STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 5

121 Easy

121 of 239

Remembering +4 | -1

According to Heisenberg uncertainty principle:

View answer and explanation

Correct option: B

\Delta x \times \Delta p \geq \frac{h}{4 \pi}

Option B (

\Delta x \times \Delta p \geq \frac{h}{4 \pi}

) is correct because the Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum is greater than or equal to

\frac{h}{4 \pi}

.

Solution approach

Correct Answer + Core Reason: $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$ . This is the mathematical expression of the Heisenberg uncertainty principle.

Step-by-Step Solution:
Step 1: Recall the Heisenberg uncertainty principle. The Heisenberg uncertainty principle states that it is impossible to determine simultaneously both the exact position and the exact momentum of a particle.
Step 2: Remember the mathematical expression of the principle: $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$ , where:
- $\Delta x$ is the uncertainty in position
- $\Delta p$ is the uncertainty in momentum
- $h$ is Planck's constant

Why other options are wrong:
- Option A: $E=m c^2$ . This is Einstein's mass-energy equivalence equation.
- Option C: $\lambda=\frac{h}{p}$ . This is de Broglie's wavelength equation.
- Option D: $\Delta x \times \Delta p=\frac{h}{6 \pi}$ . The correct constant is $\frac{h}{4 \pi}$ , not $\frac{h}{6 \pi}$ .

Key Concepts:
- Heisenberg uncertainty principle
- Uncertainty in position and momentum
- Planck's constant

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

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

122 Easy

122 of 239

Remembering +4 | -1

For a p-electron orbital angular momentum is

View answer and explanation

Correct option: A

\sqrt 2 \hbar

Option A (

\sqrt 2 \hbar

) is correct because the orbital angular momentum for a p-electron (l=1) is calculated using the formula

\sqrt{l(l+1)}\hbar

, which results in

\sqrt{2}\hbar

.

Solution approach

Correct Answer + Core Reason: $\sqrt 2 \hbar$ . The orbital angular momentum is calculated using the formula $\sqrt{l(l+1)}\hbar$ with l=1 for p-orbital.

Step-by-Step Solution:

Step 1: Recall the formula for orbital angular momentum: $L = \sqrt{l(l+1)}\hbar$ , where l is the azimuthal quantum number.

Step 2: For a p-electron, l = 1.

Step 3: Substitute l = 1 into the formula: $L = \sqrt{1(1+1)}\hbar = \sqrt{2}\hbar$ .

Why other options are wrong:
- Option B: $\hbar$ is the angular momentum for l=0 (s-orbital).
- Option C: $\sqrt 6 \hbar$ is the angular momentum for l=2 (d-orbital).
- Option D: $2\hbar$ is not the correct angular momentum for any single orbital type.

Key Concepts:
- Orbital angular momentum
- Azimuthal quantum number (l)
- p-orbital

Memory Aids:
- Remember the formula: $L = \sqrt{l(l+1)}\hbar$

Quick Tips:
- Know the l values for s, p, d, and f orbitals (0, 1, 2, 3, respectively).

123 Easy

123 of 239

Understanding +4 | -1

The KE of electron in

H{e^ + }

will be maximum in:

View answer and explanation

Correct option: C

1st orbit

Option C (1st orbit) is correct because kinetic energy is inversely proportional to the radius of the orbit, and the radius is smallest in the first orbit.

Solution approach

Correct Answer + Core Reason: 1st orbit. Kinetic energy is maximum when the radius is minimum, which occurs in the first orbit.

Step-by-Step Solution:

Step 1: Recall the formula for kinetic energy of an electron in a hydrogen-like species: $KE = \frac{{Ze^2}}{{2r_n}}$ where $r_n$ is the radius of the nth orbit.

Step 2: Recall the formula for the radius of the nth orbit: $r_n = \frac{{n^2h^2}}{{4\pi^2mZe^2}}$ .

Step 3: Observe that the kinetic energy is inversely proportional to the radius of the orbit. Therefore, the smaller the radius, the greater the kinetic energy.

Step 4: The radius is minimum for the first orbit (n=1).

Step 5: Conclude that the kinetic energy is maximum in the first orbit.

Why other options are wrong:
- Option A: 3rd orbit has a larger radius than the 1st orbit, so the kinetic energy is lower.
- Option B: 2nd orbit has a larger radius than the 1st orbit, so the kinetic energy is lower.
- Option D: In orbit with $n = \infty$ , the radius is infinite, so the kinetic energy is zero.

Key Concepts:
- Bohr's model of the atom
- Kinetic energy of an electron in an orbit
- Radius of an electron orbit

Memory Aids:
- Remember that kinetic energy is inversely proportional to the radius of the orbit.

Quick Tips:
- Use the formulas for kinetic energy and radius to compare the energy in different orbits.

124 Easy

124 of 239

Understanding +4 | -1

Which of the following configurations is correct for iron ?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the correct electronic configuration of iron (Fe), which has 26 electrons.

Solution approach

Correct Answer + Core Reason: $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2, 3 d^6$ . This is the correct electronic configuration for iron (Fe).

Step-by-Step Solution:

Step 1: Determine the atomic number of iron (Fe): The atomic number of iron is 26, meaning it has 26 electrons.

Step 2: Apply the Aufbau principle: Fill the orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d.

Step 3: Fill the orbitals with electrons: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6$ . The sum of the superscripts (number of electrons) should equal 26.

Why other options are wrong:
- Option A: $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5$ has only 23 electrons, which is not the correct number for iron.
- Option B: $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2, 3 d^5$ has only 25 electrons, which is not the correct number for iron.
- Option C: $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2, 3 d^7$ has 27 electrons, which is not the correct number for iron.

Key Concepts:
- Electronic configuration
- Aufbau principle
- Atomic number

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

Quick Tips:
- Always check that the total number of electrons in the configuration matches the atomic number.

