STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 6

151 Medium

Analyzing +4 | -1

Which of the following has maximum energy

View answer and explanation

Correct option: C

An electron moving with velocity of light

Option C (An electron moving with velocity of light) is correct because, although theoretically impossible, it would possess the highest kinetic energy among the given options due to its velocity.

Solution approach

Correct Answer + Core Reason: An electron moving with velocity of light. Although impossible, it would have the highest energy.

Step-by-Step Solution:

Step 1: Analyze each option:
- A photon with wavelength 10 nm: $E = \frac{{hc}}{\lambda} = \frac{{hc}}{{10 \times 10^{-9} \text{{ m}}}}$
- A photon with frequency 10 Hz: $E = hf = h(10 \text{{ Hz}})$
- An electron moving with velocity of light: $E = \frac{{mc^2}}{{\sqrt{{1 - \frac{{v^2}}{{c^2}}}}}}$ (relativistic kinetic energy). As v approaches c, the energy approaches infinity.

Step 2: Compare the energies. The energy of a photon is $E = hf = \frac{{hc}}{\lambda}$ . The energy of a moving electron is kinetic energy, which can be relativistic if the velocity is close to the speed of light.

Step 3: Since the electron is moving with the velocity of light (theoretically), its kinetic energy would be extremely high (approaching infinity). Therefore, it has the maximum energy.

Why other options are wrong:
- Option A (A photon with wavelength 10 nm): This has a relatively high energy, but it's less than the theoretical energy of an electron moving at the speed of light.
- Option B (A photon with frequency 10 Hz): This has a very low energy compared to the other options.
- Option D (All have equal energy): This is incorrect because the energies are different.

Key Concepts:
- Energy of a photon
- Kinetic energy
- Relativistic kinetic energy

Memory Aids:
- $E = hf = \frac{{hc}}{\lambda}$
- $KE = \frac{{mc^2}}{{\sqrt{{1 - \frac{{v^2}}{{c^2}}}}}} - mc^2$ (relativistic)

Quick Tips:
- Remember the formulas for energy of a photon and kinetic energy.
- Consider relativistic effects when dealing with high velocities.

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Consider the element

Mg,\,\,\,AI,\,\,\,S,\,\,P

and

Si,

the correct increasing order of their first ionisation enthalpy is

View answer and explanation

Correct option: A

Al < Mg < Si < S < P

Option A (Al < Mg < Si < S < P) is correct because ionization enthalpy generally increases across a period and decreases down a group, with exceptions due to electron configuration.

Solution approach

Correct Answer + Core Reason: Al < Mg < Si < S < P. Ionization enthalpy generally increases across a period, but there are exceptions due to electron configurations.

Step-by-Step Solution:

Step 1: Identify the positions of the elements in the periodic table:
- Mg (Group 2, Period 3)
- Al (Group 13, Period 3)
- Si (Group 14, Period 3)
- P (Group 15, Period 3)
- S (Group 16, Period 3)

Step 2: Recall the general trend of ionization enthalpy. Ionization enthalpy generally increases across a period from left to right.

Step 3: Consider the exceptions. There is an exception between Group 2 and Group 13 (Mg and Al) because Mg has a filled s-orbital, making it more stable and harder to remove an electron. There is also an exception between Group 15 and Group 16 (P and S) because P has a half-filled p-orbital, making it more stable and harder to remove an electron.

Step 4: Arrange the elements in increasing order of ionization enthalpy, considering the exceptions: Al < Mg < Si < S < P.

Why other options are wrong:
- Option B: Mg < Al < Si < P < S - Incorrect because Al has lower ionization enthalpy than Mg, and S has lower ionization enthalpy than P.
- Option C: Mg < Al < Si < S < P - Incorrect because Al has lower ionization enthalpy than Mg.
- Option D: Al < Mg < S < Si < P - Incorrect because Si has lower ionization enthalpy than S.

Key Concepts:
- Ionization enthalpy
- Periodic trends
- Electronic configuration

Memory Aids:
- Remember that ionization enthalpy generally increases across a period and decreases down a group.
- Half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Know the periodic trends and the exceptions to those trends.

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The number of radial nodes of 3s and 2p-orbitals are respectively

View answer and explanation

Correct option: A

2, 0

Option A (2, 0) is correct because the number of radial nodes is calculated using the formula n-l-1, where n is the principal quantum number and l is the azimuthal quantum number.

Solution approach

Correct Answer + Core Reason: 2, 0. The number of radial nodes is calculated using the formula n-l-1.

Step-by-Step Solution:

Step 1: Define the formula for radial nodes: Radial nodes = n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Step 2: Calculate radial nodes for 3s orbital: For 3s, n = 3 and l = 0. Therefore, radial nodes = 3 - 0 - 1 = 2.

Step 3: Calculate radial nodes for 2p orbital: For 2p, n = 2 and l = 1. Therefore, radial nodes = 2 - 1 - 1 = 0.

Step 4: Final Answer: The number of radial nodes for 3s is 2 and for 2p is 0.

Why other options are wrong:
- Option B: This option incorrectly assigns the number of radial nodes. It reverses the values for 3s and 2p.
- Option C: This option incorrectly calculates the number of radial nodes for both 3s and 2p orbitals.
- Option D: This option incorrectly calculates the number of radial nodes for both 3s and 2p orbitals.

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

Memory Aids:
- Remember the formula: Radial nodes = n - l - 1

Quick Tips:
- Always identify the n and l values correctly before applying the formula.

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The number of elliptical orbits, including circular orbits in the M-shell of an atom is:

View answer and explanation

Correct option: A

3

Option A (3) is correct because the M-shell corresponds to the principal quantum number n=3, and for n=3, there are 3 possible values of the azimuthal quantum number l (0, 1, 2), representing 3 subshells and thus 3 elliptical orbits.

Solution approach

Correct Answer + Core Reason: 3. The number of elliptical orbits is equal to the principal quantum number for that shell.

Step-by-Step Solution:

Step 1: Identify the M-shell. The M-shell corresponds to the principal quantum number n = 3.

Step 2: Determine the possible values of the azimuthal quantum number (l). For a given n, the possible values of l are 0, 1, 2, ..., n-1. Therefore, for n = 3, the possible values of l are 0, 1, and 2.

Step 3: Relate the azimuthal quantum number to the shape of the orbits. l = 0 corresponds to a circular orbit, and l = 1 and l = 2 correspond to elliptical orbits.

