STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 6

151 Easy

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Understanding +4 | -1

Which of the following combination of quantum number is allowed

View answer and explanation

Correct option: B

\begin{array}{cccccccccccccccccccc} n&l&m&s\\ 2&0&0&{ - 1/2} \end{array}

Option B is correct because it represents a valid set of quantum numbers following the rules:

n > 0

,

0 \le l < n

,

-l \le m \le l

, and

s = \pm \frac{1}{2}

.

Solution approach

Correct Answer + Core Reason: B. $\begin{array}{cccccccccccccccccccc} n&l&m&s\\ 2&0&0&{ - 1/2} \end{array}$ . This combination satisfies the rules for allowed quantum numbers.

Step-by-Step Solution:

Step 1: Recall the rules for quantum numbers:
- Principal quantum number ( $n$ ): $n = 1, 2, 3, ...$
- Azimuthal quantum number ( $l$ ): $0 \le l < n$
- Magnetic quantum number ( $m$ ): $-l \le m \le l$
- Spin quantum number ( $s$ ): $+1/2$ or $-1/2$

Step 2: Evaluate option A:
- $n = 3$ , $l = 2$ , $m = 1$ , $s = 0$ . $s$ must be $\pm 1/2$ , so this is incorrect.

Step 3: Evaluate option B:
- $n = 2$ , $l = 0$ , $m = 0$ , $s = -1/2$ . $l < n$ , $-l \le m \le l$ , and $s$ is allowed. This is correct.

Step 4: Evaluate option C:
- $n = 3$ , $l = -3$ , $m = -2$ , $s = +1/2$ . $l$ must be non-negative, so this is incorrect.

Step 5: Evaluate option D:
- $n = 1$ , $l = 0$ , $m = 1$ , $s = +1/2$ . $-l \le m \le l$ requires $m = 0$ when $l = 0$ , so this is incorrect.

Why other options are wrong:
- Option A: The spin quantum number ( $s$ ) must be either $+1/2$ or $-1/2$ , not 0.
- Option C: The azimuthal quantum number ( $l$ ) cannot be negative. It must be between 0 and $n-1$ .
- Option D: When $l = 0$ , the magnetic quantum number ( $m$ ) must also be 0.

Key Concepts:
- Quantum numbers: principal ( $n$ ), azimuthal ( $l$ ), magnetic ( $m$ ), and spin ( $s$ ).
- Rules for allowed values of quantum numbers.

Memory Aids:
- Remember the ranges for each quantum number: $n > 0$ , $0 \le l < n$ , $-l \le m \le l$ , $s = \pm 1/2$ .

Quick Tips:
- Always check if the given quantum numbers satisfy the fundamental rules before considering an option as correct.

152 Easy

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Arrange the following in terms of penetrating power.

\alpha

-rays,

\beta

-rays,

\gamma

-rays

View answer and explanation

Correct option: B

\alpha<\beta<\gamma

Option B (

\alpha<\beta<\gamma

) is correct because gamma rays have the highest penetrating power, followed by beta particles, and then alpha particles.

Solution approach

Correct Answer + Core Reason: $\alpha<\beta<\gamma$ . Gamma rays have the highest penetrating power, followed by beta and then alpha rays.

Step-by-Step Solution:

Step 1: Recall the properties of alpha, beta, and gamma rays.
- Alpha ( $\alpha$ ) particles are helium nuclei (2 protons and 2 neutrons) and have a +2 charge.
- Beta ( $\beta$ ) particles are high-speed electrons or positrons and have a -1 or +1 charge.
- Gamma ( $\gamma$ ) rays are high-energy photons and have no charge.

Step 2: Consider the penetrating power of each type of radiation.
- Alpha particles have the lowest penetrating power due to their large mass and charge.
- Beta particles have a higher penetrating power than alpha particles but are still less penetrating than gamma rays.
- Gamma rays have the highest penetrating power due to their high energy and lack of charge.

Step 3: Arrange the radiation in terms of increasing penetrating power.\nThe correct order is $\alpha<\beta<\gamma$ .

Why other options are wrong:
- Option A: $\alpha>\beta>\gamma$ is incorrect because alpha particles have the lowest penetrating power.
- Option C: $\alpha>\beta<\gamma$ is incorrect because alpha particles have the lowest penetrating power and beta particles have less penetrating power than gamma rays.
- Option D: $\alpha>\gamma>\beta$ is incorrect because alpha particles have the lowest penetrating power and gamma rays have the highest penetrating power.

Key Concepts:
- Alpha particles
- Beta particles
- Gamma rays
- Penetrating power

Memory Aids:
- Alpha - least penetrating, Beta - medium, Gamma - most penetrating

Quick Tips:
- Remember the relative penetrating power of alpha, beta, and gamma rays.

153 Easy

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Which quantum number is not related with Schrodinger equation

View answer and explanation

Correct option: D

Spin

Option D (Spin) is correct because the spin quantum number is not directly derived from the Schrodinger equation but is introduced to account for the intrinsic angular momentum of an electron.

Solution approach

Correct Answer + Core Reason: Spin. Spin quantum number is not derived from the Schrodinger equation.

Step-by-Step Solution:
Step 1: Recall the quantum numbers derived from the Schrodinger equation.
Step 2: The principal (n), azimuthal (l), and magnetic (ml) quantum numbers arise from solving the Schrodinger equation for the hydrogen atom.
Step 3: The spin quantum number (s) is introduced to account for the intrinsic angular momentum of the electron, which is not described by the Schrodinger equation.

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

Key Concepts:
- Quantum numbers
- Schrodinger equation
- Electron spin

Memory Aids:
- Remember that spin is an intrinsic property of the electron.

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

154 Easy

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Application +4 | -1

A photon of radiation of wavelength

4000 ~\mathop {\text{A}}\limits^ \circ

has an energy

E

. The wavelength of photon of radiation having energy

0.5 E

will be

View answer and explanation

Correct option: B

8000 ~\mathop {\text{A}}\limits^ \circ

Option B (8000 Å) is correct because energy and wavelength are inversely proportional, so halving the energy doubles the wavelength.

