STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 7

181 Medium

Application +4 | -1

The energy ∆E corresponding to intense yellow line of sodium of 

\lambda ,\;589\;\mathrm{nm}

is 

View answer and explanation

Correct option: A

2.10 eV

Option A (2.10 eV) is correct because it represents the energy calculated using the formula

E = hc/\lambda

and converting Joules to electron volts.

Solution approach

Correct Answer + Core Reason: 2.10 eV. The energy is calculated using $E = hc/\lambda$ and converting to eV.

Step-by-Step Solution:

Step 1: Convert wavelength to meters: $\lambda = 589 \,\text{nm} = 589 \times 10^{-9} \,\text{m}$

Step 2: Use the formula for energy: $E = hc/\lambda$ , where $h$ is Planck's constant ( $6.626 \times 10^{-34} \,\text{J s}$ ), and $c$ is the speed of light ( $3 \times 10^8 \,\text{m/s}$ ).

Step 3: Substitute the values: $E = (6.626 \times 10^{-34} \,\text{J s}) \times (3 \times 10^8 \,\text{m/s}) / (589 \times 10^{-9} \,\text{m})$

Step 4: Calculate the energy in Joules: $E \approx 3.37 \times 10^{-19} \,\text{J}$

Step 5: Convert Joules to electron volts: $E \,\text{(eV)} = E \,\text{(J)} / (1.602 \times 10^{-19} \,\text{J/eV})$

Step 6: $E \approx (3.37 \times 10^{-19} \,\text{J}) / (1.602 \times 10^{-19} \,\text{J/eV}) \approx 2.10 \,\text{eV}$

Step 7: Final Answer: 2.10 eV

Why other options are wrong:
- Option B: 43.37 eV: This is incorrect due to errors in calculation or incorrect units.
- Option C: 47.12 eV: This is incorrect due to errors in calculation or incorrect units.
- Option D: 2.11 kcal: This is incorrect because it's in the wrong units and has a calculation error.

Key Concepts:
- Planck's equation: $E = h\nu = hc/\lambda$
- Conversion between Joules and electron volts

Memory Aids:
- Remember the values of h and c.
- Remember the conversion factor between Joules and eV.

Quick Tips:
- Always convert the wavelength to meters before calculation.
- Pay attention to the units and convert them appropriately.

182 Medium

182 of 193Question permalink

Remembering +4 | -1

Millikan’s oil drop experiment is used to find:

View answer and explanation

Correct option: B

Electronic charge

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

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the purpose of Millikan's oil drop experiment.

Step 2: Millikan's oil drop experiment was designed to measure the elementary electric charge (e) by balancing the gravitational force on charged oil droplets with an electric force.

Why other options are wrong:
- Option A: e/m ratio of electron was determined by Thomson's experiment.
- Option C: Mass of an electron can be calculated using the e/m ratio and the electronic charge.
- Option D: Velocity of an electron is not directly measured in Millikan's experiment.

Key Concepts:
- Millikan's oil drop experiment
- Electronic charge

Memory Aids:
- Remember that Millikan's experiment measured the charge of the electron.

Quick Tips:
- Recall the purpose and results of Millikan's oil drop experiment.

183 Medium

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Application +4 | -1

Which one of the following groupings represents a collection of isoelectronic species? (At. no. Cs=55, Br=35)

View answer and explanation

Correct option: B

N ^{3-}, F ^{-}, Na ^{+}

Option B (

N ^{3-}, F ^{-}, Na ^{+}

) is correct because each of these ions has 10 electrons: N3- (7+3=10), F- (9+1=10), and Na+ (11-1=10).

Solution approach

Correct Answer + Core Reason: $N ^{3-}, F ^{-}, Na ^{+}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species:
- A. $Na ^{+}$ : 11 - 1 = 10 electrons, $Ca ^{2+}$ : 20 - 2 = 18 electrons, $Mg ^{2+}$ : 12 - 2 = 10 electrons. Not isoelectronic.
- B. $N ^{3-}$ : 7 + 3 = 10 electrons, $F ^{-}$ : 9 + 1 = 10 electrons, $Na ^{+}$ : 11 - 1 = 10 electrons. Isoelectronic.
- C. $Be$ : 4 electrons, $Al ^{3+}$ : 13 - 3 = 10 electrons, $Cl ^{-}$ : 17 + 1 = 18 electrons. Not isoelectronic.
- D. $Ca ^{2+}$ : 20 - 2 = 18 electrons, $Cs ^{+}$ : 55 - 1 = 54 electrons, $Br$ : 35 electrons. Not isoelectronic.