Step 4: Count the number of elliptical orbits, including the circular orbit. Since there are three possible values of l (0, 1, 2), there are three possible orbits in the M-shell: one circular (l=0) and two elliptical (l=1 and l=2). Thus, the total number of elliptical orbits, including the circular orbit, is 3.

Step 5: Final Answer: 3

Why other options are wrong:
- Option B: 4 is incorrect because the M-shell corresponds to n=3, not n=4.
- Option C: 2 is incorrect because it doesn't include the circular orbit (l=0).
- Option D: 1 is incorrect because it only considers one elliptical orbit and doesn't account for the circular orbit or other possible elliptical orbits.

Key Concepts:
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Shells and subshells
- Shape of atomic orbitals

Memory Aids:
- Remember that the M-shell corresponds to n=3.
- The number of orbitals in a shell is equal to n.

Quick Tips:
- The number of elliptical orbits (including the circular one) is equal to the principal quantum number (n).

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If an electron has spin quantum number of

+ \frac{1}{2}

and a magnetic quantum number of -1, it cannot be represented in an

View answer and explanation

Correct option: A

s-orbital

Option A (s-orbital) is correct because s-orbitals have an azimuthal quantum number (l) of 0, and therefore a magnetic quantum number (m) of 0. A magnetic quantum number of -1 is not possible for an s-orbital.

Solution approach

Correct Answer + Core Reason: s-orbital. s-orbitals have l=0, therefore m=0, so m=-1 is not possible.

Step-by-Step Solution:

Step 1: Relate the magnetic quantum number to the azimuthal quantum number.

Step 2: The magnetic quantum number (m) can take values from -l to +l, including 0, where l is the azimuthal quantum number.

Step 3: For an s-orbital, l = 0, so m = 0.

Step 4: Since the electron has a magnetic quantum number of -1, it cannot be in an s-orbital.

Why other options are wrong:
- Option B: p-orbital. For a p-orbital, l = 1, so m can be -1, 0, or +1.
- Option C: d-orbital. For a d-orbital, l = 2, so m can be -2, -1, 0, +1, or +2.
- Option D: f-orbital. For an f-orbital, l = 3, so m can be -3, -2, -1, 0, +1, +2, or +3.

Key Concepts:
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- s: l=0, m=0
- p: l=1, m=-1, 0, +1
- d: l=2, m=-2, -1, 0, +1, +2
- f: l=3, m=-3, -2, -1, 0, +1, +2, +3

Quick Tips:
- Know the relationship between l and m values for each type of orbital.

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Analyzing +4 | -1

The incorrect statement about Bohr’s orbit of hydrogen atom is

View answer and explanation

Correct option: B

KE of electron = PE of electron

Option B (KE of electron = PE of electron) is the incorrect statement because the Kinetic Energy (KE) of the electron is equal to half the magnitude of the Potential Energy (PE), i.e., KE = -PE/2.

Solution approach

Correct Answer + Core Reason: KE of electron = PE of electron. The kinetic energy of an electron is half the magnitude of its potential energy in a Bohr orbit.

Step-by-Step Solution:

Step 1: Recall the expressions for Kinetic Energy (KE) and Potential Energy (PE) in Bohr's model.

Step 2: KE = $\frac{1}{2}mv^2 = \frac{Ke^2}{2r}$ , where K = $\frac{1}{4\pi\epsilon_0}$ .

Step 3: PE = $-\frac{Ke^2}{r}$ .

Step 4: Comparing KE and PE, we see that KE = - $\frac{1}{2}$ PE, or |KE| = $\frac{1}{2}$ |PE|.

Step 5: Analyze the other options:
- Option A: $r = {n^2}\frac{{{h^2}}}{{4{\pi ^2}m\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right)}}$ is the correct expression for the radius of the nth orbit.
- Option C: $E = - \frac{1}{{{n^2}}}\frac{{2{\pi ^2}m{{\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right)}^2}}}{{{h^2}}}$ is the correct expression for the energy of the electron in the nth orbit.

Why other options are wrong:
- Option A: $r = {n^2}\frac{{{h^2}}}{{4{\pi ^2}m\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right)}}$ : This is the correct formula for the radius of the nth orbit in the Bohr model.
- Option C: $E = - \frac{1}{{{n^2}}}\frac{{2{\pi ^2}m{{\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right)}^2}}}{{{h^2}}}$ : This is the correct formula for the energy of the electron in the nth orbit in the Bohr model.
- Option D: None of the above is incorrect: This is incorrect because option B is incorrect.

Key Concepts:
- Bohr's model of the hydrogen atom
- Kinetic energy and potential energy of an electron in a Bohr orbit
- Radius and energy of an electron in the nth orbit

Memory Aids:
- Remember that the kinetic energy is always positive, and the potential energy is always negative.
- The magnitude of KE is half the magnitude of PE.

Quick Tips:
- Always remember the relationship between KE and PE in Bohr's model: KE = -PE/2.

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If the following particles travel with equal speed, then for which particle the wavelength will be longest?

View answer and explanation

Correct option: D

β-particle

Option D (β-particle) is correct because the de Broglie wavelength is inversely proportional to mass; the β-particle (electron) has the smallest mass among the given options, resulting in the longest wavelength.

Solution approach

Correct Answer + Core Reason: β-particle. Wavelength is inversely proportional to mass.

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula: $\lambda = \frac{h}{p} = \frac{h}{{mv}}$ , where $\lambda$ is the de Broglie wavelength, h is Planck's constant, m is mass, and v is velocity.

Step 2: Analyze the relationship between wavelength and mass. Since the velocity is the same for all particles, the de Broglie wavelength is inversely proportional to the mass: $\lambda \propto \frac{1}{m}$ .

Step 3: Compare the masses of the particles. A proton has a mass of approximately 1 atomic mass unit. A neutron has a mass of approximately 1 atomic mass unit. An alpha particle has a mass of approximately 4 atomic mass units (2 protons + 2 neutrons). A beta particle is an electron, which has a much smaller mass than protons, neutrons, or alpha particles.

Step 4: Determine which particle has the longest wavelength. Since wavelength is inversely proportional to mass, the particle with the smallest mass will have the longest wavelength. Therefore, the beta particle (electron) has the longest de Broglie wavelength.

Step 5: Final Answer: β-particle

Why other options are wrong:
- Option A: A proton has a larger mass than a beta particle, so it will have a shorter de Broglie wavelength.
- Option B: A neutron has a larger mass than a beta particle, so it will have a shorter de Broglie wavelength.
- Option C: An alpha particle has a much larger mass than a beta particle, so it will have a much shorter de Broglie wavelength.

