STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

Application +4 | -1

The ratio of ionization energy of H-and

B{e^{ + 3}}

is

View answer and explanation

Correct option: D

1:16

Option D (1:16) is correct because ionization energy is proportional to

Z^2

, where Z is the atomic number. For H, Z=1, and for Be+3, Z=4, so the ratio is 1² : 4² = 1:16.

Solution approach

Correct Answer + Core Reason: $1:16$ . Ionization energy is proportional to $Z^2$ , where Z is the atomic number.

Step-by-Step Solution:

Step 1: Recall the formula for ionization energy (IE) for hydrogen-like species:
$IE \propto \frac{Z^2}{n^2}$ , where Z is the atomic number and n is the principal quantum number.

Step 2: For hydrogen (H), Z = 1 and n = 1 (ground state). Therefore, $IE_H \propto \frac{1^2}{1^2} = 1$ .

Step 3: For $B{e^{ + 3}}$ , Z = 4 and n = 1 (ground state). Therefore, $IE_{B{e^{ + 3}}} \propto \frac{4^2}{1^2} = 16$ .

Step 4: Calculate the ratio of ionization energies:
$\frac{IE_H}{IE_{B{e^{ + 3}}}} = \frac{1}{16}$

Why other options are wrong:
- Option A: Incorrect. This would be true if both species had the same atomic number.
- Option B: Incorrect. This ratio does not reflect the correct relationship between the atomic numbers and ionization energies.
- Option C: Incorrect. This ratio does not reflect the correct relationship between the atomic numbers and ionization energies.

Key Concepts:
- Ionization Energy
- Atomic Number
- Hydrogen-like Species

Memory Aids:
- Remember that ionization energy is proportional to the square of the atomic number ( $Z^2$ ).

Quick Tips:
- Always consider the atomic number when comparing ionization energies of different species.

32 Easy

32 of 176Question permalink

Understanding +4 | -1

The de-Broglie equation applies

View answer and explanation

Correct option: C

All the material objects in motion

Option C (All the material objects in motion) is correct because the de Broglie equation applies to all matter, relating wavelength to momentum.

Solution approach

Correct Answer + Core Reason: All the material objects in motion. The de Broglie equation applies to all matter.

Step-by-Step Solution:

Step 1: State the de Broglie equation: $\lambda = \frac{h}{p}$ , where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum.

Step 2: Explain the equation: This equation relates the wavelength of a particle to its momentum. It applies to all matter, not just electrons or protons.

Step 3: Key Concept: Wave-particle duality: All matter exhibits both wave-like and particle-like properties.

Step 4: Final Answer: All the material objects in motion

Why other options are wrong:
- Option A: To protons only: Incorrect, the equation applies to all matter.
- Option B: To electrons only: Incorrect, the equation applies to all matter.
- Option D: To neutrons only: Incorrect, the equation applies to all matter.

Key Concepts:
- de Broglie equation
- Wave-particle duality
- Momentum

Memory Aids:
- de Broglie: Matter waves

Quick Tips:
- Remember that the de Broglie equation applies to all matter, regardless of its size or mass.

33 Easy

33 of 176Question permalink

Remembering +4 | -1

The quantum levels upto n=3 has:

View answer and explanation

Correct option: C

s,p,d-levels

Option C (s,p,d-levels) is correct because for n=3, the possible values of l are 0, 1, and 2, corresponding to s, p, and d orbitals, respectively.

Solution approach

Correct Answer + Core Reason: s,p,d-levels. For n=3, the possible values of l are 0, 1, and 2, corresponding to s, p, and d orbitals.

Step-by-Step Solution:

Step 1: Recall the relationship between the principal quantum number (n) and the azimuthal quantum number (l). For a given n, l can have values from 0 to n-1.

Step 2: For n=3, the possible values of l are 0, 1, and 2.

Step 3: l=0 corresponds to the s-orbital.

Step 4: l=1 corresponds to the p-orbital.

Step 5: l=2 corresponds to the d-orbital.

Step 6: Therefore, the quantum levels up to n=3 have s, p, and d levels.

Why other options are wrong:
- Option A: s and p-levels: This is incorrect because it doesn't include the d-level.
- Option B: s,p,d,f-levels: This is incorrect because f-levels start at n=4.
- Option D: s-level: This is incorrect because it only includes the s-level and misses the p and d levels.

Key Concepts:
- Quantum numbers (n, l, m_l, s)
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Atomic orbitals (s, p, d, f)

Memory Aids:
- Remember that for a given n, l can have values from 0 to n-1.
- s (l=0), p (l=1), d (l=2), f (l=3).

Quick Tips:
- Know the relationship between quantum numbers and the shapes and energies of atomic orbitals.

34 Easy

34 of 176Question permalink

Remembering +4 | -1

What is the ration of mass of an electron to the mass of a proton?

View answer and explanation

Correct option: C

1:1837

Option C (

1:1837

) is correct because the mass of an electron is approximately 1/1837th the mass of a proton.

Solution approach

Correct Answer + Core Reason: $1:1837$ . The mass of an electron is approximately 1/1837th the mass of a proton.

Step-by-Step Solution:

Step 1: Recall the mass of an electron and the mass of a proton.

Step 2: Mass of an electron ( $m_e$ ) ≈ $9.109 \times 10^{-31}$ kg

Step 3: Mass of a proton ( $m_p$ ) ≈ $1.672 \times 10^{-27}$ kg

Step 4: Calculate the ratio of the mass of an electron to the mass of a proton: $\frac{m_e}{m_p} = \frac{9.109 \times 10^{-31}}{1.672 \times 10^{-27}} \approx \frac{1}{1837}$

Why other options are wrong:
- Option A: $1:2$ : This is incorrect. The mass of a proton is much greater than twice the mass of an electron.
- Option B: $1:1$ : This is incorrect. Protons and electrons have significantly different masses.
- Option D: $1:3$ : This is incorrect. The mass of a proton is much greater than three times the mass of an electron.

