STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 3

61 Easy

61 of 239

Understanding +4 | -1

Compared with an atom of atomic weight 12 and atom ic number 6 , the atom of atomic weight 13 and atom ic number 6

View answer and explanation

Correct option: A

Contains more neutrons

Option A (Contains more neutrons) is correct because isotopes of an element have the same number of protons but different numbers of neutrons.

Solution approach

Correct Answer + Core Reason: Contains more neutrons. The atom with atomic weight 13 has one more neutron than the atom with atomic weight 12.

Step-by-Step Solution:

Step 1: Define atomic weight and atomic number.\nAtomic weight (mass number) = number of protons + number of neutrons.\nAtomic number = number of protons.

Step 2: Calculate the number of neutrons in each atom.\nFor the atom with atomic weight 12 and atomic number 6: Number of neutrons = 12 - 6 = 6.\nFor the atom with atomic weight 13 and atomic number 6: Number of neutrons = 13 - 6 = 7.

Step 3: Compare the number of neutrons.\nThe atom with atomic weight 13 has 7 neutrons, while the atom with atomic weight 12 has 6 neutrons. Therefore, the atom with atomic weight 13 contains more neutrons.

Why other options are wrong:
- Option B: Contains more electrons. Both atoms have the same atomic number (6), which means they have the same number of protons and, if neutral, the same number of electrons.
- Option C: Contains more protons. Both atoms have the same atomic number (6), which means they have the same number of protons.
- Option D: Is a different element. Both atoms have the same atomic number (6), which defines the element as carbon.

Key Concepts:
- Atomic number
- Atomic weight (mass number)
- Isotopes
- Number of neutrons

Memory Aids:
- Atomic number = number of protons
- Mass number = protons + neutrons

Quick Tips:
- Remember that isotopes have the same atomic number but different mass numbers.

62 Easy

62 of 239

Understanding +4 | -1

In which of the orbit of

\mathrm{H}{\mathrm{e}^ + },

the angular momentum of the electron in

h/2\pi ?

View answer and explanation

Correct option: A

First orbit

Option A (First orbit) is correct because the angular momentum is given by

nh/2\pi

, and for the first orbit, n=1.

Solution approach

Correct Answer + Core Reason: First orbit. Angular momentum is $nh/2\pi$ , and for the first orbit, n=1.

Step-by-Step Solution:
Step 1: Recall the formula for angular momentum: Angular momentum ( $L$ ) = $n \frac{h}{{2\pi}}$ , where $n$ is the principal quantum number.
Step 2: Set the angular momentum equal to $h/2\pi$ : $n \frac{h}{{2\pi}} = \frac{h}{{2\pi}}$
Step 3: Solve for $n$ : $n = 1$
Step 4: Relate $n$ to the orbit: $n = 1$ corresponds to the first orbit.

Why other options are wrong:
- Option B: Second orbit has $n=2$ , so the angular momentum would be $2h/2\pi$ .
- Option C: Third orbit has $n=3$ , so the angular momentum would be $3h/2\pi$ .
- Option D: Fourth orbit has $n=4$ , so the angular momentum would be $4h/2\pi$ .

Key Concepts:
- Bohr's Model
- Angular Momentum
- Principal Quantum Number

Memory Aids:
- Remember the formula for angular momentum: $L = n \frac{h}{{2\pi}}$

Quick Tips:
- Use the formula for angular momentum to determine the orbit number.

63 Easy

63 of 239

Analyzing +4 | -1

Statement (S):

{ }_{24} \mathrm{Cr}

has more paramagnetic nature than

{ }_{25} \mathrm{Mn}

.

Explanation (E):

\mathrm{Cr}

has more number of unpaired electrons than

\mathrm{Mn}

.

View answer and explanation

Correct option: C

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

Option C (Both S and E are correct and E is correct explanation of S) is correct because Chromium (Cr) has more unpaired electrons than Manganese (Mn), leading to a greater paramagnetic nature.

Solution approach

Correct Answer + Core Reason: Both S and E are correct and E is correct explanation of S. ${ }_{24} \mathrm{Cr}$ has more unpaired electrons than ${ }_{25} \mathrm{Mn}$ , hence more paramagnetic.

Step-by-Step Solution:

Step 1: Write the electronic configurations of ${ }_{24} \mathrm{Cr}$ and ${ }_{25} \mathrm{Mn}$ .

Step 2: ${ }_{24} \mathrm{Cr} = [Ar] 3d^5 4s^1$ (due to half-filled d-orbital stability)

Step 3: ${ }_{25} \mathrm{Mn} = [Ar] 3d^5 4s^2$

Step 4: Determine the number of unpaired electrons in each.

Step 5: Cr has 6 unpaired electrons, and Mn has 5 unpaired electrons.

Step 6: Paramagnetic nature is directly proportional to the number of unpaired electrons.

Step 7: Since Cr has more unpaired electrons, it is more paramagnetic than Mn.

Step 8: Final Answer: Both S and E are correct and E is the correct explanation of S.

Why other options are wrong:
- Option A: The statement (S) is correct, so this option is incorrect.
- Option B: The statement (S) is correct, so this option is incorrect.
- Option D: The explanation (E) is the correct explanation for the statement (S), so this option is incorrect.

Key Concepts:
- Electronic configuration
- Hund's rule
- Paramagnetism

Memory Aids:
- Remember that half-filled and fully-filled orbitals contribute to stability.
- Paramagnetism increases with the number of unpaired electrons.

Quick Tips:
- Write the electronic configuration carefully, considering exceptions like Cr and Cu.

64 Easy

64 of 239

Understanding +4 | -1

The number of spherical nodes in 3p orbitals is

View answer and explanation

Correct option: B

1

Option B (1) is correct because the number of spherical nodes in a 3p orbital is calculated using the formula n - l - 1, where n=3 and l=1, resulting in 3-1-1=1.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the formula for calculating the number of spherical nodes: Number of spherical nodes = n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Step 2: Identify the values of n and l for a 3p orbital. For a 3p orbital, n = 3 and l = 1 (since p corresponds to l = 1).

Step 3: Substitute the values into the formula: Number of spherical nodes = 3 - 1 - 1 = 1.

Why other options are wrong:
- Option A: 0 is incorrect. The calculation yields 1 spherical node.
- Option C: 2 is incorrect. The calculation yields 1 spherical node.
- Option D: 3 is incorrect. The calculation yields 1 spherical node.