Solution approach

Correct Answer + Core Reason: $8000 ~\mathop {\text{A}}\limits^ \circ$ . Since energy is inversely proportional to wavelength, halving the energy doubles the wavelength.

Step-by-Step Solution:

Step 1: Recall the relationship between energy and wavelength: $E = \frac{{hc}}{\lambda}$ .

Step 2: Set up the ratio: $\frac{{E_1}}{{E_2}} = \frac{{\lambda_2}}{{\lambda_1}}$ .

Step 3: Given: $E_1 = E$ , $\lambda_1 = 4000 ~\mathop {\text{A}}\limits^ \circ$ , and $E_2 = 0.5E$ .

Step 4: Substitute the values: $\frac{E}{{0.5E}} = \frac{{\lambda_2}}{{4000 ~\mathop {\text{A}}\limits^ \circ}}$ .

Step 5: Simplify: $2 = \frac{{\lambda_2}}{{4000 ~\mathop {\text{A}}\limits^ \circ}}$ .

Step 6: Solve for $\lambda_2$ : $\lambda_2 = 2 \times 4000 ~\mathop {\text{A}}\limits^ \circ = 8000 ~\mathop {\text{A}}\limits^ \circ$ .

Step 7: Final Answer: $8000 ~\mathop {\text{A}}\limits^ \circ$ .

Why other options are wrong:
- Option A (2000 Å): This is incorrect because it would mean the energy is doubled, not halved.
- Option C (4000 Å): This is incorrect because it implies the energy remains the same, which contradicts the problem statement.
- Option D (6000 Å): This is incorrect; there is no direct mathematical basis for this answer based on the given information.

Key Concepts:
- Relationship between energy and wavelength of a photon
- Inverse proportionality

Memory Aids:
- Remember that energy and wavelength are inversely proportional: $E \propto \frac{1}{\lambda}$ .

Quick Tips:
- When dealing with ratios, set up the equation carefully to avoid errors.

155 Easy

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Energy of atomic orbitals in a particular shell is in the order

View answer and explanation

Correct option: A

s < p < d < f

Option A (

s < p < d < f

) is correct because, for a given shell, the energy of atomic orbitals increases with the azimuthal quantum number (l), where s (l=0), p (l=1), d (l=2), and f (l=3).

Solution approach

Correct Answer + Core Reason: $s < p < d < f$ . Energy of atomic orbitals increases with the value of $l$ .

Step-by-Step Solution:

Step 1: Recall the relationship between orbital energy and azimuthal quantum number (l).

Step 2: For a given shell (n), the energy of an orbital increases with increasing l.

Step 3: The values of l for s, p, d, and f orbitals are 0, 1, 2, and 3, respectively.

Step 4: Therefore, the energy order is s < p < d < f.

Why other options are wrong:
- Option B: This order is incorrect; energy increases with l, not decreases.
- Option C: This order is incorrect; it mixes up the relative energies of the orbitals.
- Option D: This order is incorrect; it reverses the correct energy order.

Key Concepts:
- Energy of atomic orbitals
- Azimuthal quantum number (l)

Memory Aids:
- Remember the order: \"So People Don't Forget\" (s, p, d, f) and associate it with increasing energy.

Quick Tips:
- Understand the factors affecting the energy of atomic orbitals.

156 Easy

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Understanding +4 | -1

One require energy

{E_n}

to remove a nucleon from a nucleus and an energy

{E_e}

to remove an electron from the orbit of an atom, then:

View answer and explanation

Correct option: C

{E_n} > {E_e}

Option C is correct because the energy required to remove a nucleon from a nucleus (

{E_n}

) is much greater than the energy required to remove an electron from the orbit of an atom (

{E_e}

) due to the strong nuclear force.

Solution approach

Correct Answer + Core Reason: ${E_n} > {E_e}$ . Nuclear forces are much stronger than electromagnetic forces.

Step-by-Step Solution:

Step 1: Define nucleon removal energy ( ${E_n}$ ). This is the energy required to overcome the strong nuclear force holding nucleons (protons and neutrons) together in the nucleus.

Step 2: Define electron removal energy ( ${E_e}$ ). This is the energy required to overcome the electromagnetic force (Coulomb force) holding electrons in their orbits around the nucleus.

Step 3: Compare the strengths of nuclear and electromagnetic forces. The strong nuclear force is significantly stronger than the electromagnetic force.

Step 4: Relate force strength to removal energy. Since the nuclear force is much stronger, it requires a much greater amount of energy to remove a nucleon from the nucleus compared to the energy required to remove an electron from its orbit.

Why other options are wrong:
- Option A: The energies are not equal because the forces involved are vastly different.
- Option B: The energy required to remove a nucleon is much greater, not less than, the energy to remove an electron.
- Option D: While it's possible for the energies to be equal in very specific and unusual circumstances, generally, the nuclear removal energy is significantly greater.

Key Concepts:
- Strong nuclear force
- Electromagnetic force
- Binding energy
- Ionization energy

Memory Aids:
- Remember that the strong nuclear force is much stronger than the electromagnetic force.

Quick Tips:
- Consider the relative strengths of the forces involved when comparing removal energies.

157 Easy

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If

n = 3,

then which value of

'l'

is correct

View answer and explanation

Correct option: D

All of them

The correct answer is D (All of them) because for a given principal quantum number n=3, the azimuthal quantum number l can have values from 0 to n-1, which includes 0, 1, and 2.

Solution approach

Correct Answer + Core Reason: All of them. For $n = 3$ , $l$ can be 0, 1, or 2.

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the azimuthal quantum number (l).\nThe possible values of l range from 0 to n-1.

Step 2: Given n = 3, determine the possible values of l.\nl can be 0, 1, or 2.

Step 3: Check if the given options are within the possible values of l.\nOption A: 0 is a valid value.\nOption B: 1 is a valid value.\nOption C: 2 is a valid value.