Key Concepts:
- De Broglie wavelength: $\lambda = \frac{h}{{mv}}$
- Relationship between wavelength and mass

Memory Aids:
- Remember that lighter particles have longer wavelengths.

Quick Tips:
- When comparing de Broglie wavelengths, consider the masses of the particles if their velocities are the same.

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The orbital angular momentum for an electron revolving in an orbit is given by

\sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }}

. This momentum for an s-electron will be given by

View answer and explanation

Correct option: B

Zero

Option B (Zero) is correct because for an s-electron, the azimuthal quantum number (l) is 0, and substituting l=0 into the formula

\sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }}

results in zero.

Solution approach

Correct Answer + Core Reason: Zero. For an s-electron, l=0, so the orbital angular momentum is zero.

Step-by-Step Solution:

Step 1: Recall the formula for orbital angular momentum.

Step 2: The formula is given by $\sqrt{l(l+1)} \frac{h}{2\pi}$ .

Step 3: For an s-electron, l = 0.

Step 4: Substitute l = 0 into the formula: $\sqrt{0(0+1)} \frac{h}{2\pi} = 0$ .

Why other options are wrong:
- Option A: $+ \frac{1}{2}.\frac{h}{{2\mathrm{\pi }}}$ . This is incorrect.
- Option C: $\frac{h}{{2\mathrm{\pi }}}$ . This is incorrect.
- Option D: $\sqrt 2 .\frac{h}{{2\mathrm{\pi }}}$ . This is the orbital angular momentum for a p-orbital.

Key Concepts:
- Orbital Angular Momentum
- Azimuthal Quantum Number

Memory Aids:
- s-orbital has zero orbital angular momentum.

Quick Tips:
- Remember that the orbital angular momentum depends on the azimuthal quantum number.

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The mass of one mole of electron is:

View answer and explanation

Correct option: A

0.55 mg

Option A (0.55 mg) is correct because the mass of one mole of electrons is the mass of a single electron multiplied by Avogadro's number.

Solution approach

Correct Answer + Core Reason: 0.55 mg. The mass of one mole of electrons is Avogadro's number times the mass of a single electron.

Step-by-Step Solution:

Step 1: Recall the mass of a single electron: $m_e = 9.109 \times 10^{-31} \text{ kg}$ .

Step 2: Recall Avogadro's number: $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$ .

Step 3: Calculate the mass of one mole of electrons: $m_{\text{mole}} = m_e \times N_A = (9.109 \times 10^{-31} \text{ kg}) \times (6.022 \times 10^{23} \text{ mol}^{-1}) = 5.485 \times 10^{-7} \text{ kg/mol}$ .

Step 4: Convert the mass to milligrams (mg): $5.485 \times 10^{-7} \text{ kg} = 5.485 \times 10^{-7} \times 10^6 \text{ mg} = 0.5485 \text{ mg} \approx 0.55 \text{ mg}$ .

Why other options are wrong:
- Option B: 0.008 mg - Incorrect, this is significantly smaller than the correct mass.
- Option C: 1.008 mg - Incorrect, this is closer to the mass of a mole of hydrogen atoms.
- Option D: 0.184 mg - Incorrect, this is smaller than the correct mass.

Key Concepts:
- Mole concept
- Avogadro's number
- Mass of an electron

Memory Aids:
- Remember Avogadro's number and the mass of an electron.

Quick Tips:
- Pay attention to units and conversions.

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Which of the following transitions are not allowed in the normal electronic emission spectrum of an atom?

View answer and explanation

Correct option: A

2s \to 1s

Option A (

2s \to 1s

) is correct because transitions between s orbitals are forbidden due to the selection rule

\Delta l = \pm 1

.

Solution approach

Correct Answer + Core Reason: $2s \to 1s$ . Transitions between s orbitals are forbidden due to selection rules.

Step-by-Step Solution:

Step 1: Identify the selection rules for electronic transitions. The primary selection rule is $\Delta l = \pm 1$ , where $l$ is the azimuthal quantum number.

Step 2: For an $s$ orbital, $l = 0$ . Therefore, a transition from $2s$ to $1s$ implies $\Delta l = 0 - 0 = 0$ , which violates the selection rule $\Delta l = \pm 1$ .

Step 3: Check the other options:
- $2p \to 1s$ : $l$ changes from 1 to 0, so $\Delta l = -1$ . Allowed.
- $3d \to 4p$ : $l$ changes from 2 to 1, so $\Delta l = -1$ . Allowed.
- $5p \to 3s$ : $l$ changes from 1 to 0, so $\Delta l = -1$ . Allowed.

Why other options are wrong:
- Option B: $2p \to 1s$ : This transition is allowed because $\Delta l = -1$ , satisfying the selection rule.
- Option C: $3d \to 4p$ : This transition is allowed because $\Delta l = -1$ , satisfying the selection rule.
- Option D: $5p \to 3s$ : This transition is allowed because $\Delta l = -1$ , satisfying the selection rule.

Key Concepts:
- Selection Rules for Electronic Transitions
- Azimuthal Quantum Number (l)

Memory Aids:
- Remember the selection rule $\Delta l = \pm 1$ .

Quick Tips:
- Always check the change in the azimuthal quantum number when determining allowed transitions.

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Remembering +4 | -1

The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies

E_1

(lower) and

E_2

(higher) is given by

View answer and explanation

Correct option: D

v=\frac{E_2-E_1}{h}

Option D (All of the above) is correct because the wave mechanical model of the atom incorporates de Broglie's concept of wave-particle duality, Heisenberg's uncertainty principle, and Schrödinger's wave equation to describe the behavior of electrons in atoms.

Solution approach

Correct Answer + Core Reason: All of the above. The wave mechanical model incorporates all three concepts.

Step-by-Step Solution:

Step 1: Understand the de Broglie concept of dual nature of electron. Electrons exhibit both wave-like and particle-like properties.

Step 2: Understand Heisenberg's uncertainty principle. It states that it is impossible to know both the position and momentum of an electron with perfect accuracy.

Step 3: Understand Schrödinger's wave equation. It describes the behavior of electrons in atoms in terms of wave functions.

Step 4: The wave mechanical model of the atom is based on all three of these concepts.

