to n²/Z, and the ratio of the radii can be calculated using the given information.

Solution approach

Correct Answer + Core Reason: $\frac{x}{6} \mathrm{~nm}$ . The radius of the first orbit of Li²⁺ is calculated using the Bohr radius formula.

Step-by-Step Solution:

Step 1: Recall the formula for the radius of an orbit: $r_n = a_0 \frac{{n^2}}{Z}$ , where $a_0$ is the Bohr radius, n is the orbit number, and Z is the atomic number.

Step 2: Given the radius of the second orbit of He⁺ is x nm: $r_{\text{{He⁺, n=2}}} = a_0 \frac{{2^2}}{2} = x \text{{ nm}}$
$2a_0 = x \text{{ nm}}$ , so $a_0 = \frac{x}{2} \text{{ nm}}$

Step 3: Calculate the radius of the first orbit of Li²⁺: $r_{\text{{Li²⁺, n=1}}} = a_0 \frac{{1^2}}{3} = a_0 \frac{1}{3}$

Step 4: Substitute the value of $a_0$ : $r_{\text{{Li²⁺, n=1}}} = \frac{x}{2} \times \frac{1}{3} = \frac{x}{6} \text{{ nm}}$

Why other options are wrong:
- **Option A ( $3 \times nm$ ): Incorrect application of the Bohr radius formula.
- Option B ( $\frac{x}{2} \mathrm{~nm}$ ): Incorrect application of the Bohr radius formula.
- Option D ( $\frac{x}{4} \mathrm{~nm}$ ): Incorrect application of the Bohr radius formula.

Key Concepts:
- Bohr's model of the atom
- Radius of an orbit
- Atomic number

Memory Aids:**
- $r_n \propto \frac{{n^2}}{Z}$

Quick Tips:
- Remember that the radius of an orbit is directly proportional to $n^2$ and inversely proportional to Z.
- Use ratios to solve problems involving different atoms and orbits.

103 Easy

103 of 239

Remembering +4 | -1

Which d-orbital does not have four lobes?

View answer and explanation

Correct option: C

{d_{{z^2}}}

Option C (

{d_{{z^2}}}

) is correct because the

{d_{{z^2}}}

orbital has two lobes, while all other d-orbitals have four lobes.

Solution approach

Correct Answer + Core Reason: ${d_{{z^2}}}$ . The ${d_{{z^2}}}$ orbital has a different shape with only two lobes along the z-axis and a donut-shaped ring around the center.

Step-by-Step Solution:

Step 1: Recall the shapes of the five d-orbitals.

Step 2: The d-orbitals are ${d_{xy}}$ , ${d_{xz}}$ , ${d_{yz}}$ , ${d_{{x^2} - {y^2}}}$ , and ${d_{{z^2}}}$ .

Step 3: Visualize or recall that ${d_{xy}}$ , ${d_{xz}}$ , ${d_{yz}}$ , and ${d_{{x^2} - {y^2}}}$ each have four lobes.

Step 4: The ${d_{{z^2}}}$ orbital has two lobes along the z-axis and a ring in the xy-plane.

Step 5: Therefore, ${d_{{z^2}}}$ is the only d-orbital that does not have four lobes.

Why other options are wrong:
- Option A: ${d_{{x^2} - {y^2}}}$ has four lobes along the x and y axes.
- Option B: ${d_{xy}}$ has four lobes between the x and y axes.
- Option D: ${d_{xz}}$ has four lobes between the x and z axes.

Key Concepts:
- Atomic orbitals
- d-orbitals
- Shapes of atomic orbitals

Memory Aids:
- Remember the specific shapes of each d-orbital.

Quick Tips:
- Visualize the shapes of the orbitals to quickly identify the correct answer.

104 Easy

104 of 239

Application +4 | -1

A transition element X has configuration

\left[ {\mathrm{Ar}} \right]3{d^5}

in its +3 oxidation state. Its atomic number is:

View answer and explanation

Correct option: C

26

Option C (26) is correct because the element with the given electronic configuration in the +3 oxidation state has 26 electrons.