Key Concepts:
- Mass of electron
- Mass of proton
- Ratio

Memory Aids:
- Remember the approximate ratio: electron mass is much smaller than proton mass.

Quick Tips:
- Keep in mind the approximate ratio of the masses.

35 Easy

35 of 176Question permalink

Remembering +4 | -1

Angular momentum of an electron in an orbital is given by

View answer and explanation

Correct option: B

\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}

Option B (

\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}

) is correct because this is the formula for the magnitude of the orbital angular momentum.

Solution approach

Correct Answer + Core Reason: $\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}$ . This is the formula for angular momentum.

Step-by-Step Solution:

Step 1: Recall the formula for the angular momentum of an electron in an orbital.

Step 2: The angular momentum of an electron in an orbital is given by:
$L = \sqrt{l(l+1)} \frac{h}{2\pi}$ , where l is the azimuthal quantum number.

Step 3: Compare the given options with the formula. Option B matches the formula for angular momentum.

Why other options are wrong:
- Option A: $n\frac{h}{{2\pi }}$ represents the angular momentum in Bohr's model, not the orbital angular momentum.
- Option C: $n\frac{h}{{4\pi }}$ is not a correct formula for angular momentum.
- Option D: Option B is the correct answer.

Key Concepts:
- Angular momentum
- Azimuthal quantum number

Memory Aids:
- Remember the formula for angular momentum: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$

Quick Tips:
- Distinguish between the angular momentum in Bohr's model and the orbital angular momentum.

36 Easy

36 of 176Question permalink

Understanding +4 | -1

The specific charge for positive rays is much less than the specific charge for cathode rays. This is because:

View answer and explanation

Correct option: C

Positive rays comprise ionised atoms, whose mass is much higher

Option C (Positive rays comprise ionised atoms, whose mass is much higher) is correct because specific charge is the ratio of charge to mass, and positive rays consist of heavier ionized atoms compared to electrons in cathode rays.

Solution approach

Correct Answer + Core Reason: Positive rays comprise ionised atoms, whose mass is much higher. Specific charge is inversely proportional to mass.

Step-by-Step Solution:

Step 1: Recall the definition of specific charge: Specific charge = Charge / Mass.

Step 2: Consider the composition of positive rays and cathode rays.
- Positive rays are composed of ionized atoms or molecules.
- Cathode rays are composed of electrons.

Step 3: Compare the masses of ionized atoms/molecules and electrons.
- Ionized atoms/molecules are much heavier than electrons.

Step 4: Since specific charge is inversely proportional to mass, positive rays have a much lower specific charge than cathode rays.

Why other options are wrong:
- Option A: While positive rays are positively charged, this doesn't directly explain the difference in specific charge. It's the mass difference that matters.
- Option B: The charge on positive rays is not necessarily less; it's the mass that is significantly higher.
- Option D: The experimental method is generally accurate; the difference is due to the nature of the particles.

Key Concepts:
- Specific charge
- Positive rays
- Cathode rays
- Mass of atoms and electrons

Memory Aids:
- Specific charge = Charge / Mass

Quick Tips:
- Remember that specific charge is inversely proportional to mass.

37 Easy

37 of 176Question permalink

Remembering +4 | -1

The number of nodal planes in a

{P_X}

orbital is

View answer and explanation

Correct option: A

One

Option A (One) is correct because a

p_x

orbital has one nodal plane, which is the yz-plane.

Solution approach

Correct Answer + Core Reason: One. A ${P_X}$ orbital has one nodal plane.

Step-by-Step Solution:

Step 1: The question asks for the number of nodal planes in a ${P_X}$ orbital.

Step 2: Recall the shapes of p orbitals: They have a dumbbell shape with two lobes.

Step 3: A nodal plane is a plane where the probability of finding an electron is zero.

Step 4: For a ${P_X}$ orbital, the nodal plane is the yz-plane (the plane perpendicular to the x-axis that passes through the nucleus).

Step 5: Therefore, the number of nodal planes is one.

Why other options are wrong:
- Option B (Two): This is incorrect. A p orbital has only one nodal plane.
- Option C (Three): This is incorrect. A p orbital has only one nodal plane.
- Option D (Zero): This is incorrect. A p orbital has one nodal plane.

Key Concepts:
- Atomic orbitals
- p orbitals
- Nodal planes

Memory Aids:
- Visualize the dumbbell shape of a p orbital and the plane that separates the two lobes.

Quick Tips:
- Remember that s orbitals have zero nodal planes, p orbitals have one, d orbitals have two, and f orbitals have three.

38 Easy

38 of 176Question permalink

Analyzing +4 | -1

STATEMENT-1 : Emitted radiation will fall in visible range when an electron jump from

n=4

to

n=2

in

{H}

-atom.

STATEMENT-2 : Balmer series radiations belong to visible range for hydrogen atom only.

View answer and explanation

Correct option: A

If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3

Option A (If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3) is correct because the Balmer series corresponds to transitions to

n=2

, which results in emitted radiation falling in the visible range.

Solution approach

Correct Answer + Core Reason: If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3. The Balmer series falls in the visible range because it involves transitions to $n=2$ .

Step-by-Step Solution:

Step 1: STATEMENT-1 : Emitted radiation will fall in visible range when an electron jump from $n=4$ to $n=2$ in ${H}$ -atom. This statement is TRUE. A transition from $n=4$ to $n=2$ corresponds to a Balmer series transition, which falls in the visible region of the electromagnetic spectrum.

Step 2: STATEMENT-2 : Balmer series radiations belong to visible range for hydrogen atom only. This statement is also TRUE. The Balmer series involves transitions to the $n=2$ energy level, and the wavelengths of the emitted photons fall within the visible range.

Step 3: STATEMENT-2 explains STATEMENT-1 because the transition from $n=4$ to $n=2$ is a Balmer series transition, which is why the emitted radiation falls in the visible range.

Step 4: Final Answer: If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3

Why other options are wrong:
- Option B (If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-3): STATEMENT-2 is the correct explanation of STATEMENT-1.
- Option C (If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE): STATEMENT-2 is TRUE.
- Option D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE): STATEMENT-1 is TRUE.