Step 4: Conclude that all the given values are correct.

Why other options are wrong:
- Option A (0): While 0 is a correct value for l, it's not the only correct value.
- Option B (1): While 1 is a correct value for l, it's not the only correct value.
- Option C (2): While 2 is a correct value for l, it's not the only correct value.

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

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

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

158 Easy

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An atom having even number of electrons may be:

View answer and explanation

Correct option: C

Diamagnetic or paramagnetic

Option C (Diamagnetic or paramagnetic) is correct because atoms with an even number of electrons can have all electrons paired (diamagnetic) or some unpaired (paramagnetic).

Solution approach

Correct Answer + Core Reason: Diamagnetic or paramagnetic. Atoms with even number of electrons can be either diamagnetic or paramagnetic depending on the pairing of electrons.

Step-by-Step Solution:

Step 1: Define diamagnetic: Diamagnetic substances have all electrons paired.

Step 2: Define paramagnetic: Paramagnetic substances have unpaired electrons.

Step 3: Consider examples: Carbon (6 electrons) is paramagnetic, Magnesium (12 electrons) is diamagnetic.

Step 4: Conclude that an atom with an even number of electrons can be either diamagnetic or paramagnetic.

Why other options are wrong:
- Option A: An atom with an even number of electrons is not always diamagnetic.
- Option B: An atom with an even number of electrons is not always paramagnetic.
- Option D: The atom can be either diamagnetic or paramagnetic.

Key Concepts:
- Diamagnetism
- Paramagnetism
- Electron configuration

Memory Aids:
- Diamagnetic: Paired electrons
- Paramagnetic: Unpaired electrons

Quick Tips:
- Determine if a substance is diamagnetic or paramagnetic by looking at its electron configuration.

159 Easy

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No two electrons can have the same values of …. quantum numbers.

View answer and explanation

Correct option: D

Four

Option D (Four) is correct because, according to Pauli's exclusion principle, no two electrons in an atom can have the same set of four quantum numbers.

Solution approach

Correct Answer + Core Reason: Four. Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.

Step-by-Step Solution:

Step 1: Recall Pauli's exclusion principle.

Step 2: Pauli's exclusion principle states that no two electrons in an atom can have the same values for all four quantum numbers (n, l, ml, and ms).

Step 3: This means each electron must have a unique set of quantum numbers.

Why other options are wrong:
- Option A: One quantum number being different is not sufficient to distinguish two electrons according to Pauli's exclusion principle.
- Option B: Two quantum numbers being different is not sufficient to distinguish two electrons according to Pauli's exclusion principle.
- Option C: Three quantum numbers being different is not sufficient to distinguish two electrons according to Pauli's exclusion principle. All four must be different.

Key Concepts:
- Pauli's exclusion principle
- Quantum numbers (n, l, ml, ms)
- Electronic configuration

Memory Aids:
- Remember that Pauli's exclusion principle is about all four quantum numbers being different.

Quick Tips:
- Understand the implications of Pauli's exclusion principle for electron configurations.

160 Easy

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Which of the following is not a fundamental particle?

View answer and explanation

Correct option: D

X-rays

Option D (X-rays) is correct because X-rays are electromagnetic radiation, not a fundamental particle like electrons, protons, and neutrons.

Solution approach

Correct Answer + Core Reason: X-rays. X-rays are electromagnetic waves, not particles.

Step-by-Step Solution:

Step 1: Recall the definition of a fundamental particle.

Step 2: Fundamental particles are the basic building blocks of matter.

Step 3: Electrons, protons, and neutrons are fundamental particles.

Step 4: X-rays are electromagnetic radiation, which are waves, not particles.

Why other options are wrong:
- Option A: Electron is a fundamental particle.
- Option B: Proton is a fundamental particle.
- Option C: Neutron is a fundamental particle.

Key Concepts:
- Fundamental particles
- Electrons
- Protons
- Neutrons
- Electromagnetic radiation

Memory Aids:
- X-rays = waves

Quick Tips:
- Remember the difference between particles and waves.

161 Easy

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Application +4 | -1

Which of the following pairs of nucleides are isodiaphers ?

View answer and explanation

Correct option: D

{ }_{25}^{55} \mathrm{Mn}

and

{ }_{30}^{65} \mathrm{Zn}

Option D (

{ }_{25}^{55} \mathrm{Mn}

and

{ }_{30}^{65} \mathrm{Zn}

) is correct because isodiaphers have the same difference between the number of neutrons and protons (A-2Z).

Solution approach

Correct Answer + Core Reason: ${ }_{25}^{55} \mathrm{Mn}$ and ${ }_{30}^{65} \mathrm{Zn}$ . Isodiaphers have the same (A-2Z) value.

Step-by-Step Solution:

Step 1: Define isodiaphers: Isodiaphers have the same difference of number of neutrons and protons or $(A-2 Z)$ must be same.

Step 2: Calculate A-2Z for each option:
- A) ${ }_6^{13} \mathrm{C}$ : 13 - 2(6) = 1, ${ }_8^{16} \mathrm{O}$ : 16 - 2(8) = 0 (Not isodiaphers)
- B) ${ }_1^1 \mathrm{H}$ : 1 - 2(1) = -1, ${ }_1^2 \mathrm{H}$ : 2 - 2(1) = 0 (Not isodiaphers)
- C) ${ }_1^3 \mathrm{H}$ : 3 - 2(1) = 1, ${ }_2^4 \mathrm{He}$ : 4 - 2(2) = 0 (Not isodiaphers)
- D) ${ }_{25}^{55} \mathrm{Mn}$ : 55 - 2(25) = 5, ${ }_{30}^{65} \mathrm{Zn}$ : 65 - 2(30) = 5 (Isodiaphers)

Why other options are wrong:
- **Option A ( ${ }_6^{13} \mathrm{C}$ and ${ }_8^{16} \mathrm{O}$ ): The (A-2Z) values are different.
- Option B ( ${ }_1^1 \mathrm{H}$ and ${ }_1^2 \mathrm{H}$ ): The (A-2Z) values are different.
- Option C ( ${ }_1^3 \mathrm{H}$ and ${ }_2^4 \mathrm{He}$ ): The (A-2Z) values are different.