Solution approach

Correct Answer + Core Reason: 26. ${\mathrm{\;}_{26}}\mathrm{F}{\mathrm{e}^{3 + }}$ has $3{d^5}$ configuration

Step-by-Step Solution:

Step 1: The element X has a +3 oxidation state with the configuration $\left[ {\mathrm{Ar}} \right]3{d^5}$ .

Step 2: The electronic configuration of Ar is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$ , which accounts for 18 electrons.

Step 3: In the +3 oxidation state, the element has lost 3 electrons. Therefore, the neutral atom has 3 more electrons than the $\left[ {\mathrm{Ar}} \right]3{d^5}$ configuration.

Step 4: The number of electrons in $\left[ {\mathrm{Ar}} \right]3{d^5}$ is $18 + 5 = 23$ .

Step 5: The neutral atom has $23 + 3 = 26$ electrons.

Step 6: Therefore, the atomic number of the element is 26.

Why other options are wrong:
- Option A: 22 is incorrect. If the atomic number were 22, the +3 ion would have 19 electrons, which does not match the given configuration.
- Option B: 25 is incorrect. If the atomic number were 25, the +3 ion would have 22 electrons, which does not match the given configuration.
- Option D: 19 is incorrect. If the atomic number were 19, the +3 ion would have 16 electrons, which does not match the given configuration.

Key Concepts:
- Electronic configuration
- Oxidation state
- Atomic number

Memory Aids:
- Remember that a positive oxidation state means the element has lost electrons.

Quick Tips:
- Always account for the electrons lost or gained when dealing with ions.

105 Easy

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Understanding +4 | -1

Two electrons occupying the same orbital are distinguished by:

View answer and explanation

Correct option: D

Spin quantum number

Option D (Spin quantum number) is correct because the spin quantum number differentiates two electrons occupying the same orbital, as they must have opposite spins (+1/2 and -1/2) according to the Pauli Exclusion Principle.

Solution approach

Correct Answer + Core Reason: Spin quantum number. Electrons in the same orbital must have different spin quantum numbers.

Step-by-Step Solution:

Step 1: Recall the Pauli Exclusion Principle.

Step 2: The Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers (n, l, ml, ms).

Step 3: Electrons occupying the same orbital have the same principal quantum number (n), azimuthal quantum number (l), and magnetic quantum number (ml).

Step 4: Therefore, the only quantum number that can be different for two electrons in the same orbital is the spin quantum number (ms).

Step 5: The spin quantum number can have two values: +1/2 and -1/2.

Why other options are wrong:
- Option A: Principal quantum number (n) is the same for electrons in the same orbital.
- Option B: Magnetic quantum number (ml) is the same for electrons in the same orbital.
- Option C: Azimuthal quantum number (l) is the same for electrons in the same orbital.

Key Concepts:
- Quantum numbers
- Pauli Exclusion Principle
- Spin quantum number

Memory Aids:
- Remember the Pauli Exclusion Principle and that electrons in the same orbital must have opposite spins.

Quick Tips:
- Understand the meaning of each quantum number and how they relate to the properties of electrons in atoms.

106 Easy

106 of 239

Remembering +4 | -1

The minimum energy required to eject an electron from an atom is called :

View answer and explanation

Correct option: D

Work function

Option D (Work function) is correct because the work function is defined as the minimum energy required to eject an electron from an atom.

Solution approach

Correct Answer + Core Reason: Work function. The work function is the minimum energy needed to remove an electron.

Step-by-Step Solution:

Step 1: Define the problem: The question asks for the term that represents the minimum energy to eject an electron.

Step 2: Recall the definition of work function: The work function ( $W$ ) is the minimum energy required to remove an electron from a solid to a point immediately outside the solid surface.

Step 3: Relate to photoelectric effect: In the photoelectric effect, $h\nu = W + KE_{max}$ , where $h\nu$ is the energy of the incident photon, $W$ is the work function, and $KE_{max}$ is the maximum kinetic energy of the emitted electron. When $KE_{max} = 0$ , $h\nu = W$ .