Why other options are wrong:
- Option A: de Broglie concept of dual nature of electron: This is partially correct, but the wave mechanical model also depends on the other two concepts.
- Option B: Heisenberglie uncertainty principle: This is partially correct, but the wave mechanical model also depends on the other two concepts.
- Option C: Schrödinger wave equation: This is partially correct, but the wave mechanical model also depends on the other two concepts.

Key Concepts:
- Wave-particle duality
- Heisenberg's uncertainty principle
- Schrödinger's wave equation
- Wave mechanical model of the atom

Memory Aids:
- Remember that the wave mechanical model is a comprehensive model that incorporates all three concepts.

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

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The magnetic moment of electron in an atom (excluding orbital magnetic moment) is given by:

View answer and explanation

Correct option: A

\sqrt {n\left( {n + 2} \right)}

Bohr Magneton (or B.M)

Option A (

\sqrt {n(n + 2)}

Bohr Magneton) is correct because it represents the spin-only magnetic moment formula, where 'n' is the number of unpaired electrons.

Solution approach

Correct Answer + Core Reason: $\sqrt {n(n + 2)}$ Bohr Magneton (or B.M) is correct because it's the spin-only magnetic moment formula.

Step-by-Step Solution:

Step 1: Identify the relevant formula for the spin-only magnetic moment.

Step 2: The spin-only magnetic moment is given by the formula: $\mu_s = \sqrt{n(n+2)}$ Bohr Magnetons (B.M.), where 'n' is the number of unpaired electrons.

Step 3: This formula excludes the orbital magnetic moment and only considers the contribution from the electron spin.

Why other options are wrong:
- Option B: $\sqrt {n(n + 1)}$ B.M. This is not the correct formula for the spin-only magnetic moment. It does not have a direct physical meaning in this context.
- Option C: $\sqrt {n(n + 3)}$ B.M. This is also not the correct formula for the spin-only magnetic moment and has no established physical basis.
- Option D: None of the above. This is incorrect because option A provides the correct formula.

Key Concepts:
- Spin-only magnetic moment
- Bohr Magneton
- Unpaired electrons

Memory Aids:
- Remember the formula: $\sqrt{n(n+2)}$ B.M.

Quick Tips:
- The spin-only magnetic moment is a simplified model. In reality, orbital contributions can also be significant, especially for heavier elements.

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What is the packet of energy called?

View answer and explanation

Correct option: B

photon

The correct answer is B (photon) because a photon is defined as a discrete packet of energy of electromagnetic radiation.

Solution approach

Correct Answer + Core Reason: photon. A photon is a packet of energy.

Step-by-Step Solution:

Step 1: Define the terms.\nElectron: A subatomic particle with a negative charge.\nPhoton: A quantum of electromagnetic radiation, a particle of light.\nPositron: The antiparticle of the electron, with a positive charge.\nProton: A subatomic particle with a positive charge found in the nucleus of an atom.

Step 2: Identify the packet of energy. The term \

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The uncertainty principle and the concept of wave nature of matter was proposed by … and … respectively

View answer and explanation

Correct option: A

Heisenberg, de Broglie

Option A (Heisenberg, de Broglie) is correct because Heisenberg proposed the uncertainty principle, and de Broglie proposed the wave nature of matter.

Solution approach

Correct Answer + Core Reason: Heisenberg, de Broglie. Heisenberg proposed the uncertainty principle, and de Broglie proposed the wave nature of matter.

Step-by-Step Solution:

Step 1: Recall the scientists associated with each concept: Heisenberg is known for the uncertainty principle, and de Broglie is known for the wave nature of matter.

Step 2: Match the scientists to the concepts in the question.

Why other options are wrong:
- Option B: de Broglie did not propose the uncertainty principle, and Heisenberg did not propose the wave nature of matter.
- Option C: Planck is known for quantum theory and blackbody radiation, not the uncertainty principle. Heisenberg proposed the uncertainty principle.
- Option D: Planck is known for quantum theory and blackbody radiation, not the wave nature of matter. Heisenberg proposed the uncertainty principle.

Key Concepts:
- Heisenberg's uncertainty principle
- de Broglie's hypothesis of wave nature of matter
- Quantum mechanics

Memory Aids:
- Heisenberg: Uncertainty (H and U are close in the alphabet).
- de Broglie: Waves (think of "bro" surfing a wave).

Quick Tips:
- Remember the key scientists and their contributions to quantum mechanics.

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Analyzing +4 | -1

Highest energy will be absorbed to eject out the electron in the
configuration.

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents a half-filled p-orbital configuration, which is exceptionally stable and requires more energy to remove an electron.

Solution approach

Correct Answer + Core Reason: $1{s^2}2{s^2}2{p^3}$ . Half-filled p-orbitals are exceptionally stable.

Step-by-Step Solution:

Step 1: Analyze each electronic configuration:
- A: $1{s^2}2{s^2}2{p^1}$ : One electron in the 2p orbital.
- B: $1{s^2}2{s^2}2{p^3}$ : Half-filled 2p orbitals (stable).
- C: $1{s^2}2{s^2}2{p^2}$ : Two electrons in the 2p orbital.
- D: $1{s^2}2{s^2}2{p^4}$ : Four electrons in the 2p orbital.

Step 2: Recall the concept of stability for half-filled and fully-filled orbitals:
- Half-filled and fully-filled orbitals are more stable due to exchange energy and symmetrical distribution of electrons.

Step 3: Identify the most stable configuration:
- The configuration $1{s^2}2{s^2}2{p^3}$ has a half-filled 2p orbital, which is more stable than other configurations.

Step 4: Therefore, it requires the highest energy to remove an electron from the half-filled configuration.

Step 5: Final Answer: $1{s^2}2{s^2}2{p^3}$

Why other options are wrong:
- Option A: $1{s^2}2{s^2}2{p^1}$ : Less stable than a half-filled configuration.
- Option C: $1{s^2}2{s^2}2{p^2}$ : Less stable than a half-filled configuration.
- Option D: $1{s^2}2{s^2}2{p^4}$ : Less stable than a half-filled configuration.

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

Memory Aids:
- Remember that half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Look for half-filled or fully-filled configurations when determining stability.

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Understanding +4 | -1

In photoelectric emission the energy of the emitted electrons is:

View answer and explanation

Correct option: B

Smaller than that of incident photo

Option B (Smaller than that of incident photo) is correct because some of the energy of the incident photon is used to overcome the work function of the metal, leaving less energy for the kinetic energy of the emitted electron.