125 Easy

125 of 239

Application +4 | -1

The orbital angular momentum of an electron in the 3d-orbital is

View answer and explanation

Correct option: B

\sqrt 6 \frac{h}{{2\pi }}

Option B (

\sqrt 6 \frac{h}{{2\pi }}

) is correct because it is calculated using the formula for orbital angular momentum, which depends on the azimuthal quantum number (l).

Solution approach

Correct Answer + Core Reason: $\sqrt 6 \frac{h}{{2\pi }}$ . The orbital angular momentum is calculated using the formula involving the azimuthal quantum number.

Step-by-Step Solution:

Step 1: Recall the formula for orbital angular momentum: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$ , where l is the azimuthal quantum number.

Step 2: Identify the value of l for a 3d-orbital: For a d-orbital, l = 2.

Step 3: Substitute the value of l into the formula: $L = \sqrt{2(2+1)} \frac{h}{2\pi} = \sqrt{2(3)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$

Step 4: Simplify: The orbital angular momentum is $\sqrt{6} \frac{h}{2\pi}$

Why other options are wrong:
- **Option A ( $\sqrt 5 \frac{h}{{2\pi }}$ ): This is incorrect because it uses an incorrect value for l or miscalculates the expression.
- Option C ( $\frac{{\sqrt 6 }}{{2\pi }}$ ): This is incorrect because it is missing the h term in the formula.
- Option D ( $\frac{{3h}}{{2\pi }}$ ): This is incorrect because it uses an incorrect value for l and miscalculates the expression.

Key Concepts:
- Orbital Angular Momentum
- Azimuthal Quantum Number (l)
- Atomic Orbitals

Memory Aids:**
- $L = \sqrt{l(l+1)} \frac{h}{2\pi}$

Quick Tips:
- Remember the values of l for different orbitals (s=0, p=1, d=2, f=3).

126 Easy

126 of 239

Analyzing +4 | -1

Which of the following element will have the same number of electrons in

s

and as well as

p

-type of orbitals?

View answer and explanation

Correct option: B

\operatorname{Mg}(Z=12)

Option B is correct because Magnesium (Mg) has the electronic configuration

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

, which has an equal number of electrons in s-orbitals (6) and p-orbitals (6).

Solution approach

Correct Answer + Core Reason: $\operatorname{Mg}(Z=12)$ . Magnesium has 6 electrons in s-orbitals and 6 electrons in p-orbitals.

Step-by-Step Solution:
Step 1: Write the electronic configuration for each element.
Step 2: Count the number of electrons in s and p orbitals for each element.
Step 3: For Mg (Z=12): $1s^2 2s^2 2p^6 3s^2$ . s-electrons = 2 + 2 + 2 = 6, p-electrons = 6.
Step 4: For Fe (Z=26): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$ . s-electrons = 2 + 2 + 2 + 2 = 8, p-electrons = 6 + 6 = 12.
Step 5: For Ne (Z=10): $1s^2 2s^2 2p^6$ . s-electrons = 2 + 2 = 4, p-electrons = 6.
Step 6: For Ar (Z=18): $1s^2 2s^2 2p^6 3s^2 3p^6$ . s-electrons = 2 + 2 + 2 = 6, p-electrons = 6 + 6 = 12.
Step 7: Identify the element with the same number of s and p electrons: Mg.

Why other options are wrong:
- Option A: Fe has 8 s-electrons and 12 p-electrons.
- Option C: Ne has 4 s-electrons and 6 p-electrons.
- Option D: Ar has 6 s-electrons and 12 p-electrons.

Key Concepts:
- Electronic configuration
- Orbitals (s, p, d, f)

Memory Aids:
- Aufbau principle

Quick Tips:
- Write the complete electronic configuration to avoid errors.

127 Easy

127 of 239

Understanding +4 | -1

The ratio of charge and mass would be greter for

View answer and explanation

Correct option: B

Electron

Option B (Electron) is correct because the charge-to-mass ratio is significantly higher for an electron due to its very small mass.

Solution approach

Correct Answer + Core Reason: Electron. The charge-to-mass ratio is greatest for the electron due to its small mass.

Step-by-Step Solution:

Step 1: Recall the charge and mass of each particle:
- Proton: Charge = +1, Mass ≈ 1 amu
- Electron: Charge = -1, Mass ≈ 1/1837 amu
- Neutron: Charge = 0, Mass ≈ 1 amu
- Alpha particle: Charge = +2, Mass ≈ 4 amu

Step 2: Calculate the charge-to-mass ratio (e/m) for each particle:
- Proton: e/m ≈ 1/1
- Electron: e/m ≈ 1/(1/1837) = 1837
- Neutron: e/m = 0/1 = 0
- Alpha particle: e/m ≈ 2/4 = 0.5

Step 3: Compare the charge-to-mass ratios.

Step 4: The electron has the highest charge-to-mass ratio.

Why other options are wrong:
- Option A: Proton has a lower charge-to-mass ratio compared to the electron.
- Option C: Neutron has a charge of 0, so its charge-to-mass ratio is 0.
- Option D: Alpha particle has a lower charge-to-mass ratio compared to the electron.

Key Concepts:
- Charge-to-Mass Ratio
- Subatomic Particles

Memory Aids:
- Electrons are very light, so they have a high charge-to-mass ratio.

Quick Tips:
- Remember the relative masses and charges of subatomic particles.

128 Easy

128 of 239

Understanding +4 | -1

Which of the following set of quantum number is possible?

View answer and explanation

Correct option: C

n=4, I=3, m=-3, s=\frac{1}{2}

The correct answer is C (

n=4, I=3, m=-3, s=\frac{1}{2}

) because it satisfies all the rules for quantum numbers: n > 0, 0 ≤ l < n, -l ≤ m ≤ +l, and s = ±1/2.

Solution approach

Correct Answer + Core Reason: $n=4, I=3, m=-3, s=\frac{1}{2}$ . This set of quantum numbers is possible because it adheres to the rules governing their values.