Step 4: Conclusion: Therefore, the minimum energy required is the work function.

Why other options are wrong:
- Option A: Kinetic energy is the energy of motion, not the minimum energy required to eject an electron.
- Option B: Electrical energy is a form of energy related to electric fields and charges, not the specific energy for electron ejection.
- Option C: Chemical energy is the energy stored in chemical bonds, not the energy required for electron ejection.

Key Concepts:
- Work function
- Photoelectric effect
- Energy levels

Memory Aids:
- Work function: Think of it as the "work" needed to "functionally" remove an electron.

Quick Tips:
- Remember the photoelectric effect equation: $h\nu = W + KE_{max}$

107 Easy

107 of 239

Understanding +4 | -1

Which of the following energy level may bring absorption of photon but never emission of photon?

View answer and explanation

Correct option: C

1 s

Option C (

1s

) is correct because the 1s energy level is the lowest and thus cannot emit a photon, only absorb.

Solution approach

Correct Answer + Core Reason: $1 s$ . The $1s$ electronic level is the lowest energy level, allowing absorption but not emission of a photon.

Step-by-Step Solution:

Step 1: Understand the concept of energy levels and photon absorption/emission.

Step 2: Recall that absorption involves moving to a higher energy level, and emission involves moving to a lower energy level.

Step 3: Recognize that the $1s$ level is the ground state (lowest energy level).

Step 4: An electron in the $1s$ level can absorb a photon to move to a higher energy level (e.g., $2s$ , $2p$ , $3d$ ).

Step 5: However, an electron in the $1s$ level cannot emit a photon because there is no lower energy level to transition to.

Why other options are wrong:
- Option A: $3d$ can both absorb and emit photons.
- Option B: $2p$ can both absorb and emit photons.
- Option D: $2s$ can both absorb and emit photons.

Key Concepts:
- Energy levels
- Photon absorption
- Photon emission
- Ground state

Memory Aids:
- Absorption: electron goes UP in energy level.
- Emission: electron goes DOWN in energy level.

Quick Tips:
- The lowest energy level can only absorb, not emit.

108 Easy

108 of 239

Understanding +4 | -1

The quantum number not obtained from the schrodinger wave equation is

View answer and explanation

Correct option: D

s

The correct answer is D (s) because the spin quantum number (s) is not derived directly from solving the Schrodinger wave equation.

Solution approach

Correct Answer + Core Reason: s. The spin quantum number is not obtained directly from the Schrodinger wave equation.

Step-by-Step Solution:

Step 1: Recall the Schrodinger wave equation. The Schrodinger equation provides solutions that describe the energy and spatial distribution of electrons in atoms.

Step 2: Identify the quantum numbers derived from the Schrodinger equation. Solving the Schrodinger equation yields three quantum numbers: the principal quantum number (n), the azimuthal or angular momentum quantum number (l), and the magnetic quantum number (m).

Step 3: Recognize the origin of the spin quantum number. The spin quantum number (s) arises from relativistic quantum mechanics (specifically, the Dirac equation) and is not a direct solution of the non-relativistic Schrodinger equation.

Why other options are wrong:
- Option A (n): The principal quantum number (n) is obtained from the Schrodinger equation and describes the energy level of the electron.
- **Option B ( $l$ ):** The azimuthal quantum number ( $l$ ) is obtained from the Schrodinger equation and describes the shape of the electron's orbital.
- Option C (m): The magnetic quantum number (m) is obtained from the Schrodinger equation and describes the orientation of the electron's orbital in space.

Key Concepts:
- Schrodinger wave equation
- Quantum numbers (n, l, m, s)
- Spin quantum number

Memory Aids:
- Remember that n, l, and m are solutions to the Schrodinger equation, while s is a separate concept related to electron spin.

Quick Tips:
- Understand the origin of each quantum number to answer similar questions.

