31 Year NEET Previous Questions: Structure of Atom - 1 Free MCQ Practice

• For He⁺ (Z = 2) and Li²⁺ (Z = 3), we calculate the energy and radius for the first orbit (n = 1).
• Energy for He⁺:
  • E_n(He⁺) = - (Z² × R_H) / n²
  • = - (2² × 2.18 × 10^-18) / 1²
  • = - 8.72 × 10^-18 J
• Radius for He⁺:
  • r_n(He⁺) = (n² × a₀) / Z
  • = (1² × 52.9) / 2
  • = 26.4 pm
• Energy for Li²⁺:
  • E_n(Li²⁺) = - (Z² × R_H) / n²
  • = - (3² × 2.18 × 10^-18) / 1²
  • = - 19.62 × 10^-18 J
• Radius for Li²⁺:
  • r_n(Li²⁺) = (n² × a₀) / Z
  • = (1² × 52.9) / 3
  • = 17.6 pm

Conclusion
E_n(Li²⁺) = -19.62 × 10^-18 J and r_n(Li²⁺) = 17.6 pm & E_n(He⁺) = -8.72 × 10^-18 J and r_n(He⁺) = 26.4 pm

For the transition n = 2 → n = 3:

• Energy difference:
  E₁ = -13.6 × (1/3² - 1/2²)
  E₁ = -13.6 × (1/9 - 1/4) = -13.6 × (-5/36) = 13.6 × 5/36
• Corresponding wavelength:
  λ₁ = hc/E₁
  λ₁ is inversely proportional to E₁.

For the transition n = 4 → n = 6:

• Energy difference:
  E₂ = -13.6 × (1/6² - 1/4²)
  E₂ = -13.6 × (1/36 - 1/16) = -13.6 × (-5/144) = 13.6 × 5/144
• Corresponding wavelength:
  λ₂ = hc/E₂
  λ₂ is inversely proportional to E₂.

Ratio of wavelengths:

• λ₁/λ₂ = E₂/E₁
• E₁ = 13.6 × 5/36, E₂ = 13.6 × 5/144
• λ₁/λ₂ = (5/144) / (5/36) = 36/144 = 1/4

Therefore, the ratio of the wavelengths is 1/4.

The energy levels of an electron in a hydrogen-like ion (an atom or ion with only one electron) can be quantified using the formula:

where:
E_n is the energy of the electron in the nth energy level,
Z is the atomic number (number of protons) of the ion,
13.6 eV is the ionization energy of hydrogen,
n is the principal quantum number (the energy level).
Since we need to compare this across different ions in different energy states, let's plug in some numbers:
For He⁺ (Helium ion):
Z = 2 (as helium has 2 protons)
n = 1 for ground state

However, the problem gives the energy in joules, and it's given a constant x.
So,  x = 54.4 eV (converted to joules as needed).
Next, for the
Be³⁺ ion:
Z = 4 (as beryllium has 4 protons)
n = 2

Since initially x = 54.4 eV (or its equivalent in joules) for at He⁺ at n = 1, and now we have the same energy for  at Be³⁺ at n = 2, the energy level in joules would also equate to x, just at a different energy state and ion.
So, the answer is:
Option A: −x.

Statement I: The Balmer spectral line for H atom with the lowest energy appears at 5/36 R_H cm^-1 (R_H = Rydberg constant).

• The Balmer series corresponds to the transitions in the hydrogen atom where the electron falls from higher energy levels to the second energy level (n = 2).
• The wavelength of the Balmer spectral line for the lowest energy transition (from n = 3 to n = 2) is given by the Rydberg formula: 1/λ = RH * (1/2² - 1/3²) This simplifies to: 1/λ = R_H * (1/4 - 1/9) = 5/36 R_H cm^-1.

Thus, Statement I is correct.
Statement II: When the temperature of a black body increases, the maxima of the curve (intensity versus wavelength) shifts towards shorter wavelengths.

• According to Wien's displacement law, the wavelength at which the intensity of radiation is maximum for a black body is inversely proportional to the temperature. This means that as the temperature increases, the peak of the emission spectrum shifts toward shorter wavelengths.
• Thus, Statement II is correct.

Since both statements are correct, the correct answer is (c) Both Statement I and Statement II are correct.

Correct Answer: I > II > III > IV
Explanation: The energy of electrons depends on n+ℓn+ℓ. For the given electrons:

• I: n=4,ℓ=2→n+ℓ=6n=4,ℓ=2→n+ℓ=6
• II: n=3,ℓ=2→n+ℓ=5n=3,ℓ=2→n+ℓ=5
• III: n=4,ℓ=1→n+ℓ=5n=4,ℓ=1→n+ℓ=5
• IV: n=3,ℓ=1→n+ℓ=4n=3,ℓ=1→n+ℓ=4
• Order: I (highest n+ℓn+ℓ) > II (same n+ℓn+ℓ as III, but lower nn) > III > IV (lowest n+ℓn+ℓ).