Step 2: Identify the grouping where all species have the same number of electrons.

Why other options are wrong:
- Option A: $Na ^{+}, Ca ^{2+}, Mg ^{2+}$ - Incorrect, they have different numbers of electrons (10, 18, 10).
- Option C: $Be , Al ^{3+}, Cl ^{-}$ - Incorrect, they have different numbers of electrons (4, 10, 18).
- Option D: $Ca ^{2+}, Cs ^{+}, Br$ - Incorrect, they have different numbers of electrons (18, 54, 35).

Key Concepts:
- Isoelectronic species
- Number of electrons in ions

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

Quick Tips:
- Calculate the number of electrons for each ion carefully.

184 Medium

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Analyzing +4 | -1

Which of the following options is incorrect regarding Bohr’s model of an atom

View answer and explanation

Correct option: C

Ratio of magnitude of KE to that of PE of first Bohr orbit in ‘H’ will be in the ratio 1:2

Option D is incorrect because the speed of an electron in the nth shell decreases as n increases, so the order should be reversed.

Solution approach

Correct Answer + Core Reason: Ratio of magnitude of KE to that of PE of first Bohr orbit in ‘H’ will be in the ratio 1:2. This statement is correct, hence not incorrect.

Step-by-Step Solution:

Step 1: Analyze option A: Ionization energy increases with increasing nuclear charge for isoelectronic species.

Step 2: Analyze option B: Angular momentum is quantized as $mvr = n\frac{h}{{2\pi }}$ , so it increases with n.

Step 3: Analyze option C: KE = $\frac{1}{2}mv^2$ and PE = $-\frac{{kZe^2}}{r}$ . For Bohr orbit, KE = - $\frac{1}{2}$ PE. Thus, |KE|:|PE| = 1:2.

Step 4: Analyze option D: Speed of electron in nth orbit is given by $v \propto \frac{Z}{n}$ . Therefore, as n increases, v decreases.

Step 5: Conclude that option D is incorrect because the speed decreases with increasing n.

Why other options are wrong:
- Option A: The I.E. order is correct because as the nuclear charge increases for isoelectronic species, the ionization energy increases.
- Option B: The angular momentum order is correct because angular momentum is directly proportional to the shell number n.
- Option C: The ratio of KE to PE is correct as KE = -1/2 PE in Bohr's model.
- Option D: This option is incorrect, and hence the correct answer to the question.

Key Concepts:
- Bohr's model postulates
- Ionization energy trends
- Angular momentum quantization
- Kinetic and potential energy relationships in Bohr's model

Memory Aids:
- Remember that velocity decreases with increasing shell number: v ∝ 1/n
- KE = -1/2 PE in Bohr's model

Quick Tips:
- Always remember the relationships between energy, velocity, and shell number in Bohr's model.

185 Medium

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Analyzing +4 | -1

Statement - I: For Hydrogen spectrum all Balmer lines in the emission spectrum have longer
wavelengths than that of any line of Lyman series.

Statement - II: Transition energy is always greater when electrons come to an orbit with
lower orbit number than an electron come to an orbit with higher orbit number, irrespective to
higher orbit from which electron is coming.

View answer and explanation

Correct option: A

If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I

Option A (If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I) is correct because the transition energy determines the wavelength, and transitions to lower orbits have higher energy and thus shorter wavelengths.

Solution approach

Correct Answer + Core Reason: If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I. Balmer series involves transitions to n=2, while Lyman series involves transitions to n=1. Transitions to n=1 have higher energy and shorter wavelengths.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Balmer series involves transitions to n=2, and Lyman series involves transitions to n=1. Since transitions to n=1 release more energy, Lyman series has shorter wavelengths than Balmer series. Therefore, Statement-I is true.

Step 2: Analyze Statement-II: Transition energy is greater when electrons come to an orbit with a lower orbit number. This is because the energy difference between orbits increases as you get closer to the nucleus. Therefore, Statement-II is true.

Step 3: Determine if Statement-II explains Statement-I: Yes, Statement-II explains why Lyman series has shorter wavelengths than Balmer series. The higher transition energy in the Lyman series (transitions to n=1) results in shorter wavelengths.