109 Easy

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Remembering +4 | -1

The shape of an orbital is given by the quantum number

View answer and explanation

Correct option: B

l

Option B (

l

) is correct because the shape of an orbital is determined by the azimuthal quantum number (

l

).

Solution approach

Correct Answer + Core Reason: $l$ . The azimuthal quantum number determines the shape of an orbital.

Step-by-Step Solution:

Step 1: Define quantum numbers.

Step 2: Recall the role of each quantum number:
- n: principal quantum number (energy level)
- l: azimuthal quantum number (shape)
- m: magnetic quantum number (orientation)
- s: spin quantum number (spin)

Step 3: The shape of an orbital is determined by the azimuthal quantum number (l).

Why other options are wrong:
- Option A: $n$ determines the energy level and size of the orbital, not the shape.
- Option C: $m$ determines the spatial orientation of the orbital, not the shape.
- Option D: $s$ determines the spin of the electron, not the shape of the orbital.

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

Memory Aids:
- \"l\" for \"shape\" (azimuthal quantum number)

Quick Tips:
- Know the role of each quantum number in describing an electron in an atom.

110 Easy

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Remembering +4 | -1

The velocity of a photon is:

View answer and explanation

Correct option: A

Independent of its wavelength

Option A (Independent of its wavelength) is correct because the velocity of a photon in a vacuum is a constant, the speed of light (c), and is independent of its wavelength or frequency.

Solution approach

Correct Answer + Core Reason: Independent of its wavelength. The velocity of a photon is constant (speed of light).

Step-by-Step Solution:

Step 1: Recall the properties of photons.

Step 2: Photons are massless particles that travel at the speed of light in a vacuum.

Step 3: The speed of light is a constant, approximately $3 \times 10^8 \text{{ m/s}}$ , and is independent of the photon's wavelength or frequency.

Why other options are wrong:
- Option B: The velocity of a photon does not depend on its wavelength in a vacuum. While wavelength and frequency are related to energy, they don't change the speed of the photon itself.
- Option C: The velocity of a photon does not depend on its source. All photons travel at the speed of light in a vacuum.
- Option D: The velocity of a photon is not equal to the square of its amplitude. Amplitude is related to the intensity of light, not its speed.

Key Concepts:
- Photons
- Speed of light
- Electromagnetic radiation

Memory Aids:
- Remember that all photons travel at the same speed in a vacuum.

Quick Tips:
- Understanding the fundamental properties of photons helps in answering such questions.

111 Easy

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Understanding +4 | -1

The atom with lowest atomic number that has a ground state (EC) of

(n-1) d^6 n s^2

is in the

View answer and explanation

Correct option: C

4th period

Option C (4th period) is correct because the electronic configuration

(n-1) d^6 n s^2

indicates a transition metal, and the d-orbitals are first filled in the 4th period (n=4).

Solution approach

Correct Answer + Core Reason: 4th period. The electronic configuration $(n-1) d^6 n s^2$ indicates a transition metal in the 4th period.

Step-by-Step Solution:

Step 1: Analyze the electronic configuration. The electronic configuration is given as $(n-1) d^6 n s^2$ .

Step 2: Determine the value of n. Since the d-orbital is $(n-1)d$ , the d-orbital being filled is one energy level lower than the ns orbital. The first d-orbital to be filled is the 3d orbital, which occurs when n = 4 (4s orbital is filled after 3d).

Step 3: Identify the period. The period number corresponds to the principal quantum number (n). Therefore, the element is in the 4th period.

Why other options are wrong:
- Option A (2nd period): The 2nd period does not have d-orbitals.
- Option B (3rd period): The 3rd period does not have d-orbitals being filled in the ground state.
- Option D (5th period): While the 5th period does have d-orbitals, the lowest atomic number with the given configuration is in the 4th period (Iron).

Key Concepts:
- Electronic configuration
- Transition metals
- Periodicity
- Principal quantum number

Memory Aids:
- d-orbitals start filling from the 4th period.

Quick Tips:
- Remember the order of filling orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, ...

112 Easy

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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.