Key Concepts:
- Spherical nodes
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Atomic orbitals (s, p, d, f)

Memory Aids:
- Remember the formula: Spherical nodes = n - l - 1.
- s: l=0, p: l=1, d: l=2, f: l=3

Quick Tips:
- Always remember the formula for calculating spherical nodes. Make sure to correctly identify the values of n and l for the given orbital.

65 Easy

65 of 239

Understanding +4 | -1

\mathrm{Co}^{3+}, \mathrm{Mn}^{+}, \mathrm{Fe}^{2+}

and

\mathrm{Cr}

are isoelectronic (24 electrons) in each. Thus, EC are identical in

View answer and explanation

Correct option: A

\mathrm{Mn}^{+}, \mathrm{Cr}

Option A is correct because only Mn+ and Cr have the same electronic configuration after accounting for the charge and the exception in Cr's configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mn}^{+}, \mathrm{Cr}$ . Both have the same electronic configuration: [Ar] 4s1 3d5 or [Ar] 3d6.

Step-by-Step Solution:
Step 1: Write the electronic configurations of the neutral atoms: Co, Mn, Fe, and Cr.
Step 2: Write the electronic configurations of the ions: $\mathrm{Co}^{3+}$ , $\mathrm{Mn}^{+}$ , $\mathrm{Fe}^{2+}$ .
Step 3: Compare the electronic configurations to find the isoelectronic species (24 electrons each).
Step 4: Identify which of the isoelectronic species have identical electronic configurations.
Step 5: Co(Z=27) = [Ar] 4s2 3d7, Co3+ = [Ar] 3d6\nMn(Z=25) = [Ar] 4s2 3d5, Mn+ = [Ar] 4s1 3d5\nFe(Z=26) = [Ar] 4s2 3d6, Fe2+ = [Ar] 3d6\nCr(Z=24) = [Ar] 4s1 3d5\nMn+ and Cr have identical EC.

Why other options are wrong:
- Option B: $\mathrm{Co}^{3+}$ has a different electronic configuration ([Ar] 3d6) than $\mathrm{Mn}^{+}$ and $\mathrm{Cr}$ .
- Option C: $\mathrm{Fe}^{2+}$ has a different electronic configuration ([Ar] 3d6) than $\mathrm{Cr}$ .
- Option D: Not all of the listed combinations have identical electronic configurations.

Key Concepts:
- Electronic configuration
- Isoelectronic species
- Exceptions to Aufbau principle (Chromium)

Memory Aids:
- Remember the electronic configurations of common elements and ions.

Quick Tips:
- Always write out the electronic configurations to compare them accurately.

66 Easy

66 of 239

Remembering +4 | -1

The shape of

s

-orbital is

View answer and explanation

Correct option: B

Spherical

Option B (Spherical) is correct because the s-orbital is spherically symmetrical around the nucleus.

Solution approach

Correct Answer + Core Reason: Spherical. The s-orbital has a spherical shape.

Step-by-Step Solution:
Step 1: Recall the shapes of atomic orbitals.
Step 2: The s-orbital is spherically symmetrical.
Step 3: The p-orbital is dumbbell-shaped, and the d-orbital has more complex shapes.

Why other options are wrong:
- Option A: Pyramidal is not the shape of any standard atomic orbital.
- Option C: Tetrahedral is not the shape of any standard atomic orbital.
- Option D: Dumb-bell shaped is the shape of a p-orbital, not an s-orbital.

Key Concepts:
- Atomic orbitals
- Shapes of orbitals

Memory Aids:
- Remember that s-orbitals are spherical.

Quick Tips:
- Know the shapes of s, p, and d orbitals.

67 Easy

67 of 239

Remembering +4 | -1

The probability of finding the electron in the orbital is

View answer and explanation

Correct option: B

90-95%

Option B (90-95%) is correct because the probability of finding an electron within the defined space of an orbital is conventionally accepted to be within this range.

Solution approach

Correct Answer + Core Reason: 90-95%. The probability of finding an electron in an orbital is defined within this range.

Step-by-Step Solution:

Step 1: Recall the definition of an atomic orbital. An orbital represents a region of space around the nucleus where there is a high probability of finding an electron.

Step 2: Understand that the probability is not 100% because the electron's position cannot be determined with absolute certainty due to the Heisenberg Uncertainty Principle.

Step 3: Recognize that the 90-95% probability is a commonly accepted convention for defining the boundary of an orbital.

Why other options are wrong:
- Option A (100%): The electron's location cannot be determined with 100% certainty due to quantum mechanics.
- Option C (70-80%): This range is lower than the standard convention for defining an orbital.
- Option D (50-60%): This range is significantly lower than the standard convention for defining an orbital.

Key Concepts:
- Atomic Orbitals
- Probability Density
- Heisenberg Uncertainty Principle

Memory Aids:
- Think of orbitals as "fuzzy" regions where electrons are likely to be found, not as fixed paths.

Quick Tips:
- Remember the conventional probability range (90-95%) for defining the spatial extent of an orbital.

68 Easy

68 of 239

Analyzing +4 | -1

Assertion: Nuclear binding energy per nucleon is in the order

{ }_4^9 {Be}>{ }_3^7 {Li}>{ }_2^4 {He}

.

Reason: Binding energy per nucleon increases linearly with difference in number of neutrons and protons.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

The correct answer is D (If both the Assertion and Reason are incorrect) because the binding energy per nucleon order is reversed, and the reason provided is not the correct explanation.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. The assertion is incorrect because the order of binding energy per nucleon is reversed, and the reason is incorrect because binding energy does not increase linearly with the difference in the number of neutrons and protons.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states that the nuclear binding energy per nucleon is in the order ${ }_4^9 {Be}>{ }_3^7 {Li}>{ }_2^4 {He}$ . This is incorrect.

Step 2: Correct the order of binding energy per nucleon.\nThe correct order is ${ }_2^4 {He}>{ }_3^7 {Li}>{ }_4^9 {Be}$ .

Step 3: Analyze the reason.\nThe reason states that binding energy per nucleon increases linearly with the difference in the number of neutrons and protons. This is incorrect.

Step 4: Understand the factors affecting binding energy.\nBinding energy depends on the mass defect and the number of nucleons. It's not a simple linear relationship with the neutron-proton difference.

Why other options are wrong:
- Option A: Both Assertion and Reason are incorrect.
- Option B: Both Assertion and Reason are incorrect.
- Option C: Both Assertion and Reason are incorrect.

