STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy

1 Easy

1 of 239

Application +4 | -1

Which of the following is paramagnetic?

View answer and explanation

Correct option: A

\mathrm{Ni}^{++}

Option A (

\mathrm{Ni}^{++}

) is correct because it has unpaired electrons in its d-orbital configuration, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ni}^{++}$ . $\mathrm{Ni}^{++}$ has unpaired electrons in its d-orbital configuration, making it paramagnetic.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $\mathrm{Ni}^{++}$ .\nNi has an atomic number of 28. Its electronic configuration is $[Ar]3d^84s^2$ . $\mathrm{Ni}^{++}$ loses 2 electrons, so its electronic configuration is $[Ar]3d^8$ .

Step 2: Determine the number of unpaired electrons in $\mathrm{Ni}^{++}$ .\nIn the $3d^8$ configuration, there are 2 unpaired electrons.

Step 3: Determine the electronic configuration of $\mathrm{Cu}^{+}$ .\nCu has an atomic number of 29. Its electronic configuration is $[Ar]3d^{10}4s^1$ . $\mathrm{Cu}^{+}$ loses 1 electron, so its electronic configuration is $[Ar]3d^{10}$ .

Step 4: Determine the number of unpaired electrons in $\mathrm{Cu}^{+}$ .\nIn the $3d^{10}$ configuration, all electrons are paired, so there are 0 unpaired electrons.

Step 5: Determine the electronic configuration of $\mathrm{Zn}^{++}$ .\nZn has an atomic number of 30. Its electronic configuration is $[Ar]3d^{10}4s^2$ . $\mathrm{Zn}^{++}$ loses 2 electrons, so its electronic configuration is $[Ar]3d^{10}$ .

Step 6: Determine the number of unpaired electrons in $\mathrm{Zn}^{++}$ .\nIn the $3d^{10}$ configuration, all electrons are paired, so there are 0 unpaired electrons.

Step 7: Paramagnetic substances have unpaired electrons. Therefore, $\mathrm{Ni}^{++}$ is paramagnetic.

Why other options are wrong:
- **Option B ( $\mathrm{Cu}^{+}$ ):** This is incorrect because $\mathrm{Cu}^{+}$ has a $d^{10}$ configuration with no unpaired electrons, making it diamagnetic.
- **Option C ( $\mathrm{Zn}^{++}$ ):** This is incorrect because $\mathrm{Zn}^{++}$ has a $d^{10}$ configuration with no unpaired electrons, making it diamagnetic.
- Option D (All): This is incorrect because only $\mathrm{Ni}^{++}$ is paramagnetic.

Key Concepts:
- Electronic configuration
- Paramagnetism
- Diamagnetism
- Unpaired electrons

Memory Aids:
- Paramagnetic: Para = beside, magnetic = unpaired electrons are beside each other, creating a magnetic field.
- Diamagnetic: Dia = against, magnetic = all electrons are paired, so they oppose an external magnetic field.

Quick Tips:
- Determine the electronic configuration of the ion.
- Count the number of unpaired electrons.
- If there are unpaired electrons, the substance is paramagnetic. If all electrons are paired, the substance is diamagnetic.

2 Easy

2 of 239

Remembering +4 | -1

Quantum number which defines the orientation of orbital present in a subshell is

View answer and explanation

Correct option: C

Magnetic quantum number (m)

Option C (Magnetic quantum number (m)) is correct because the magnetic quantum number defines the orientation of an orbital in space.

Solution approach

Correct Answer + Core Reason: Magnetic quantum number (m). The magnetic quantum number defines the spatial orientation of an orbital.

Step-by-Step Solution:
Step 1: Identify the question type: This is a conceptual question about quantum numbers.
Step 2: Recall the definitions of quantum numbers:
- Principal quantum number (n): Energy level
- Azimuthal quantum number (l): Shape of the orbital
- Magnetic quantum number (m): Orientation of the orbital in space
- Spin quantum number (s): Spin of the electron
Step 3: Select the correct option: The magnetic quantum number (m) describes the orientation of an orbital.

Why other options are wrong:
- Option A: Principal quantum number (n) describes the energy level of the electron.
- Option B: Azimuthal quantum number (l) describes the shape of the orbital (s, p, d, f).
- Option D: Spin quantum number (s) describes the intrinsic angular momentum of the electron (spin up or spin down).

Key Concepts:
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- Magnetic = Orientation (think of a compass pointing in a direction)

Quick Tips:
- Remember the roles of each quantum number: n (energy), l (shape), m (orientation), s (spin).

3 Easy

3 of 239

Understanding +4 | -1

Maximum number of electrons associated with

\ell = 1,\,\,m = 0

in magnesium atom is

View answer and explanation

Correct option: B

2

The correct answer is B (2) because only the two electrons in the 2p orbital with m=0 satisfy the given conditions in Magnesium's electronic configuration.

Solution approach

Correct Answer + Core Reason: 2. Only two electrons in the 2p orbital have $\ell = 1$ and $m = 0$ .

Step-by-Step Solution:

Step 1: Write the electronic configuration of Magnesium (Mg): $1s^2 2s^2 2p^6 3s^2$ .

Step 2: Identify the orbitals with $\ell = 1$ : These are the p orbitals.

Step 3: Identify the p orbitals with $m = 0$ : In the $2p^6$ configuration, there are three p orbitals ( $2p_x, 2p_y, 2p_z$ ) with $m = -1, 0, +1$ respectively. The $2p_z$ orbital has $m = 0$ and can hold 2 electrons.

Step 4: Count the number of electrons in the $2p$ orbital with $m=0$ : There are 2 electrons.

Why other options are wrong:
- Option A (1): This is incorrect because each orbital can hold a maximum of 2 electrons.
- Option C (6): This is the total number of electrons in the 2p orbitals, but only 2 have m=0.
- Option D (8): This is incorrect and doesn't relate to the electronic configuration of Mg.

Key Concepts:
- Electronic configuration
- Quantum numbers ( $\ell$ and $m$ )
- Atomic orbitals

Memory Aids:
- Remember the relationship between $\ell$ and the type of orbital (s, p, d, f).
- Remember the possible values of $m$ for a given $\ell$ .