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Application +4 | -1

According to de-Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass

100 \mathrm{~g}

does not move like a wave when it is thrown by a bowler at a speed of

100 \mathrm{~km} / \mathrm{h}

. Calculate the wavelength of the ball.

View answer and explanation

Correct option: C

238.5 \times 10^{-36} \mathrm{~m}

Option C (

238.5 \times 10^{-36} \mathrm{~m}

) is correct because it is calculated using the de Broglie wavelength formula with the given mass and velocity.

Solution approach

Correct Answer + Core Reason: $238.5 \times 10^{-36} \mathrm{~m}$ . This is calculated using the de Broglie wavelength equation.

Step-by-Step Solution:

Step 1: Convert the velocity from km/h to m/s.\nv = 100 km/h = $100 \times \frac{1000 m}{3600 s} = \frac{1000}{36} m/s \approx 27.78 m/s$

Step 2: Convert the mass from grams to kilograms.\nm = 100 g = 0.1 kg

Step 3: Use the de Broglie wavelength equation: $\lambda = \frac{h}{mv}$ \nwhere h is Planck's constant ( $6.626 \times 10^{-34} Js$ ), m is the mass, and v is the velocity.

Step 4: Substitute the values.
$\lambda = \frac{6.626 \times 10^{-34} Js}{0.1 kg \times \frac{1000}{36} m/s} = \frac{6.626 \times 10^{-34} Js}{2.778 kg \cdot m/s} \approx 2.385 \times 10^{-34} m$
$\lambda = 238.5 \times 10^{-36} m$

Why other options are wrong:
- Option A: Incorrect exponent. The correct exponent is -36, not -34.
- Option B: Incorrect coefficient and exponent. The correct coefficient is 238.5 and the exponent is -36.
- Option D: Incorrect coefficient. The correct coefficient is 238.5, not 2385.

Key Concepts:
- de Broglie Wavelength
- Planck's Constant

Memory Aids:
- Remember the de Broglie equation: $\lambda = \frac{h}{mv}$

Quick Tips:
- Make sure to use consistent units (SI units) when applying the de Broglie equation.

114 Easy

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Remembering +4 | -1

The number of protons, neutrons and electrons in

{}_{71}^{175}\mathrm{Lu}

, respectively, are

View answer and explanation

Correct option: A

\mathrm{71,}\,\,\mathrm{104}\,\,\mathrm{and}\,\,\mathrm{71}

Option A (71, 104, and 71) is correct because the number of protons and electrons equals the atomic number, and the number of neutrons is the mass number minus the atomic number.

Solution approach

Correct Answer + Core Reason: $\mathrm{71}\,\,\mathrm{104}\,\,\mathrm{and}\,\,\mathrm{71}$ . Number of protons = atomic number, number of neutrons = mass number - atomic number, number of electrons = number of protons in a neutral atom.

Step-by-Step Solution:
Step 1: Identify the given information: The element is ${}_{71}^{175}\mathrm{Lu}$ .
Step 2: Determine the number of protons: The atomic number is 71, so there are 71 protons.
Step 3: Determine the number of electrons: In a neutral atom, the number of electrons is equal to the number of protons, so there are 71 electrons.
Step 4: Determine the number of neutrons: The mass number is 175, so the number of neutrons is 175 - 71 = 104.

Why other options are wrong:
- Option B: This option incorrectly assigns the number of neutrons as 71 and the number of protons as 104.
- Option C: This option incorrectly assigns the number of electrons as 104.
- Option D: This option incorrectly assigns the number of protons as 175.

Key Concepts:
- Atomic number
- Mass number
- Protons, neutrons, and electrons

Memory Aids:
- APE (Atomic number = Protons = Electrons) in a neutral atom
- MAN (Mass number - Atomic number = Neutrons)

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

115 Easy

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Analyzing +4 | -1

Which statement relating to the spectrum of H atom is false?