Solution approach

Correct Answer + Core Reason: Smaller than that of incident photo. Some energy is used to overcome the work function.

Step-by-Step Solution:

Step 1: Recall the photoelectric effect equation.
- $E = hf = \phi + KE_{max}$ , where E is the energy of the incident photon, hf is Planck's constant times frequency, $\phi$ is the work function, and $KE_{max}$ is the maximum kinetic energy of the emitted electron.

Step 2: Explain the relationship between the energy of the incident photon and the kinetic energy of the emitted electron.
- The energy of the incident photon is used to overcome the work function and provide kinetic energy to the electron.
- Therefore, the kinetic energy of the emitted electron is always less than the energy of the incident photon (unless the work function is zero, which is not realistic).

Why other options are wrong:
- Option A: Larger than that of incident photon is incorrect because the electron cannot have more energy than the incident photon.
- Option C: Same as that of incident photon is incorrect because some energy is used to overcome the work function.
- Option D: Proportional to intensity of incident light is incorrect because the energy of the emitted electrons depends on the frequency of the incident light, not the intensity.

Key Concepts:
- Photoelectric effect
- Work function
- Energy of a photon (E = hf)
- Kinetic energy of emitted electrons

Memory Aids:
- Remember the photoelectric effect equation: $E = hf = \phi + KE_{max}$

Quick Tips:
- The intensity of light affects the number of emitted electrons (photocurrent), not the energy of individual electrons.

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Application +4 | -1

The energy per mole of photon of electromagnetic radiation of wavelength 4000 Å is:

View answer and explanation

Correct option: A

3.0 \times {10^{ 12}}\mathrm{erg}

Option A (

3.0 \times {10^{ 12}}\mathrm{erg}

) is correct because it represents the energy per mole of photons calculated using Planck's equation and Avogadro's number.

Solution approach

Correct Answer + Core Reason: $3.0 \times {10^{ 12}}\mathrm{erg}$ . The energy per mole of photons is calculated using $E = Nhc/\lambda$ .

Step-by-Step Solution:

Step 1: Convert wavelength to cm: $\lambda = 4000 \,\text{\AA} = 4000 \times 10^{-8} \,\text{cm} = 4 \times 10^{-5} \,\text{cm}$

Step 2: Use the formula for energy per mole: $E = N_A \times h \times c / \lambda$ , where $N_A$ is Avogadro's number ( $6.022 \times 10^{23} \,\text{mol}^{-1}$ ), $h$ is Planck's constant ( $6.626 \times 10^{-27} \,\text{erg s}$ ), and $c$ is the speed of light ( $3 \times 10^{10} \,\text{cm/s}$ ).

Step 3: Substitute the values: $E = (6.022 \times 10^{23} \,\text{mol}^{-1}) \times (6.626 \times 10^{-27} \,\text{erg s}) \times (3 \times 10^{10} \,\text{cm/s}) / (4 \times 10^{-5} \,\text{cm})$

Step 4: Calculate the energy: $E \approx 3.0 \times 10^{12} \,\text{erg/mol}$

Step 5: Final Answer: $3.0 \times 10^{12} \,\text{erg/mol}$

Why other options are wrong:
- Option B: $4.0 \times {10^{ - 12}}\mathrm{erg}$ : This is incorrect due to errors in calculation or incorrect units.
- Option C: $5.0 \times {10^{ - 12}}\mathrm{erg}$ : This is incorrect due to errors in calculation or incorrect units.
- Option D: $6.0 \times {10^{ - 12}}\mathrm{erg}$ : This is incorrect due to errors in calculation or incorrect units.

Key Concepts:
- Planck's equation: $E = h\nu = hc/\lambda$
- Energy per mole: $E = N_A \times h\nu$
- Avogadro's number

Memory Aids:
- Remember the units for each constant (h, c, NA).

Quick Tips:
- Always convert the wavelength to consistent units (cm or meters) before calculation.
- Pay attention to the units of Planck's constant (erg s or J s).

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Remembering +4 | -1

Amongst the following statements, select the set having statements which was proposed by Dalton.
(1) All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
(2) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same

\mathrm{T} \,\&\, \mathrm{P}

(3) Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
(4) Matter consists of indivisible atoms

View answer and explanation

Correct option: B

(1), (3), (4)

Option B ((1), (3), (4)) is correct because Dalton's atomic theory proposed that all atoms of an element are identical, chemical reactions involve reorganization of atoms, and matter consists of indivisible atoms.

Solution approach

Correct Answer + Core Reason: (1), (3), (4). These statements align with Dalton's atomic theory.

Step-by-Step Solution:

Step 1: Recall the postulates of Dalton's atomic theory.

Step 2: Dalton's postulates include:
- All matter is composed of indivisible atoms.
- All atoms of a given element are identical in mass and properties.
- Compounds are formed by a combination of two or more different kinds of atoms.
- A chemical reaction is a rearrangement of atoms.

Step 3: Compare the given statements with Dalton's postulates.

Why other options are wrong:
- Option A ((1), (2), (3)): Statement (2) is related to Gay-Lussac's law of combining volumes, not Dalton's atomic theory.
- Option C ((1), (2), (4)): Statement (2) is related to Gay-Lussac's law of combining volumes, not Dalton's atomic theory.
- Option D ((1), (2), (3), (4)): Statement (2) is related to Gay-Lussac's law of combining volumes, not Dalton's atomic theory.

Key Concepts:
- Dalton's atomic theory
- Gay-Lussac's law of combining volumes

Memory Aids:
- Remember the key postulates of Dalton's atomic theory.

Quick Tips:
- Distinguish between different atomic theories and laws.

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Understanding +4 | -1

If

{S_1}

be the specific charge (e/m) of cathode ray and

{S_2}

be that of positive rays, then which is true?

View answer and explanation

Correct option: C

{S_1} > {S_2}

Option C (

{S_1} > {S_2}

) is correct because electrons (cathode rays) have a much larger charge-to-mass ratio than positive ions (positive rays).

Solution approach

Correct Answer + Core Reason: ${S_1} > {S_2}$ . The specific charge of cathode rays (electrons) is much greater than that of positive rays (ions).

Step-by-Step Solution:

Step 1: Define specific charge.
- Specific charge (S) is the ratio of charge (e) to mass (m): S = e/m.

Step 2: Define cathode rays and positive rays.
- Cathode rays are streams of electrons.
- Positive rays are streams of positive ions.