Step-by-Step Solution:

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

Step 2: Evaluate each option based on these rules:
- Option A: $n=4, l=2, m=-2, s=-2$ . Incorrect because s can only be +1/2 or -1/2.
- Option B: $n=4, I=4, m=0, s=\frac{1}{2}$ . Incorrect because l must be less than n (l < n).
- Option C: $n=4, I=3, m=-3, s=\frac{1}{2}$ . Correct because it satisfies all the rules: n=4, l<n, -l<=m<=l, and s is either +1/2 or -1/2.
- Option D: $n=4, I=0, m=0, s=0$ . Incorrect because s can only be +1/2 or -1/2.

Why other options are wrong:
- **Option A ( $n=4, l=2, m=-2, s=-2$ ): The spin quantum number (s) can only be +1/2 or -1/2, not -2.
- Option B ( $n=4, I=4, m=0, s=\frac{1}{2}$ ): The azimuthal quantum number (l) must be less than the principal quantum number (n). Here, l = n, which is not allowed.
- Option D ( $n=4, I=0, m=0, s=0$ ): The spin quantum number (s) can only be +1/2 or -1/2, not 0.

Key Concepts:
- Quantum numbers (n, l, m, s)
- Rules for quantum numbers

Memory Aids:
- Remember the allowed ranges for each quantum number: n > 0, 0 ≤ l < n, -l ≤ m ≤ +l, and s = ±1/2.

Quick Tips:**
- Always check if the given set of quantum numbers adheres to the rules before selecting the answer.

129 Easy

129 of 239

Understanding +4 | -1

Bohr’s model can explain

View answer and explanation

Correct option: B

Spectrum of an atom or ion containing one electron only

Option B (Spectrum of an atom or ion containing one electron only) is correct because Bohr's model is based on the hydrogen atom, which has only one electron, and can be extended to other one-electron species.

Solution approach

Correct Answer + Core Reason: Spectrum of an atom or ion containing one electron only. Bohr's model is based on the hydrogen atom.

Step-by-Step Solution:

Step 1: Recall the postulates of Bohr's model.

Step 2: Understand that Bohr's model is based on the hydrogen atom, which has only one electron.

Step 3: Recognize that Bohr's model can be extended to other one-electron species, such as He+ and Li2+.

Why other options are wrong:
- Option A: Bohr's model can be applied to other one-electron species, not just hydrogen.
- Option C: Bohr's model cannot explain the spectrum of hydrogen molecules, which have multiple electrons.
- Option D: The solar spectrum is complex and involves many elements, so Bohr's model cannot explain it.

Key Concepts:
- Bohr's model
- Atomic spectra
- One-electron species

Memory Aids:
- Bohr's model is for one-electron systems.

Quick Tips:
- Know the limitations of Bohr's model.

130 Easy

130 of 239

Remembering +4 | -1

Rutherford's experiment which established the nuclear model of the atom used a beam of

View answer and explanation

Correct option: D

Helium nuclei, which impinged on a metal foil and got scattered

Option D (Helium nuclei, which impinged on a metal foil and got scattered) is correct because Rutherford's experiment used alpha particles (helium nuclei) to bombard a gold foil.

Solution approach

Correct Answer + Core Reason: Helium nuclei, which impinged on a metal foil and got scattered. Rutherford's experiment used alpha particles (helium nuclei) to bombard a gold foil.

Step-by-Step Solution:

Step 1: Rutherford's Experiment: Describe the setup of Rutherford's gold foil experiment.

Step 2: Alpha Particles: Alpha particles (helium nuclei, $^4_2He^{2+}$ ) were used as the projectile.

Step 3: Gold Foil: A thin gold foil was used as the target.

Step 4: Scattering: The scattering pattern of the alpha particles was observed to deduce the structure of the atom.

Why other options are wrong:
- **Option A ( $\beta$ -particles which impinged on a metal foil and got absorbed): Rutherford used alpha particles, not beta particles.
- Option B ( $\gamma$ -rays which impinged on a metal foil and ejected electrons): This describes the photoelectric effect, not Rutherford's experiment.
- Option C (Helium atoms, which impinged on a metal foil and got scattered):** Rutherford used helium *nuclei* (alpha particles), not neutral helium atoms.

Key Concepts:
- Rutherford's Gold Foil Experiment
- Alpha Particles
- Nuclear Model of the Atom

Memory Aids:
- Remember that Rutherford \"scattered alpha particles\" to discover the nucleus.

Quick Tips:
- Rutherford's experiment led to the discovery of the nucleus and the nuclear model of the atom.

131 Easy

131 of 239

Remembering +4 | -1

Electron behaves both as a particle and a wave. This was proposed by

View answer and explanation

Correct option: C

de-Broglie

Option C (de-Broglie) is correct because de-Broglie proposed the wave-particle duality of electrons.

Solution approach

Correct Answer + Core Reason: de-Broglie. de-Broglie proposed the wave-particle duality of electrons.

Step-by-Step Solution:

Step 1: Recall the historical experiments and theories related to the nature of electrons.

Step 2: de-Broglie proposed that matter (including electrons) exhibits wave-particle duality. The de-Broglie wavelength is given by $\lambda = \frac{h}{{mv}}$ , where h is Planck's constant, m is mass, and v is velocity.

Step 3: Conclude that de-Broglie is the correct answer.

Why other options are wrong:
- Option A (Heisenberg): Heisenberg is known for the Uncertainty Principle, not the wave-particle duality.
- Option B (Gilbert N. Lewis): Gilbert N. Lewis is known for his work on chemical bonding and Lewis structures, not the wave-particle duality.
- Option D (L. Rutherford): Rutherford is known for his gold foil experiment and the discovery of the nucleus, not the wave-particle duality.

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

Memory Aids:
- de-Broglie = Dual nature

Quick Tips:
- Remember the key contributions of famous scientists in physics and chemistry.

132 Easy

132 of 239

Remembering +4 | -1

The planck’s constant has a unit of :

View answer and explanation

Correct option: C

Angular momentum

Option C (Angular momentum) is correct because Planck's constant has units of energy multiplied by time, which is equivalent to the units of angular momentum.