View answer and explanation

Correct option: C

The spectral lines are closer together at longer wavelengths

Option C (The spectral lines are closer together at longer wavelengths) is false because the spectral lines in the hydrogen atom spectrum converge at shorter wavelengths (higher energies), not longer wavelengths.

Solution approach

Correct Answer + Core Reason: The spectral lines are closer together at longer wavelengths. Spectral lines converge at shorter wavelengths.

Step-by-Step Solution:

Step 1: Recall the Rydberg formula for the wavelengths of spectral lines in the hydrogen atom spectrum:
$\frac{1}{\lambda} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ \nwhere $R_H$ is the Rydberg constant, $n_1$ and $n_2$ are integers with $n_2 > n_1$ .

Step 2: Analyze the energy levels.\nThe energy levels in the hydrogen atom become closer together as $n$ increases. This means that the energy difference between adjacent levels decreases as $n$ increases.

Step 3: Relate energy difference to wavelength.\nThe energy of a photon is given by $E = hf = \frac{hc}{\lambda}$ , where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Therefore, the wavelength is inversely proportional to the energy difference.

Step 4: Determine the trend of spectral line spacing.\nSince the energy levels become closer together as $n$ increases, the energy difference between transitions decreases for larger $n$ values. This corresponds to longer wavelengths. Therefore, the spectral lines become closer together at shorter wavelengths (higher energies), not longer wavelengths.

Step 5: Analyze the given statements.
- A: The lines can be defined by quantum number - True. The Rydberg formula uses quantum numbers $n_1$ and $n_2$ .
- B: The lines of longest wavelength in the Balmer series corresponds to the transition between n=3 and n=2 levels - True. The Balmer series has $n_1 = 2$ , and the smallest energy difference (longest wavelength) is for $n_2 = 3$ .
- C: The spectral lines are closer together at longer wavelengths - False. The spectral lines are closer together at shorter wavelengths.
- D: A continuum occurs at $n = \infty$ - True. As $n$ approaches infinity, the energy levels become continuous, leading to a continuum.

Why other options are wrong:
- Option A: The spectral lines are indeed defined by quantum numbers, so this statement is true.
- Option B: The longest wavelength in the Balmer series corresponds to the transition between n=3 and n=2, so this statement is true.
- Option D: A continuum does occur at n = infinity, so this statement is true.

Key Concepts:
- Hydrogen Atom Spectrum
- Rydberg Formula
- Energy Levels
- Quantum Numbers

Memory Aids:
- Remember the Rydberg formula and the relationship between energy, frequency, and wavelength.

Quick Tips:
- Understand the energy level diagram of the hydrogen atom.
- Remember that energy and frequency are directly proportional, while wavelength is inversely proportional to energy.

116 Easy

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Understanding +4 | -1

The angular momentum of an electron in a given stationary state can be expressed as

m_e v r=n \frac{h}{2 \pi}

. Based on this expression an electron can move only in those orbits for which its angular momentum is

View answer and explanation

Correct option: B

Integral multiple of

\frac{h}{2 \pi}

Option B (Integral multiple of

\frac{h}{2 \pi}

) is correct because Bohr's quantization condition states that the angular momentum of an electron is an integral multiple of

\frac{h}{2 \pi}

.

Solution approach

Correct Answer + Core Reason: Integral multiple of $\frac{h}{2 \pi}$ . Bohr's quantization condition states that angular momentum is quantized.

Step-by-Step Solution:

Step 1: Recall Bohr's quantization condition: $m_e v r = n \frac{h}{2 \pi}$ , where n is an integer.

Step 2: Interpret the equation: The angular momentum ( $m_e v r$ ) is equal to an integer (n) times $\frac{h}{2 \pi}$ .

Step 3: Conclude that the angular momentum is an integral multiple of $\frac{h}{2 \pi}$ .

Why other options are wrong:
- Option A: Equal to $n$ is incorrect because the angular momentum is equal to $n \frac{h}{2 \pi}$ , not just n.
- Option C: Multiple of $n$ is incorrect because it doesn't specify the correct factor of $\frac{h}{2 \pi}$ .
- Option D: Equal to $\frac{h}{2 \pi}$ only is incorrect because the angular momentum can be any integral multiple of $\frac{h}{2 \pi}$ , not just equal to it.