Key Concepts:
- Hydrogen Atomic Spectra
- Balmer Series
- Visible Spectrum
- Electronic Transitions

Memory Aids:
- Remember that the Balmer series corresponds to transitions to $n=2$ and falls in the visible range.

Quick Tips:
- Know the different series in the hydrogen spectrum and their corresponding regions of the electromagnetic spectrum.

39 Easy

39 of 176Question permalink

Remembering +4 | -1

Splitting of spectral lines when atoms are subjected to strong electric field is called:

View answer and explanation

Correct option: B

Stark effect

Option B (Stark effect) is correct because the splitting of spectral lines in a strong electric field is known as the Stark effect.

Solution approach

Correct Answer + Core Reason: Stark effect. This is the splitting of spectral lines in an electric field.

Step-by-Step Solution:

Step 1: Recall the definitions of Zeeman and Stark effects.

Step 2: Define Zeeman effect: Splitting of spectral lines in a magnetic field.

Step 3: Define Stark effect: Splitting of spectral lines in an electric field.

Step 4: Identify the correct effect. Since the question mentions a strong electric field, the correct answer is the Stark effect.

Why other options are wrong:
- Option A: Zeeman effect is the splitting of spectral lines in a magnetic field, not an electric field.
- Option C: Decay refers to radioactive decay, which is not related to the splitting of spectral lines.
- Option D: Disintegration is a general term for breaking apart, not specifically related to spectral lines.

Key Concepts:
- Stark effect
- Zeeman effect

Memory Aids:
- Stark = Electric field, Zeeman = Magnetic field

Quick Tips:
- Remember the definitions of Stark and Zeeman effects to distinguish between them.

40 Easy

40 of 176Question permalink

Application +4 | -1

The Bohr's energy equation for H atom reveals that the energy level of a shell is given by

E = - 13.58/{n^2}\;\mathrm{eV}.

The smallest amount that an H-atom will absorb, if in ground state is:

View answer and explanation

Correct option: D

10.19 eV

Option D (10.19 eV) is correct because it represents the energy difference between the ground state (n=1) and the first excited state (n=2) of the hydrogen atom.

Solution approach

Correct Answer + Core Reason: 10.19 eV. This is the energy difference between n=1 and n=2 levels.

Step-by-Step Solution:

Step 1: Calculate the energy of the ground state (n=1): $E_1 = -\frac{{13.58}}{{1^2}} = -13.58 \text{{ eV}}$

Step 2: Calculate the energy of the first excited state (n=2): $E_2 = -\frac{{13.58}}{{2^2}} = -\frac{{13.58}}{4} = -3.395 \text{{ eV}}$

Step 3: Calculate the energy difference between the two levels: $\Delta E = E_2 - E_1 = -3.395 - (-13.58) = 10.185 \text{{ eV}}$

Step 4: Round to two decimal places: $\Delta E \approx 10.19 \text{{ eV}}$

Why other options are wrong:
- Option A: 1.0 eV: This is incorrect and doesn't represent the energy difference between any two levels.
- Option B: 3.39 eV: This is the energy of the second energy level (n=2), not the energy difference.
- Option C: 6.79 eV: This is incorrect and doesn't represent the energy difference between any two levels.

Key Concepts:
- Bohr's energy equation
- Energy levels in hydrogen atom

Memory Aids:
- Remember the formula: $E_n = -\frac{{13.6}}{{n^2}} \text{{ eV}}$

Quick Tips:
- Always calculate the energy difference between the initial and final states to find the energy absorbed or emitted.

41 Easy

41 of 176Question permalink

Analyzing +4 | -1

The number of molecules or ions from the following, which do not have odd number of electrons are _____
(A)

\mathrm{NO}_{2}

\quad

(B)

\mathrm{ICl}_{4}^{-}

\quad

(C)

\mathrm{BrF}_{3}

\quad

(D)

\mathrm{ClO}_{2}

\quad

(E)

\mathrm{NO}_{2}^{+}

\quad

(F)

\mathrm{NO}

View answer and explanation

Correct option: A

3

The correct answer is 3.

ICl_4^-

,

BrF_3

, and

NO_2^+

have an even number of electrons. The others have an odd number of electrons.

Solution approach

Correct Answer + Core Reason: 3. $ICl_4^-$ , $BrF_3$ , and $NO_2^+$ have an even number of electrons.

Step-by-Step Solution:

Step 1: Calculate the number of valence electrons for each molecule/ion.
(A) $NO_2$ : 5 (N) + 2 * 6 (O) = 17 (odd)
(B) $ICl_4^-$ : 7 (I) + 4 * 7 (Cl) + 1 (charge) = 7 + 28 + 1 = 36 (even)
(C) $BrF_3$ : 7 (Br) + 3 * 7 (F) = 7 + 21 = 28 (even)
(D) $ClO_2$ : 7 (Cl) + 2 * 6 (O) = 7 + 12 = 19 (odd)
(E) $NO_2^+$ : 5 (N) + 2 * 6 (O) - 1 (charge) = 5 + 12 - 1 = 16 (even)
(F) $NO$ : 5 (N) + 6 (O) = 11 (odd)

Step 2: Count the number of molecules/ions with an even number of electrons.
$ICl_4^-$ , $BrF_3$ , and $NO_2^+$ have an even number of electrons. That's 3.

Why other options are wrong:
- Option A: Incorrect count of molecules with even number of electrons.
- Option B: Incorrect count of molecules with even number of electrons.
- Option C: Incorrect count of molecules with even number of electrons.

Key Concepts:
- Valence electrons
- Odd and even number of electrons

Memory Aids:
- Sum the valence electrons of all atoms in the molecule/ion.
- Add 1 for each negative charge and subtract 1 for each positive charge.

Quick Tips:
- Focus on valence electrons only.

42 Easy

42 of 176Question permalink

Remembering +4 | -1

The best metal to be used for photoemission is:

View answer and explanation

Correct option: C

Cesium

Option C (Cesium) is correct because Cesium has the lowest work function among the given options, making it the most efficient for photoemission.