Key Concepts:
- Isotopes, Isobars, Isotones, Isodiaphers

Memory Aids:
- Isodiaphers: Same difference between neutrons and protons (A-2Z).

Quick Tips:**
- Calculate (A-2Z) for each nuclide and compare.

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Understanding +4 | -1

The electronic configuration of element with atomic number 24 is

View answer and explanation

Correct option: D

1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1

Option D (

1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1

) is correct because Chromium (atomic number 24) has an exception to Hund's rule, leading to a half-filled d-orbital.

Solution approach

Correct Answer + Core Reason: $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1$ . Chromium has a half-filled d-orbital for stability.

Step-by-Step Solution:

Step 1: Recall the Aufbau principle and Hund's rule.

Step 2: Write the expected electronic configuration for Chromium (Z=24): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^4$

Step 3: Apply the exception to Hund's rule: Due to the relatively small energy difference between the 4s and 3d orbitals, one electron from the 4s orbital is promoted to the 3d orbital to achieve a more stable half-filled d-orbital configuration.

Step 4: Write the actual electronic configuration for Chromium: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$

Step 5: Conclusion: The correct electronic configuration is $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1$ .

Why other options are wrong:
- **Option A ( $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^4, 4 s^2$ ): This is the expected configuration without considering the exception for stability.
- Option B ( $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^{10}$ ): Incorrect. This configuration does not correspond to Chromium.
- Option C ( $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^6$ ): Incorrect. This configuration does not correspond to Chromium.

Key Concepts:
- Electronic configuration
- Aufbau principle
- Hund's rule
- Exceptions to Hund's rule (half-filled and fully-filled orbitals)

Memory Aids:
- Remember that Chromium and Copper are common exceptions to Hund's rule.

Quick Tips:**
- Be aware of the exceptions to the Aufbau principle and Hund's rule for elements like Chromium and Copper.

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The electronic configuration of silver atom in ground state is:

View answer and explanation

Correct option: C

\left[ {\mathrm{Kr}} \right]4{d^{10}},5{s^1}

Option C (

[\mathrm{Kr}]

4{d¹⁰},5{s¹}) is correct because silver's electronic configuration has a completely filled d-orbital and a single electron in the s-orbital for stability.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Kr}} \right]4{d^{10}},5{s^1}$ . Completely filled orbitals are extra stable.

Step-by-Step Solution:

Step 1: Determine the atomic number of silver (Ag): Z = 47

Step 2: Write the expected electronic configuration based on Aufbau principle:
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^9$

Step 3: Consider the stability of completely filled and half-filled orbitals: A more stable configuration is achieved by moving one electron from the 5s orbital to the 4d orbital, resulting in a completely filled 4d orbital.

Step 4: Write the actual electronic configuration:
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^{10}$

Step 5: Express the configuration in noble gas notation: The electronic configuration of Krypton (Kr) is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$ . Therefore, the electronic configuration of silver can be written as $\left[ {\mathrm{Kr}} \right]4{d^{10}},5{s^1}$ .

Why other options are wrong:
- Option A: $\left[ {\mathrm{Ar}} \right]3{d^{10}},4{s^1}$ : Argon has an atomic number of 18, which is too low to be the preceding noble gas for silver. Also, the 3d and 4s orbitals are filled before the 4p orbital.
- Option B: $\left[ {\mathrm{Xr}} \right]4{f^{14}},5{d^{10}},6{s^1}$ : Xenon has an atomic number of 54, which is higher than silver's. Silver does not have electrons in the 4f orbital in its ground state.
- Option D: $\left[ {\mathrm{Kr}} \right]4{d^9},5{s^2}$ : This configuration violates the stability rule. Having a completely filled d-orbital (d¹⁰) is more stable than having a filled s-orbital (s²) and an incomplete d-orbital (d⁹).

Key Concepts:
- Electronic configuration
- Aufbau principle
- Hund's rule
- Stability of completely filled and half-filled orbitals
- Noble gas notation

Memory Aids:
- Remember that elements like Cr and Cu also show exceptions to Hund's rule for increased stability.

Quick Tips:
- When determining electronic configurations, always consider the stability gained by having completely filled or half-filled orbitals.

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Analyzing +4 | -1

Assertion : Total number of orbitals associated with principal quantum number

n=3

is 6 .

Reason : Number of orbitals in a shell equals to

2 n

.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D is correct. The assertion is false because the total number of orbitals with n=3 is 9 (1s + 3p + 5d), and the reason is false because the number of orbitals in a shell is n².

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The number of orbitals for n=3 is 9, and the number of orbitals in a shell is $n^2$ .

Step-by-Step Solution:

Step 1: Analyze the assertion: Total number of orbitals associated with principal quantum number $n=3$ is 6.

Step 2: Calculate the number of orbitals for $n=3$ : For $n=3$ , the possible values of $l$ are 0, 1, and 2. These correspond to the 3s, 3p, and 3d orbitals, respectively. The number of orbitals for each value of $l$ is $2l+1$ . So, we have:
- $l=0$ (3s): $2(0)+1 = 1$ orbital
- $l=1$ (3p): $2(1)+1 = 3$ orbitals
- $l=2$ (3d): $2(2)+1 = 5$ orbitals
Total number of orbitals = $1 + 3 + 5 = 9$
Therefore, the assertion is false.

Step 3: Analyze the reason: Number of orbitals in a shell equals to $2n$ .

Step 4: Determine the correct formula for the number of orbitals in a shell: The number of orbitals in a shell is given by $n^2$ . For example, when $n=3$ , the number of orbitals is $3^2 = 9$ .
Therefore, the reason is false.

Step 5: Conclusion: Both the assertion and the reason are false.