Quick Tips:
- Write out the electronic configuration first.
- Identify the orbitals that match the given quantum numbers.

4 Easy

4 of 239

Understanding +4 | -1

The nucleus of the element having atomic number 25 and atomic weight 55 will contain

View answer and explanation

Correct option: A

25 protons and 30 neutrons

Option A (25 protons and 30 neutrons) is correct because the number of protons equals the atomic number, and the number of neutrons is the difference between the atomic weight and the atomic number.

Solution approach

Correct Answer + Core Reason: 25 protons and 30 neutrons. The number of protons equals the atomic number, and the number of neutrons is the atomic weight minus the atomic number.

Step-by-Step Solution:

Step 1: Identify the atomic number (Z) and atomic weight (A): Z = 25, A = 55.

Step 2: Number of protons = Atomic number = 25.

Step 3: Number of neutrons = Atomic weight - Atomic number = 55 - 25 = 30.

Step 4: Therefore, the nucleus contains 25 protons and 30 neutrons.

Why other options are wrong:
- Option B (25 neutrons and 30 protons): This reverses the number of protons and neutrons, which is incorrect.
- Option C (55 protons): This is incorrect because the atomic weight represents the total number of protons and neutrons, not just protons.
- Option D (55 neutrons): This is incorrect because the atomic weight represents the total number of protons and neutrons, not just neutrons.

Key Concepts:
- Atomic Number
- Atomic Weight
- Protons, Neutrons, and Electrons

Memory Aids:
- A = Z + N (Atomic weight = Atomic number + Number of neutrons)

Quick Tips:
- Remember that the atomic number defines the element.

5 Easy

5 of 239

Understanding +4 | -1

Which is not permissible subshell?

View answer and explanation

Correct option: A

2d

Option A (2d) is correct because the d subshell starts from the third energy level (n=3), so 2d is not a permissible subshell.

Solution approach

Correct Answer + Core Reason: 2d. The d subshell corresponds to $l=2$ , and for $n=2$ , the possible values of $l$ are 0 and 1.

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the azimuthal quantum number (l).

Step 2: The possible values of l range from 0 to n-1.

Step 3: For n=2, the possible values of l are 0 and 1, corresponding to the s and p subshells, respectively.

Step 4: Therefore, a 2d subshell is not possible.

Why other options are wrong:
- Option B: 4f is permissible because for n=4, l can be 0, 1, 2, or 3. l=3 corresponds to the f subshell.
- Option C: 6p is permissible because for n=6, l can be 0, 1, 2, 3, 4, or 5. l=1 corresponds to the p subshell.
- Option D: 3s is permissible because for n=3, l can be 0, 1, or 2. l=0 corresponds to the s subshell.

Key Concepts:
- Quantum Numbers (n and l)
- Atomic Orbitals
- Subshells (s, p, d, f)

Memory Aids:
- Remember the relationship: 0 ≤ l ≤ n-1

Quick Tips:
- Know the allowed values of quantum numbers.

6 Easy

6 of 239

Analyzing +4 | -1

Statement (S): H-atom has only one electron in its orbit, but several spectral lines are noticed

Explanation (E): The

\mathrm{H}

-spectra is observed in

\mathrm{H}_2

gas

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

Option A is incorrect. The explanation is wrong, it should be that the H-spectra is observed in a sample of H atoms, not H2 gas. The spectral lines arise from electronic transitions within the H atom.

Solution approach

Correct Answer + Core Reason: S is correct but E is wrong. The statement is correct, but the explanation is incorrect.

Step-by-Step Solution:

Step 1: Understand the statement: H-atom has only one electron, but its spectrum shows several spectral lines. This is because the electron can be excited to different energy levels and then transition back to lower energy levels, emitting photons of different energies (wavelengths).

Step 2: Understand the explanation: The $\mathrm{H}$ -spectra is observed in a sample of H atoms, not $\mathrm{H}_2$ gas. The spectral lines arise from electronic transitions within the H atom.

Step 3: Evaluate the statement: The statement is correct.

Step 4: Evaluate the explanation: The explanation is incorrect. The H-spectra is observed in a sample of H atoms, not $\mathrm{H}_2$ gas.

Why other options are wrong:
- Option B: The statement is correct, so this option is incorrect.
- Option C: The explanation is incorrect, so this option is incorrect.
- Option D: The explanation is incorrect, so this option is incorrect.

Key Concepts:
- Atomic Spectra
- Electronic Transitions
- Energy Levels

Memory Aids:
- Remember that spectral lines arise from electronic transitions between energy levels.

Quick Tips:
- The hydrogen spectrum is a result of the quantized energy levels of the electron in the hydrogen atom.

7 Easy

7 of 239

Application +4 | -1

When the value of azimuthal quantum number is 3, the maximum and minimum values of spin multiplicity are

View answer and explanation

Correct option: B

8,1

Option B (8,1) is correct because when the azimuthal quantum number is 3 (f orbital), the maximum spin multiplicity is 8 (7 unpaired electrons) and the minimum is 1 (all paired).

Solution approach

Correct Answer + Core Reason: 8,1. The maximum spin multiplicity occurs with 7 unpaired electrons, and the minimum occurs when all electrons are paired except for one.

Step-by-Step Solution:

Step 1: Identify the azimuthal quantum number: $l = 3$ corresponds to an f orbital.

Step 2: Determine the maximum number of unpaired electrons in an f orbital: An f orbital can hold a maximum of 7 unpaired electrons.

Step 3: Calculate the maximum spin multiplicity: $S = \frac{7}{2}$ , so the maximum spin multiplicity is $2S + 1 = 2(\frac{7}{2}) + 1 = 8$ .

Step 4: Determine the minimum number of unpaired electrons. In an f orbital, the minimum number of unpaired electrons can be 0 or 1. Since the question asks for spin multiplicity, we consider the case with 1 unpaired electron.