Key Concepts:
- Nuclear binding energy
- Binding energy per nucleon
- Mass defect

Memory Aids:
- Higher binding energy per nucleon means greater stability.

Quick Tips:
- Remember that ${ }_2^4 {He}$ has a relatively high binding energy per nucleon.

69 Easy

69 of 239

Remembering +4 | -1

Mass of neutron is … times the mass of electron

View answer and explanation

Correct option: A

1840

Option A (1840) is correct because the mass of a neutron is approximately 1840 times the mass of an electron.

Solution approach

Correct Answer + Core Reason: 1840. The mass of a neutron is approximately 1840 times the mass of an electron.

Step-by-Step Solution:

Step 1: Identify the given information.\nMass of neutron ( $m_n$ ) = $1.675 \times 10^{-27}$ kg\nMass of electron ( $m_e$ ) = $9.108 \times 10^{-31}$ kg

Step 2: Calculate the ratio of the mass of a neutron to the mass of an electron.\nRatio = $\frac{m_n}{m_e} = \frac{1.675 \times 10^{-27}}{9.108 \times 10^{-31}}$

Step 3: Perform the calculation.\nRatio ≈ 1840

Step 4: Final Answer
1840

Why other options are wrong:
- Option B (1480): This is an incorrect approximation of the ratio of neutron to electron mass.
- Option C (2000): This is also an incorrect approximation; the actual ratio is closer to 1840.
- Option D (None of these): This is incorrect because option A is the correct answer.

Key Concepts:
- Mass of neutron
- Mass of electron
- Ratio and proportion

Memory Aids:
- Remember the approximate ratio of neutron to electron mass is around 1840.

Quick Tips:
- Use scientific notation to simplify calculations involving very small or very large numbers.

70 Easy

70 of 239

Analyzing +4 | -1

Statement (S): Dipositive zinc ion exhibits paramagnetism due to the loss of two electrons from

3 d

-orbitals of neutral atom.

Explanation (E): Paramagnetism is due to the presence of unpaired electron.

View answer and explanation

Correct option: B

S is wrong but E is correct

Option B (S is wrong but E is correct) is correct because dipositive zinc ion is diamagnetic due to the complete filling of d-orbitals, and paramagnetism is indeed due to unpaired electrons.

Solution approach

Correct Answer + Core Reason: S is wrong but E is correct. Zinc ion is diamagnetic, and paramagnetism is due to unpaired electrons.

Step-by-Step Solution:

Step 1: Write the electronic configuration of Zn: [Ar] 3d10 4s2.

Step 2: Write the electronic configuration of Zn2+: [Ar] 3d10. The two electrons are lost from the 4s orbital.

Step 3: Since the 3d orbitals are completely filled, there are no unpaired electrons. Therefore, Zn2+ is diamagnetic, not paramagnetic.

Step 4: Paramagnetism is indeed due to the presence of unpaired electrons.

Step 5: Therefore, statement S is wrong, and explanation E is correct.

Why other options are wrong:
- Option A (S is correct but E is wrong): S is not correct, and E is correct.
- Option C (Both S and E are correct and E is correct explanation of S): S is not correct.
- Option D (Both S and E are correct but E is not correct explanation of S): S is not correct.

Key Concepts:
- Paramagnetism
- Diamagnetism
- Electronic Configuration
- Transition Metal Ions

Memory Aids:
- Filled or half-filled orbitals contribute to stability and affect magnetic properties.

Quick Tips:
- Write the electronic configuration to determine the presence of unpaired electrons.

71 Easy

71 of 239

Understanding +4 | -1

Statement (S): All the

d

-orbitals are identical in shape.

Explanation (E): All the

p

-orbitals are identical in shape.

View answer and explanation

Correct option: B

S is wrong but E is correct

Option B (S is wrong but E is correct) is correct because not all d-orbitals have the same shape, while all p-orbitals do have the same dumbbell shape, differing only in orientation.

Solution approach

Correct Answer + Core Reason: S is wrong but E is correct. d-orbitals have different shapes, p-orbitals have the same shape.

Step-by-Step Solution:

Step 1: Recall the shapes of d-orbitals.\nThe five d-orbitals are $d_{xy}$ , $d_{yz}$ , $d_{xz}$ , $d_{x^2-y^2}$ , and $d_{z^2}$ . The $d_{z^2}$ orbital has a different shape compared to the other four d-orbitals. Therefore, not all d-orbitals are identical in shape.

Step 2: Recall the shapes of p-orbitals.\nThe three p-orbitals are $p_x$ , $p_y$ , and $p_z$ . All three p-orbitals have the same dumbbell shape, but they are oriented along different axes (x, y, and z).

Step 3: Evaluate the statement and explanation.
- Statement (S): All the $d$ -orbitals are identical in shape. This is incorrect.
- Explanation (E): All the $p$ -orbitals are identical in shape. This is correct.

Why other options are wrong:
- Option A: S is not correct, and E is correct.
- Option C: Both S and E are not correct.
- Option D: Both S and E are not correct.

Key Concepts:
- Atomic orbitals
- d-orbitals
- p-orbitals
- Shapes of orbitals

Memory Aids:
- Remember the shapes of s, p, and d orbitals.

Quick Tips:
- Visualize the shapes of atomic orbitals to answer such questions.

72 Easy

72 of 239

Remembering +4 | -1

Heisenberg’s uncertainty principle rules out the exact simultaneous measurement of

View answer and explanation

Correct option: D

Position and momentum

Option D (Position and momentum) is correct because Heisenberg's uncertainty principle states that it is impossible to simultaneously know the exact position and momentum of a particle.

Solution approach

Correct Answer + Core Reason: Position and momentum cannot be simultaneously measured exactly due to Heisenberg's uncertainty principle.

Step-by-Step Solution:
Step 1: Recall Heisenberg's uncertainty principle.
Step 2: Heisenberg's uncertainty principle states that the more accurately we know the position of a particle, the less accurately we can know its momentum, and vice versa.
Step 3: Mathematically, this is expressed as $\Delta x \Delta p \ge \frac{h}{{4\pi}}$ , where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant.

Why other options are wrong:
- Option A: Probability and intensity are not directly related by the uncertainty principle.
- Option B: Energy and velocity are not the pair directly addressed by the most common form of the uncertainty principle, although energy and time are.
- Option C: Charge density and radius are not directly related by the uncertainty principle.

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

Memory Aids:
- Uncertainty = Position & Momentum

Quick Tips:
- Remember the mathematical expression of Heisenberg's uncertainty principle.