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

Key Concepts:
- Quantum Numbers
- Atomic Orbitals
- Electronic Configuration

Memory Aids:
- Number of orbitals in a shell = $n^2$

Quick Tips:
- Remember the relationship between $n$ , $l$ , and the number of orbitals.
- For each value of $n$ , $l$ ranges from 0 to $n-1$ .

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Pauli's exclusion principle states that

View answer and explanation

Correct option: D

All the four quantum numbers of twoelectrons in an atom cannot be equal

Option D (All the four quantum numbers of twoelectrons in an atom cannot be equal) is correct because Pauli's exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers.

Solution approach

Correct Answer + Core Reason: All the four quantum numbers of twoelectrons in an atom cannot be equal. This is the statement of Pauli's exclusion principle.

Step-by-Step Solution:

Step 1: State Pauli's exclusion principle. Pauli's exclusion principle states that no two electrons in the same atom can have the same set of all four quantum numbers (n, l, ml, and ms).

Step 2: Explain the meaning of the principle. This means that each electron in an atom must have a unique set of quantum numbers.

Why other options are wrong:
- Option A: Nucleus of an atom contains no negative charge is incorrect; the nucleus contains protons and neutrons.
- Option B: Electrons move in circular orbits around the nucleus is incorrect; this is a simplified Bohr model, not related to Pauli's principle.
- Option C: Electrons occupy orbitals of lowest energy is incorrect; this is Hund's rule and the Aufbau principle, not Pauli's principle.

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

Memory Aids:
- Pauli's principle: No two electrons can have the same four quantum numbers.

Quick Tips:
- Remember the statement of Pauli's exclusion principle.

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Possible number of orientations of a subshell is:

View answer and explanation

Correct option: C

2l + 1

Option C (

2l + 1

) is correct because the number of possible orientations of a subshell is determined by the magnetic quantum number (ml), which ranges from -l to +l, including 0, giving a total of 2l + 1 orientations.

Solution approach

Correct Answer + Core Reason: $2l + 1$ . The number of orientations of a subshell is given by this formula.

Step-by-Step Solution:

Step 1: Recall the quantum numbers.

Step 2: Understand that the number of possible orientations of a subshell is determined by the magnetic quantum number (ml).

Step 3: Remember that the magnetic quantum number (ml) can take values from -l to +l, including 0, where l is the azimuthal quantum number.

Step 4: Calculate the number of possible orientations. The number of possible orientations is 2l + 1.

Why other options are wrong:
- Option A: $l$ is incorrect because it represents the azimuthal quantum number, not the number of orientations.
- Option B: $n$ is incorrect because it represents the principal quantum number, not the number of orientations.
- Option D: ${n^2}$ is incorrect because it is related to the number of orbitals in a shell, not the number of orientations of a subshell.

Key Concepts:
- Quantum numbers
- Magnetic quantum number (ml)
- Azimuthal quantum number (l)
- Subshells
- Atomic orbitals

Memory Aids:
- Remember that the number of orientations is 2l + 1.

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

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The speed of the cathode rays is:

View answer and explanation

Correct option: B

Less than light

Option B (Less than light) is correct because cathode rays are streams of electrons, which have mass and cannot travel at the speed of light.

Solution approach

Correct Answer + Core Reason: Cathode rays travel at a speed less than light.

Step-by-Step Solution:

Step 1: Identify the nature of cathode rays: Cathode rays are streams of electrons.

Step 2: Recall the speed of light and the factors affecting the speed of particles.

Step 3: Electrons have mass, and no particle with mass can travel at the speed of light.

Step 4: Therefore, the speed of cathode rays is less than the speed of light.

Step 5: Final Answer: Less than light

Why other options are wrong:
- Option A: Equal to light is incorrect because particles with mass cannot travel at the speed of light.
- Option C: Greater than light is incorrect because nothing can travel faster than light according to the theory of relativity.
- Option D: May be less than, greater than or equal to light is incorrect because cathode rays cannot travel at or greater than the speed of light.

Key Concepts:
- Cathode rays
- Speed of light
- Theory of relativity

Memory Aids:
- Massive particles cannot reach the speed of light.

Quick Tips:
- Remember that only massless particles (like photons) can travel at the speed of light.

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Two electron in an atm of an element cannot have:

View answer and explanation

Correct option: D

An identical set of quantum numbers

Option D (An identical set of quantum numbers) is correct because the Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.

Solution approach

Correct Answer + Core Reason: An identical set of quantum numbers. The Pauli Exclusion Principle forbids two electrons in the same atom from having the same set of quantum numbers.

Step-by-Step Solution:

Step 1: Recall the Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers (n, l, ml, s).

Step 2: Analyze the options:
- A: The same principal quantum number (n) is possible for multiple electrons in an atom.
- B: The same azimuthal quantum number (l) is possible for multiple electrons in an atom.
- C: The same magnetic quantum number (ml) is possible for multiple electrons in an atom.
- D: An identical set of quantum numbers (n, l, ml, s) is not possible for any two electrons in an atom.

Why other options are wrong:
- Option A (The same principle quantum number): Electrons in the same energy level can have the same principal quantum number.
- Option B (The same azimuthal quantum number): Electrons in the same subshell can have the same azimuthal quantum number.
- Option C (The same magnetic quantum number): Electrons in the same orbital can have the same magnetic quantum number.
- Option D (An identical set of quantum numbers): This violates the Pauli Exclusion Principle.

Key Concepts:
- Pauli Exclusion Principle
- Quantum Numbers (n, l, ml, s)
- Atomic Structure

Memory Aids:
- Pauli: "No identical twins in the electron world!"

Quick Tips:
- Remember the four quantum numbers and what they represent: n (energy level), l (subshell shape), ml (orbital orientation), s (electron spin).

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The maximum number of atomic orbitals associated with a principal quantum number 5 is:

View answer and explanation

Correct option: D

25

Option D (25) is correct because the maximum number of atomic orbitals associated with a principal quantum number n is n², and for n=5, it is 5² = 25.