Why other options are wrong:
- Option B: If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I: This is incorrect because Statement-II does explain Statement-I.
- Option C: If Statement-I is true but Statement-II is false: This is incorrect because both statements are true.
- Option D: If Statement-I is false but Statement-II is true: This is incorrect because Statement-I is also true.

Key Concepts:
- Hydrogen spectrum
- Lyman series
- Balmer series
- Transition energy
- Wavelength

Memory Aids:
- Remember that higher energy transitions result in shorter wavelengths (E = hc/λ).
- Lyman series involves transitions to the ground state (n=1).
- Balmer series involves transitions to n=2.

Quick Tips:
- Understand the relationship between energy levels, transitions, and wavelengths in the hydrogen spectrum.

186 Medium

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Remembering +4 | -1

\alpha

-particles can be detected by using

View answer and explanation

Correct option: C

Zinc sulphide screen

Option C (Zinc sulphide screen) is correct because alpha particles cause scintillations (small flashes of light) when they strike a zinc sulphide screen, making them detectable.

Solution approach

Correct Answer + Core Reason: Zinc sulphide screen. Alpha particles cause scintillations on a zinc sulphide screen.

Step-by-Step Solution:

Step 1: Recall the properties of alpha particles and methods of detection.

Step 2: Alpha particles are positively charged and can cause ionization and excitation of atoms.

Step 3: Zinc sulphide is a material that emits light (scintillation) when struck by ionizing radiation such as alpha particles.

Why other options are wrong:
- Option A (Thin aluminium sheet): Aluminium can be used to stop alpha particles, but not for detection.
- Option B (Barium sulphate): Barium sulphate is used in medical imaging, not for detecting alpha particles.
- Option D (Gold foil): Gold foil was used in Rutherford's experiment to scatter alpha particles, but not for direct detection.

Key Concepts:
- Alpha particles
- Scintillation
- Zinc sulphide

Memory Aids:
- Remember that zinc sulphide screens are used to detect alpha particles due to the scintillation effect.

Quick Tips:
- Know the common methods for detecting different types of radiation.

187 Medium

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Understanding +4 | -1

The statements, which is/are correct:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all three statements regarding the number of nodes (total, radial, and angular) in an orbital are accurate.

Solution approach

Correct Answer + Core Reason: All of the above. All the statements about nodes are correct.

Step-by-Step Solution:

Step 1: Define total nodes.

Step 2: Total number of nodes = n - 1

Step 3: Define radial nodes.

Step 4: Number of radial nodes = n - l - 1

Step 5: Define angular nodes.

Step 6: Number of angular nodes = l

Step 7: Verify the correctness of each statement.

Why other options are wrong:
- Option A: Number of total nodes in an orbital $= n - 1$ . This is a correct statement, but not the complete answer.
- Option B: Number of radial nodes in an orbital $= n - l - 1$ . This is a correct statement, but not the complete answer.
- Option C: Number of angular nodes in an orbital $= l$ . This is a correct statement, but not the complete answer.

Key Concepts:
- Nodes
- Radial Nodes
- Angular Nodes
- Quantum Numbers

Memory Aids:
- Total nodes = n-1
- Angular nodes = l
- Radial nodes = n-l-1

Quick Tips:
- Remember the formulas for calculating the number of different types of nodes.

188 Medium

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Application +4 | -1

The frequency of first line of Balmer series in hydrogen atom is

{v_0}

. The frequency of corresponding line emitted by singly ionised helium atom is :

View answer and explanation

Correct option: B

4{v_0}

The correct answer is B (

4{v_0}

) because the frequency of the spectral line is proportional to the square of the atomic number (Z²). For He+, Z=2, so the frequency is 4 times that of hydrogen.

Solution approach

Correct Answer + Core Reason: $4{v_0}$ . The frequency is proportional to $Z^2$ , and for He+, Z=2.

Step-by-Step Solution:

Step 1: Recall the Rydberg formula for the frequency of a spectral line:
$\nu = RZ^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})$ \nwhere:
$\nu$ is the frequency\nR is the Rydberg constant\nZ is the atomic number\nn1 and n2 are the principal quantum numbers of the initial and final energy levels.

Step 2: For the first line of the Balmer series, n1 = 2 and n2 = 3.