Key Concepts:
- Bohr's model
- Quantization of angular momentum

Memory Aids:
- Remember Bohr's quantization condition: $m_e v r = n \frac{h}{2 \pi}$ .

Quick Tips:
- Understand the postulates of Bohr's model.

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Understanding +4 | -1

Which electron transition in a hydrogen atom requires the largest amount of energy?

View answer and explanation

Correct option: A

From n=1 to n=2

Option A (From n=1 to n=2) is correct because the energy difference between n=1 and n=2 is the largest in the hydrogen atom spectrum.

Solution approach

Correct Answer + Core Reason: From n=1 to n=2. The energy difference is largest for transitions involving the ground state (n=1).

Step-by-Step Solution:

Step 1: Recall the energy levels of the hydrogen atom. The energy of an electron in the nth orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}$ eV.

Step 2: Calculate the energy difference for each transition:
- From n=1 to n=2: $\Delta E = E_2 - E_1 = -13.6(\frac{1}{2^2} - \frac{1}{1^2}) = -13.6(\frac{1}{4} - 1) = 10.2$ eV
- From n=2 to n=3: $\Delta E = E_3 - E_2 = -13.6(\frac{1}{3^2} - \frac{1}{2^2}) = -13.6(\frac{1}{9} - \frac{1}{4}) = 1.89$ eV
- From $n = \infty$ to n=1: $\Delta E = E_1 - E_{\infty} = -13.6(\frac{1}{1^2} - 0) = -13.6$ eV (magnitude is 13.6 eV)
- From n=3 to n=5: $\Delta E = E_5 - E_3 = -13.6(\frac{1}{5^2} - \frac{1}{3^2}) = -13.6(\frac{1}{25} - \frac{1}{9}) = 0.97$ eV

Step 3: Compare the energy differences. The largest energy difference is for the transition from n=1 to n=2.

Why other options are wrong:
- Option B: From n=2 to n=3 requires less energy than from n=1 to n=2.
- Option C: From $n = \infty$ to n=1 requires more energy in magnitude, but the question asks for the largest *amount* of energy, implying a positive value (absorption). While this transition releases the most energy, the transition from n=1 to n=2 *requires* the most energy to be absorbed.
- Option D: From n=3 to n=5 requires less energy than from n=1 to n=2.

Key Concepts:
- Hydrogen atom spectrum
- Energy levels
- Electronic transitions

Memory Aids:
- Energy levels are closer together as n increases.

Quick Tips:
- Remember the formula for energy levels in the hydrogen atom.

118 Easy

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Understanding +4 | -1

Which of the following expression represent the wave number of spectral lines in Balmer series (if

n

is the principal quantum number of higher energy level)?

View answer and explanation

Correct option: C

\frac{R(n-2)(n+2)}{4 n^2}

The correct answer is C. The wave number for the Balmer series is given by

\frac{R(n-2)(n+2)}{4 n^2}

, derived from the Rydberg formula.

Solution approach

Correct Answer + Core Reason: $\frac{R(n-2)(n+2)}{4 n^2}$ . This formula correctly represents the wave number of spectral lines in the Balmer series.

Step-by-Step Solution:

Step 1: Identify the problem: Find the expression for the wave number of spectral lines in the Balmer series.

Step 2: Recall the Rydberg formula: $\bar{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ , where $R$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the higher energy level.

Step 3: For the Balmer series, $n_1 = 2$ . So, $\bar{v} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) = R \left( \frac{1}{4} - \frac{1}{n^2} \right)$

Step 4: Simplify the expression: $\bar{v} = R \left( \frac{n^2 - 4}{4n^2} \right)$

Step 5: Factor the numerator: $\bar{v} = R \left( \frac{(n-2)(n+2)}{4n^2} \right) = \frac{R(n-2)(n+2)}{4n^2}$

Why other options are wrong:
- Option A: This is incorrect because it simplifies to $\frac{R(n^2-4)}{n^2}$ which is not the correct Balmer series formula.
- Option B: This is incorrect because it simplifies to $\frac{R(n-2)(n+2)}{n^2}$ which is not the correct Balmer series formula.
- Option D: This is incorrect because it has $(n-1)(n+1)$ in the numerator, which is not related to the Balmer series.