Solution approach

Correct Answer + Core Reason: Angular momentum. Planck's constant has units of angular momentum.

Step-by-Step Solution:

Step 1: Recall the formula relating energy and frequency: $E = h\nu$ , where $E$ is energy, $h$ is Planck's constant, and $\nu$ is frequency.

Step 2: The units of energy are Joules (J), and the units of frequency are Hertz (Hz) or $s^{-1}$ .

Step 3: Therefore, the units of Planck's constant are $J \cdot s$ .

Step 4: Angular momentum is given by $L = I\omega$ , where $I$ is the moment of inertia and $\omega$ is the angular velocity.

Step 5: The units of moment of inertia are $kg \cdot m^2$ , and the units of angular velocity are $rad/s$ or $s^{-1}$ .

Step 6: Therefore, the units of angular momentum are $kg \cdot m^2 \cdot s^{-1}$ .

Step 7: Since $J = kg \cdot m^2 \cdot s^{-2}$ , then $J \cdot s = kg \cdot m^2 \cdot s^{-1}$ , which are the same units as angular momentum.

Why other options are wrong:
- Option A: Work has units of energy (Joules), not $J \cdot s$ .
- Option B: Energy has units of Joules, not $J \cdot s$ .
- Option D: Linear momentum has units of $kg \cdot m/s$ , not $J \cdot s$ .

Key Concepts:
- Planck's constant
- Energy
- Frequency
- Angular momentum

Memory Aids:
- Planck's constant: \"Energy times time.\"

Quick Tips:
- Remember the units of Planck's constant and how they relate to energy and angular momentum.

133 Easy

133 of 239

Remembering +4 | -1

The shape of orbital with

l = 1

and

m = 0

is

View answer and explanation

Correct option: B

Dumbell

Option B (Dumbell) is correct because an orbital with

l = 1

corresponds to a p-orbital, which has a dumbbell shape.

Solution approach

Correct Answer + Core Reason: Dumbell. The shape of an orbital with $l = 1$ is a dumbbell, which corresponds to a p-orbital.

Step-by-Step Solution:

Step 1: Relate the quantum number $l$ to orbital shape:
- $l = 0$ : s-orbital (spherical)
- $l = 1$ : p-orbital (dumbbell)
- $l = 2$ : d-orbital (clover-leaf)
- $l = 3$ : f-orbital (more complex shape)

Step 2: Identify the shape for $l = 1$ : The shape is dumbbell.

Why other options are wrong:
- Option A: Spherical corresponds to an s-orbital ( $l = 0$ ).
- Option C: Clover-leaf corresponds to a d-orbital ( $l = 2$ ).
- Option D: Doughnut shape is not a standard description of atomic orbitals.

Key Concepts:
- Atomic orbitals
- Quantum numbers ( $l$ , $m$ )
- Shapes of s, p, and d orbitals

Memory Aids:
- $l = 0$ : s-orbital (spherical)
- $l = 1$ : p-orbital (dumbbell)

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

134 Easy

134 of 239

Analyzing +4 | -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Loss of electron from hydrogen atom results in nucleus of

\sim 1.5 \times 10^{-3} \, \mathrm{pm}

size.

Reason R: Proton

\left(\mathrm{H}^{+}\right)

always exists in combined form.

In the light of the above statements, choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: D

Both A and R are correct but R is NOT the
correct explanation of A

The correct answer is D because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. The size of the nucleus is indeed around

1.5 \times 10^{-3} \, \mathrm{pm}

, and protons always exist in combined form, but these two facts are unrelated.

Solution approach

Correct Answer + Core Reason: Both A and R are correct but R is NOT the correct explanation of A. The size of the nucleus is approximately $1.5 \times 10^{-3} \, \mathrm{pm}$ , and protons exist in combined form, but these are unrelated facts.

Step-by-Step Solution:

Step 1: Analyze Assertion A: Loss of an electron from a hydrogen atom results in a nucleus of size $\sim 1.5 \times 10^{-3} \, \mathrm{pm}$ . This statement is correct. The size of the nucleus is on the order of $10^{-15} \, \mathrm{m}$ , which is $1.5 \times 10^{-3} \, \mathrm{pm}$ .

Step 2: Analyze Reason R: Proton $(\mathrm{H}^{+})$ always exists in combined form. This statement is also correct. A free proton is highly reactive and will always associate with other molecules or ions.

Step 3: Determine if R is the correct explanation for A. The size of the nucleus is an intrinsic property and is not related to the fact that protons exist in combined form. Therefore, R is not the correct explanation for A.

Why other options are wrong:
- Option A: Both A and R are correct and R is the correct explanation of A - Incorrect because R does not explain A.
- Option B: A is correct but R is not correct - Incorrect because both A and R are correct.
- Option C: A is not correct but R is correct - Incorrect because both A and R are correct.

Key Concepts:
- Nuclear size
- Existence of protons
- Assertion and Reason analysis

Memory Aids:
- Remember the approximate size of the nucleus.
- Protons are highly reactive and exist in combined form.

Quick Tips:
- Carefully analyze each statement (Assertion and Reason) independently before considering their relationship.

135 Easy

135 of 239

Application +4 | -1

If there are five radial nodes, when what can be the correct representation of the orbital for

n = 8

View answer and explanation

Correct option: A

8d

Answer A (8d) is correct because the number of radial nodes is given by n - l - 1, and solving for l gives the d orbital.

Solution approach

Correct Answer + Core Reason: 8d. The orbital is determined by the number of radial nodes and the principal quantum number.

Step-by-Step Solution:

Step 1: Identify the given information: $n = 8$ , number of radial nodes = 5.

Step 2: Use the formula for radial nodes: $\text{Radial nodes} = n - l - 1$ .

Step 3: Substitute the given values: $5 = 8 - l - 1$ .

Step 4: Solve for $l$ : $l = 8 - 1 - 5 = 2$ .

Step 5: Determine the orbital based on the value of $l$ : $l = 0$ corresponds to s, $l = 1$ corresponds to p, $l = 2$ corresponds to d, $l = 3$ corresponds to f.