Solution approach

Correct Answer + Core Reason: 25. The number of atomic orbitals is $n^2$ .

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the number of atomic orbitals.

Step 2: The number of atomic orbitals for a given principal quantum number n is $n^2$ .

Step 3: Given n = 5, the number of atomic orbitals is $5^2 = 25$ .

Why other options are wrong:
- Option A (9): This would be correct for n = 3, not n = 5.
- Option B (12): This is not a valid formula for calculating the number of atomic orbitals.
- Option C (16): This would be correct for n = 4, not n = 5.

Key Concepts:
- Principal quantum number (n)
- Atomic orbitals

Memory Aids:
- $n^2$ = number of orbitals

Quick Tips:
- Remember the formula $n^2$ for calculating the number of atomic orbitals.

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Which of the given radiations is having high penetrating power and not affected by electrical and magnetic field?

View answer and explanation

Correct option: C

Gamma rays

Option C (Gamma rays) is correct because gamma rays have high penetrating power and are not affected by electric or magnetic fields due to being uncharged photons.

Solution approach

Correct Answer + Core Reason: Gamma rays possess high penetrating power and are unaffected by electric and magnetic fields.

Step-by-Step Solution:

Step 1: Identify the properties of each type of radiation.

Step 2: Alpha rays are positively charged particles (Helium nuclei) and have low penetrating power. They are affected by electric and magnetic fields.

Step 3: Beta rays are negatively charged particles (electrons or positrons) and have moderate penetrating power. They are affected by electric and magnetic fields.

Step 4: Gamma rays are high-energy photons (electromagnetic radiation) and have high penetrating power. They are not affected by electric and magnetic fields because they are uncharged.

Step 5: Neutrons are neutral particles and have high penetrating power, but they are not typically classified as \"radiations\" in the same context as alpha, beta, and gamma rays, which are products of radioactive decay. While neutrons can penetrate materials, the question is specifically about radiations.

Step 6: Compare the penetrating power and charge of each radiation type.

Step 7: Conclude that gamma rays have the highest penetrating power and are not affected by electric or magnetic fields.

Why other options are wrong:
- Option A: Alpha rays are charged and have low penetrating power.
- Option B: Beta rays are charged and have moderate penetrating power.
- Option D: Neutrons are neutral and have high penetrating power, but the question is about radiations from radioactive decay, where gamma rays are the correct answer.

Key Concepts:
- Alpha, beta, and gamma radiation
- Penetrating power of radiation
- Effect of electric and magnetic fields on charged particles

Memory Aids:
- Alpha particles are like Helium nuclei (heavy and charged).
- Beta particles are like electrons (light and charged).
- Gamma rays are like light (uncharged and energetic).

Quick Tips:
- Remember the order of penetrating power: Alpha < Beta < Gamma.
- Charged particles are affected by electric and magnetic fields, while uncharged particles are not.

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The energy of an electron in

{n}^{\text {th }}

orbit of hydrogen atom is

View answer and explanation

Correct option: C

\frac{13.6}{n^2} \mathrm{eV}

Option C (

\frac{{13.6}}{{n^2}} \mathrm{{eV}}

) is correct because the energy of an electron in the nth orbit of a hydrogen atom is given by the formula

E = -13.6/n^2

eV.

Solution approach

Correct Answer + Core Reason: $\frac{{13.6}}{{n^2}} \mathrm{{eV}}$ . The energy of an electron in the nth orbit of a hydrogen atom is given by $E = -13.6/n^2$ eV.

Step-by-Step Solution:

Step 1: Recall the formula for the energy of an electron in the nth orbit of a hydrogen atom.

Step 2: The formula is: $E = -13.6 \frac{{Z^2}}{{n^2}} \text{{ eV}}$ , where Z is the atomic number and n is the principal quantum number.

Step 3: For hydrogen, Z = 1.

Step 4: Therefore, $E = -13.6 \frac{{1^2}}{{n^2}} \text{{ eV}} = -\frac{{13.6}}{{n^2}} \text{{ eV}}$

Why other options are wrong:
- Option A: $\frac{{13.6}}{{n^4}} \mathrm{{eV}}$ is incorrect. The energy is inversely proportional to $n^2$ , not $n^4$ .
- Option B: $\frac{{13.6}}{{n^3}} \mathrm{{eV}}$ is incorrect. The energy is inversely proportional to $n^2$ , not $n^3$ .
- Option D: $\frac{{13.6}}{{n}} \mathrm{{eV}}$ is incorrect. The energy is inversely proportional to $n^2$ , not $n$ .

Key Concepts:
- Bohr's model of the hydrogen atom.
- Energy levels of hydrogen atom.

Memory Aids:
- Remember the formula: $E = -13.6/n^2$ eV.

Quick Tips:
- The energy is always negative, indicating that the electron is bound to the nucleus.

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The correct designation of an electron with

n = 4,l = 3,m = 2,

and

s = 1/2

is:

View answer and explanation

Correct option: B

4f

Option B (4f) is correct because n=4 and l=3 corresponds to the 4f subshell.

Solution approach

Correct Answer + Core Reason: 4f. n=4 and l=3 corresponds to the 4f subshell.

Step-by-Step Solution:

Step 1: Understand the relationship between quantum numbers and atomic orbitals.

Step 2: The principal quantum number (n) indicates the energy level or shell.

Step 3: The azimuthal quantum number (l) indicates the shape of the orbital (subshell):
- 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 4: Given n = 4 and l = 3, the electron is in the 4f subshell.

Why other options are wrong:
- Option A: 3d corresponds to n = 3 and l = 2.
- Option C: 5p corresponds to n = 5 and l = 1.
- Option D: 6s corresponds to n = 6 and l = 0.

Key Concepts:
- Quantum Numbers
- Atomic Orbitals
- Electronic Configuration

Memory Aids:
- Remember the values of l for different subshells (s, p, d, f).