Step 3: For Hydrogen (H), Z = 1. Therefore, the frequency ${v_0}$ is:
${v_0} = R(1)^2(\frac{1}{2^2} - \frac{1}{3^2}) = R(\frac{1}{4} - \frac{1}{9}) = R(\frac{5}{36})$

Step 4: For singly ionized Helium (He+), Z = 2. Therefore, the frequency $\nu_{He+}$ is:
$\nu_{He+} = R(2)^2(\frac{1}{2^2} - \frac{1}{3^2}) = 4R(\frac{1}{4} - \frac{1}{9}) = 4R(\frac{5}{36})$

Step 5: Compare the frequencies:
$\nu_{He+} = 4R(\frac{5}{36}) = 4{v_0}$

Why other options are wrong:
- Option A: $2{v_0}$ : Incorrect. The frequency is proportional to Z², not Z.
- Option C: ${v_0}/2$ : Incorrect. The frequency increases with Z.
- Option D: ${v_0}/4$ : Incorrect. The frequency increases with Z.

Key Concepts:
- Rydberg formula
- Balmer series
- Atomic number

Memory Aids:
- Remember the Rydberg formula: $\nu = RZ^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})$

Quick Tips:
- The frequency of spectral lines is proportional to the square of the atomic number.
- For the Balmer series, the electron transitions to n=2.

189 Medium

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Remembering +4 | -1

Number of waves in fourth orbit

View answer and explanation

Correct option: A

4

Option A (4) is correct because the number of waves in the nth orbit is equal to n. Therefore, in the fourth orbit, there are 4 waves.

Solution approach

Correct Answer + Core Reason: 4. The number of waves in the fourth orbit is 4.

Step-by-Step Solution:

Step 1: The question asks for the number of waves in the fourth orbit.

Step 2: Recall the relationship between the orbit number and the number of waves: The number of waves in the nth orbit is equal to n.

Step 3: Apply this relationship to the fourth orbit.

Step 4: For the fourth orbit, the number of waves is 4.

Why other options are wrong:
- Option B (5): This is incorrect. The number of waves is equal to the orbit number, not one more than the orbit number.
- Option C (0): This is incorrect. There are waves in the orbit.
- Option D (1): This is incorrect. The number of waves is equal to the orbit number.

Key Concepts:
- Bohr's model
- de Broglie's hypothesis
- Wave nature of electrons

Memory Aids:
- Remember that the number of waves is directly equal to the orbit number.

Quick Tips:
- The number of waves is always the same as the orbit number.

190 Medium

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Understanding +4 | -1

In photoelectric effect, at which frequency electron will be ejected with certain kinetic energy

\left(v_0=\right.

threshold frequency).

View answer and explanation

Correct option: A

v>v_0

Option A is correct. For electron ejection with kinetic energy, the frequency of incident light must be greater than the threshold frequency.

Solution approach

Correct Answer + Core Reason: $v>v_0$ . The frequency of incident light must exceed the threshold frequency for electron ejection with kinetic energy.

Step-by-Step Solution:

Step 1: Define threshold frequency ( $v_0$ ): The minimum frequency required to eject an electron.

Step 2: Understand the photoelectric equation: $E = h\nu = h\nu_0 + KE$ , where $E$ is the energy of the photon, $h\nu_0$ is the work function, and $KE$ is the kinetic energy of the ejected electron.

Step 3: Condition for electron ejection with kinetic energy: $h\nu > h\nu_0$ , which implies $v > v_0$ .

Step 4: Conclusion: The frequency of incident light ( $v$ ) must be greater than the threshold frequency ( $v_0$ ).

Why other options are wrong:
- Option B: $v_0>v$ : Incorrect because the frequency of incident light must be greater than the threshold frequency for electron ejection.
- Option C: $v_0 \geq v$ : Incorrect because the frequency of incident light must be strictly greater than the threshold frequency for electron ejection with kinetic energy.
- Option D: $v \geq v_0$ : Incorrect because the frequency of incident light must be strictly greater than the threshold frequency for electron ejection with kinetic energy.

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Photoelectric equation

Memory Aids:
- Frequency must be greater than threshold for electron ejection.

Quick Tips:
- Remember the relationship between frequency, threshold frequency, and kinetic energy in the photoelectric effect.

191 Medium

191 of 193Question permalink

Understanding +4 | -1

Which are in the ascending order of wavelength?

View answer and explanation

Correct option: B

Lyman limit, Balmer limit, Paschen limit in the hydrogen spectrum

Option B (Lyman limit, Balmer limit, Paschen limit in the hydrogen spectrum) is correct because the wavelengths of the series limits increase in the order Lyman < Balmer < Paschen.