Statement I is True: The energy levels in both the He⁺ ion and the H atom are related by the formula for the energy of the nth level of a hydrogen-like atom:

• Energy for an electron in level n is given by: E = -13.6 eV * Z² / n²
• For He⁺ (Z = 2) in the n = 2 state, the energy is: E = -13.6 eV * 2² / 2² = -3.4 eV
• For H (Z = 1) in the n = 1 state, the energy is: E = -13.6 eV
• These energies are indeed comparable when considering the difference in nuclear charge. Therefore, Statement I is correct.

Statement II is False: According to the Heisenberg Uncertainty Principle, it is not possible to simultaneously determine both the exact position and momentum of an electron. The more precisely one is measured, the less precisely the other can be determined. Hence, Statement II is incorrect.

To solve this, let's break it down step by step:

• The element's atomic number is either 26 (Fe) or 27 (Co).
• The problem states that the isotope has 19.23% more neutrons than protons.

Now, for any given isotope, the number of protons is the atomic number (Z), and the number of neutrons can be calculated as the mass number (A) minus the atomic number (Z).
So, the number of neutrons = A - Z.
Let’s denote the number of neutrons as N and the number of protons as P, where P = Z (atomic number), and N = A - Z (mass number - atomic number).
From the given information, the number of neutrons is 19.23% more than the number of protons. This means:
N = P + 0.1923P, which simplifies to:
A - Z = Z + 0.1923Z
Thus:
A = Z + 1.1923Z A = 2.1923Z
Now, let's calculate the mass number for both Fe and Co.

• For Fe (Z = 26): A = 2.1923 × 26 ≈ 57.0, so the mass number is approximately 57. This corresponds to the isotope ⁵⁷Fe.
• For Co (Z = 27): A = 2.1923 × 27 ≈ 59.2, so the mass number is approximately 59. This is closer to 60, which corresponds to the isotope 60Co.

But the correct answer, based on the information provided, is (b) ⁵⁷Fe because the percentage difference matches closer to this isotope.
Therefore, the correct answer is (b) ⁵⁷Fe.

To match List I (Quantum Numbers) with List II (Information provided), we need to understand what each quantum number represents:
1. Principal Quantum Number (n): This quantum number determines the size and energy level of the orbital. An increase in n implies a higher energy level and a larger orbital size.
2. Azimuthal Quantum Number (ℓ): (not directly listed but related to m_l because m_l depends on ℓ): This quantum number defines the shape of the orbital. Different values of ℓ correspond to different shapes (s, p, d, f, etc.).
3. Magnetic Quantum Number (m_l): This quantum number describes the orientation of the orbital in space. It can take values from −ℓ to ℓ, where each value corresponds to a specific orientation of the orbital.
4. Spin Quantum Number (m_s): This quantum number specifies the orientation of the spin of the electron. The two possible values are + 1/2 and −1/2, representing the two possible spin orientations (up or down).
Matching the given options:
A. m_l - should be matched with "Orientation of orbital" (III).
B. m_s - should be matched with "Orientation of spin of electron" (IV).
C. I - This is likely intended to be ℓ, and would be matched with "Shape of orbital" (I).
D. n - relates to "Size of orbital" (II), as it affects the distance from the nucleus along with the energy level.
Using the understanding above, we can identify the correct matches:
A-III (Orientation of orbital)
B-IV (Orientation of spin of electron)
C-I (Shape of orbital)
D-II (Size of orbital)
Therefore, the correct answer is:
Option B: A-III, B-IV, C-I, D-II

Quantum numbers:

1. n (Principal quantum number): Determines the energy level or shell. It can take positive integer values (1, 2, 3, ...).
2. l (Azimuthal quantum number): Determines the subshell (shape of the orbital). It can take values from 0 to n-1.
3. mₗ (Magnetic quantum number): Determines the orientation of the orbital. It can take values from -l to +l, including 0.
4. mₛ (Spin quantum number): Determines the spin of the electron. It can take values of +1/2 or -1/2.

Analyzing each option:
(a) n = 4, l = 3, mₗ = -3, -2, -1, 0, +1, +2, +3, mₛ = -1/2:

• n = 4, so l can be 0, 1, 2, or 3 (valid for l = 3).
• mₗ can range from -l to +l, so for l = 3, mₗ can be -3, -2, -1, 0, +1, +2, +3 (valid).
• mₛ = -1/2 is valid for electron spin.
• This set of quantum numbers is correct.