Quick Tips:
- Know the relationship between quantum numbers and the shape and energy of atomic orbitals.

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S-I: There is one nodal regions in 3s-orbital.

S-II: There is one nodal plane in 3s-orbital.

View answer and explanation

Correct option: B

Both S-I and S-II are incorrect

Option B (Both S-I and S-II are incorrect) is correct because a 3s-orbital has two spherical nodes (nodal regions) and no nodal planes.

Solution approach

Correct Answer + Core Reason: Both S-I and S-II are incorrect. A 3s orbital has two radial nodes (spherical) and no nodal planes.

Step-by-Step Solution:

Step 1: Understand the concept of nodal regions and nodal planes.

Step 2: Nodal regions (or radial nodes) are spherical surfaces where the probability of finding an electron is zero.

Step 3: Nodal planes are flat planes where the probability of finding an electron is zero.

Step 4: For a 3s orbital, the number of radial nodes is given by $n - l - 1$ , where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. For a 3s orbital, $n = 3$ and $l = 0$ . Therefore, the number of radial nodes is $3 - 0 - 1 = 2$ .

Step 5: The number of nodal planes is given by $l$ . For a 3s orbital, $l = 0$ . Therefore, the number of nodal planes is 0.

Step 6: Statement I says there is one nodal region, which is incorrect. There are two nodal regions.

Step 7: Statement II says there is one nodal plane, which is incorrect. There are zero nodal planes.

Why other options are wrong:
- Option A: Both S-I and S-II are correct: This is incorrect because both statements are false.
- Option C: S-I is correct and S-II is incorrect: This is incorrect because S-I is also incorrect.
- Option D: S-I is incorrect and S-II is correct: This is incorrect because S-II is also incorrect.

Key Concepts:
- Atomic orbitals
- Nodal regions (radial nodes)
- Nodal planes
- Quantum numbers

Memory Aids:
- Remember the formula for radial nodes: $n - l - 1$ .
- Remember that s orbitals have no nodal planes.

Quick Tips:
- Know the shapes and nodal properties of s, p, and d orbitals.
- Practice calculating the number of radial nodes and nodal planes for different orbitals.

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Which set has the same number of unpaired electrons in their ground state

View answer and explanation

Correct option: D

N,P,V

The correct answer is N, P, V because Nitrogen, Phosphorus, and Vanadium each have 3 unpaired electrons in their ground state electron configurations.

Solution approach

Correct Answer + Core Reason: N,P,V because each of these elements has 3 unpaired electrons in their ground state.

Step-by-Step Solution:

Step 1: Write the electronic configurations of each element/ion in the options:
- Cl-: [Ne] 3s2 3p6 (0 unpaired electrons)
- Fe+3: [Ar] 3d5 (5 unpaired electrons)
- Cr+3: [Ar] 3d3 (3 unpaired electrons)
- N+: [He] 2s2 2p1 (2 unpaired electrons)
- Mg+2: [Ne] (0 unpaired electrons)
- Al: [Ne] 3s2 3p1 (1 unpaired electron)
- Na: [Ne] 3s1 (1 unpaired electron)
- P: [Ne] 3s2 3p3 (3 unpaired electrons)
- Cl: [Ne] 3s2 3p5 (1 unpaired electron)
- N: [He] 2s2 2p3 (3 unpaired electrons)
- V: [Ar] 4s2 3d3 (3 unpaired electrons)

Step 2: Identify the set with the same number of unpaired electrons:
- N, P, and V each have 3 unpaired electrons.

Why other options are wrong:
- Option A: Cl- has 0, Fe+3 has 5, and Cr+3 has 3 unpaired electrons. They are not equal.
- Option B: N+ has 2, Mg+2 has 0, and Al has 1 unpaired electron. They are not equal.
- Option C: Na has 1, P has 3, and Cl has 1 unpaired electron. They are not equal.

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

Memory Aids:
- Hund's rule states that electrons will individually occupy each orbital within a subshell before doubling up in any one orbital.

Quick Tips:
- Write out the electronic configuration to easily determine the number of unpaired electrons.

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Spin angular momentum for electron

View answer and explanation

Correct option: A

\sqrt {S(s + 1)} \frac{h}{{2\pi }}

Option A (

\sqrt {S(s + 1)} \frac{h}{{2\pi }}

) is correct because this is the formula for the spin angular momentum of an electron.

Solution approach

Correct Answer + Core Reason: $\sqrt {S(s + 1)} \frac{h}{{2\pi }}$ . This is the standard formula for spin angular momentum.

Step-by-Step Solution:
Step 1: Recall the formula for spin angular momentum.
Step 2: The formula is given by $\sqrt {S(s + 1)} \frac{h}{{2\pi }}$ , where S is the spin quantum number.

Why other options are wrong:
- Option B: $\sqrt {2S(s + 1)} \frac{h}{{2\pi }}$ is not the correct formula.
- Option C: $\sqrt {S(s + 2)} \frac{h}{{2\pi }}$ is not the correct formula.
- Option D: Option A is the correct formula, so this is incorrect.

Key Concepts:
- Spin angular momentum
- Spin quantum number

Memory Aids:
- Remember the formula directly.

Quick Tips:
- Know the formulas for different types of angular momentum.

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The principle, which gives a way to fill the electrons in the available energy level is:

View answer and explanation

Correct option: C

Aufbau principle

Option C (Aufbau principle) is correct because the Aufbau principle dictates the filling of electrons in available energy levels.

Solution approach

Correct Answer + Core Reason: Aufbau principle. The Aufbau principle governs the filling of electrons in available energy levels.

Step-by-Step Solution:

Step 1: Recall the definitions of Hund's rule, Pauli's exclusion principle, and the Aufbau principle.

Step 2: Hund's rule states that electrons will individually occupy each orbital within a subshell before doubling up in any one orbital.

Step 3: Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.

Step 4: The Aufbau principle states that electrons first fill the lowest energy levels available before occupying higher energy levels.