Step 6: Since $l = 2$ , the orbital is d. Therefore, the correct representation is 8d.

Why other options are wrong:
- Option B (8s): This is incorrect because for 8s, $l = 0$ , and the number of radial nodes would be $8 - 0 - 1 = 7$ , not 5.
- Option C (8p): This is incorrect because for 8p, $l = 1$ , and the number of radial nodes would be $8 - 1 - 1 = 6$ , not 5.
- Option D (8f): This is incorrect because for 8f, $l = 3$ , and the number of radial nodes would be $8 - 3 - 1 = 4$ , not 5.

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

Memory Aids:
- Remember the relationship between $l$ and the orbital type: s (0), p (1), d (2), f (3).

Quick Tips:
- Always remember the formula for radial nodes: $n - l - 1$ .
- Know the relationship between the azimuthal quantum number ( $l$ ) and the shape of the orbital.

136 Easy

136 of 239

Application +4 | -1

The number of unpaired electrons in the

{Fe}^{2+}

ion is

View answer and explanation

Correct option: B

4

Option B (4) is correct because the electronic configuration of Fe2+ has 4 unpaired electrons in the 3d orbitals.

Solution approach

Correct Answer + Core Reason: 4. The number of unpaired electrons in Fe2+ is determined by its electronic configuration.

Step-by-Step Solution:

Step 1: Write the electronic configuration of Fe: Fe (26) = [Ar] 3d6 4s2

Step 2: Write the electronic configuration of Fe2+: Fe2+ = [Ar] 3d6

Step 3: Fill the 3d orbitals according to Hund's rule: d orbitals have 5 sub-orbitals. Fill each sub-orbital with one electron first, then pair up.\nSo, 3d6 will have 4 unpaired electrons.

Why other options are wrong:
- Option A (0): This would be the case if all electrons were paired, which is not true for Fe2+.
- Option C (6): This would be the case if all 6 d electrons were unpaired, which violates Hund's rule.
- Option D (3): This is an incorrect number of unpaired electrons for the 3d6 configuration.

Key Concepts:
- Electronic configuration
- Hund's rule
- Unpaired electrons

Memory Aids:
- Remember Hund's rule: electrons fill orbitals individually before pairing up.

Quick Tips:
- Always write the electronic configuration first.
- Use Hund's rule to determine the number of unpaired electrons.

137 Easy

137 of 239

Understanding +4 | -1

If the uncertainty in the position of electron is zero, the uncertainty in its momentum would be

View answer and explanation

Correct option: D

Infinite

Option D (Infinite) is correct because, according to the Heisenberg uncertainty principle, if the uncertainty in position is zero, the uncertainty in momentum must be infinite.

Solution approach

Correct Answer + Core Reason: Infinite. According to the Heisenberg uncertainty principle, if the uncertainty in position is zero, the uncertainty in momentum is infinite.

Step-by-Step Solution:

Step 1: State the Heisenberg uncertainty principle. The Heisenberg uncertainty principle states that the product of the uncertainties in position ( $\Delta x$ ) and momentum ( $\Delta p$ ) is greater than or equal to $\frac{h}{4\pi}$ : $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ , where h is Planck's constant.

Step 2: Apply the given condition. Given that the uncertainty in the position of the electron is zero, i.e., $\Delta x = 0$ .

Step 3: Calculate the uncertainty in momentum. Substituting $\Delta x = 0$ into the Heisenberg uncertainty principle: $0 \cdot \Delta p \ge \frac{h}{4\pi}$ . This implies that $\Delta p \ge \frac{h}{0}$ , which means $\Delta p$ approaches infinity.

Why other options are wrong:
- Option A (Zero): If the uncertainty in position is zero, the uncertainty in momentum cannot be zero according to the Heisenberg uncertainty principle.
- **Option B (Greater than $\frac{h}{4 \pi}$ ):** While the product of uncertainties must be greater than or equal to $\frac{h}{4 \pi}$ , when the uncertainty in position is zero, the uncertainty in momentum must be infinite.
- **Option C (Less than $\frac{h}{4 \pi}$ ):** The product of uncertainties cannot be less than $\frac{h}{4 \pi}$ according to the Heisenberg uncertainty principle.

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

Memory Aids:
- $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$

Quick Tips:
- Remember the Heisenberg uncertainty principle and its implications.

138 Easy

138 of 239

Remembering +4 | -1

The unit of wavelength (nm) is equal to:

View answer and explanation

Correct option: A

10 Å

Option A (10 Å) is correct because 1 nanometer (nm) is equal to 10 Angstroms (Å).

Solution approach

Correct Answer + Core Reason: 1 nm is equal to 10 Å.

Step-by-Step Solution:

Step 1: Define nanometer (nm). 1 nm = $1 \times 10^{-9}$ m

Step 2: Define Angstrom (Å). 1 Å = $1 \times 10^{-10}$ m

Step 3: Convert nm to Å. To convert nm to Å, divide the value in nm by the conversion factor:
$1 nm = \frac{{1 \times 10^{-9} m}}{{1 \times 10^{-10} m/Å}} = 10 Å$

Why other options are wrong:
- Option B: 100 Å is incorrect. 1 nm is not equal to 100 Å.
- Option C: 1000 Å is incorrect. 1 nm is not equal to 1000 Å.
- Option D: 55 Å is incorrect. 1 nm is not equal to 55 Å.

Key Concepts:
- Units of length
- Nanometer (nm)
- Angstrom (Å)
- Unit conversion

Memory Aids:
- 1 nm = 10 Å

Quick Tips:
- Remember the conversion factor between nm and Å.

139 Easy

139 of 239

Application +4 | -1

An atom has the electronic configuration of

1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^{10}, 4 s^2 4 p^5

. Its atomic mass is 80 . Its atomic number and the number of neutrons in its nucleus shall be:

View answer and explanation

Correct option: A

35 and 45

Option A (35 and 45) is correct because the electronic configuration indicates 35 electrons, which is the atomic number, and the difference between the atomic mass (80) and atomic number (35) gives the number of neutrons (45).