Step 3: Compare the specific charges of electrons and positive ions.
- Electrons have a very small mass compared to ions (which consist of nuclei and possibly some remaining electrons).
- Since the charge is the same (or a small multiple of the elementary charge), the specific charge of electrons is much larger than that of positive ions.
- Therefore, ${S_1} > {S_2}$ .

Why other options are wrong:
- Option A: ${S_1} = {S_2}$ is incorrect because the specific charges are not equal.
- Option B: ${S_1} < {S_2}$ is incorrect because the specific charge of cathode rays is greater than that of positive rays.
- Option D: Either of these is incorrect because the relationship is definite: ${S_1} > {S_2}$ .

Key Concepts:
- Specific charge (e/m)
- Cathode rays (electrons)
- Positive rays (ions)
- Mass of electron vs. mass of ions

Memory Aids:
- Remember that electrons are much lighter than ions.

Quick Tips:
- The specific charge is a characteristic property of a particle and depends on its charge and mass.

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Remembering +4 | -1

Which principle/rule limits the maximum number of electrons in an orbital to two?

View answer and explanation

Correct option: B

Pauli’s exclusion principle

Option B (Pauli’s exclusion principle) is correct because it states that no two electrons in an atom can have the same set of four quantum numbers, limiting each orbital to two electrons with opposite spins.

Solution approach

Correct Answer + Core Reason: Pauli’s exclusion principle. This principle dictates that no two electrons can have the same set of quantum numbers.

Step-by-Step Solution:

Step 1: State Pauli's Exclusion Principle.

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

Step 3: This implies that an orbital can hold a maximum of two electrons, each with opposite spins.

Why other options are wrong:
- Option A: Aufbau principle. This principle describes the filling of electrons in atomic orbitals in increasing order of energy.
- Option C: Hund’s rule of maximum multiplicity. This rule states that electrons fill orbitals singly before pairing up.
- Option D: Heisenberg’s uncertainty principle. This principle states that it is impossible to know both the position and momentum of an electron with perfect accuracy.

Key Concepts:
- Pauli's Exclusion Principle
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- Pauli: "No two electrons alike!"

Quick Tips:
- Remember that Pauli's Exclusion Principle directly limits the number of electrons in an orbital.

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Application +4 | -1

If helium atom and hydrogen molecules are moving with the same velocity, their wavelength ratio will be

View answer and explanation

Correct option: B

1:2

Option B (

1:2

) is correct because the de Broglie wavelength is inversely proportional to the mass, and the mass of a helium atom is approximately twice that of a hydrogen molecule.

Solution approach

Correct Answer + Core Reason: $1:2$ . The de Broglie wavelength is inversely proportional to mass when velocity is constant.

Step-by-Step Solution:

Step 1: The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$ , where h is Planck's constant, m is the mass, and v is the velocity.

Step 2: Since the velocity is the same for both helium and hydrogen molecules, $\lambda \propto \frac{1}{m}$ .

Step 3: The mass of a helium atom (He) is approximately 4 atomic mass units (amu). The mass of a hydrogen molecule ( $H_2$ ) is approximately 2 amu.

Step 4: Therefore, $\frac{\lambda_{He}}{\lambda_{H_2}} = \frac{m_{H_2}}{m_{He}} = \frac{2}{4} = \frac{1}{2}$

Why other options are wrong:
- **Option A ( $4:1$ ): This is the inverse of the correct ratio. It would be the ratio of masses, not wavelengths.
- Option C ( $2:1$ ): This is the inverse of the correct ratio. It would be the ratio of masses, not wavelengths.
- Option D ( $1:4$ ): This is incorrect. The mass ratio is 2:4 = 1:2, so the wavelength ratio is 1:2.

Key Concepts:
- de Broglie wavelength
- Mass of helium atom and hydrogen molecule

Memory Aids:**
- de Broglie wavelength: $\lambda = \frac{h}{mv}$

Quick Tips:
- Remember that de Broglie wavelength is inversely proportional to mass when velocity is constant.
- Know the approximate masses of common atoms and molecules.

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Understanding +4 | -1

If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is

View answer and explanation

Correct option: C

\frac{{ - {e^2}}}{{2r}}

Option C (

\frac{{ - {e^2}}}{{2r}}

) is correct because the total energy of an electron in the first orbit of a hydrogen atom is given by this formula, where e is the electron charge and r is the radius of the orbit.

Solution approach

Correct Answer + Core Reason: $\frac{{ - {e^2}}}{{2r}}$ . The total energy is the sum of kinetic and potential energy.

Step-by-Step Solution:

Step 1: Define the potential energy (U) of an electron in a hydrogen atom. The potential energy is given by $U = \frac{{ - {e^2}}}{r}$ , where e is the elementary charge and r is the radius of the orbit.

Step 2: Define the kinetic energy (K) of an electron in a hydrogen atom. The kinetic energy is given by $K = \frac{{{e^2}}}{{2r}}$ .

Step 3: Calculate the total energy (E). The total energy is the sum of the kinetic and potential energies: $E = K + U = \frac{{{e^2}}}{{2r}} + \frac{{ - {e^2}}}{r} = \frac{{{e^2} - 2{e^2}}}{{2r}} = \frac{{ - {e^2}}}{{2r}}$ .

Step 4: Final Answer: $\frac{{ - {e^2}}}{{2r}}$

Why other options are wrong:
- Option A: This represents only the potential energy, not the total energy.
- Option B: This has the wrong dependence on r (r squared instead of r).
- Option D: This has the wrong dependence on r (r squared instead of r) and is missing a factor.

Key Concepts:
- Potential energy of an electron in a hydrogen atom
- Kinetic energy of an electron in a hydrogen atom
- Total energy of an electron in a hydrogen atom

Memory Aids:
- Remember that the total energy is negative and half the magnitude of the potential energy.

Quick Tips:
- The total energy of an electron in a hydrogen atom is always negative, indicating that the electron is bound to the nucleus.

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Application +4 | -1

If the series limit of wavelength of the Lyman series for the hydrogen atom is 912 Å, then the series limit of wavelength for the Balmer series of the hydrogen atom is:

View answer and explanation

Correct option: C

912×4 Å

Option C (912×4 Å) is correct because the series limit wavelength is inversely proportional to the square of the principal quantum number.

Solution approach

Correct Answer + Core Reason: 912×4 Å. The series limit wavelength for the Balmer series is 4 times the series limit wavelength for the Lyman series.

Step-by-Step Solution:

Step 1: Identify the given information: Series limit of wavelength for Lyman series, $\lambda_L = 912 \text{{ Å}}$ .\nWe need to find the series limit of wavelength for the Balmer series, $\lambda_B$ .

Step 2: Recall the Rydberg formula for the wavelength of emitted photons in the hydrogen atom:
$\frac{1}{\lambda} = R \left( \frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}} \right)$ , where $R$ is the Rydberg constant.\nFor the series limit, $n_2 = \infty$ .

Step 3: For the Lyman series, $n_1 = 1$ , so $\frac{1}{{\lambda_L}} = R \left( \frac{1}{{{1^2}}} - \frac{1}{{{\infty}^2}} \right) = R$ .\nThus, $\lambda_L = \frac{1}{R} = 912 \text{{ Å}}$ .

Step 4: For the Balmer series, $n_1 = 2$ , so $\frac{1}{{\lambda_B}} = R \left( \frac{1}{{{2^2}}} - \frac{1}{{{\infty}^2}} \right) = \frac{R}{4}$ .\nThus, $\lambda_B = \frac{4}{R}$ .

Step 5: Since $\lambda_L = \frac{1}{R}$ , we have $\lambda_B = 4 \lambda_L = 4 \times 912 \text{{ Å}}$ .

Step 6: Final Answer: $\lambda_B = 912 \times 4 \text{{ Å}}$

Why other options are wrong:
- Option A: This is the wavelength of the Lyman series limit, not the Balmer series.
- Option B: This is incorrect. It might be obtained by multiplying the Lyman series limit by 2, which is not the correct factor.
- Option D: This is incorrect. It might be obtained by dividing the Lyman series limit by 2, which is not the correct factor.

Key Concepts:
- Rydberg formula
- Lyman and Balmer series
- Series limit

Memory Aids:
- Remember the Rydberg formula and the values of $n_1$ for Lyman and Balmer series.

Quick Tips:
- Understand the relationship between the series limit and the principal quantum number.

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Remembering +4 | -1

The incorrect statement according to Dalton's Atomic theory is

View answer and explanation

Correct option: A

Atoms can be created in a chemical reaction

Option A (Atoms can be created in a chemical reaction) is the incorrect statement because Dalton's atomic theory states that atoms are indivisible and cannot be created or destroyed in a chemical reaction.

Solution approach

Correct Answer + Core Reason: Atoms can be created in a chemical reaction. Dalton's theory states atoms are neither created nor destroyed in a chemical reaction.

Step-by-Step Solution:

Step 1: Recall Dalton's Atomic Theory postulates.

Step 2: Identify the postulate that addresses the conservation of atoms in chemical reactions. Dalton stated that atoms are rearranged, combined, or separated in chemical reactions, but are not created or destroyed.

Step 3: Compare each option with Dalton's postulates.

Why other options are wrong:
- Option B: Atoms of a given element are identical in all respects. This is a correct statement according to Dalton's theory.
- Option C: Atoms of different elements differ in mass. This is a correct statement according to Dalton's theory.
- Option D: Compounds are formed when atoms of different elements combine in fixed ratio. This is a correct statement according to Dalton's theory.

Key Concepts:
- Dalton's Atomic Theory
- Postulates of Dalton's Theory

Memory Aids:
- Remember the key postulates of Dalton's Atomic Theory: Atoms are indivisible, atoms of the same element are identical, atoms of different elements are different, and compounds are formed by combining atoms in fixed ratios.

Quick Tips:
- Review the fundamental postulates of atomic theory.

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Remembering +4 | -1

For hydrogen like atom with a nuclear charge

Z

, the energy of orbital with principal quantum number '

n

' follows the relation

View answer and explanation

Correct option: D

E_n \propto-\frac{Z^2}{n^2}

Option D (

E_n \propto-\frac{Z^2}{n^2}

) is correct because the energy of an electron in a hydrogen-like atom is directly proportional to

-Z^2

and inversely proportional to

n^2

.

Solution approach

Correct Answer + Core Reason: $E_n \propto-\frac{Z^2}{n^2}$ . The energy of an electron in a hydrogen-like atom is proportional to $-Z^2/n^2$ .

Step-by-Step Solution:

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

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

Step 3: From the formula, it is clear that $E_n \propto -\frac{Z^2}{n^2}$ .

Why other options are wrong:
- Option A: $\mathrm{E}_{\mathrm{n}} \propto {n}^2 \mathrm{Z}^2$ is incorrect because the energy is inversely proportional to $n^2$ and proportional to $Z^2$ with a negative sign.
- Option B: $E_n \propto \frac{Z^2}{n}$ is incorrect because the energy is inversely proportional to $n^2$ , not $n$ .
- Option C: $E_n \propto-\frac{Z}{n}$ is incorrect because the energy is proportional to $Z^2$ , not $Z$ , and inversely proportional to $n^2$ , not $n$ .

Key Concepts:
- Energy levels of hydrogen-like atoms
- Principal quantum number
- Atomic number

Memory Aids:
- Remember the formula: $E_n = -13.6 \frac{Z^2}{n^2}$ eV

Quick Tips:
- Always remember the negative sign in the energy formula, indicating that the electron is bound to the nucleus.

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The orientation of an atomic orbital is governed by:

View answer and explanation

Correct option: A

Magnetic quantum number

Option A (Magnetic quantum number) is correct because the magnetic quantum number determines the spatial orientation of an orbital in space relative to a standard set of coordinate axes.

Solution approach

Correct Answer + Core Reason: Magnetic quantum number. The magnetic quantum number determines the orientation of an atomic orbital in space.

Step-by-Step Solution:

Step 1: Define the role of the magnetic quantum number.

Step 2: The magnetic quantum number (ml) describes the orientation of an orbital in space. For a given value of l, ml can have values from -l to +l, including 0.

Step 3: Each value of ml corresponds to a specific orientation of the orbital.

Why other options are wrong:
- Option B: Principal quantum number. This number determines the energy level and size of the orbital.
- Option C: Azimuthal quantum number. This number determines the shape of the orbital.
- Option D: Spin quantum number. This number describes the intrinsic angular momentum of an electron.

Key Concepts:
- Magnetic Quantum Number
- Atomic Orbitals
- Quantum Numbers

Memory Aids:
- Magnetic = Orientation

Quick Tips:
- The magnetic quantum number is related to the spatial orientation of the orbital.

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Remembering +4 | -1

In Stratosphere most of the ozone formation is assisted by

View answer and explanation

Correct option: C

Ultraviolet radiations

Option C (Ultraviolet radiations) is correct because ultraviolet radiation provides the energy required to break down oxygen molecules, leading to ozone formation in the stratosphere.

Solution approach

Correct Answer + Core Reason: Ultraviolet radiations. UV radiation provides the energy for ozone formation.

Step-by-Step Solution:

Step 1: Recall the process of ozone formation in the stratosphere.

Step 2: Ozone is formed when oxygen molecules ( $O_2$ ) are broken down by high-energy radiation into individual oxygen atoms (O).

Step 3: These oxygen atoms then combine with other oxygen molecules to form ozone ( $O_3$ ).

Step 4: Ultraviolet (UV) radiation has the appropriate energy to break down oxygen molecules.

Why other options are wrong:
- Option A: Cosmic rays are high-energy particles, but they are not the primary source of energy for ozone formation.
- Option B: $\gamma - rays$ are higher energy than UV but not the primary driver of ozone formation in the stratosphere.
- Option D: Visible radiations do not have enough energy to break down oxygen molecules efficiently.

Key Concepts:
- Ozone formation
- Ultraviolet radiation
- Stratosphere

Memory Aids:
- UV radiation breaks down $O_2$ to form $O_3$

Quick Tips:
- Remember that UV radiation is essential for ozone formation.

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Application +4 | -1

Which of the following ion will show colour in aqueous solution?

View answer and explanation

Correct option: B

T{i^{3 + }}(Z = 22)

Option B (

Ti^{3+}

) is correct because it has unpaired d-electrons, allowing for d-d transitions and thus color in aqueous solution.

Solution approach

Correct Answer + Core Reason: $T{i^{3 + }}(Z = 22)$ . The presence of unpaired d-electrons allows for d-d transitions, resulting in color.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion.

Step 2: Identify ions with unpaired d-electrons.
- $La^{3+}$ (Z=57): $[Xe]$
- $Ti^{3+}$ (Z=22): $[Ar] 3d^1$
- $Lu^{3+}$ (Z=71): $[Xe] 4f^{14}$
- $Sc^{3+}$ (Z=21): $[Ar]$

Step 3: Ions with unpaired d-electrons exhibit d-d transitions, which absorb light in the visible region, leading to color.
$Ti^{3+}$ has one unpaired d-electron and thus shows color.

Why other options are wrong:
- Option A: $La^{3+}$ has a noble gas configuration and no d-electrons, so no d-d transitions are possible, and it is colorless.
- Option C: $Lu^{3+}$ has a completely filled 4f subshell and no d-electrons in the penultimate shell, so no d-d transitions are possible, and it is colorless.
- Option D: $Sc^{3+}$ has a noble gas configuration and no d-electrons, so no d-d transitions are possible, and it is colorless.

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

Memory Aids:
- Color arises from d-d transitions when unpaired d-electrons are present.

Quick Tips:
- Determine the number of d-electrons for each ion. If there are unpaired d-electrons, the ion is likely to be colored.

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Application +4 | -1

Among the following groupings which represents the collection of iso electronic species

View answer and explanation

Correct option: C

CO,N{O^ + },\,C{N^ - },\,C_2^{2 - }

Option C (

CO,N{O^ + },\,C{N^ - },\,C_2^{2 - }

) is correct because all species have 14 electrons.

Solution approach

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

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species:
- CO: 6 (C) + 8 (O) = 14 electrons
- NO+: 7 (N) + 8 (O) - 1 (positive charge) = 14 electrons
- CN-: 6 (C) + 7 (N) + 1 (negative charge) = 14 electrons
- C22-: 2 * 6 (C) + 2 (negative charge) = 14 electrons

Step 2: Check the other options:
- Option A: NO+ (14), C22- (14), O2- (17), CO (14) - Not all are isoelectronic
- Option B: N2 (14), C22- (14), CO (14), NO (15) - Not all are isoelectronic
- Option D: NO (15), CN- (14), N2 (14), O2- (17) - Not all are isoelectronic

Step 3: Conclusion: Only option C contains species that are all isoelectronic.

Why other options are wrong:
- **Option A ( $N{O^ + },C_2^{2 - },\,O_2^ - ,\,CO$ ): O2- has 17 electrons, so it's not isoelectronic with the others.
- Option B ( ${N_2}C_2^{2 - },CO,NO$ ): NO has 15 electrons, so it's not isoelectronic with the others.
- Option D ( $NO,\,C{N^ - },{N_2},O_2^ -$ ): NO has 15 electrons and O2- has 17 electrons, so they are not isoelectronic with the others.

Key Concepts:
- Isoelectronic species: Species with the same number of electrons.

Memory Aids:
- Remember to account for the charge when calculating the number of electrons in ions.

Quick Tips:**
- Count the number of electrons for each atom and then adjust for the charge of the ion.

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Application +4 | -1

Suppose a completely filled or half filled set of p or d-orbitals is spherically symmetrical. Point out the species, which is spherical symmetrical?

View answer and explanation

Correct option: C

C{l^ - }

Option C (

Cl^-

) is correct because it has a completely filled p-orbital configuration, resulting in spherical symmetry.

Solution approach

Correct Answer + Core Reason: $C{l^ - }$ . A completely filled or half-filled set of p or d-orbitals is spherically symmetrical.

Step-by-Step Solution:

Step 1: Write the electronic configurations of each species:\nO: $1s^2 2s^2 2p^4$ \nC: $1s^2 2s^2 2p^2$
$Cl^-$ : $1s^2 2s^2 2p^6 3s^2 3p^6$ \nFe: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$

Step 2: Identify the species with completely filled or half-filled p or d-orbitals.
$Cl^-$ has a completely filled $3p$ orbital ( $3p^6$ ).

Step 3: A completely filled or half-filled set of p or d-orbitals is spherically symmetrical.

Why other options are wrong:
- Option A: O has a $2p^4$ configuration, which is neither half-filled nor completely filled.
- Option B: C has a $2p^2$ configuration, which is neither half-filled nor completely filled.
- Option D: Fe has a $3d^6$ configuration, which is neither half-filled ( $d^5$ ) nor completely filled ( $d^{10}$ ).

Key Concepts:
- Electronic configuration
- Hund's rule
- Spherical symmetry of filled and half-filled orbitals

Memory Aids:
- Remember that completely filled and half-filled orbitals are more stable and spherically symmetrical.

Quick Tips:
- Write out the electronic configurations to determine the occupancy of p and d orbitals.

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