Step 5: Calculate the minimum spin multiplicity: $S = \frac{1}{2}$ , so the minimum spin multiplicity is $2S + 1 = 2(\frac{1}{2}) + 1 = 2$ . However, the question implies a minimum value of 1, which would occur if all electrons are paired (S=0), giving 2S+1 = 1. In this case, we consider the case where all electrons are paired, giving a spin multiplicity of 1.

Step 6: Final Answer: The maximum spin multiplicity is 8, and the minimum spin multiplicity is 1.

Why other options are wrong:
- Option A (1,8): This reverses the maximum and minimum values, which is incorrect.
- Option C (6,1): The maximum spin multiplicity is not 6. This would correspond to 5 unpaired electrons, which is not the maximum for an f orbital.
- Option D (7,0): The minimum spin multiplicity cannot be 0. Even if all electrons are paired, the spin multiplicity is 1.

Key Concepts:
- Azimuthal quantum number (l)
- Spin quantum number (s)
- Spin multiplicity (2S+1)
- Hund's rule

Memory Aids:
- Spin multiplicity formula: 2S+1
- f orbital can hold up to 7 unpaired electrons

Quick Tips:
- Remember that the maximum spin multiplicity occurs when all electrons are unpaired.
- The minimum spin multiplicity is 1 when all electrons are paired.

8 Easy

8 of 239

Remembering +4 | -1

The orientation of an atomic orbital is governed by

View answer and explanation

Correct option: C

m

Option C (m) is correct because the orientation of an atomic orbital in space is governed by the magnetic quantum number (m).

Solution approach

Correct Answer + Core Reason: m. The magnetic quantum number determines the spatial orientation of an orbital.

Step-by-Step Solution:
Step 1: Define the magnetic quantum number (m).
Step 2: The magnetic quantum number (m) describes the orientation of an atomic orbital in space relative to a standard set of coordinate axes.
Step 3: The values of m range from -l to +l, including 0, where l is the azimuthal quantum number.

Why other options are wrong:
- Option A: n is incorrect because n (principal quantum number) determines the energy level and size of the orbital.
- Option B: $l$ is incorrect because $l$ (azimuthal quantum number) determines the shape of the orbital.
- Option D: s is incorrect because s (spin quantum number) describes the intrinsic angular momentum of an electron.

Key Concepts:
- Quantum numbers
- Magnetic quantum number
- Atomic orbitals

Memory Aids:
- m = orientation

Quick Tips:
- Remember the roles of the four quantum numbers: n, l, m, and s.

9 Easy

9 of 239

Remembering +4 | -1

The energy of different orbitals in an atom or ion having only one electron depends on

View answer and explanation

Correct option: A

n

only

Option A (

n

only) is correct because for a one-electron system, the energy of an orbital depends only on the principal quantum number (

n

).

Solution approach

Correct Answer + Core Reason: $n$ only. In a one-electron system, energy depends only on the principal quantum number.

Step-by-Step Solution:

Step 1: Recall the factors affecting orbital energy.

Step 2: Consider the case of a one-electron system.

Step 3: Recognize that in a one-electron system, the energy depends only on the principal quantum number ( $n$ ).

Why other options are wrong:
- Option B: $n$ and $l$ are important for multi-electron systems but not for one-electron systems.
- Option C: $n, l$ and $m$ are important for multi-electron systems but not for one-electron systems.
- Option D: $n, l, m$ and $s$ are important for multi-electron systems but not for one-electron systems.

Key Concepts:
- Quantum numbers
- Atomic orbitals

Memory Aids:
- Remember that energy depends only on $n$ for one-electron systems.

Quick Tips:
- Understand the factors affecting orbital energy in different systems.

10 Easy

10 of 239

Understanding +4 | -1

Statement (S): Number of waves in an orbit of atom is equal to number of that orbit.

Explanation (E): Number of waves in an orbit is derived by

\frac{2 \pi r_n}{\lambda}

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

Option C is correct because the number of waves in an orbit is equal to the orbit number, and this is derived from the relationship between the circumference of the orbit and the wavelength of the electron.

Solution approach

Correct Answer + Core Reason: Both S and E are correct and E is correct explanation of S. The number of waves in an orbit is equal to the orbit number, and this is derived from the relationship between the circumference of the orbit and the wavelength of the electron.

Step-by-Step Solution:

Step 1: Understand the de Broglie wavelength concept. The de Broglie wavelength, $\lambda$ , of an electron is related to its momentum by $\lambda = \frac{h}{mu}$ , where $h$ is Planck's constant, $m$ is the mass of the electron, and $u$ is its velocity.

Step 2: Relate the circumference of the orbit to the wavelength. According to Bohr's quantization condition, the angular momentum of an electron in an orbit is quantized: $mur_n = \frac{nh}{2\pi}$ , where $r_n$ is the radius of the $n$ -th orbit.

Step 3: Express the circumference of the orbit in terms of the wavelength. The circumference of the $n$ -th orbit is $2\pi r_n$ . The number of waves in the orbit is given by $\frac{2\pi r_n}{\lambda}$ .

Step 4: Substitute the expressions for $r_n$ and $\lambda$ . From Bohr's quantization condition, $r_n = \frac{nh}{2\pi mu}$ . Substituting this into the expression for the number of waves, we get $\frac{2\pi (\frac{nh}{2\pi mu})}{\lambda} = \frac{nh}{mu\lambda}$ . Since $\lambda = \frac{h}{mu}$ , we have $\frac{nh}{mu(\frac{h}{mu})} = n$ .

Step 5: Conclusion: The number of waves in the $n$ -th orbit is equal to $n$ , the orbit number. Thus, both the statement and the explanation are correct, and the explanation correctly explains the statement.

Why other options are wrong:
- Option A: S is correct but E is wrong: E is correct, so this option is incorrect.
- Option B: S is wrong but E is correct: S is correct, so this option is incorrect.
- Option D: Both S and E are correct but E is not correct explanation of S: E is the correct explanation of S, so this option is incorrect.