(b) n = 5, l = 2, mₗ = -2, -1, +1, +2, mₛ = +1/2:

• n = 5, so l can be 0, 1, 2, 3, or 4 (valid for l = 2).
• mₗ can range from -l to +l, so for l = 2, mₗ can be -2, -1, 0, +1, +2 (valid).
• mₛ = +1/2 is valid for electron spin.
• This set of quantum numbers is correct.

(c) n = 4, l = 2, mₗ = -2, -1, 0, +1, +2, mₛ = -1/2:

• n = 4, so l can be 0, 1, 2, or 3 (valid for l = 2).
• mₗ can range from -l to +l, so for l = 2, mₗ can be -2, -1, 0, +1, +2 (valid).
• mₛ = -1/2 is valid for electron spin.
• This set of quantum numbers is correct.

(d) n = 5, l = 3, mₗ = -3, -2, -1, 0, +1, +2, +3, mₛ = +1/2:

• n = 5, so l can be 0, 1, 2, 3, or 4 (valid for l = 3).
• mₗ can range from -l to +l, so for l = 3, mₗ can be -3, -2, -1, 0, +1, +2, +3 (valid).
• mₛ = +1/2 is valid for electron spin.
• This set of quantum numbers is correct.

Conclusion: Option (b) is the incorrect set of quantum numbers because it incorrectly specifies that mₗ can take values from -2 to +2 for l = 2, but it includes -2, -1, +1, +2. There is no zero value included for mₗ. Therefore, Option (b) is the incorrect one.
The correct answer is (b).

Statement I: The value of wave function, Ψ, depends upon the coordinates of the electron in the atom.
This is true. The wave function (Ψ) represents the state of an electron in an atom and its value depends on the position coordinates of the electron, i.e., the spatial coordinates (x, y, z) of the electron in space. The wave function provides information about the probability amplitude of finding an electron at a specific location.
Statement II: The probability of finding an electron at a point within an atom is proportional to the orbital wave function.
This is false. The probability of finding an electron at a point is proportional to the square of the wave function (Ψ²), not directly proportional to Ψ itself. The square of the wave function, |Ψ|², gives the probability density of finding an electron at a specific point in space.
Thus, the correct answer is (a) Statement I is True but Statement II is False.

It is statement based question.
Statements B, C & E are correct.
(B) Mass of the electron is  9.10939×10^−-31 kg
(C) All the isotopes of given elements show same chemical properties.
(E) Dalton's atomic theory, regarded the atom as an ultimate particle of matter.

Values of n_m(magnetic quantum number) for given azimuthal quantum number can be calculated as following

In an atom, all the five 3d orbitals are equal in energy in free state i.e., degenerate.
The shape of d_{x² − y²} is different then shape of dz²

 The size of orbital depends on principal quantum number 'n' therefore all the five 3d orbitals are different in size when compared to the respective 4d orbitals.
Shape of orbitals depends on azimuthal quantum number 'I' therefore shapes of 4d orbitals are similar to the respective 3d orbitals.

Correct option: D

Aufbau principle requires that electrons occupy orbitals in order of increasing energy (lower-energy subshells are filled first). The given filling order satisfies this requirement.

Hund's rule states that for degenerate orbitals electrons occupy them singly with parallel spins as far as possible before pairing. The distribution of electrons across the degenerate p orbitals follows this rule and then pairs when required.

Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers, so each orbital can hold at most two electrons with opposite spins. The electronic distribution does not violate this principle.

Because all three rules are satisfied, the configuration violates none of them; therefore option D is correct.

Energy of electromagnetic radiation (E)

Isoelectronic species have some number of electrons.

As per Aufbau principle (n + l rule), higher is the value of (n + l), greater will be the energy. When (n + l) values are same then which orbital have higher value of n will have more energy.
4d = 4 + 2 = 6
5p = 5 + 1 = 6
5f = 5 + 3 = 8
6p = 6 + 1 = 7
Hence, the correct order of decreasing energy will
be : 5f > 6p > 5p > 4d.

• Balmer series lies in visible region.
• Paschen series lies in infra red region.
• Brackett series lies in infra red region.
• Lyman series lies in ultraviolet region.

In degnerate orbital all unpaired electrons show same spin. So the correct configuration of N atom is

In the case of hydrogen type atoms, energy depends on the principal quantum number only. Therefore 2-s orbital will have energy equal to 2-p orbital.

For n = 3 and l = 1, the subshell is 3p and a particular 3p orbital can accodomodate only 2 electrons.

d_{x² - y²} and d_z² orbitals have electron density along the axes. 

For any two electrons occupying the same orbital values of n, l and ml are same but ms is different [ + 1/2 and - 1/2]

Angular momentum of electron in 'd‘ orbital =

Number of d-electrons in Fe²⁺ = 6
Number of p-electrons in Cl = 11 

Ti (22) ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²
∴ Order of increasing energy is 3s, 3p, 4s, 3d