Solution approach

Correct Answer + Core Reason: Lyman limit, Balmer limit, Paschen limit in the hydrogen spectrum. Wavelengths of series limits increase in this order.

Step-by-Step Solution:

Step 1: The question asks for the ascending order of wavelength among the given options.

Step 2: Consider the wavelengths of the series limits in the hydrogen spectrum:
- Lyman series limit: shortest wavelength (UV region)
- Balmer series limit: intermediate wavelength (Visible region)
- Paschen series limit: longer wavelength (Infrared region)

Step 3: Therefore, the ascending order of wavelength is Lyman limit < Balmer limit < Paschen limit.

Why other options are wrong:
- Option A: ${H_\alpha },{H_\beta },{H_\gamma } \ldots$ lines in Balmer series of hydrogen atom is incorrect because the wavelengths decrease in this order.
- Option C: Blue, violet, yellow, red colours in solar spectrum is incorrect because the ascending order of wavelength is violet, blue, yellow, red.
- Option D: None of the above is incorrect because option B is correct.

Key Concepts:
- Hydrogen spectrum
- Lyman series
- Balmer series
- Paschen series
- Wavelength

Memory Aids:
- Lyman < Balmer < Paschen (wavelength)

Quick Tips:
- Remember the relative wavelengths of the different series in the hydrogen spectrum.

192 Medium

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Understanding +4 | -1

Which element possess non-spherical shells?

View answer and explanation

Correct option: B

B

Option B (B) is correct because Boron (B) has the electronic configuration

1s^22s^22p^1

, and the presence of electrons in the p-orbital results in non-spherical shells.

Solution approach

Correct Answer + Core Reason: B. Boron has electrons in p-orbitals, which are non-spherical.

Step-by-Step Solution:

Step 1: Write the electronic configurations of each element.

Step 2: He: $1s^2$ \nB: $1s^2 2s^2 2p^1$ \nBe: $1s^2 2s^2$ \nLi: $1s^2 2s^1$

Step 3: Identify the element with electrons in p-orbitals.

Step 4: Boron has one electron in the 2p orbital, which is non-spherical.

Why other options are wrong:
- Option A: He. Helium only has electrons in the s orbital, which is spherical.
- Option C: Be. Beryllium only has electrons in the s orbital, which is spherical.
- Option D: Li. Lithium only has electrons in the s orbital, which is spherical.

Key Concepts:
- Electronic Configuration
- Atomic Orbitals
- Shape of Orbitals

Memory Aids:
- s orbitals are spherical, p orbitals are dumbbell-shaped.

Quick Tips:
- The presence of p, d, or f orbitals leads to non-spherical shapes.

193 Medium

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Understanding +4 | -1

Number of electrons in

- \mathrm{CON}{\mathrm{H}_2}

are

View answer and explanation

Correct option: A

24

Option A (24) is correct because -CONH₂ has 1 carbon (6 electrons), 1 oxygen (8 electrons), 1 nitrogen (7 electrons), and 2 hydrogens (2 electrons), totaling 6 + 8 + 7 + 2 + 1 (for the negative charge) = 24 electrons.

Solution approach

Correct Answer + Core Reason: 24. Count the electrons from each atom in the molecule.

Step-by-Step Solution:

Step 1: Identify the atoms in the molecule $-\mathrm{CON}{\mathrm{H}_2}$ .
- Carbon (C): 1 atom
- Oxygen (O): 1 atom
- Nitrogen (N): 1 atom
- Hydrogen (H): 2 atoms

Step 2: Determine the number of electrons in each atom.
- C: 6 electrons
- O: 8 electrons
- N: 7 electrons
- H: 1 electron

Step 3: Calculate the total number of electrons in the molecule.
- Total electrons = (1 × 6) + (1 × 8) + (1 × 7) + (2 × 1) + 1 (for the negative charge) = 6 + 8 + 7 + 2 + 1 = 24 electrons.

Why other options are wrong:
- Option B: 20 is incorrect; the correct calculation yields 24.
- Option C: 22 is incorrect; the correct calculation yields 24.
- Option D: 18 is incorrect; the correct calculation yields 24.

Key Concepts:
- Counting electrons in molecules
- Atomic structure

Memory Aids:
- Remember the number of electrons in common atoms (C, H, N, O).

Quick Tips:
- Carefully count the number of atoms of each element and their corresponding electrons.

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