Key Concepts:
- de Broglie wavelength
- Bohr's quantization condition
- Wave-particle duality

Memory Aids:
- Remember the Bohr quantization condition: $mur = \frac{nh}{2\pi}$
- de Broglie's equation: $\lambda = \frac{h}{p} = \frac{h}{mu}$

Quick Tips:
- Always relate the circumference of the orbit to the de Broglie wavelength to find the number of waves in an orbit.
- Remember that the number of waves must be an integer.

11 Easy

11 of 239

Remembering +4 | -1

Normally, the time taken in the transition is :

View answer and explanation

Correct option: D

{10^{ - 8}}\mathrm{sec}

Option D (

{10^{ - 8}}\mathrm{sec}

) is correct because electronic transitions in atoms typically occur on the order of nanoseconds.

Solution approach

Correct Answer + Core Reason: ${10^{ - 8}}\mathrm{sec}$ . Electronic transitions are very fast, typically occurring in nanoseconds.

Step-by-Step Solution:

Step 1: Recall the nature of electronic transitions in atoms.

Step 2: Understand that these transitions involve the movement of electrons between energy levels.

Step 3: Recognize that these processes are very rapid due to the small size and high speed of electrons.

Step 4: Consider the typical timescales for such transitions, which are on the order of nanoseconds ( $\sim 10^{-9}$ s) to picoseconds ( $\sim 10^{-12}$ s).

Step 5: Choose the option that best represents this timescale, which is ${10^{ - 8}}\mathrm{sec}$ .

Why other options are wrong:
- Option A (Zero): Transitions cannot occur in zero time.
- Option B (1 sec): This is far too long for an electronic transition.
- **Option C ( ${10^{ - 5}}\mathrm{sec}$ ): This is also too long for an electronic transition; it's more typical of some chemical reactions.

Key Concepts:
- Electronic transitions
- Energy levels in atoms
- Time scales of atomic processes

Memory Aids:
- Remember that electronic processes are generally very fast.

Quick Tips:**
- Keep in mind the typical timescales for different physical and chemical processes.

12 Easy

12 of 239

Understanding +4 | -1

The energy of the electron in a hydrogen atom has a negative sign because

View answer and explanation

Correct option: A

the energy of the electron in the atom is lower than the energy of a free electron at rest

Option A (the energy of the electron in the atom is lower than the energy of a free electron at rest) is correct because the negative sign indicates that the electron is in a bound state, requiring energy to be removed from the atom.

Solution approach

Correct Answer + Core Reason: the energy of the electron in the atom is lower than the energy of a free electron at rest. The negative sign indicates a bound state.

Step-by-Step Solution:

Step 1: Understand the concept of energy levels in an atom.
- Electrons in an atom have quantized energy levels.
- The energy of an electron in an atom is defined relative to a free electron at rest (infinitely far from the nucleus), which is assigned an energy of zero.

Step 2: Explain the negative sign of the electron's energy.
- The negative sign indicates that the electron is in a bound state, meaning it is attracted to the nucleus and requires energy to be removed from the atom to become a free electron.
- The energy of the electron in the atom is lower (more negative) than the energy of a free electron at rest (zero).

Why other options are wrong:
- Option B: A free electron at rest is infinitely far away from the nucleus, not very close.
- Option C: The energy value assigned to a free electron at rest is zero, not infinity.
- Option D: Options B and C are incorrect.

Key Concepts:
- Energy levels in an atom
- Bound state
- Ionization energy

Memory Aids:
- Negative energy means the electron is bound to the nucleus.

Quick Tips:
- Remember that the energy of an electron in an atom is always negative relative to a free electron.

13 Easy

13 of 239

Remembering +4 | -1

Electromagnetic radiation with minimum wavelength is

View answer and explanation

Correct option: C

X-ray

Option C (X-ray) is correct because X-rays have the shortest wavelength among the given electromagnetic radiations.

Solution approach

Correct Answer + Core Reason: X-rays have the minimum wavelength among the given options.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum.

Step 2: The electromagnetic spectrum, in order of decreasing wavelength (increasing frequency), is: Radio waves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.

Step 3: Comparing the given options, X-rays have the shortest wavelength.

Why other options are wrong:
- Option A: Ultra-violet has a longer wavelength than X-rays.
- Option B: Radio waves have the longest wavelength in the electromagnetic spectrum.
- Option D: Infra-red has a longer wavelength than X-rays.

Key Concepts:
- Electromagnetic spectrum
- Wavelength and frequency

Memory Aids:
- Remember the order of the electromagnetic spectrum (e.g., ROYGBIV for visible light). Radio waves are longest, Gamma rays are shortest.

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

14 Easy

14 of 239

Remembering +4 | -1

Which of the following scientists explained his model on the basis of centrifugal force?

View answer and explanation

Correct option: D

Rutherford

Option D (Rutherford) is correct because Rutherford's model of the atom proposed that electrons orbit the nucleus, with the centrifugal force balancing the electrostatic force of attraction.

Solution approach

Correct Answer + Core Reason: Rutherford. Rutherford explained his model on the basis of centrifugal force balancing the electrostatic force of attraction.

Step-by-Step Solution:

Step 1: Recall Rutherford's model of the atom.

Step 2: Understand that Rutherford proposed that electrons revolve around the nucleus in specific orbits.

Step 3: Recognize that the centrifugal force due to the revolution of electrons balances the electrostatic force of attraction between the electrons and the nucleus.

Step 4: Conclude that Rutherford's model explained the atom based on centrifugal force.

Why other options are wrong:
- Option A: Thomson proposed the plum pudding model.
- Option B: Dalton proposed the atomic theory.
- Option C: Millikan conducted the oil drop experiment to determine the charge of an electron.

Key Concepts:
- Rutherford's model of the atom
- Centrifugal force
- Electrostatic force

Memory Aids:
- Rutherford: \"R\" for \"Revolution,\" reminding you of electrons revolving around the nucleus.

Quick Tips:
- Remember the key features of different atomic models.

15 Easy

15 of 239

Remembering +4 | -1

The number of waves in an orbit are

View answer and explanation

Correct option: B

n

Option B (

n

) is correct because the number of waves in an orbit is equal to the principal quantum number,

n

.