73 Easy

73 of 239

Understanding +4 | -1

In hydrogen spectrum least energetic transition of electrons are found in:

View answer and explanation

Correct option: D

Pfund series

Option D (Pfund series) is correct because the Pfund series corresponds to transitions to the n=5 energy level, which results in the lowest energy transitions in the hydrogen spectrum compared to Lyman, Balmer, and Bracket series.

Solution approach

Correct Answer + Core Reason: Pfund series. Pfund series has the smallest energy transitions.

Step-by-Step Solution:

Step 1: Recall the Hydrogen Spectrum Series.\nThe hydrogen spectrum consists of several series of lines, each corresponding to transitions to a specific energy level:
- Lyman series: Transitions to n = 1
- Balmer series: Transitions to n = 2
- Paschen series: Transitions to n = 3
- Brackett series: Transitions to n = 4
- Pfund series: Transitions to n = 5

Step 2: Understand the relationship between energy and transition level.\nThe energy of a transition is inversely proportional to the square of the final energy level. Higher the final energy level, lower the energy of the transition.
$\Delta E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ , where $n_1$ is the final energy level and $n_2$ is the initial energy level.

Step 3: Compare the energy of transitions for each series.\nSince the Pfund series corresponds to transitions to n = 5, it has the lowest energy transitions compared to the other series.

Why other options are wrong:
- Option A: Lyman series: Transitions to n = 1, which have the highest energy.
- Option B: Balmer series: Transitions to n = 2, which have higher energy than Pfund.
- Option C: Bracket series: Transitions to n = 4, which have higher energy than Pfund.

Key Concepts:
- Hydrogen spectrum
- Energy levels
- Rydberg formula
- Lyman, Balmer, Paschen, Brackett, and Pfund series

Memory Aids:
- Remember the order of the series: Lyman, Balmer, Paschen, Brackett, Pfund (in increasing order of final energy level).

Quick Tips:
- The series with the lowest final energy level (n) will have the highest energy transitions, and vice versa.

74 Easy

74 of 239

Understanding +4 | -1

If the value of azimuthal quantum number is 3 , the possible values of magnetic quantum number would be:

View answer and explanation

Correct option: C

0, \pm 1, \pm 2, \pm 3

Option C (

0, \pm 1, \pm 2, \pm 3

) is correct because for an azimuthal quantum number of 3, the magnetic quantum number can range from -3 to +3, including 0.

Solution approach

Correct Answer + Core Reason: $0, \pm 1, \pm 2, \pm 3$ . The magnetic quantum number ranges from -l to +l, including 0.

Step-by-Step Solution:

Step 1: Recall the relationship between the azimuthal quantum number (l) and the magnetic quantum number (m).
- The magnetic quantum number (m) can take integer values from -l to +l, including 0.

Step 2: Determine the possible values of m when l = 3.
- When l = 3, the possible values of m are -3, -2, -1, 0, +1, +2, +3.

Step 3: Express the possible values of m in the given format.
- The possible values of m are 0, $\pm 1$ , $\pm 2$ , $\pm 3$ .

Step 4: Conclude that the possible values of the magnetic quantum number are 0, $\pm 1$ , $\pm 2$ , $\pm 3$ .

Why other options are wrong:
- Option A (0,1,2,3): This option only includes positive values and 0, but the magnetic quantum number can also be negative.
- Option B (0,-1,-2,-3): This option only includes negative values and 0, but the magnetic quantum number can also be positive.
- **Option D ( $\pm 1, \pm 2, \pm 3$ ): This option excludes 0, which is a valid value for the magnetic quantum number.

Key Concepts:
- Quantum numbers
- Azimuthal quantum number (l)
- Magnetic quantum number (m)

Memory Aids:
- Remember that m ranges from -l to +l, including 0.

Quick Tips:**
- Use the value of l to determine the possible values of m.

75 Easy

75 of 239

Analyzing +4 | -1

An ion has 18 electrons in the outermost shell, it is

View answer and explanation

Correct option: A

{Cu}^{+}

Option A (

{Cu}^{+}

) is correct because

{Cu}^{+}

has 28 electrons, and its electronic configuration is

1s^22s^22p^63s^23p^63d^{10}

, which has 18 electrons in the outermost shell (3s, 3p, and 3d).

Solution approach

Correct Answer + Core Reason: ${Cu}^{+}$ . ${Cu}^{+}$ has 28 electrons with 18 electrons in its outermost shell.

Step-by-Step Solution:

Step 1: Determine the number of electrons for each ion.
- ${Cu}^{+} = 29 - 1 = 28$ electrons
- ${Th}^{4+} = 90 - 4 = 86$ electrons
- ${Cs}^{+} = 55 - 1 = 54$ electrons
- ${K}^{+} = 19 - 1 = 18$ electrons

Step 2: Write the electronic configuration for each ion.
- ${Cu}^{+} = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}$
- ${Th}^{4+} = [Rn] 5f^0 6d^0 7s^0$
- ${Cs}^{+} = [Xe]$
- ${K}^{+} = 1s^2 2s^2 2p^6 3s^2 3p^6$

Step 3: Identify the ion with 18 electrons in the outermost shell.
- ${Cu}^{+}$ has 18 electrons in the 3rd shell (3s, 3p, 3d).

Why other options are wrong:
- Option B: ${Th}^{4+}$ does not have 18 electrons in its outermost shell.
- Option C: ${Cs}^{+}$ does not have 18 electrons in its outermost shell.
- Option D: ${K}^{+}$ has 8 electrons in its outermost shell (3s and 3p).

Key Concepts:
- Electronic Configuration
- Ion Formation
- Shell Structure

Memory Aids:
- Remember the order of filling electrons in orbitals: Aufbau principle.

Quick Tips:
- Write the electronic configuration to determine the number of electrons in the outermost shell.

76 Easy

76 of 239

Understanding +4 | -1

Which one of the following sets of quantum numbers is possible

View answer and explanation

Correct option: C

n=1, l=0, m_1=0, m_s=+1 / 2

Option C (

n=1, l=0, m_1=0, m_s=+1 / 2

) is correct because it adheres to all the quantum number rules:

l

must be between 0 and

n-1

,

m_l

must be between

-l

and

+l

, and

m_s

must be either +1/2 or -1/2.

Solution approach

Correct Answer + Core Reason: $n=1, l=0, m_1=0, m_s=+1 / 2$ . This set of quantum numbers is valid because it follows all the rules for allowed values of each quantum number.