Why other options are wrong:
- Option A: Hund's rule deals with the filling of orbitals within a subshell, not the filling of energy levels in general.
- Option B: Pauli's exclusion principle states that no two electrons can have the same set of quantum numbers, not the order of filling energy levels.
- Option D: The correct answer is one of the options provided.

Key Concepts:
- Aufbau principle
- Hund's rule
- Pauli's exclusion principle

Memory Aids:
- Aufbau = Building up (of electron configuration).

Quick Tips:
- Understand the difference between the three principles.

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The number of electrons in the nucleus of

{C^{12}}

is

View answer and explanation

Correct option: C

0

Option C (0) is correct because the nucleus of an atom contains protons and neutrons, but not electrons.

Solution approach

Correct Answer + Core Reason: 0. Electrons are not present in the nucleus.

Step-by-Step Solution:

Step 1: Recall the structure of an atom. An atom consists of a nucleus containing protons and neutrons, surrounded by electrons in orbitals.

Step 2: Identify the location of electrons. Electrons are located outside the nucleus in electron orbitals.

Step 3: Conclude that the number of electrons in the nucleus is 0.

Why other options are wrong:
- Option A: 6 is the number of protons in the nucleus of ${C^{12}}$ , not the number of electrons in the nucleus.
- Option B: 12 is the mass number of ${C^{12}}$ , not the number of electrons in the nucleus.
- Option D: 3 is not a relevant number related to the composition of the nucleus of ${C^{12}}$ .

Key Concepts:
- Atomic structure
- Nucleus (protons and neutrons)
- Electrons

Memory Aids:
- Electrons orbit the nucleus, they are not inside it.

Quick Tips:
- Remember the basic structure of an atom: protons and neutrons in the nucleus, electrons orbiting the nucleus.

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Which one is not in agreement with Bohr's model of the atom?

View answer and explanation

Correct option: D

Heisenberg’s uncertainty principle

Option D (Heisenberg’s uncertainty principle) is correct because Bohr's model violates the Heisenberg uncertainty principle by defining both position and momentum of an electron simultaneously.

Solution approach

Correct Answer + Core Reason: Heisenberg’s uncertainty principle. Bohr's model violates the uncertainty principle.

Step-by-Step Solution:

Step 1: Recall the postulates of Bohr's model.\nBohr's model assumes that electrons move in fixed orbits with definite energies and radii.

Step 2: State Heisenberg's uncertainty principle.\nHeisenberg's uncertainty principle states that it is impossible to simultaneously determine the exact position and momentum of an electron.
$\Delta x \Delta p \geq \frac{h}{{4\pi}}$

Step 3: Explain why Bohr's model is not in agreement with Heisenberg's uncertainty principle.\nBohr's model defines both the position (radius of the orbit) and momentum (related to the velocity) of the electron simultaneously, which violates Heisenberg's uncertainty principle.

Step 4: Consider the other options.
- Line spectra of the hydrogen atom: Bohr's model successfully explained the line spectra of hydrogen.
- Pauli's exclusion principle: This principle is related to the filling of orbitals and is not directly contradicted by Bohr's model.
- Planck's theory: Bohr's model incorporated Planck's quantum theory.

Why other options are wrong:
- Option A: Bohr's model explains the line spectra of the hydrogen atom.
- Option B: Pauli's exclusion principle is not directly contradicted by Bohr's model.
- Option C: Bohr's model incorporated Planck's quantum theory.

Key Concepts:
- Bohr's model
- Heisenberg's uncertainty principle
- Quantum mechanics

Memory Aids:
- Bohr's model has limitations and is superseded by quantum mechanics.

Quick Tips:
- Remember that Bohr's model is a simplified model and has limitations.

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A metallic ion

{M^{2 + }}

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

View answer and explanation

Correct option: A

30

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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Which among the following species have the same number of electrons in its outermost as well as penultimate shell?

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{a}^{2 + }}

Option D (

\mathrm{C}{\mathrm{a}^{2 + }}

) is correct because it has 8 electrons in both its outermost and penultimate shells.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{a}^{2 + }}$ . Calcium ion has 8 electrons in both its outermost and penultimate shells after losing 2 electrons.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each species:
- Mg: 1s² 2s² 2p⁶ 3s² => Mg²⁺: 1s² 2s² 2p⁶ (2, 8)
- O: 1s² 2s² 2p⁴ => O²⁻: 1s² 2s² 2p⁶ (2, 8)
- F: 1s² 2s² 2p⁵ => F⁻: 1s² 2s² 2p⁶ (2, 8)
- Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² => Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ (2, 8, 8)

Step 2: Identify the outermost and penultimate shells for each ion:
- Mg²⁺: Outermost = 2nd shell (8 electrons), Penultimate = 1st shell (2 electrons)
- O²⁻: Outermost = 2nd shell (8 electrons), Penultimate = 1st shell (2 electrons)
- F⁻: Outermost = 2nd shell (8 electrons), Penultimate = 1st shell (2 electrons)
- Ca²⁺: Outermost = 3rd shell (8 electrons), Penultimate = 2nd shell (8 electrons)

Step 3: Determine which species has the same number of electrons in both shells:
- Only Ca²⁺ has 8 electrons in both its outermost and penultimate shells.

Why other options are wrong:
- Option A: $\mathrm{M}{\mathrm{g}^{2 + }}$ has 8 electrons in the outermost shell and 2 in the penultimate.
- Option B: ${\mathrm{O}^{2 - }}$ has 8 electrons in the outermost shell and 2 in the penultimate.
- Option C: ${\mathrm{F}^ - }$ has 8 electrons in the outermost shell and 2 in the penultimate.

Key Concepts:
- Electronic configuration
- Outermost and penultimate shells
- Ion formation

Memory Aids:
- Remember the order of filling electron shells (1s, 2s, 2p, 3s, 3p, 4s, etc.).

Quick Tips:
- Write out the full electronic configuration to easily identify the number of electrons in each shell.
- Remember that positive ions lose electrons, and negative ions gain electrons.

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