Solution approach

Correct Answer + Core Reason: 35 and 45. The atomic number is determined from the electronic configuration, and the number of neutrons is the difference between the atomic mass and atomic number.

Step-by-Step Solution:

Step 1: Determine the atomic number (Z) from the electronic configuration:
$1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^{10}, 4s^2 4p^5$ \nZ = 2 + 2 + 6 + 2 + 6 + 10 + 2 + 5 = 35.

Step 2: Identify the atomic mass (A):\nGiven, A = 80.

Step 3: Calculate the number of neutrons (n):\nn = A - Z = 80 - 35 = 45.

Why other options are wrong:
- Option B (45 and 35): This is incorrect because it reverses the atomic number and the number of neutrons.
- Option C (40 and 40): This is incorrect because the electronic configuration does not correspond to an atomic number of 40.
- Option D (30 and 50): This is incorrect because the electronic configuration does not correspond to an atomic number of 30.

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

Memory Aids:
- Z = number of protons = number of electrons (in a neutral atom)
- A = number of protons + number of neutrons

Quick Tips:
- Always count the number of electrons in the electronic configuration to determine the atomic number.

140 Easy

140 of 239

Understanding +4 | -1

Statement (S): The location and momentum of an electron in an orbital are complementry to each other.

Explanation (E): The statement is against Heisenberg uncertainty principle.

View answer and explanation

Correct option: A

S is correct but E is wrong

Option A (S is correct but E is wrong) is correct because while the statement reflects the uncertainty principle, the explanation misinterprets its meaning.

Solution approach

Correct Answer + Core Reason: S is correct but E is wrong. The statement is related to the Heisenberg uncertainty principle, but the explanation is incorrect.

Step-by-Step Solution:

Step 1: Understand the Heisenberg uncertainty principle.

Step 2: The Heisenberg uncertainty principle states that it is impossible to determine both the position and momentum of an electron (or any particle) with perfect accuracy simultaneously.

Step 3: The statement (S) reflects this principle: the more accurately we know the position, the less accurately we know the momentum, and vice versa.

Step 4: The explanation (E) says the statement is *against* the Heisenberg uncertainty principle, which is incorrect. The statement *is* an illustration of the principle.

Step 5: Final Answer: S is correct but E is wrong

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

Key Concepts:
- Heisenberg uncertainty principle
- Position and momentum
- Quantum mechanics

Memory Aids:
- You can't know both position and momentum precisely.

Quick Tips:
- Understand the implications of the Heisenberg uncertainty principle.

141 Easy

141 of 239

Understanding +4 | -1

The ground state electronic configuration of

F{e^{3 + }}

is

View answer and explanation

Correct option: C

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

Option C is correct because Fe3+ has lost 3 electrons from Fe, resulting in a 3d5 configuration.

Solution approach

Correct Answer + Core Reason: $\left[ {Ar} \right]3{d^5}$ . $F{e^{3 + }}$ has lost 3 electrons from Fe, resulting in a 3d5 configuration.

Step-by-Step Solution:
Step 1: Write the electronic configuration of Fe: $[Ar] 4s^2 3d^6$ .
Step 2: When Fe loses 3 electrons to form $F{e^{3 + }}$ , the electrons are removed from the 4s orbital first, then from the 3d orbital.
Step 3: Therefore, the electronic configuration of $F{e^{3 + }}$ is $[Ar] 3d^5$ .

Why other options are wrong:
- Option A: Incorrect because it shows 3 electrons in the d orbital and 2 in the s orbital, which is not the correct configuration for $F{e^{3 + }}$ .
- Option B: Incorrect because it shows 6 electrons in the d orbital and 2 in the s orbital, which is the configuration of Fe, not $F{e^{3 + }}$ .
- Option D: Incorrect because it shows 4 electrons in the d orbital and 1 in the s orbital, which is not the correct configuration for $F{e^{3 + }}$ .

Key Concepts:
- Electronic configuration
- Ion formation
- Hund's rule

Memory Aids:
- Electrons are removed from the outermost shell first.

Quick Tips:
- Remember to remove electrons from the s-orbital before the d-orbital when forming cations.

142 Easy

142 of 239

Remembering +4 | -1

Which is not basic postulate of Dalton’s atomic theory?

View answer and explanation

Correct option: C

Atoms of all elements are alike, including their masses

The correct answer is C (Atoms of all elements are alike, including their masses) because Dalton's atomic theory states that atoms of the *same* element are alike, not atoms of *all* elements.

Solution approach

Correct Answer + Core Reason: Atoms of all elements are alike, including their masses. Dalton's theory postulates that atoms of the same element are identical, not atoms of different elements.

Step-by-Step Solution:

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

Step 2: Identify the statement that contradicts Dalton's original postulates.

Step 3: Dalton's theory states that atoms of a given element are identical in mass and properties, but atoms of different elements are different.

Why other options are wrong:
- Option A (Atoms are neither created nor destroyed in a chemical reaction): This is a correct postulate of Dalton's atomic theory related to the conservation of atoms in chemical reactions.
- Option B (In a given compound, the relative number and kinds of atoms are constant): This is a correct postulate of Dalton's atomic theory related to the law of definite proportions.
- Option D (Each element is composed of extremely small particles called atoms): This is a correct postulate of Dalton's atomic theory defining the fundamental unit of an element.

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

Memory Aids:
- Remember Dalton's theory focuses on atoms of the *same* element being alike.

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

143 Easy

143 of 239

Remembering +4 | -1

The electron is

View answer and explanation

Correct option: B

\beta

-ray particle

Option B (

\beta

-ray particle) is correct because an electron is a type of beta particle, which is a high-energy, high-speed electron or positron emitted during radioactive decay.

Solution approach

Correct Answer + Core Reason: $\beta$ -ray particle. An electron is a beta particle.

Step-by-Step Solution:

Step 1: Recall the properties of electrons and beta particles.

Step 2: Beta particles are high-energy electrons or positrons emitted during radioactive decay.