Step-by-Step Solution:

Step 1: Review 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 (ml): ml = -l, -l+1, ..., 0, ..., l-1, l
- Spin quantum number (ms): ms = +1/2 or -1/2

Step 2: Evaluate option A: $n=3, l=3, m_1=-3, m_s=+1 / 2$ . Since l must be less than n, l cannot be 3 when n is 3. This option is incorrect.

Step 3: Evaluate option B: $n=2, l=1, m_1=2, m_s=-1 / 2$ . Since ml must be between -l and +l, ml cannot be 2 when l is 1. This option is incorrect.

Step 4: Evaluate option C: $n=1, l=0, m_1=0, m_s=+1 / 2$ . Since l = 0, ml must be 0. ms can be +1/2 or -1/2. This option is correct.

Step 5: Evaluate option D: $n=3, l=2, m_1=3, m_s=-1 / 2$ . Since ml must be between -l and +l, ml cannot be 3 when l is 2. This option is incorrect.

Why other options are wrong:
- **Option A ( $n=3, l=3, m_1=-3, m_s=+1 / 2$ ):** The value of $l$ cannot be equal to $n$ . For $n=3$ , the possible values of $l$ are 0, 1, and 2.
- **Option B ( $n=2, l=1, m_1=2, m_s=-1 / 2$ ):** The value of $m_l$ must be between $-l$ and $+l$ . For $l=1$ , the possible values of $m_l$ are -1, 0, and 1.
- **Option D ( $n=3, l=2, m_1=3, m_s=-1 / 2$ ):** The value of $m_l$ must be between $-l$ and $+l$ . For $l=2$ , the possible values of $m_l$ are -2, -1, 0, 1, and 2.

Key Concepts:
- Quantum numbers (n, l, ml, ms)
- Rules for allowed values of quantum numbers

Memory Aids:
- Remember the ranges for each quantum number: n > 0, 0 <= l < n, -l <= ml <= l, ms = +/- 1/2

Quick Tips:
- Always check if the given quantum numbers satisfy the rules for allowed values.

77 Easy

77 of 239

Understanding +4 | -1

Which of the following particle is not deflected in the magnetic field?

View answer and explanation

Correct option: C

Neutron

Option C (Neutron) is correct because neutrons are neutral particles and are not affected by magnetic fields.

Solution approach

Correct Answer + Core Reason: Neutron. Neutrons are neutral particles and are not affected by magnetic fields.

Step-by-Step Solution:

Step 1: Recall the properties of charged particles in a magnetic field.

Step 2: Charged particles (electrons, protons, deuterons) experience a force in a magnetic field, causing them to deflect.

Step 3: The force on a charged particle in a magnetic field is given by $F = qvB\sin{\theta}$ , where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field.

Step 4: Neutrons have no charge ( $q = 0$ ), so they experience no force and are not deflected.

Why other options are wrong:
- Option A: Electron is deflected because it has a negative charge.
- Option B: Proton is deflected because it has a positive charge.
- Option C: Neutron is the correct answer as it is not deflected because it has no charge.
- Option D: Deuteron is deflected because it has a positive charge.

Key Concepts:
- Magnetic force on a charged particle
- Properties of subatomic particles

Memory Aids:
- No charge, no force in a magnetic field.

Quick Tips:
- Remember that only charged particles are affected by magnetic fields.

78 Easy

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Remembering +4 | -1

Which wavelength falls in a X-rays region?

View answer and explanation

Correct option: C

1Å

Option C (1Å) is correct because X-rays have wavelengths in the range of approximately 0.01 to 10 Angstroms.

Solution approach

Correct Answer + Core Reason: 1Å falls in the X-rays region.

Step-by-Step Solution:

Step 1: Identify the question: The question asks which wavelength falls in the X-rays region.

Step 2: Recall the electromagnetic spectrum and the wavelengths of X-rays.

Step 3: X-rays have wavelengths in the range of approximately 0.01 to 10 Angstroms.

Why other options are wrong:
- Option A: 10,000 Å is in the ultraviolet region.
- Option B: 1000 Å is in the ultraviolet region.
- Option D: ${10^{ - 2}}$ is not a valid unit of wavelength.

Key Concepts:
- Electromagnetic spectrum
- X-rays
- Wavelength

Memory Aids:
- X-rays have short wavelengths.

Quick Tips:
- Remember the approximate wavelengths of different regions of the electromagnetic spectrum.

79 Easy

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Application +4 | -1

The maximum number of electrons that can be identified with the quantum number

\ell=2

for

\mathrm{Fe}^{3+}

ion will be

View answer and explanation

Correct option: C

5

Option C (5) is correct because

\ell=2

corresponds to the d-orbital, and

\mathrm{Fe}^{3+}

has 5 electrons in its 3d orbital.

Solution approach

Correct Answer + Core Reason: 5. The number of electrons in the d-orbital of $\mathrm{Fe}^{3+}$ is 5.

Step-by-Step Solution:

Step 1: Identify the electronic configuration of Fe: $\mathrm{Fe} = [Ar] 3d^6 4s^2$ .

Step 2: Determine the electronic configuration of $\mathrm{Fe}^{3+}$ : $\mathrm{Fe}^{3+} = [Ar] 3d^5$ .

Step 3: The quantum number $\ell = 2$ corresponds to the d-orbital.

Step 4: Count the number of electrons in the 3d orbital of $\mathrm{Fe}^{3+}$ : There are 5 electrons.

Why other options are wrong:
- Option A: This is incorrect. The number of electrons is not 3.
- Option B: This is incorrect. The number of electrons is not 4.
- Option D: This is incorrect. The number of electrons is not 6.

Key Concepts:
- Electronic configuration
- Quantum numbers
- d-orbital

Memory Aids:
- Remember that $\ell = 0, 1, 2, 3$ corresponds to s, p, d, f orbitals respectively.

Quick Tips:
- Write the electronic configuration first to determine the number of electrons in each orbital.

80 Easy

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Application +4 | -1

The set of quantum numbers

n=3, l=0, m=0, s=-1 / 2

belongs to the element

View answer and explanation

Correct option: A

{Mg}

Option A (Mg) is correct because the given quantum numbers (n=3, l=0, m=0, s=-1/2) correspond to the 3s orbital, which is the valence electron configuration of Magnesium.

Solution approach

Correct Answer + Core Reason: ${Mg}$ . The quantum numbers correspond to the 3s orbital, which is the valence electron configuration of Magnesium.

Step-by-Step Solution:

Step 1: Identify the orbital based on the quantum numbers. n = 3 indicates the third energy level, and l = 0 indicates an s orbital.

Step 2: Determine the electronic configuration corresponding to the given quantum numbers. The quantum numbers n=3, l=0, m=0, s=-1/2 correspond to a 3s electron.

Step 3: Determine the element with the 3s electron in its electronic configuration. Magnesium (Mg) has the electronic configuration $1s^2 2s^2 2p^6 3s^2$ . The last electron fills the 3s orbital.

Why other options are wrong:
- Option B: ${Na}$ : Sodium (Na) has the electronic configuration $1s^2 2s^2 2p^6 3s^1$ . It has only one electron in the 3s orbital, not a filled 3s orbital as suggested by the given quantum numbers.
- Option C: ${Ne}$ : Neon (Ne) has the electronic configuration $1s^2 2s^2 2p^6$ . It does not have any electrons in the 3s orbital.
- Option D: $F$ : Fluorine (F) has the electronic configuration $1s^2 2s^2 2p^5$ . It does not have any electrons in the 3s orbital.

Key Concepts:
- Quantum numbers
- Electronic configuration
- Atomic structure

Memory Aids:
- Remember the relationship between quantum numbers and orbitals: n (principal quantum number), l (azimuthal quantum number, l=0 for s, l=1 for p, l=2 for d), m (magnetic quantum number), s (spin quantum number).

Quick Tips:
- Always determine the orbital based on the quantum numbers first.
- Then, determine the electronic configuration corresponding to the orbital.
- Finally, identify the element with that electronic configuration.

81 Easy

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Remembering +4 | -1

Which of the following is Heisenberg uncertainty principle?

View answer and explanation

Correct option: A

\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}

Option A (

\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}}

) is correct because it represents the correct mathematical expression of Heisenberg's uncertainty principle.

Solution approach

Correct Answer + Core Reason: $\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}$ is the correct expression of Heisenberg's uncertainty principle.

Step-by-Step Solution:

Step 1: Recall Heisenberg's uncertainty principle.

Step 2: Understand that the principle states that it is impossible to simultaneously determine the exact position and momentum of a particle.

Step 3: The mathematical expression is given by $\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}$ , where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant.

Why other options are wrong:
- Option B: $\Delta x.\Delta p = \frac{h}{{4\mathrm{\pi }}}$ implies a definite equality, which is not the case in the uncertainty principle.
- Option C: $\Delta x.\Delta p \le \frac{h}{{4\mathrm{\pi }}}$ contradicts the principle, as the product of uncertainties must be greater than or equal to $\frac{h}{{4\mathrm{\pi }}}$ .
- Option D: $\Delta x.\Delta p < \frac{h}{{4\mathrm{\pi }}}$ is also incorrect for the same reason as option C.

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

Memory Aids:
- Remember the inequality sign ( $\ge$ ) in the expression.

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

82 Easy

82 of 239

Remembering +4 | -1

Which one of the following is considered as the main postulate of Bohr’s model of atom

View answer and explanation

Correct option: D

Angular momentum of electron is an integral multiple of

\frac{h}{{2\pi }}

Option D is correct because Bohr's postulate states that the angular momentum of an electron is quantized and is an integral multiple of

\frac{h}{{2\pi }}

.

Solution approach

Correct Answer + Core Reason: Angular momentum of electron is an integral multiple of $\frac{h}{{2\pi }}$ . This is a fundamental postulate of Bohr's atomic model.

Step-by-Step Solution:

Step 1: Recall Bohr's postulates regarding the quantization of angular momentum.

Step 2: Bohr's postulate states that the angular momentum (L) of an electron in an atom is quantized and can only take on specific values that are integer multiples of $\frac{h}{{2\pi }}$ , where h is Planck's constant.

Step 3: Mathematically, this is expressed as: $L = n \frac{h}{{2\pi}}$ , where n is an integer (n = 1, 2, 3, ...).

Step 4: Therefore, the angular momentum of the electron is an integral multiple of $\frac{h}{{2\pi}}$ .

Why other options are wrong:
- Option A: Protons being present in the nucleus is part of Rutherford's model and the nuclear model of the atom, but not a main postulate of Bohr's model.
- Option B: Electrons revolving around the nucleus is part of Rutherford's model, but Bohr's model adds the quantization of energy levels and angular momentum.
- Option C: Balancing of centrifugal and electrostatic forces is a necessary condition for stable orbits, but not the main postulate of Bohr's model. The quantization of angular momentum is the key distinguishing feature.

Key Concepts:
- Bohr's Atomic Model
- Quantization of Angular Momentum
- Planck's Constant

Memory Aids:
- Remember that Bohr's model introduces the idea of quantized energy levels and angular momentum.

Quick Tips:
- Focus on the key postulates of each atomic model to differentiate them.

83 Easy

83 of 239

Application +4 | -1

The electrons identified by 517 ‘n’ and ‘l’

(i)n = 4,\,l = 1,\,(ii)\,n = 4,\,l = 0,\,(iii)\,n = 3,\,l = 2\,,\,(iv)\,n = 3,\,l = 1

, Can be placed in order of increasing from lowest to highest is

View answer and explanation

Correct option: A

iv < ii < iii < i

Option A (iv < ii < iii < i) is correct because it correctly orders the orbitals based on increasing energy using the n+l rule and Hund's rule.

Solution approach

Correct Answer + Core Reason: iv < ii < iii < i. The energy of an orbital increases with increasing n+l value.

Step-by-Step Solution:

Step 1: Determine the n+l value for each orbital:
(i) n=4, l=1; n+l = 5
(ii) n=4, l=0; n+l = 4
(iii) n=3, l=2; n+l = 5
(iv) n=3, l=1; n+l = 4

Step 2: Compare the n+l values. Orbitals (ii) and (iv) have the same n+l value (4), and orbitals (i) and (iii) have the same n+l value (5).

Step 3: For orbitals with the same n+l value, the orbital with the lower n value has lower energy. Therefore, orbital (iv) has lower energy than orbital (ii), and orbital (iii) has lower energy than orbital (i).

Step 4: Arrange the orbitals in increasing order of energy: (iv) < (ii) < (iii) < (i).

Why other options are wrong:
- Option B: ii < iv < i < iii: Incorrectly places orbital (iv) with higher energy than (ii).
- Option C: i < iii < ii < iv: Completely incorrect order.
- Option D: iii < i < iv < ii: Incorrectly places orbitals (iii) and (i) with lower energy than (ii) and (iv).