Solution approach

Correct Answer + Core Reason: $n$ . The number of waves in an orbit=n.

Step-by-Step Solution:
Step 1: Recall Bohr's model: According to Bohr's model, electrons revolve around the nucleus in specific orbits.
Step 2: Relate to de Broglie wavelength: The circumference of an orbit is an integral multiple of the de Broglie wavelength, i.e., $2\pi r = n\lambda$ .
Step 3: Number of waves: The number of waves in an orbit is equal to $n$ .

Why other options are wrong:
- Option A: ${n^2}$ is not the number of waves in an orbit.
- Option C: $n - 1$ is not the number of waves in an orbit.
- Option D: $n - 2$ is not the number of waves in an orbit.

Key Concepts:
- Bohr's model
- de Broglie wavelength
- Principal quantum number

Memory Aids:
- The number of waves is directly equal to the principal quantum number $n$ .

Quick Tips:
- Remember the relationship between the orbit circumference and the de Broglie wavelength.

16 Easy

16 of 239

Remembering +4 | -1

When the azimuthal quantum number has a value of

l=1

, the shape of the orbital is:

View answer and explanation

Correct option: C

dumb-bell

Option C (dumb-bell) is correct because when the azimuthal quantum number l = 1, it corresponds to a p-orbital, which has a dumb-bell shape.

Solution approach

Correct Answer + Core Reason: dumb-bell. l=1 corresponds to p-orbital which is dumb-bell shaped.

Step-by-Step Solution:

Step 1: Define azimuthal quantum number (l). The azimuthal quantum number (l) describes the shape of an atomic orbital.

Step 2: Relate l to orbital shape. When l = 0, the orbital is s-orbital and is spherically symmetrical. When l = 1, the orbital is a p-orbital and has a dumb-bell shape. When l = 2, the orbital is a d-orbital and has a more complex shape.

Step 3: Conclusion. Therefore, when l = 1, the shape of the orbital is dumb-bell.

Why other options are wrong:
- Option A: unsymmetrical is a vague description and not specific to l=1.
- Option B: spherically symmetrical corresponds to l = 0 (s-orbital).
- Option D: complicated is a general term and not specific to l=1. d-orbitals (l=2) have more complicated shapes.

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

Memory Aids:
- l = 0: s-orbital (spherical), l = 1: p-orbital (dumb-bell), l = 2: d-orbital (more complex).

Quick Tips:
- Know the relationship between the azimuthal quantum number and the shape of the orbital.

17 Easy

17 of 239

Remembering +4 | -1

The electron possesses wave properties was shown experimentally by:

View answer and explanation

Correct option: C

Davission and germer

Option C (Davission and germer) is correct because the Davisson-Germer experiment demonstrated the wave nature of electrons through diffraction patterns.

Solution approach

Correct Answer + Core Reason: Davission and germer. The Davisson-Germer experiment demonstrated the wave nature of electrons.

Step-by-Step Solution:

Step 1: Recall the Davisson-Germer experiment.

Step 2: The Davisson-Germer experiment involved scattering electrons off a nickel crystal and observing diffraction patterns, which confirmed the wave nature of electrons as predicted by de Broglie.

Step 3: Identify the correct experimental verification.

Why other options are wrong:
- Option A: Bohr: Bohr's model explained the quantized energy levels of electrons in atoms but did not directly demonstrate the wave nature of electrons.
- Option B: de Broglie: de Broglie proposed the wave-particle duality of matter, but the Davisson-Germer experiment provided experimental evidence.
- Option D: Schrödinger: Schrödinger developed the wave equation to describe the behavior of electrons in atoms, but the Davisson-Germer experiment provided experimental evidence for the wave nature of electrons.

Key Concepts:
- Wave-particle duality
- Davisson-Germer experiment
- de Broglie hypothesis

Memory Aids:
- Remember that the Davisson-Germer experiment confirmed de Broglie's hypothesis.

Quick Tips:
- Know the key experiments that led to the development of quantum mechanics.

18 Easy

18 of 239

Understanding +4 | -1

For which of the following species, Bohr’s theory doesn’t apply

View answer and explanation

Correct option: D

N{a^ + }

Option D is correct because Bohr's theory is only applicable to hydrogen-like species (species with only one electron), and Na+ has more than one electron.

Solution approach

Correct Answer + Core Reason: $N{a^ + }$ . Bohr's theory applies only to single-electron species.

Step-by-Step Solution:

Step 1: Recall Bohr's Theory: Bohr's theory is applicable to species with only one electron and a nucleus.

Step 2: Analyze the options:
- A. H (1 electron) - Applicable
- B. He+ (1 electron) - Applicable
- C. Li2+ (1 electron) - Applicable
- D. Na+ (10 electrons) - Not Applicable

Step 3: Conclusion: Na+ has 10 electrons, so Bohr's theory doesn't apply.

Why other options are wrong:
- Option A: H has only one electron, so Bohr's theory applies.
- Option B: He+ has only one electron, so Bohr's theory applies.
- Option C: Li2+ has only one electron, so Bohr's theory applies.

Key Concepts:
- Bohr's Theory
- Hydrogen-like species
- Electronic configuration

Memory Aids:
- Bohr's model works for single electron systems only.

Quick Tips:
- Check the number of electrons before applying Bohr's theory.

19 Easy

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Understanding +4 | -1

In order to designate an orbital n in an atom, the number of quantum number required are:

View answer and explanation

Correct option: C

3

Option C (3) is correct because three quantum numbers (n, l, and m) are required to define an orbital in an atom.

Solution approach

Correct Answer + Core Reason: 3. To designate an orbital, n, l, and m are required.

Step-by-Step Solution:

Step 1: The principal quantum number (n) describes the energy level of the electron.

Step 2: The azimuthal quantum number (l) describes the shape of the orbital (s, p, d, f). It ranges from 0 to n-1.

Step 3: The magnetic quantum number (m) describes the orientation of the orbital in space. It ranges from -l to +l, including 0.

Step 4: To uniquely define an orbital, we need n, l, and m.

Why other options are wrong:
- Option A: 1 is incorrect because only the principal quantum number is not sufficient to define an orbital.
- Option B: 2 is incorrect because the principal and azimuthal quantum numbers are not sufficient to define an orbital.
- Option D: 4 is incorrect because the spin quantum number (s) describes the spin of the electron, not the orbital itself.

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

Memory Aids:
- Remember that n describes the energy level, l describes the shape, and m describes the orientation of the orbital.

Quick Tips:
- Understand the meaning of each quantum number and what it describes.

20 Easy

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Remembering +4 | -1

Millikan performed an experiment method to determine which of the following?

View answer and explanation

Correct option: B

Charge of the electron

Option B (Charge of the electron) is correct because Millikan's oil drop experiment was designed to measure the elementary electric charge.

Solution approach

Correct Answer + Core Reason: Charge of the electron. Millikan's oil drop experiment determined the charge of an electron.

Step-by-Step Solution:

Step 1: Recall Millikan's oil drop experiment. This experiment involved observing charged oil droplets between two electrodes.

Step 2: Identify the purpose of the experiment. The purpose was to determine the charge of an electron.

Step 3: Confirm the answer. Millikan's experiment measured the charge of the electron.

Why other options are wrong:
- Option A (Mass of the electron): Incorrect. Millikan's experiment did not directly measure the mass of the electron.
- **Option C (\ ${e} / {m}$ ratio of electron): Incorrect. The e/m ratio was determined by Thomson's experiment, not Millikan's.
- Option D (Both (1) and (2)): Incorrect. Millikan only determined the charge of the electron.

Key Concepts:
- Millikan's oil drop experiment
- Charge of an electron

Memory Aids:
- Remember that Millikan used oil drops to find the charge of an electron.

Quick Tips:**
- Know the key experiments and their contributions to atomic theory.

21 Easy

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Analyzing +4 | -1

Statement (S): Electronic configuration of

{ }_{23} \mathrm{V}^{3+}

ion is

[\mathrm{Ar}]^{18} 3 d^2

and not

[\mathrm{Ar}]^{18} 3 d^0 4 s^2.

Explanation (E):

\mathrm{V}^{3+}

ion is diamagnetic in nature.

View answer and explanation

Correct option: A

S is correct but E is wrong

Option A is correct because V3+ has the electronic configuration [Ar]3d², and it is paramagnetic, not diamagnetic.

Solution approach

Correct Answer + Core Reason: S is correct but E is wrong. The electronic configuration of V3+ is [Ar]3d², and it is paramagnetic.

Step-by-Step Solution:

Step 1: Write the electronic configuration of V (Vanadium). V has an atomic number of 23, so its electronic configuration is [Ar] 3d³ 4s².

Step 2: Determine the electronic configuration of V3+. When V loses 3 electrons to form V3+, it loses the 4s electrons first, followed by one 3d electron. Therefore, the electronic configuration of V3+ is [Ar] 3d².

Step 3: Evaluate the statement (S). The statement that the electronic configuration of V3+ is [Ar] 3d² is correct.

Step 4: Determine the magnetic nature of V3+. V3+ has two unpaired electrons in the 3d orbitals, making it paramagnetic.

Step 5: Evaluate the explanation (E). The explanation that V3+ is diamagnetic is wrong.

Step 6: Conclude that the statement is correct, but the explanation is wrong.

Why other options are wrong:
- Option B: The statement is correct.
- Option C: The explanation is wrong.
- Option D: The explanation is wrong.

Key Concepts:
- Electronic configuration
- Ion formation
- Paramagnetism and diamagnetism

Memory Aids:
- Remember that electrons are removed from the outermost shell first when forming cations.
- Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired.

Quick Tips:
- Write the electronic configuration of the neutral atom first, then remove electrons to form the ion.

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Remembering +4 | -1

How many neutrons are present in tritium nucleus?

View answer and explanation

Correct option: A

2

Option A (2) is correct because tritium, being an isotope of hydrogen with a mass number of 3, has 1 proton and therefore 2 neutrons (3 - 1 = 2).

Solution approach

Correct Answer + Core Reason: Tritium has 2 neutrons.

Step-by-Step Solution:

Step 1: Recall the composition of tritium. Tritium is $^3H_1$ .

Step 2: Determine the number of protons. The atomic number (1) indicates 1 proton.

Step 3: Determine the number of neutrons. The mass number (3) minus the number of protons (1) equals the number of neutrons: 3 - 1 = 2.

Why other options are wrong:
- Option B: 3 is the mass number of tritium, not the number of neutrons.
- Option C: 1 is the number of protons in tritium, not the number of neutrons.
- Option D: 0 neutrons would mean it is protium, not tritium.

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

Memory Aids:
- Mass number = Protons + Neutrons

Quick Tips:
- Remember that the number of neutrons = mass number - atomic number.

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Application +4 | -1

Calculate number of electron for which

n=4

and

{S}=+\frac{1}{2}

View answer and explanation

Correct option: A

16

Option A (16) is correct because for n=4, there are 16 electrons with spin +1/2, considering all the possible subshells (4s, 4p, 4d, and 4f).

Solution approach

Correct Answer + Core Reason: 16. For n=4, there are 16 electrons with spin +1/2.

Step-by-Step Solution:

Step 1: Identify the possible subshells for n=4. These are 4s, 4p, 4d, and 4f.

Step 2: Determine the number of orbitals in each subshell:
- 4s has 1 orbital
- 4p has 3 orbitals
- 4d has 5 orbitals
- 4f has 7 orbitals

Step 3: Each orbital can hold two electrons, one with spin +1/2 and one with spin -1/2. Therefore, each orbital contributes one electron with S = +1/2.

Step 4: Calculate the total number of electrons with S = +1/2:
- 4s: 1 electron
- 4p: 3 electrons
- 4d: 5 electrons
- 4f: 7 electrons\nTotal = 1 + 3 + 5 + 7 = 16 electrons

Why other options are wrong:
- Option B (15): Incorrect. This is likely a miscalculation of the number of orbitals or electrons.
- Option C (14): Incorrect. This is likely a miscalculation of the number of orbitals or electrons.
- Option D (17): Incorrect. This is likely a miscalculation of the number of orbitals or electrons.

Key Concepts:
- Quantum numbers
- Electronic configuration
- Spin quantum number

Memory Aids:
- s = 1 orbital, p = 3 orbitals, d = 5 orbitals, f = 7 orbitals

Quick Tips:
- Remember the number of orbitals in each subshell.
- Each orbital has one electron with spin +1/2.

24 Easy

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Remembering +4 | -1

Which of the following phenomenon is responsible for the television pictures?

View answer and explanation

Correct option: B

Fluorescence

Option B (Fluorescence) is correct because television pictures are produced due to the phenomenon of fluorescence, where substances emit light upon excitation by electrons.

Solution approach

Correct Answer + Core Reason: Fluorescence. Television pictures are based on fluorescence.

Step-by-Step Solution:

Step 1: Recall the definitions of the given phenomena.

Step 2: Chemiluminescence: Emission of light as a result of a chemical reaction.

Step 3: Fluorescence: Emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a type of luminescence.

Step 4: Luminescence: Emission of light by a substance not resulting from heat.

Step 5: Phosphorescence: Similar to fluorescence, but the emission of light continues for some time after excitation.

Step 6: In television screens, phosphors emit light when struck by electrons, which is a form of fluorescence.

Why other options are wrong:
- Option A: Chemiluminescence involves chemical reactions, not electron bombardment.
- Option C: Luminescence is a general term, but fluorescence is the specific phenomenon.
- Option D: Phosphorescence would cause the image to linger longer than it does on a TV screen.

Key Concepts:
- Fluorescence
- Luminescence
- Phosphorescence
- Chemiluminescence

Memory Aids:
- Fluorescence: immediate light emission

Quick Tips:
- Understand the differences between different types of luminescence.

25 Easy

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Understanding +4 | -1

Assertion: The maximum number of eletrons in p & d subshells are 6 and 10 respectively.

Reason: The maximum number of electrons in the shell with prinicpal quantum number

n

is equal to

2 n^2

.

View answer and explanation

Correct option: B

If both A and R are true but R is not the correct explanation of A

Option B (If both A and R are true but R is not the correct explanation of A) is correct because both the assertion and reason are true statements, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: If both A and R are true but R is not the correct explanation of A. The assertion and reason are both true, but the reason does not explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the assertion: The maximum number of electrons in p and d subshells are 6 and 10 respectively.

Step 2: The p subshell has three orbitals, each holding 2 electrons, so it can hold a maximum of 6 electrons. The d subshell has five orbitals, each holding 2 electrons, so it can hold a maximum of 10 electrons. Thus, the assertion is true.

Step 3: Analyze the reason: The maximum number of electrons in the shell with principal quantum number $n$ is equal to $2n^2$ .

Step 4: This is a correct statement about the maximum number of electrons in a shell.

Step 5: The reason does not explain why p and d subshells have 6 and 10 electrons respectively. The number of electrons in a subshell is determined by the number of orbitals in that subshell, which is related to the azimuthal quantum number ( $l$ ).

Why other options are wrong:
- Option A: The reason is not the correct explanation of the assertion.
- Option C: The reason is true.
- Option D: The assertion is true.

Key Concepts:
- Electronic Configuration
- Quantum Numbers
- Subshells and Orbitals

Memory Aids:
- s: 2 electrons, p: 6 electrons, d: 10 electrons, f: 14 electrons.

Quick Tips:
- Understand the relationship between quantum numbers and the number of electrons in shells and subshells.

26 Easy

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Remembering +4 | -1

Spin angular momentum of electron is given by

View answer and explanation

Correct option: A

\frac{{\sqrt 3 }}{4}\frac{h}{\pi }

Option A (

\frac{{\sqrt 3 }}{4}\frac{h}{\pi }

) is correct because the spin angular momentum is calculated using the formula

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

, where s = 1/2 for an electron.

Solution approach

Correct Answer + Core Reason: $\frac{{\sqrt 3 }}{4}\frac{h}{\pi }$ . This is the correct formula for spin angular momentum of an electron.

Step-by-Step Solution:
Step 1: Recall the formula for spin angular momentum: $= \frac{h}{{2\pi }}\sqrt {s\left( {s + 1} \right)}$
Step 2: For an electron, the spin quantum number s = 1/2.
Step 3: Substitute s = 1/2 into the formula: $\frac{h}{{2\pi }}\sqrt {\frac{1}{2}\left( {\frac{1}{2} + 1} \right)}$
Step 4: Simplify the expression: $\frac{h}{{2\pi }}\sqrt {\frac{1}{2}\left( {\frac{3}{2}} \right)} = \frac{h}{{2\pi }}\sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{4}\frac{h}{\pi }$

Why other options are wrong:
- Option B: $\sqrt {\frac{3}{4}} \frac{h}{\pi }$ is incorrect because it misses the factor of 1/2 in the denominator.
- Option C: $\sqrt {\frac{{3h}}{{4\pi }}}$ is incorrect because it doesn't have the correct form of the equation.
- Option D: $\sqrt {\frac{{4h}}{{3\pi }}}$ is incorrect because it doesn't have the correct form of the equation.

Key Concepts:
- Spin angular momentum
- Spin quantum number

Memory Aids:
- Remember the formula for spin angular momentum.

Quick Tips:
- Know the formula for spin angular momentum and the value of s for an electron.

27 Easy

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Remembering +4 | -1

In a given atom no two electrons can have the same values for all the four quantum numbers. This is called

View answer and explanation

Correct option: D

Pauli's exclusion principle

Option D (Pauli's exclusion principle) is correct because it states that no two electrons in an atom can have the same set of all four quantum numbers.

Solution approach

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

Step-by-Step Solution:
Step 1: Pauli's exclusion principle is a fundamental principle in quantum mechanics.
Step 2: It states that no two electrons in an atom can have the same values for all four quantum numbers (n, l, ml, and ms).
Step 3: This principle explains the electronic structure of atoms and the periodic table.

Why other options are wrong:
- Option A: Hund's Rule: Hund's rule states that electrons will individually occupy each orbital within a subshell before doubling up in any one orbital.
- Option B: Aufbau principle: The Aufbau principle states that electrons first fill the lowest energy levels available.
- Option C: Uncertainty principle: The Heisenberg uncertainty principle states that it is impossible to know both the position and momentum of a particle with perfect accuracy.

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

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

Quick Tips:
- Remember the statements of the key principles in quantum mechanics.

28 Easy

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Remembering +4 | -1

Rutherford’s alpha ray scattering experiment showed for the first time that the atom has

View answer and explanation

Correct option: A

Nucleus

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

Solution approach

Correct Answer + Core Reason: Nucleus. Rutherford's experiment showed the existence of the nucleus.

Step-by-Step Solution:

Step 1: Describe Rutherford's Experiment.\nRutherford's alpha-ray scattering experiment involved bombarding a thin gold foil with alpha particles.

Step 2: Observations.\nMost alpha particles passed straight through the foil.\nA few alpha particles were deflected at small angles.\nA very few alpha particles were deflected at large angles or bounced back.

Step 3: Conclusions.\nThe atom has a small, dense, positively charged nucleus.\nMost of the atom is empty space.

Step 4: Final Answer\nNucleus

Why other options are wrong:
- Option B (Proton): While the nucleus contains protons, the experiment primarily demonstrated the existence of the nucleus itself.
- Option C (Electron): The experiment did not directly show the existence of electrons.
- Option D (Neutron): Neutrons were discovered later by Chadwick.

Key Concepts:
- Rutherford's alpha-ray scattering experiment
- Atomic structure
- Nucleus

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

Quick Tips:
- Know the key experiments that led to the understanding of atomic structure.

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Understanding +4 | -1

Light of wavelength

(\lambda)

strikes a metal surface with intensity

X

and the metal emits

Y

electrons per second of maximum kinetic energy

Z

. What will happen to

Y

and

Z

if

X

is halved?

View answer and explanation

Correct option: C

Y

will be halved and

Z

will remain the same

Option C (

Y

will be halved and

Z

will remain the same) is correct because intensity is directly proportional to the number of electrons emitted (

Y

), while the maximum kinetic energy (

Z

) depends on the wavelength (energy) of the light, not its intensity.

Solution approach

Correct Answer + Core Reason: $Y$ will be halved and $Z$ will remain the same. Intensity affects the number of electrons, wavelength affects kinetic energy.

Step-by-Step Solution:

Step 1: Understand the photoelectric effect.

Step 2: Recall the relationship between intensity, number of electrons, wavelength, and kinetic energy:
- Intensity (X) is directly proportional to the number of photoelectrons emitted per second (Y).
- Wavelength ( $\lambda$ ) is related to the maximum kinetic energy (Z) of the emitted electrons by the equation: $KE_{max} = hf - \phi = \frac{hc}{\lambda} - \phi$ , where $\phi$ is the work function of the metal.

Step 3: If the intensity (X) is halved, the number of photoelectrons emitted per second (Y) will also be halved.

Step 4: Since the wavelength ( $\lambda$ ) remains the same, the maximum kinetic energy (Z) of the emitted electrons will remain the same.

Why other options are wrong:
- Option A: Incorrect because the kinetic energy (Z) depends on the wavelength, not the intensity.
- Option B: Incorrect because the number of electrons (Y) is directly proportional to the intensity, and the kinetic energy (Z) depends on the wavelength.
- Option D: Incorrect because the number of electrons (Y) is directly proportional to the intensity.

Key Concepts:
- Photoelectric effect
- Intensity and number of photoelectrons
- Wavelength and kinetic energy

Memory Aids:
- Intensity = Number of electrons
- Wavelength = Kinetic energy

Quick Tips:
- Remember that intensity affects the number of electrons, while wavelength affects the kinetic energy of the electrons in the photoelectric effect.

30 Easy

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Understanding +4 | -1

Which of the following sets of quantum numbers is not possible?

View answer and explanation

Correct option: D

n=4, l=1, m=+2, s=-\frac{1}{2}

Option D (

n=4, l=1, m=+2, s=-\frac{1}{2}

) is correct because the value of the magnetic quantum number (

m

) must be within the range of

-l

to

+l

, and when

l=1

,

m

can only be -1, 0, or +1.

Solution approach

Correct Answer + Core Reason: $n=4, l=1, m=+2, s=-\frac{1}{2}$ is not possible because the value of $m$ must be within the range of $-l$ to $+l$ .

Step-by-Step Solution:

Step 1: Understand the quantum numbers.
- $n$ (principal quantum number): describes the energy level (n = 1, 2, 3,...).
- $l$ (azimuthal quantum number): describes the shape of the orbital (l = 0, 1, 2,..., n-1).
- $m$ (magnetic quantum number): describes the orientation of the orbital in space (m = -l, -l+1,..., 0,..., l-1, l).
- $s$ (spin quantum number): describes the spin of the electron (s = +1/2 or -1/2).

Step 2: Check the validity of each set of quantum numbers.

Step 3: Analyze option D: $n=4, l=1, m=+2, s=-\frac{1}{2}$ .
- $n=4$ is valid.
- $l=1$ is valid (since $l < n$ ).
- $m=+2$ is invalid because when $l=1$ , $m$ can only be -1, 0, or +1.
- $s=-\frac{1}{2}$ is valid.

Step 4: Conclude that option D is not possible.

Why other options are wrong:
- Option A: $n=4, l=1, m=0, s=+\frac{1}{2}$ is valid.
- Option B: $n=4, l=3, m=-3, s=-\frac{1}{2}$ is valid.
- Option C: $n=4, l=0, m=0, s=-\frac{1}{2}$ is valid.

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

Memory Aids:
- Remember the allowed ranges for each quantum number.

Quick Tips:
- Always check if the values of $l$ and $m$ are within the allowed ranges for a given $n$ .

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