Key Concepts:
- Rydberg Formula
- Balmer Series
- Wave Number
- Atomic Spectra

Memory Aids:
- Remember that the Balmer series corresponds to transitions to the $n=2$ energy level.

Quick Tips:
- Memorize the Rydberg formula and the values of $n_1$ for different spectral series (Lyman, Balmer, Paschen, etc.).

119 Easy

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Application +4 | -1

Which of the following pairs of ions have the same electronic configuration?

View answer and explanation

Correct option: B

\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}

Option B (

\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}

) is correct because both ions have the same electronic configuration of [Ar]3d⁵.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}$ . Both ions have the same electronic configuration.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each ion:
- $\mathrm{Cr}^{3+}$ : $[Ar] 3d^3$
- $\mathrm{Fe}^{3+}$ : $[Ar] 3d^5$
- $\mathrm{Mn}^{2+}$ : $[Ar] 3d^5$
- $\mathrm{Co}^{3+}$ : $[Ar] 3d^6$
- $\mathrm{Sc}^{3+}$ : Not provided

Step 2: Compare the electronic configurations:
- $\mathrm{Fe}^{3+}$ and $\mathrm{Mn}^{2+}$ both have the electronic configuration $[Ar] 3d^5$ .

Why other options are wrong:
- Option A: $\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$ have different electronic configurations.
- Option C: $\mathrm{Fe}^{3+}$ and $\mathrm{Co}^{3+}$ have different electronic configurations.
- Option D: Not provided in the options.

Key Concepts:
- Electronic configuration of ions

Memory Aids:
- Remember the order of filling electrons in orbitals.

Quick Tips:
- Write the electronic configuration of each ion to compare them.

120 Easy

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Application +4 | -1

Hydrogen atoms are excited to

n = 4

state, In the spectrum of emitted radiation, number of lines in the
UV and visible regions are resp.

View answer and explanation

Correct option: D

3,2

Option D (

3,2

) is correct because when hydrogen atoms are excited to n=4, there are 3 Lyman series lines (UV region) and 2 Balmer series lines (visible region).

Solution approach

Correct Answer + Core Reason: $3,2$ . The number of lines in UV and visible regions are determined by the transitions to n=1 and n=2 respectively.

Step-by-Step Solution:

Step 1: Hydrogen atoms are excited to $n = 4$ state.

Step 2: Total number of lines emitted: $\sum_{i=1}^{n-1} i = \sum_{i=1}^{3} i = 3 + 2 + 1 = 6$ .

Step 3: Number of lines in UV region (Lyman series, transitions to $n=1$ ): $4 \rightarrow 1$ , $3 \rightarrow 1$ , $2 \rightarrow 1$ . So, there are 3 lines.

Step 4: Number of lines in visible region (Balmer series, transitions to $n=2$ ): $4 \rightarrow 2$ , $3 \rightarrow 2$ . So, there are 2 lines.

Step 5: Number of lines in the infrared region (Paschen series, transitions to $n=3$ ): $4 \rightarrow 3$ . So, there is 1 line.

Step 6: Final Answer: $3,2$

Why other options are wrong:
- Option A: Incorrect number of lines in UV and visible regions.
- Option B: Incorrect number of lines in UV and visible regions.
- Option C: Incorrect number of lines in UV and visible regions.

Key Concepts:
- Hydrogen spectrum
- Lyman series (UV region)
- Balmer series (Visible region)
- Paschen series (Infrared region)

Memory Aids:
- Lyman: UV, Balmer: Visible, Paschen: IR

Quick Tips:
- Remember the series and their corresponding regions of the electromagnetic spectrum.

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