Key Concepts:
- n+l rule
- Energy of orbitals
- Hund's rule

Memory Aids:
- Aufbau principle: Electrons fill orbitals in order of increasing energy.

Quick Tips:
- Remember that the n+l rule is a guideline, and exceptions may exist for orbitals with very close energies.

84 Easy

84 of 239

Understanding +4 | -1

STATEMENT-1 : Half-filled and fully-filled degenerate orbitals are more stable.

STATEMENT-2 : Extra stability is due to the symmetrical distribution of electrons and exchange energy.

View answer and explanation

Correct option: A

If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-4

Option A is correct because both statements are true, and Statement-2 correctly explains why half-filled and fully-filled orbitals are more stable.

Solution approach

Correct Answer + Core Reason: If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-4. Extra stability arises from symmetrical electron distribution and exchange energy.

Step-by-Step Solution:
Step 1: Statement-1 asserts the stability of half-filled and fully-filled degenerate orbitals. This is a well-established fact in chemistry.
Step 2: Statement-2 explains this stability by referencing symmetrical electron distribution and exchange energy. Symmetrical distribution minimizes electron-electron repulsion, and exchange energy arises from electrons with parallel spins exchanging positions within the orbital, leading to a more stable configuration.
Step 3: The explanation in Statement-2 directly supports and explains the observation in Statement-1.

Why other options are wrong:
- Option B: While both statements are true, Statement-2 *is* the correct explanation for Statement-1.
- Option C: Statement-1 is true, so this option is incorrect.
- Option D: Statement-1 is true, so this option is incorrect.

Key Concepts:
- Hund's Rule
- Exchange Energy
- Stability of Half-filled and Fully-filled Orbitals
- Electronic Configuration

Memory Aids:
- Remember that symmetrical configurations are generally more stable.
- Half-filled and fully-filled orbitals are exceptions to simple filling rules due to exchange energy.

Quick Tips:
- When evaluating stability, consider both electron-electron repulsion and exchange energy.

85 Easy

85 of 239

Remembering +4 | -1

Maximum number of electrons in a subshell is given by

View answer and explanation

Correct option: B

2\left( {2l + 1} \right)

Option B (

2\left( {2l + 1} \right)

) is correct because it represents the total number of electrons that can occupy a subshell with a given azimuthal quantum number l.

Solution approach

Correct Answer + Core Reason: $2\left( {2l + 1} \right)$ . The number of orbitals in a subshell is $(2l+1)$ , and each orbital can hold 2 electrons.

Step-by-Step Solution:

Step 1: Define the azimuthal quantum number, $l$ .

Step 2: The number of orbitals in a subshell is given by $2l + 1$ .

Step 3: Each orbital can hold a maximum of 2 electrons.

Step 4: Therefore, the maximum number of electrons in a subshell is $2(2l + 1)$ .

Why other options are wrong:
- Option A: $\left( {2l + 1} \right)$ is incorrect. This represents the number of orbitals in a subshell, not the number of electrons.
- Option C: $\left( {2l + 1} \right)2$ is incorrect. This is mathematically equivalent to option A, just written differently, and still represents the number of orbitals multiplied by 2 incorrectly.
- Option D: $2\left( {2l + 1} \right)2$ is incorrect. This would mean each orbital can hold 4 electrons, which is not true.

Key Concepts:
- Quantum numbers (azimuthal quantum number)
- Atomic structure
- Electron configuration

Memory Aids:
- Remember that each orbital can hold 2 electrons.
- The number of orbitals is $2l+1$ .

Quick Tips:
- Know the relationship between quantum numbers and the number of orbitals and electrons.

86 Easy

86 of 239

Analyzing +4 | -1

Ions which have the same electronic configuration are those of

View answer and explanation

Correct option: C

Potassium and calcium

Option C is correct. K⁺ and Ca²⁺ both have the same electronic configuration: 1s²2s²2p⁶3s²3p⁶, which is the electronic configuration of Argon.

Solution approach

Correct Answer + Core Reason: Potassium and calcium. Both ions have the same number of electrons and thus the same electronic configuration.

Step-by-Step Solution:
Step 1: Write the electronic configuration of each ion.
Step 2: K⁺: [Ar] (1s²2s²2p⁶3s²3p⁶)
Step 3: Ca²⁺: [Ar] (1s²2s²2p⁶3s²3p⁶)
Step 4: Li⁺: 1s²
Step 5: Na⁺: [Ne] (1s²2s²2p⁶)
Step 6: Cl⁻: [Ar] (1s²2s²2p⁶3s²3p⁶)
Step 7: O²⁻: [Ne] (1s²2s²2p⁶)
Step 8: Compare the electronic configurations.

Why other options are wrong:
- Option A: Lithium and sodium have different numbers of electrons.
- Option B: Sodium and potassium have different numbers of electrons.
- Option D: Oxygen and chlorine have different numbers of electrons.

Key Concepts:
- Electronic configuration of ions
- Isoelectronic species

Memory Aids:
- Noble gas configurations are stable.

Quick Tips:
- Count the number of electrons in each ion.

87 Easy

87 of 239

Remembering +4 | -1

For a Bohr atom angular momentum M of the electron is (n=0,1,2,…)

View answer and explanation

Correct option: D

\frac{{nh}}{{2\mathrm{\pi }}}

Option D (

\frac{{nh}}{{2\mathrm{\pi }}}}

) is correct because the angular momentum of an electron in a Bohr atom is quantized and given by this formula.

Solution approach

Correct Answer + Core Reason: $\frac{{nh}}{{2\mathrm{\pi }}}}$ Angular momentum is quantized in Bohr's model and is an integer multiple of $\frac{{h}}{{2\mathrm{\pi }}}}$

Step-by-Step Solution:

Step 1: Recall Bohr's quantization condition for angular momentum.

Step 2: The angular momentum (L) of an electron in a Bohr orbit is given by: $L = mvr = n\frac{{h}}{{2\pi}}$ , where n is the principal quantum number (n = 1, 2, 3, ...), h is Planck's constant, m is the mass of the electron, v is its velocity, and r is the radius of the orbit.

Step 3: Therefore, the angular momentum M is equal to L, so $M = \frac{{nh}}{{2\pi}}$

Why other options are wrong:
- Option A: $\frac{{n{h^2}}}{{4\mathrm{\pi }}}}$ : This is incorrect because the angular momentum is directly proportional to h, not $h^2$ .
- Option B: $\frac{{{n^2}{h^2}}}{{4\mathrm{\pi }}}}\;$ : This is incorrect because the angular momentum is directly proportional to n and h, not $n^2$ and $h^2$ .
- Option C: $\frac{{\sqrt {\mathrm{\pi }{h^2}} }}{{4\mathrm{\pi }}}$ : This option has no physical basis in Bohr's model for angular momentum.

Key Concepts:
- Bohr's model of the atom
- Quantization of angular momentum
- Planck's constant

Memory Aids:
- Remember that angular momentum is quantized in units of h/2π.

Quick Tips:
- Always remember the formula for angular momentum in Bohr's model: $L = n\frac{{h}}{{2\pi}}$

88 Easy

88 of 239

Remembering +4 | -1

Mass of a proton in

\mathrm{Kg}

is

View answer and explanation

Correct option: D

1.66 \times 10^{-27}

Option D (

1.66 \times 10^{-27}

) is correct because the mass of a proton is approximately

1.66 \times 10^{-27}

kg.

Solution approach

Correct Answer + Core Reason: $1.66 \times 10^{-27}$ . The mass of a proton is approximately $1.66 \times 10^{-27}$ kg.

Step-by-Step Solution:

Step 1: Recall the mass of a proton in kilograms.

Step 2: The mass of a proton is approximately $1.6726219 \times 10^{-27}$ kg, which can be rounded to $1.66 \times 10^{-27}$ kg.

Why other options are wrong:
- **Option A ( $1.66 \times 10^{-24}$ ): This value is too large.
- Option B ( $6.022 \times 10^{-23}$ ): This is related to Avogadro's number and not the mass of a proton.
- Option C ( $6.022 \times 10^{-27}$ ): This value is too small and also related to Avogadro's number.

Key Concepts:
- Mass of a proton
- Fundamental constants

Memory Aids:**
- Remember the approximate mass of a proton: $1.66 \times 10^{-27}$ kg.

Quick Tips:
- Know the values of fundamental constants like the mass of a proton, neutron, and electron.

89 Easy

89 of 239

Application +4 | -1

If threshold wavelength

\left( {{\lambda ^0}} \right)

for ejection of electron from metal is 330 nm. Then work function for the photo electric emission is

View answer and explanation

Correct option: C

6 \times {10^{ - 19}}J

Option C (

6 \times {10^{ - 19}}J

) is correct because it is the work function calculated using the threshold wavelength and Planck's constant.

Solution approach

Correct Answer + Core Reason: $6 \times {10^{ - 19}}J$ . The work function is calculated using the threshold wavelength.

Step-by-Step Solution:

Step 1: Identify the given information: Threshold wavelength $\lambda_0 = 330 \text{{ nm}} = 330 \times 10^{-9} \text{{ m}}$

Step 2: Recall the formula for work function: $W_0 = h \nu_0 = \frac{{hc}}{{\lambda_0}}$ , where $h = 6.626 \times 10^{-34} \text{{ J s}}$ and $c = 3 \times 10^8 \text{{ m/s}}$

Step 3: Substitute the values into the formula:
$W_0 = \frac{{(6.626 \times 10^{-34} \text{{ J s}})(3 \times 10^8 \text{{ m/s}})}}{{330 \times 10^{-9} \text{{ m}}}}$

Step 4: Calculate the work function:
$W_0 = \frac{{19.878 \times 10^{-26}}}{{330 \times 10^{-9}}} \text{{ J}}$
$W_0 = 0.060236 \times 10^{-17} \text{{ J}}$
$W_0 = 6.0236 \times 10^{-19} \text{{ J}}$

Step 5: Round to the nearest value: $W_0 \approx 6 \times 10^{-19} \text{{ J}}$

Why other options are wrong:
- **Option A ( $1.2 \times {10^{ - 18}}J$ ): This is incorrect. This value is twice the correct work function and may be due to a calculation error.
- Option B ( $1.2 \times {10^{ - 20}}J$ ): This is incorrect. This value is much smaller than the correct work function and may be due to an incorrect exponent in the calculation.
- Option D ( $6 \times {10^{ - 12}}J$ ): This is incorrect. This value is much larger than the correct work function and may be due to an incorrect exponent in the calculation.

Key Concepts:
- Photoelectric effect
- Work function
- Threshold wavelength
- Planck's constant

Memory Aids:**
- $W_0 = h \nu_0 = \frac{{hc}}{{\lambda_0}}$

Quick Tips:
- Remember to convert the wavelength to meters before calculating the work function.
- Pay attention to the units and make sure they are consistent.

90 Easy

90 of 239

Application +4 | -1

Hydrogen atoms are excited to

n = 4

state, In the spectrum of emitted radiation, number of lines in the
UV and visible regions are resp.

View answer and explanation

Correct option: D

3,2

Option D (

3,2

) is correct because when hydrogen atoms are excited to n=4, there are 3 Lyman series lines (UV region) and 2 Balmer series lines (visible region).

Solution approach

Correct Answer + Core Reason: $3,2$ . The number of lines in UV and visible regions are determined by the transitions to n=1 and n=2 respectively.

Step-by-Step Solution:

Step 1: Hydrogen atoms are excited to $n = 4$ state.

Step 2: Total number of lines emitted: $\sum_{i=1}^{n-1} i = \sum_{i=1}^{3} i = 3 + 2 + 1 = 6$ .

Step 3: Number of lines in UV region (Lyman series, transitions to $n=1$ ): $4 \rightarrow 1$ , $3 \rightarrow 1$ , $2 \rightarrow 1$ . So, there are 3 lines.

Step 4: Number of lines in visible region (Balmer series, transitions to $n=2$ ): $4 \rightarrow 2$ , $3 \rightarrow 2$ . So, there are 2 lines.

Step 5: Number of lines in the infrared region (Paschen series, transitions to $n=3$ ): $4 \rightarrow 3$ . So, there is 1 line.

Step 6: Final Answer: $3,2$

Why other options are wrong:
- Option A: Incorrect number of lines in UV and visible regions.
- Option B: Incorrect number of lines in UV and visible regions.
- Option C: Incorrect number of lines in UV and visible regions.

Key Concepts:
- Hydrogen spectrum
- Lyman series (UV region)
- Balmer series (Visible region)
- Paschen series (Infrared region)

Memory Aids:
- Lyman: UV, Balmer: Visible, Paschen: IR

Quick Tips:
- Remember the series and their corresponding regions of the electromagnetic spectrum.

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