Step 3: Therefore, an electron is a $\beta$ -ray particle.

Why other options are wrong:
- Option A: $\alpha$ -ray particles are helium nuclei (2 protons and 2 neutrons).
- Option C: Hydrogen ion is a proton ( $H^+$ ).
- Option D: Positron is the antiparticle of the electron, with a positive charge.

Key Concepts:
- Beta decay
- Alpha particles
- Electrons
- Positrons

Memory Aids:
- Remember that beta particles are electrons.

Quick Tips:
- Know the composition of alpha and beta particles.

144 Easy

144 of 239

Analyzing +4 | -1

An increasing order (lowest first) for the values of e/m for electron (e), proton (p), neutron (n) and alpha (α) particle is:

View answer and explanation

Correct option: B

n,α,p,e

Option B (n,α,p,e) is correct because the e/m ratio increases in the order: neutron (0), alpha, proton, and electron due to their respective charges and masses.

Solution approach

Correct Answer + Core Reason: n,α,p,e. The e/m ratio increases in the order: neutron (0), alpha, proton, and electron.

Step-by-Step Solution:

Step 1: Define e/m ratio: The charge-to-mass ratio (e/m) is the ratio of the electric charge (e) to the mass (m) of a particle.

Step 2: Analyze each particle:
- Neutron (n): Charge = 0, Mass ≈ 1 amu. e/m = 0
- Alpha particle (α): Charge = +2, Mass ≈ 4 amu. e/m = 2/4 = 0.5
- Proton (p): Charge = +1, Mass ≈ 1 amu. e/m = 1/1 = 1
- Electron (e): Charge = -1, Mass ≈ 1/1836 amu. e/m ≈ 1836

Step 3: Compare the e/m ratios: 0 < 0.5 < 1 < 1836. Therefore, the increasing order is n, α, p, e.

Step 4: Conclusion: The increasing order of e/m values is neutron < alpha particle < proton < electron.

Why other options are wrong:
- Option A (e,p,n,α): This is incorrect as it places the electron with the lowest e/m, which is wrong.
- Option C (n,p,e,α): This is incorrect as the alpha particle has a lower e/m ratio than both the proton and electron.
- Option D (n,p,α,e): This is incorrect as it places the alpha particle with a lower e/m ratio than the proton but higher than the electron, which is wrong.

Key Concepts:
- Charge-to-mass ratio (e/m)
- Properties of subatomic particles (electron, proton, neutron, alpha particle)

Memory Aids:
- Remember that neutrons are neutral (zero charge).
- Electrons have a very small mass compared to protons and neutrons.

Quick Tips:
- Approximate the masses of protons and neutrons as 1 amu and the mass of an electron as negligible when comparing e/m ratios.
- Remember the charge of an alpha particle is +2 and its mass is approximately 4 amu.

145 Easy

145 of 239

Understanding +4 | -1

Which of the following electron transition in H-atom will require the energy equivalent to its ionisation energy

View answer and explanation

Correct option: D

From

n = 1

to

n = \, \propto

Option D is correct because ionization energy is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞).

Solution approach

Correct Answer + Core Reason: From $n = 1$ to $n = \propto$ . Ionization energy is the energy to remove an electron from n=1 to infinity.

Step-by-Step Solution:
Step 1: Recall the definition of ionization energy.
Step 2: Ionization energy is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞).
Step 3: $I.E = - {E_1} = 13.6\,eV$

Why other options are wrong:
- Option A: This transition requires less energy than ionization energy.
- Option B: This transition requires less energy than ionization energy.
- Option C: This transition requires less energy than ionization energy.

Key Concepts:
- Ionization energy
- Energy levels in hydrogen atom

Memory Aids:
- Ionization = removal to infinity

Quick Tips:
- Remember the definition of ionization energy.

146 Easy

146 of 239

Application +4 | -1

An electron moves around the nucleus in a circle of radius

r

. Assuming that the uncertainty of the momentum of electron is of the same order as momentum itself, the momentum of the electron would be

View answer and explanation

Correct option: A

\frac{h}{4 \pi r}

Option A (

\frac{h}{4 \pi r}

) is correct because it is derived from the Heisenberg uncertainty principle, relating the uncertainty in position and momentum.

Solution approach

Correct Answer + Core Reason: $\frac{h}{4 \pi r}$ . The momentum is estimated using the Heisenberg uncertainty principle, assuming the uncertainty in position is the radius of the circle.

Step-by-Step Solution:

Step 1: State the Heisenberg uncertainty principle:
$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$

Step 2: Identify the uncertainty in position:
$\Delta x = r$ (radius of the circle)

Step 3: Assume the uncertainty in momentum is of the same order as the momentum itself:
$\Delta p \approx p$

Step 4: Substitute the values into the uncertainty principle:\nr · p = $\frac{h}{4 π}$ $\n\n**Step 5:** Solve for the momentum (p):\np = \frac{h}{4 \pi r}$

Final Answer: p = $\frac{h}{4 \pi r}$

Why other options are wrong:
- **Option B ( $2 \pi h r$ ): This is incorrect; it does not follow from the uncertainty principle.
- Option C ( $\frac{h}{2 \pi r}$ ): This is incorrect; it does not follow from the uncertainty principle.
- Option D ( $4 \pi {hr}$ ): This is incorrect; it does not follow from the uncertainty principle.

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

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

Quick Tips:
- Understand the relationship between uncertainty in position and momentum.
- Apply the uncertainty principle correctly.

147 Easy

147 of 239

Remembering +4 | -1

According to Bohr's postulates which quantity can take up only discrete values:

View answer and explanation

Correct option: B

Angular momentum

Option B (Angular momentum) is correct because, according to Bohr's postulates, the angular momentum of an electron in an orbit is quantized and can only take discrete values.

Solution approach

Correct Answer + Core Reason: Angular momentum. Angular momentum $= n.\frac{{h}}{{{{2\pi }}}};$ where n is integer and thus discrete value.