Solution approach

Correct Answer + Core Reason: Cesium. Cesium has the lowest work function among the given alkali metals, making it the most suitable for photoemission.

Step-by-Step Solution:

Step 1: Define Work Function: The work function is the minimum energy required to remove an electron from the surface of a solid.

Step 2: Photoemission and Work Function: For photoemission to occur, the energy of the incident photon must be greater than or equal to the work function of the metal. A lower work function means that less energy is required to eject an electron.

Step 3: Comparing Alkali Metals: Among the alkali metals (Potassium, Sodium, Cesium, Lithium), Cesium has the lowest ionization energy and, therefore, the lowest work function. This is because Cesium's outermost electron is farthest from the nucleus and experiences the least attraction.

Why other options are wrong:
- Option A: Potassium. Potassium has a higher work function than Cesium.
- Option B: Sodium. Sodium has a higher work function than Cesium.
- Option D: Lithium. Lithium has the highest work function among the given alkali metals.

Key Concepts:
- Photoemission
- Work function
- Ionization energy
- Alkali metals

Memory Aids:
- Cesium (Cs) is at the bottom of Group 1 (alkali metals) and has the lowest ionization energy.

Quick Tips:
- Remember that metals with lower work functions are more efficient for photoemission.

43 Easy

43 of 176Question permalink

Application +4 | -1

For a metal ion, the calculated magnetic moment is 4.90 BM. This metal ion has____number of unpaired electrons.

View answer and explanation

Correct option: A

4

Option A (4) is correct because the magnetic moment is related to the number of unpaired electrons by the formula

\mu = \sqrt{{n(n + 2)}}

BM, where n is the number of unpaired electrons.

Solution approach

Correct Answer + Core Reason: 4. The magnetic moment is related to the number of unpaired electrons.

Step-by-Step Solution:

Step 1: The formula for the magnetic moment ( $\mu$ ) is given by: $\mu = \sqrt{{n(n + 2)}}$ BM, where n is the number of unpaired electrons.

Step 2: Given that $\mu = 4.90$ BM, we need to find the value of n.

Step 3: Substitute the value of $\mu$ into the formula: $4.90 = \sqrt{{n(n + 2)}}$

Step 4: Square both sides: $(4.90)^2 = n(n + 2)$ => $24.01 = n^2 + 2n$

Step 5: Rearrange the equation: $n^2 + 2n - 24.01 = 0$

Step 6: Since we are looking for an integer value for n, we can approximate 24.01 to 24. Then, we can test integer values of n to see which one satisfies the equation $\mu = \sqrt{{n(n + 2)}}$ \nIf n=4, $\mu = \sqrt{{4(4 + 2)}} = \sqrt{{24}} \approx 4.9$

Step 7: Therefore, the number of unpaired electrons is 4.

Why other options are wrong:
- Option B: If n=0, $\mu = 0$ , which is not 4.90 BM.
- Option C: Duplicated option. If n=0, $\mu = 0$ , which is not 4.90 BM.
- Option D: Duplicated option. If n=0, $\mu = 0$ , which is not 4.90 BM.

Key Concepts:
- Magnetic moment
- Unpaired electrons
- Formula: $\mu = \sqrt{{n(n + 2)}}$ BM

Memory Aids:
- Remember the formula for magnetic moment.

Quick Tips:
- Use the magnetic moment formula to find the number of unpaired electrons.

44 Easy

44 of 176Question permalink

Remembering +4 | -1

Which of the subshell has double dumb-bell shape?

View answer and explanation

Correct option: C

d

Option C (d) is correct because d-orbitals, except for

d_{z^2}

, have a double dumbbell shape.

Solution approach

Correct Answer + Core Reason: d. d-orbitals (except $d_{z^2}$ ) have a double dumbbell shape.

Step-by-Step Solution:

Step 1: Recall the shapes of s, p, d, and f orbitals.

Step 2: s-orbitals are spherical.

Step 3: p-orbitals are dumbbell-shaped.

Step 4: d-orbitals (except $d_{z^2}$ ) have a double dumbbell shape.

Step 5: f-orbitals have more complex shapes.

Step 6: Therefore, the subshell with a double dumbbell shape is the d subshell.

Why other options are wrong:
- Option A: s: s-orbitals are spherical, not double dumbbell-shaped.
- Option B: p: p-orbitals are dumbbell-shaped, not double dumbbell-shaped.
- Option D: f: f-orbitals have more complex shapes, not double dumbbell-shaped.

Key Concepts:
- Atomic orbitals (s, p, d, f)
- Shapes of atomic orbitals

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

Quick Tips:
- Know the shapes of the common atomic orbitals (s, p, d).

45 Easy

45 of 176Question permalink

Remembering +4 | -1

Which d-orbital does not have four lobes?

View answer and explanation

Correct option: C

{d_{{z^2}}}

Option C is correct because the dz² orbital has two lobes along the z-axis and a donut-shaped ring in the xy-plane, not four lobes.

Solution approach

Correct Answer + Core Reason: ${d_{{z^2}}}$ . The $d_{z^2}$ orbital does not have four lobes.

Step-by-Step Solution:

Step 1: Recall the shapes of the d-orbitals.
- $d_{x^2-y^2}$ : Has four lobes along the x and y axes.
- $d_{xy}$ : Has four lobes between the x and y axes.
- $d_{xz}$ : Has four lobes between the x and z axes.
- $d_{yz}$ : Has four lobes between the y and z axes.
- $d_{z^2}$ : Has two lobes along the z-axis and a donut-shaped ring in the xy-plane.

Step 2: Identify the d-orbital that does not have four lobes.
- The $d_{z^2}$ orbital has a different shape compared to the other d-orbitals; it has two lobes along the z-axis and a donut-shaped ring in the xy-plane, not four lobes.

Why other options are wrong:
- Option A: ${d_{{x^2} - {y^2}}}$ has four lobes.
- Option B: ${d_{xy}}$ has four lobes.
- Option D: ${d_{xz}}$ has four lobes.

Key Concepts:
- d-orbitals
- Shapes of atomic orbitals

Memory Aids:
- Visualize the shapes of the d-orbitals to remember their lobe arrangements.

Quick Tips:
- Remember that the $d_{z^2}$ orbital is unique in its shape compared to the other four d-orbitals.

46 Easy

46 of 176Question permalink

Understanding +4 | -1

Which one is not true for the cathode rays?

View answer and explanation

Correct option: D

They are electromagnetic waves

Option D (They are electromagnetic waves) is correct because cathode rays are streams of electrons, which are particles, not electromagnetic waves.

Solution approach

Correct Answer + Core Reason: They are electromagnetic waves. Cathode rays are streams of electrons, not EM waves.

Step-by-Step Solution:

Step 1: Recall the properties of cathode rays.

Step 2: Cathode rays are streams of electrons.

Step 3: Electrons are particles with mass and charge.

Step 4: Electromagnetic waves are disturbances in electric and magnetic fields.

Step 5: Cathode rays are not electromagnetic waves.

Why other options are wrong:
- Option A: Cathode rays have kinetic energy because they are moving electrons.
- Option B: Cathode rays cause fluorescence when they strike certain substances.
- Option C: Cathode rays travel in straight lines in the absence of external fields.

Key Concepts:
- Cathode rays
- Electrons
- Electromagnetic waves

Memory Aids:
- Cathode rays are streams of electrons.

Quick Tips:
- Remember that cathode rays are particles, not waves.

47 Easy

47 of 176Question permalink

Understanding +4 | -1

Which wave property is directly proportional to energy of electromagnetic radiation:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because energy of electromagnetic radiation is directly proportional to frequency, wave number, and indirectly proportional to wavelength, but since velocity is constant, it is also directly proportional.

Solution approach

Correct Answer + Core Reason: All of these. Energy is directly proportional to frequency, wave number, and velocity (since velocity is constant).

Step-by-Step Solution:

Step 1: Recall the relationship between energy and frequency of electromagnetic radiation: $E = hf$ , where E is energy, h is Planck's constant, and f is frequency.

Step 2: Recall the relationship between frequency, wavelength, and velocity: $c = f\lambda$ , where c is the speed of light and λ is the wavelength. Therefore, $f = c/\lambda$ .

Step 3: Wave number is defined as the reciprocal of wavelength: $\bar{\nu} = 1/\lambda$ . Since $f = c \bar{\nu}$ , energy is also directly proportional to wave number.

Step 4: Since velocity is constant, it is also directly proportional to energy.

Why other options are wrong:
- Option A (Velocity): While the velocity of electromagnetic radiation is constant in a given medium, it is still related to the energy through frequency and wavelength.
- Option B (Frequency): Energy is directly proportional to frequency, but it's not the only correct answer.
- Option C (Wave number): Energy is directly proportional to wave number, but it's not the only correct answer.

Key Concepts:
- Energy of electromagnetic radiation: $E = hf$
- Relationship between frequency, wavelength, and velocity: $c = f\lambda$
- Wave number: $\bar{\nu} = 1/\lambda$

Memory Aids:
- Remember the formula $E = hf$ and $c = f\lambda$ .

Quick Tips:
- Understand the relationships between energy, frequency, wavelength, and wave number.

48 Easy

48 of 176Question permalink

Understanding +4 | -1

Which of the following atoms has same number of protons and neutrons in its nucleus?

View answer and explanation

Correct option: B

Deuterium

Option B (Deuterium) is correct because Deuterium (²H) has 1 proton and 1 neutron in its nucleus.

Solution approach

Correct Answer + Core Reason: Deuterium. Deuterium has 1 proton and 1 neutron.

Step-by-Step Solution:

Step 1: Recall the composition of each atom's nucleus:
- Carbon: Usually 6 protons and 6 neutrons (¹²C).
- Deuterium: 1 proton and 1 neutron (²H).
- Tritium: 1 proton and 2 neutrons (³H).
- Nitrogen: Usually 7 protons and 7 neutrons (¹⁴N).

Step 2: Identify the atom with the same number of protons and neutrons.

Step 3: Deuterium has 1 proton and 1 neutron.

Why other options are wrong:
- Option A: Carbon has 6 protons and usually 6 neutrons.
- Option C: Tritium has 1 proton and 2 neutrons.
- Option D: Nitrogen has 7 protons and usually 7 neutrons.

Key Concepts:
- Atomic structure
- Protons and neutrons
- Isotopes

Memory Aids:
- Deuterium is an isotope of hydrogen with 1 proton and 1 neutron.

Quick Tips:
- Remember the composition of common isotopes.

49 Easy

49 of 176Question permalink

Analyzing +4 | -1

C a

has atomic no. 20 and atomic weight 40 . Which of the following statements is not correct about

{Ca}

atom

View answer and explanation

Correct option: C

The number of protons is half of the number of neutrons

Option C (The number of protons is half of the number of neutrons) is correct because Calcium has 20 protons and 20 neutrons, so the number of protons equals the number of neutrons.

Solution approach

Correct Answer + Core Reason: The number of protons is half of the number of neutrons. Calcium has an equal number of protons and neutrons.

Step-by-Step Solution:

Step 1: Determine the number of protons, neutrons, and electrons in a Calcium atom:
- Atomic number (number of protons) = 20
- Number of electrons = Number of protons = 20 (for a neutral atom)
- Number of neutrons = Atomic weight - Atomic number = 40 - 20 = 20
- Number of nucleons = Atomic weight = 40

Step 2: Evaluate each statement:
- A: The number of electrons is the same as the number of neutrons (20 = 20) - Correct
- B: The number of nucleons is double of the number of electrons (40 = 2 * 20) - Correct
- C: The number of protons is half of the number of neutrons (20 = 20/2) - Incorrect, the number of protons is equal to the number of neutrons.
- D: The number of nucleons is double of the atomic number (40 = 2 * 20) - Correct

Step 3: Identify the incorrect statement: Statement C is incorrect.

Step 4: Final Answer: The number of protons is half of the number of neutrons

Why other options are wrong:
- Option A: The number of electrons is same as the number of neutrons: This statement is correct for Calcium.
- Option B: The number of nucleons is double of the number of electrons: This statement is correct for Calcium.
- Option D: The number of nucleons isdouble of the atomic number: This statement is correct for Calcium.

Key Concepts:
- Atomic number, mass number, number of protons, neutrons, and electrons

Memory Aids:
- Remember that the atomic number is the number of protons.
- Remember that the mass number is the number of protons plus neutrons.

Quick Tips:
- Always remember that in a neutral atom, the number of protons equals the number of electrons.

50 Easy

50 of 176Question permalink

Understanding +4 | -1

In an atom wave mechanics suggests that electrons:

View answer and explanation

Correct option: C

Form diffused cloud around the nucleus

Option C (Form diffused cloud around the nucleus) is correct because wave mechanics describes electrons as probability distributions, forming electron clouds rather than fixed orbits.

Solution approach

Correct Answer + Core Reason: Form diffused cloud around the nucleus. Wave mechanics describes electrons as existing in probability distributions around the nucleus, forming electron clouds.

Step-by-Step Solution:

Step 1: Recall the basic postulates of wave mechanics (quantum mechanics). In wave mechanics, electrons are not considered to be moving in fixed orbits but are described by wave functions.

Step 2: Explain the concept of electron clouds. The square of the wave function gives the probability density of finding an electron at a particular point in space. This probability density forms a diffused cloud around the nucleus.

Step 3: Conclusion: Therefore, in an atom, wave mechanics suggests that electrons form a diffused cloud around the nucleus.

Why other options are wrong:
- Option A (Move around the nucleus in circular orbits): This is the Bohr model, which is an outdated model of the atom. Wave mechanics replaced this model.
- Option B (Move around the nucleus in elliptical orbits): This is the Sommerfeld model, which is also an outdated model of the atom. While it improved upon the Bohr model, it was still superseded by wave mechanics.
- Option D (None of the above): Option C is correct, so this option is incorrect.

Key Concepts:
- Wave Mechanics (Quantum Mechanics)
- Electron Cloud
- Atomic Orbitals

Memory Aids:
- Think of the electron as a fuzzy cloud, not a tiny ball orbiting the nucleus.

Quick Tips:
- Understand the difference between the Bohr model and the wave mechanical model of the atom.

51 Easy

51 of 176Question permalink

Application +4 | -1

If velocity of an electron in first Bohr orbit of Hydrogen atom is

x

, its velocity in 3rd orbit will be

View answer and explanation

Correct option: A

\frac{x}{3}

Option A (

\frac{x}{3}

) is correct because the velocity of an electron in the nth Bohr orbit is inversely proportional to n.

Solution approach

Correct Answer + Core Reason: $\frac{x}{3}$ . The velocity of an electron in the nth Bohr orbit is inversely proportional to n.

Step-by-Step Solution:

Step 1: Recall the formula for the velocity of an electron in the nth Bohr orbit: $v_n = \frac{v_1}{n}$ , where $v_1$ is the velocity in the first Bohr orbit and n is the orbit number.

Step 2: Given that the velocity in the first Bohr orbit (n=1) is x, so $v_1 = x$ .

Step 3: We want to find the velocity in the 3rd orbit (n=3), so $v_3 = \frac{v_1}{3} = \frac{x}{3}$ .

Why other options are wrong:
- Option B: $3x$ is incorrect because the velocity decreases as the orbit number increases.
- Option C: $9x$ is incorrect because the velocity decreases as the orbit number increases.
- Option D: $\frac{x}{9}$ is incorrect because the velocity is inversely proportional to n, not $n^2$ .

Key Concepts:
- Bohr's model
- Velocity of an electron in the nth orbit

Memory Aids:
- Remember that the velocity of an electron in the nth Bohr orbit is inversely proportional to n.

Quick Tips:
- Use the formula $v_n = \frac{v_1}{n}$ to quickly calculate the velocity in any Bohr orbit.

52 Easy

52 of 176Question permalink

Remembering +4 | -1

X-rays were discovered by :

View answer and explanation

Correct option: B

Roentgen

Option B is correct because Wilhelm Conrad Roentgen discovered X-rays in 1895.

Solution approach

Correct Answer + Core Reason: Roentgen. Wilhelm Conrad Roentgen discovered X-rays in 1895.

Step-by-Step Solution:

Step 1: Recall the history of X-ray discovery.

Step 2: Identify the scientist associated with the discovery of X-rays.

Step 3: Roentgen is the correct answer.

Why other options are wrong:
- Option A: Becquerel discovered radioactivity.
- Option C: Mme. Curie studied radioactivity.
- Option D: Van Laue contributed to X-ray diffraction.

Key Concepts:
- History of physics
- Discovery of X-rays

Memory Aids:
- Roentgen discovered X-rays.

Quick Tips:
- Remember key scientific discoveries and their discoverers.

53 Easy

53 of 176Question permalink

Remembering +4 | -1

A photoelectric cell is a device, which :

View answer and explanation

Correct option: A

Converts light into electricity

Option A (Converts light into electricity) is correct because a photoelectric cell's primary function is to convert light energy into electrical energy.

Solution approach

Correct Answer + Core Reason: Converts light into electricity. Photoelectric cells are designed to convert light energy into electrical energy.

Step-by-Step Solution:

Step 1: Define Photoelectric Cell: A photoelectric cell is a device that utilizes the photoelectric effect to generate an electric current when exposed to light.

Step 2: Functionality: When light (photons) strikes the surface of a suitable material (typically a metal), electrons are emitted. These emitted electrons can be collected to create an electric current.

Why other options are wrong:
- Option B: Converts electricity into light. This describes a light-emitting diode (LED), not a photoelectric cell.
- Option C: Stores lights. No device directly stores light. Some materials can exhibit phosphorescence or fluorescence, but they don't store light in the same way a battery stores electricity.
- Option D: Stores electricity. This describes a capacitor or battery, not a photoelectric cell.

Key Concepts:
- Photoelectric effect
- Conversion of energy

Memory Aids:
- Photo = Light, Electric = Electricity. Photoelectric cell converts light to electricity.

Quick Tips:
- Remember the basic principle of the photoelectric effect: light causes the emission of electrons.

54 Easy

54 of 176Question permalink

Understanding +4 | -1

Minimum de Broglie wavelength is associated with

View answer and explanation

Correct option: D

S{O_2}

molecule

Option D (

S{O_2}

molecule) is correct because, for the same velocity, heavier particles have shorter de Broglie wavelengths.

Solution approach

Correct Answer + Core Reason: $S{O_2}$ molecule. Minimum de Broglie wavelength is associated with the largest mass.

Step-by-Step Solution:

Step 1: State the de Broglie equation: $\lambda = \frac{h}{p} = \frac{h}{{mv}}$ , where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity.

Step 2: Analyze the equation: For a given velocity, the de Broglie wavelength is inversely proportional to the mass. Therefore, the larger the mass, the smaller the wavelength.

Step 3: Determine the masses of the given particles/molecules:
- Electron: Very small mass
- Proton: Intermediate mass
- $CO_2$ : Molar mass = 12 + 2(16) = 44 g/mol
- $SO_2$ : Molar mass = 32 + 2(16) = 64 g/mol

Step 4: Compare the masses: $SO_2$ has the largest mass.

Step 5: Final Answer: $S{O_2}$ molecule

Why other options are wrong:
- Option A: Electron: Electrons have the smallest mass, so they will have the largest de Broglie wavelength.
- Option B: Proton: Protons have a larger mass than electrons, but smaller than the molecules.
- Option C: $C{O_2}$ molecule: $CO_2$ has a smaller mass than $SO_2$ , so it will have a larger de Broglie wavelength than $SO_2$ .

Key Concepts:
- de Broglie equation
- Mass and wavelength relationship

Memory Aids:
- Larger mass = Shorter wavelength

Quick Tips:
- Remember that de Broglie wavelength is inversely proportional to mass. When comparing different particles, focus on their masses.

55 Easy

55 of 176Question permalink

Application +4 | -1

The radius of second Bohr orbit of H-atom is

View answer and explanation

Correct option: C

0.2116 nm

Option C (0.2116 nm) is correct because the radius of the nth Bohr orbit is given by

r_n = n^2 * a_0

, where

a_0

is the Bohr radius (0.0529 nm). For the second orbit (n=2),

r_2 = 2^2 * 0.0529 nm = 0.2116 nm

.

Solution approach

Correct Answer + Core Reason: 0.2116 nm. The radius of the second Bohr orbit is calculated using the formula $r_n = n^2 a_0$ .

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n = n^2 a_0$ , where $a_0$ is the Bohr radius (0.0529 nm).

Step 2: For the second Bohr orbit, n = 2.

Step 3: Substitute n = 2 into the formula: $r_2 = 2^2 \times 0.0529 nm = 4 \times 0.0529 nm = 0.2116 nm$ .

Why other options are wrong:
- Option A: 0.053 nm is the Bohr radius ( $a_0$ ) itself, which is the radius of the first Bohr orbit (n=1).
- Option B: 0.106 nm is incorrect. It might be a calculation error.
- Option D: 0.42 nm is incorrect. It might be a calculation error.

Key Concepts:
- Bohr's model
- Radius of the nth Bohr orbit
- Bohr radius ( $a_0$ )

Memory Aids:
- Remember the formula for the radius of the nth Bohr orbit: $r_n = n^2 a_0$ .

Quick Tips:
- Use the formula to quickly calculate the radius of any Bohr orbit.

56 Easy

56 of 176Question permalink

Understanding +4 | -1

The increasing order (lowest first) of the value of

\frac{e}{m}

for electron (e), proton (p),neutron (n) and alpha particle (α) will be

View answer and explanation

Correct option: A

n,\alpha ,p,e

Option A (

n,\alpha ,p,e

) is correct because the e/m ratio increases as the mass decreases, with the neutron having no charge and therefore a ratio of 0.

Solution approach

Correct Answer + Core Reason: $n,\alpha ,p,e$ . The e/m ratio is directly proportional to charge and inversely proportional to mass.

Step-by-Step Solution:

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

Step 2: Consider the neutron (n). The neutron has no charge (e=0), so its e/m ratio is 0.

Step 3: Consider the alpha particle ( $\alpha$ ). An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. Its charge is +2e and its mass is approximately 4 atomic mass units.

Step 4: Consider the proton (p). The proton has a charge of +e and a mass of approximately 1 atomic mass unit.

Step 5: Consider the electron (e). The electron has a charge of -e and a mass much smaller than a proton (approximately 1/1836 atomic mass units).

Step 6: Compare the e/m ratios. Since the neutron has no charge, it has the lowest e/m ratio (0). The electron has the smallest mass, so it has the highest e/m ratio. The alpha particle has a charge of 2e and a mass of 4, so its e/m ratio is half that of a proton. Therefore, the increasing order is neutron < alpha particle < proton < electron.

Step 7: Final Answer: $n,\alpha ,p,e$

Why other options are wrong:
- Option B: This is incorrect because it places the electron before the other particles, but the electron has the highest e/m ratio.
- Option C: This is incorrect because it places the electron before the alpha particle, but the electron has a higher e/m ratio.
- Option D: This is incorrect because it places the alpha particle before the proton, but the alpha particle has a lower e/m ratio.

Key Concepts:
- e/m ratio
- Charge and mass of subatomic particles

Memory Aids:
- Remember that the neutron is neutral and has no charge.
- The electron is much lighter than protons, neutrons, and alpha particles.

Quick Tips:
- When comparing e/m ratios, consider both the charge and the mass of the particles.

57 Easy

57 of 176Question permalink

Understanding +4 | -1

The KE of electron in

H{e^ + }

will be maximum in:

View answer and explanation

Correct option: C

1st orbit

Option C (1st orbit) is correct because the kinetic energy of an electron in a hydrogen-like species is inversely proportional to the square of the principal quantum number (n), so it's maximum at n=1.

Solution approach

Correct Answer + Core Reason: 1st orbit. The KE of electron in $H{e^ + }$ will be maximum in the 1st orbit.

Step-by-Step Solution:

Step 1: Recall the relationship between KE and orbit number: The kinetic energy (KE) of an electron in a hydrogen-like species is given by $KE = \frac{{kZe^2}}{{2r}}$ , where k is Coulomb's constant, Z is the atomic number, e is the electron charge, and r is the radius of the orbit.

Step 2: Relate radius to principal quantum number: The radius of the orbit is proportional to $n^2$ , where n is the principal quantum number: $r \propto n^2$ .

Step 3: Combine the relationships: Therefore, $KE \propto \frac{1}{{n^2}}$ .

Step 4: Determine the orbit with maximum KE: The KE is maximum when n is minimum, which is n = 1 (1st orbit).

Step 5: Final Answer: The KE of the electron in $H{e^ + }$ will be maximum in the 1st orbit.

Why other options are wrong:
- Option A: The KE is lower in the 3rd orbit compared to the 1st orbit.
- Option B: The KE is lower in the 2nd orbit compared to the 1st orbit.
- Option D: As n approaches infinity, the KE approaches zero.

Key Concepts:
- Kinetic energy of electron in an orbit
- Bohr's theory
- Principal quantum number (n)

Memory Aids:
- KE is inversely proportional to $n^2$ .

Quick Tips:
- Remember the relationship between energy, radius, and principal quantum number in Bohr's model.

58 Easy

58 of 176Question permalink

Remembering +4 | -1

Transition from

n = 4,5,6

to n=3 in hydrogen spectrum gives:

View answer and explanation

Correct option: B

Paschen series

Option B (Paschen series) is correct because the Paschen series corresponds to transitions from higher energy levels to n=3.

Solution approach

Correct Answer + Core Reason: Paschen series. This series corresponds to transitions to n=3.

Step-by-Step Solution:

Step 1: Recall the definitions of the different series in the hydrogen spectrum:
- 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: Identify the series corresponding to transitions to n=3. The Paschen series corresponds to transitions from $n = 4, 5, 6$ to $n = 3$ .

Why other options are wrong:
- Option A: Lyman series corresponds to transitions to n=1.
- Option C: Balmer series corresponds to transitions to n=2.
- Option D: Pfund series corresponds to transitions to n=5.

Key Concepts:
- Hydrogen spectrum
- Lyman series
- Balmer series
- Paschen series
- Brackett series
- Pfund series

Memory Aids:
- Remember the series and their corresponding final energy levels.

Quick Tips:
- Memorize the series and their corresponding final energy levels in the hydrogen spectrum.

59 Easy

59 of 176Question permalink

Application +4 | -1

If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is:

View answer and explanation

Correct option: D

{10^{12}}h

Option D (

{10^{12}}h

) is correct because the momentum of a photon is given by

p = \frac{h}{\lambda}

, where

\lambda

is the wavelength. Given

\lambda = 0.01 \AA = 0.01 \times 10^{-10} m = 10^{-12} m

, so

p = \frac{h}{10^{-12}} = 10^{12}h

.

Solution approach

Correct Answer + Core Reason: ${10^{12}}h$ . The momentum of a photon is inversely proportional to its wavelength.

Step-by-Step Solution:

Step 1: Recall the formula for the momentum of a photon: $p = \frac{h}{\lambda}$

Step 2: Convert the wavelength to meters: $\lambda = 0.01 \AA = 0.01 \times 10^{-10} \text{{ m}} = 10^{-12} \text{{ m}}$

Step 3: Substitute the wavelength into the formula: $p = \frac{h}{10^{-12}}$

Step 4: Simplify the expression: $p = 10^{12}h$

Why other options are wrong:
- Option A: ${10^{ - 2}}h$ : This is incorrect because it uses the wavelength directly without taking the inverse.
- Option B: h: This is incorrect because it assumes the wavelength is 1, which is not the case.
- Option C: ${10^2}h$ : This is incorrect because it uses an incorrect power of 10.

Key Concepts:
- Momentum of a photon
- Planck's constant
- Wavelength

Memory Aids:
- Remember the formula: $p = \frac{h}{\lambda}$

Quick Tips:
- Always convert the wavelength to meters before using the formula.

60 Easy

60 of 176Question permalink

Remembering +4 | -1

In the Schrödinger wave equation, ψ represents:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because ψ represents the wave function, which describes the behavior of an electron in an atom and can also be interpreted as an amplitude function related to the probability of finding the electron in a specific region of space, thus defining orbitals.

Solution approach

Correct Answer + Core Reason: All of these. $\psi$ represents the wave function, amplitude function, and defines orbitals.

Step-by-Step Solution:

Step 1: Recall the meaning of $\psi$ in the context of quantum mechanics.

Step 2: $\psi$ is the wave function, which describes the quantum state of a particle.

Step 3: $\psi$ can also be interpreted as an amplitude function, where the square of the amplitude ( $|\psi|^2$ ) gives the probability density of finding the particle at a particular location.

Step 4: The spatial distribution of the wave function defines the shape and properties of atomic orbitals.

Step 5: Therefore, $\psi$ represents all of the given options.

Why other options are wrong:
- Option A: Orbitals: While $\psi$ is related to orbitals, it is not solely orbitals. It also represents the wave function and amplitude function.
- Option B: Wave function: While $\psi$ is the wave function, it also represents the amplitude function and defines orbitals.
- Option C: Amplitude function: While $\psi$ is the amplitude function, it also represents the wave function and defines orbitals.

Key Concepts:
- Wave function ( $\psi$ )
- Amplitude function
- Atomic orbitals
- Quantum mechanics

Memory Aids:
- Remember that $\psi$ is a fundamental concept in quantum mechanics and has multiple interpretations.

Quick Tips:
- Understand the different interpretations of the wave function.

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