Step-by-Step Solution:
Step 1: Recall Bohr's postulates.
Step 2: Bohr's postulate states that the angular momentum of an electron is quantized: $L = n \frac{h}{2\pi}$ , where n is an integer.
Step 3: Quantized means that the angular momentum can only take discrete values.

Why other options are wrong:
- Option A: Kinetic energy is quantized, but not directly stated as a primary postulate regarding discrete values in Bohr's model like angular momentum.
- Option C: Momentum is related to kinetic energy and is not directly quantized in Bohr's original postulates in the same explicit way as angular momentum.
- Option D: Potential energy is also quantized, but angular momentum is the fundamental quantity that is directly postulated to have discrete values.

Key Concepts:
- Bohr's postulates
- Quantization of angular momentum
- Atomic structure

Memory Aids:
- Bohr's model: Angular momentum is the key quantized property.

Quick Tips:
- Remember the formula for quantized angular momentum: $L = n \frac{h}{2\pi}$ .

148 Easy

148 of 239

Remembering +4 | -1

The total number of orbitals possible for principle quantum number n is

View answer and explanation

Correct option: B

\,\,{n^{2\,\,\,}}

Option B (

n^{2\,\,\}

) is correct because the total number of orbitals possible for a principal quantum number n is

n^2

.

Solution approach

Correct Answer + Core Reason: $\,{n^{2\,\,\}}$ . The total number of orbitals possible for principal quantum number n is equal to ${n^2}$

Step-by-Step Solution:

Step 1: The question asks for the total number of orbitals for a given principal quantum number n.

Step 2: Recall the relationship between the principal quantum number (n) and the number of orbitals.

Step 3: For a given n, the number of orbitals is $n^2$ .

Why other options are wrong:
- Option A: $n$ is the principal quantum number itself, not the number of orbitals.
- Option C: $2n$ is related to the number of electrons in a subshell, not the total number of orbitals.
- Option D: $2n^2$ is the maximum number of electrons that can occupy a shell with principal quantum number n.

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

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

Quick Tips:
- Remember the relationships between quantum numbers and the properties of atomic orbitals.

149 Easy

149 of 239

Understanding +4 | -1

The number of d-electrons retained in

\mathrm{F}{\mathrm{e}^{2 + }}

(At. No. Fe=26) ions is

View answer and explanation

Correct option: D

6

Option D (6) is correct because

\mathrm{F}{\mathrm{e}^{2 + }}

has lost two electrons from the 4s orbital, leaving 6 electrons in the 3d orbitals.

Solution approach

Correct Answer + Core Reason: 6. $\mathrm{F}{\mathrm{e}^{2 + }}$ has 6 d-electrons.

Step-by-Step Solution:
Step 1: Write the electronic configuration of Fe: ${\mathrm{\;}_{26}}\mathrm{Fe} = [Ar]3{d^6}4{s^2}$
Step 2: Determine the electronic configuration of $\mathrm{F}{\mathrm{e}^{2 + }}$ . It loses 2 electrons from the 4s orbital: $\mathrm{F}{\mathrm{e}^{2 + }} = [Ar]3{d^6}$
Step 3: Count the number of d-electrons: 6

Why other options are wrong:
- Option A: 3 is incorrect. It doesn't account for the correct number of d electrons.
- Option B: 4 is incorrect. It doesn't account for the correct number of d electrons.
- Option C: 5 is incorrect. It doesn't account for the correct number of d electrons.

Key Concepts:
- Electronic configuration
- Transition metal ions

Memory Aids:
- Remember to remove electrons from the s-orbital before the d-orbital when forming cations.

Quick Tips:
- Practice writing electronic configurations for ions.

150 Easy

150 of 239

Application +4 | -1

Which one of the following sets of ions represents a collection of iso electronic species

View answer and explanation

Correct option: B

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

Option B (

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

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

Solution approach

Correct Answer + Core Reason: $C{N^-}, C_2^{2 - },{N_2}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.

Step 2: For $SO_4^{2-}$ : 16 (S) + 4*8 (O) + 2 = 16 + 32 + 2 = 50 electrons\nFor $PO_4^{3-}$ : 15 (P) + 4*8 (O) + 3 = 15 + 32 + 3 = 50 electrons\nFor $BO_3^{3-}$ : 5 (B) + 3*8 (O) + 3 = 5 + 24 + 3 = 32 electrons

Step 3: For $CN^-$ : 6 (C) + 7 (N) + 1 = 14 electrons\nFor $C_2^{2-}$ : 2*6 (C) + 2 = 12 + 2 = 14 electrons\nFor $N_2$ : 2*7 (N) = 14 electrons

Step 4: For $O_2$ : 2*8 (O) = 16 electrons\nFor $F_2$ : 2*9 (F) = 18 electrons\nFor $N_2$ : 2*7 (N) = 14 electrons

Step 5: For $CO_3^{2-}$ : 6 (C) + 3*8 (O) + 2 = 6 + 24 + 2 = 32 electrons\nFor $NO_3^-$ : 7 (N) + 3*8 (O) + 1 = 7 + 24 + 1 = 32 electrons\nFor $ClO_4^-$ : 17 (Cl) + 4*8 (O) + 1 = 17 + 32 + 1 = 50 electrons

Step 6: Identify the set with the same number of electrons.

Step 7: $CN^-, C_2^{2-}, N_2$ all have 14 electrons.

Why other options are wrong:
- Option A: $SO_4^{2-}, PO_4^{3-}, BO_3^{3-}$ have different numbers of electrons (50, 50, and 32 respectively).
- Option C: ${O_2},{F_2},{N_2}$ have different numbers of electrons (16, 18, and 14 respectively).
- Option D: $CO_3^{ - 2},NO_3^ - ,ClO_4^ -$ have different numbers of electrons (32, 32, and 50 respectively).

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

Memory Aids:
- Remember to add electrons for negative charges and subtract for positive charges.

Quick Tips:
- Practice counting electrons quickly for common ions and molecules.

Your Progress

Structure of atom practice · Easy

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice