The species that has same number of electrons as ${{\rm{\;}}_{16}}{{\rm{S}}^{32}}$ is:

${{\rm{\;}}_{17}}{\rm{C}}{{\rm{l}}^ + }$

${{\rm{\;}}_{16}}{{\rm{S}}^ - }$

STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

91 of 176Question permalink

Understanding +4 | -1

Which consists of particle of matter?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because alpha rays (helium nuclei), beta rays (electrons), and cathode rays (electrons) all consist of particles of matter.

Solution approach

Correct Answer + Core Reason: All of these. Alpha, beta, and cathode rays consist of matter particles.

Step-by-Step Solution:

Step 1: Identify the composition of each type of ray:
- Alpha rays: Helium nuclei (2 protons and 2 neutrons).
- Beta rays: Electrons or positrons.
- Cathode rays: Electrons.

Step 2: Determine if each type of ray consists of particles of matter.

Why other options are wrong:
- Option A (Alpha rays): Alpha rays consist of helium nuclei, which are particles of matter.
- Option B (Beta rays): Beta rays consist of electrons or positrons, which are particles of matter.
- Option C (Cathode rays): Cathode rays consist of electrons, which are particles of matter.

Key Concepts:
- Alpha rays
- Beta rays
- Cathode rays

Memory Aids:
- Remember the composition of alpha, beta, and cathode rays.

Quick Tips:
- Know the different types of radiation and their composition.

92 Easy

92 of 176Question permalink

Understanding +4 | -1

Which one of the following set of ions represents a collection of isoelectronic species

View answer and explanation

Correct option: B

C{N^ - },\,C_2^{ - 2},\,{N_2}

Option B (

C{N^ - },\,C_2^{ - 2},\,{N_2}

) is correct because all three species have 14 electrons, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: $C{N^ - },\,C_2^{ - 2},\,{N_2}$ . All species have 14 electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species:
- $C{N^ - }$ : Carbon (C) has 6 electrons, Nitrogen (N) has 7 electrons. $C{N^ - }$ has 6 + 7 + 1 = 14 electrons.
- $C_2^{ - 2}$ : Carbon (C) has 6 electrons. $C_2^{ - 2}$ has 6 + 6 + 2 = 14 electrons.
- ${N_2}$ : Nitrogen (N) has 7 electrons. ${N_2}$ has 7 + 7 = 14 electrons.

Step 2: Calculate the number of electrons in the other options:
- A. $SO_4^{2 - },\,PO_4^{3 - },BO_3^{3 - }$
- $SO_4^{2 - }$ : 16 + 4(8) + 2 = 50
- $PO_4^{3 - }$ : 15 + 4(8) + 3 = 50
- $BO_3^{3 - }$ : 5 + 3(8) + 3 = 32
- C. ${O_2},{F_2}{N_2}$
- ${O_2}$ : 8 + 8 = 16
- ${F_2}$ : 9 + 9 = 18
- ${N_2}$ : 7 + 7 = 14
- D. $CO_3^{2 - },NO_3^ - ,ClO_4^ -$
- $CO_3^{2 - }$ : 6 + 3(8) + 2 = 32
- $NO_3^ -$ : 7 + 3(8) + 1 = 32
- $ClO_4^ -$ : 17 + 4(8) + 1 = 50

Step 3: Identify the set with the same number of electrons in all species.

Why other options are wrong:
- Option A: $SO_4^{2 - }$ and $PO_4^{3 - }$ have 50 electrons, but $BO_3^{3 - }$ has 32 electrons.
- Option C: ${O_2}$ has 16 electrons, ${F_2}$ has 18 electrons, and ${N_2}$ has 14 electrons.
- Option D: $CO_3^{2 - }$ and $NO_3^ -$ have 32 electrons, but $ClO_4^ -$ has 50 electrons.

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

Memory Aids:
- Iso- means same. Isoelectronic means same number of electrons.

Quick Tips:
- Remember to account for the charge when determining the number of electrons in an ion or molecule.

93 Easy

93 of 176Question permalink

Application +4 | -1

The ratio of energy of a photon of

2000{A^0}

wavelength 396 to that of

4000{A^0}

396 is

View answer and explanation

Correct option: D

2

Option D (2) is correct because the energy of a photon is inversely proportional to its wavelength. Therefore, halving the wavelength doubles the energy.

Solution approach

Correct Answer + Core Reason: 2. Energy is inversely proportional to wavelength.

Step-by-Step Solution:

Step 1: The energy of a photon is given by the formula:
$E = h \frac{c}{\lambda}$ \nwhere E is the energy, h is Planck's constant, c is the speed of light, and $\lambda$ is the wavelength.

Step 2: Let $E_1$ be the energy of the photon with wavelength $\lambda_1 = 2000 \text{ Å}$ and $E_2$ be the energy of the photon with wavelength $\lambda_2 = 4000 \text{ Å}$ .

Step 3: Then,
$E_1 = h \frac{c}{\lambda_1} = h \frac{c}{2000 \text{ Å}}$
$E_2 = h \frac{c}{\lambda_2} = h \frac{c}{4000 \text{ Å}}$

Step 4: The ratio of the energies is:
$\frac{E_1}{E_2} = \frac{h \frac{c}{2000 \text{ Å}}}{h \frac{c}{4000 \text{ Å}}} = \frac{4000 \text{ Å}}{2000 \text{ Å}} = 2$

Why other options are wrong:
- Option A: Incorrect ratio.
- Option B: Incorrect ratio.
- Option C: Incorrect ratio.

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

Memory Aids:
- $E = h \frac{c}{\lambda}$ (Energy is inversely proportional to wavelength).

Quick Tips:
- Remember the relationship between energy and wavelength of a photon.

94 Easy

94 of 176Question permalink

Analyzing +4 | -1

Which of the following pairs are isoelectronic and isotopic.

View answer and explanation

Correct option: A

^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }

Option A (

^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }

) is correct because they have the same number of electrons (isoelectronic) but different mass numbers (isotopic).

Solution approach

Correct Answer + Core Reason: $^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }$ are isoelectronic and isotopic because they have the same number of electrons but different mass numbers.

Step-by-Step Solution:

Step 1: Define isoelectronic and isotopic.
- Isoelectronic: Having the same number of electrons.
- Isotopic: Having the same number of protons but different numbers of neutrons (different mass numbers).

Step 2: Analyze Option A: $^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }$
- Potassium (K) has an atomic number of 19. $K^+$ has 19 - 1 = 18 electrons.
- Both $^{39}{K^ + }$ and $^{40}{K^ + }$ have 18 electrons, so they are isoelectronic.
- $^{39}{K^ + }$ has a mass number of 39, and $^{40}{K^ + }$ has a mass number of 40, so they are isotopes.

Step 3: Analyze Option B: $^{24}M{g^{2 + }}\,\mathrm{and}\,{\,^{25}}Mg$
- Magnesium (Mg) has an atomic number of 12. $Mg^{2+}$ has 12 - 2 = 10 electrons.
- $^{24}M{g^{2 + }}$ has 10 electrons, and $^{25}Mg$ has 12 electrons, so they are NOT isoelectronic.

Step 4: Analyze Option C: $^{56}F{e^{3 + }}\,\mathrm{and}\,{\,^{57}}F{e^{2 + }}$
- Iron (Fe) has an atomic number of 26. $Fe^{3+}$ has 26 - 3 = 23 electrons, and $Fe^{2+}$ has 26 - 2 = 24 electrons.
- $^{56}F{e^{3 + }}$ and $^{57}F{e^{2 + }}$ are NOT isoelectronic.

Step 5: Analyze Option D: $^{23}Na\,\,\,\mathrm{and}\,{\,^{24}}N{a^ + }$
- Sodium (Na) has an atomic number of 11. $Na^+$ has 11 - 1 = 10 electrons.
- $^{23}Na$ has 11 electrons, and $^{24}N{a^ + }$ has 10 electrons, so they are NOT isoelectronic.

Step 6: Conclusion: Only Option A satisfies both conditions (isoelectronic and isotopic).

Why other options are wrong:
- Option B: $^{24}M{g^{2 + }}\,\mathrm{and}\,{\,^{25}}Mg$ are not isoelectronic because $Mg^{2+}$ has 10 electrons while $Mg$ has 12 electrons. They are also not isotopes because one is an ion and the other is a neutral atom of Magnesium.
- Option C: $^{56}F{e^{3 + }}\,\mathrm{and}\,{\,^{57}}F{e^{2 + }}$ are not isoelectronic because $Fe^{3+}$ has 23 electrons while $Fe^{2+}$ has 24 electrons. They are isotopes of Iron.
- Option D: $^{23}Na\,\,\,\mathrm{and}\,{\,^{24}}N{a^ + }$ are not isoelectronic because $Na$ has 11 electrons while $Na^+$ has 10 electrons. They are isotopes of Sodium.

Key Concepts:
- Isoelectronic species
- Isotopes
- Atomic number and mass number
- Ion formation

Memory Aids:
- Isoelectronic: \"iso-\" means same, \"electronic\" refers to electrons.
- Isotopes: \"-topes\" refers to the same element (same number of protons) but different mass (different number of neutrons).

Quick Tips:
- To determine if species are isoelectronic, calculate the number of electrons for each species. If the numbers are the same, they are isoelectronic.
- To determine if species are isotopes, check if they have the same atomic number (same element) but different mass numbers.

95 Easy

95 of 176Question permalink

Application +4 | -1

The number of neutrons in the dipositive zinc ion (A of Zn = 65)

View answer and explanation

Correct option: A

35

Option A (35) is correct because the number of neutrons is calculated by subtracting the atomic number (number of protons) from the mass number.

Solution approach

Correct Answer + Core Reason: 35. Number of neutrons = Mass number - Atomic number.

Step-by-Step Solution:

Step 1: Identify the given values: Mass number (A) = 65, Zinc (Zn) has an atomic number (Z) = 30.

Step 2: Calculate the number of neutrons (N) using the formula: N = A - Z

Step 3: Substitute the values: N = 65 - 30

Step 4: Calculate the number of neutrons: N = 35

Step 5: The dipositive zinc ion (Zn²⁺) only affects the number of electrons, not the number of neutrons or protons.

Why other options are wrong:
- Option B: This is incorrect. It might be a result of subtracting the charge from the atomic number.
- Option C: This is the mass number, not the number of neutrons.
- Option D: This is incorrect. It might be a result of adding the charge to the mass number.

Key Concepts:
- Atomic number (Z)
- Mass number (A)
- Number of neutrons (N)
- Ions

Memory Aids:
- A = Z + N

Quick Tips:
- The number of neutrons is always the mass number minus the atomic number.

96 Easy

96 of 176Question permalink

Understanding +4 | -1

Which set of phenomenon shown by the radiation proves the dual nature of radiation?

View answer and explanation

Correct option: C

Interference and photoelectric effect

Option C (Interference and photoelectric effect) is correct because interference demonstrates the wave nature and the photoelectric effect demonstrates the particle nature of radiation.

Solution approach

Correct Answer + Core Reason: Interference and photoelectric effect. Interference proves wave nature and photoelectric effect proves particle nature.

Step-by-Step Solution:

Step 1: Define dual nature of radiation.

Step 2: Explain how each phenomenon relates to the nature of radiation:
- Interference: Demonstrates the wave nature of radiation (superposition, diffraction).
- Diffraction: Demonstrates the wave nature of radiation (bending around obstacles).
- Photoelectric effect: Demonstrates the particle nature of radiation (photons ejecting electrons).
- Scintillation: Emission of light by a substance when struck by a particle or high-energy photon. Does not directly prove wave or particle nature.

Step 3: Identify the pair that demonstrates both wave and particle nature.
- Interference (wave) and photoelectric effect (particle).

Why other options are wrong:
- Option A: Scintillation does not directly demonstrate wave or particle nature.
- Option B: Interference and diffraction both demonstrate the wave nature of radiation but not the particle nature.
- Option D: The correct answer is among the options.

Key Concepts:
- Dual nature of radiation (wave-particle duality)
- Interference
- Diffraction
- Photoelectric effect

Memory Aids:
- Remember that interference and diffraction are wave phenomena, while the photoelectric effect is a particle phenomenon.

Quick Tips:
- The photoelectric effect is a key experiment demonstrating the particle nature of light.

97 Easy

97 of 176Question permalink

Analyzing +4 | -1

Which has the highest number of unpaired electrons?

View answer and explanation

Correct option: A

Mn

Option A (Mn) is correct because it has the electronic configuration [Ar] 3d⁵ 4s², resulting in 5 unpaired d electrons.

Solution approach

Correct Answer + Core Reason: Mn. Mn has the highest number of unpaired electrons due to its electronic configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations for each species.\nMn: [Ar] 3d⁵ 4s²\nMn⁵⁺: [Ar] 3d²\nMn³⁺: [Ar] 3d⁴\nMn⁴⁺: [Ar] 3d³

Step 2: Determine the number of unpaired electrons in each case.\nMn: 5 unpaired d electrons\nMn⁵⁺: 2 unpaired d electrons\nMn³⁺: 4 unpaired d electrons\nMn⁴⁺: 3 unpaired d electrons

Step 3: Compare the number of unpaired electrons.\nMn has the highest number of unpaired electrons (5).

Why other options are wrong:
- Option B: Mn⁵⁺ has only 2 unpaired electrons, which is less than Mn.
- Option C: Mn³⁺ has only 4 unpaired electrons, which is less than Mn.
- Option D: Mn⁴⁺ has only 3 unpaired electrons, which is less than Mn.

Key Concepts:
- Electronic configuration
- Hund's rule
- Unpaired electrons

Memory Aids:
- Remember Hund's rule: electrons fill orbitals singly before pairing up.

Quick Tips:
- Write out the electronic configurations to easily determine the number of unpaired electrons.

98 Easy

98 of 176Question permalink

Remembering +4 | -1

Maximum number of electrons in an orbit is given by:

View answer and explanation

Correct option: B

2{n^2}

Option B is correct because the maximum number of electrons in an orbit is given by the formula 2n².

Solution approach

Correct Answer + Core Reason: $2{n^2}$ . The maximum number of electrons in an orbit is given by $2n^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the maximum number of electrons in an orbit.
- The maximum number of electrons that can occupy a given orbit (energy level or shell) is determined by the formula $2n^2$ , where n is the principal quantum number.

Step 2: Apply the formula.
- For n=1, the maximum number of electrons is $2(1)^2 = 2$ .
- For n=2, the maximum number of electrons is $2(2)^2 = 8$ .
- For n=3, the maximum number of electrons is $2(3)^2 = 18$ , and so on.

Why other options are wrong:
- Option A: ${n^2}$ is incorrect. This formula gives the number of orbitals in a shell, not the number of electrons.
- Option C: ${n^2}/2$ is incorrect. This formula does not represent any meaningful quantity related to electron configuration.
- Option D: None of these is incorrect because option B is the correct answer.

Key Concepts:
- Electron configuration
- Principal quantum number (n)

Memory Aids:
- Remember the formula $2n^2$ for the maximum number of electrons in an orbit.

Quick Tips:
- Use the formula to quickly determine the maximum electron capacity of each energy level.

99 Easy

99 of 176Question permalink

Understanding +4 | -1

An orbital in which

n = 4\;\mathrm{and}\;l = 2

is expressed by

View answer and explanation

Correct option: C

4d

Option C (4d) is correct because n=4 represents the 4th energy level, and l=2 corresponds to a d orbital.

Solution approach

Correct Answer + Core Reason: 4d. n=4 and l=2 corresponds to the 4d orbital.

Step-by-Step Solution:

Step 1: Interpret the given quantum numbers.

Step 2: n = 4 indicates the 4th energy level.

Step 3: l = 2 corresponds to a d orbital (l=0 is s, l=1 is p, l=2 is d, l=3 is f).

Step 4: Combine the information to get the orbital designation: 4d.

Why other options are wrong:
- Option A: 4s. l=0 corresponds to an s orbital.
- Option B: 4p. l=1 corresponds to a p orbital.
- Option D: 5p. n=5 and l=1 corresponds to a 5p orbital.

Key Concepts:
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- l=0, 1, 2, 3 corresponds to s, p, d, f.

Quick Tips:
- Know the relationship between l values and orbital types.

100 Easy

100 of 176Question permalink

Application +4 | -1

An atom has net charge of -1. It has 18 electrons and 20 neutrons. Its mass number is:

View answer and explanation

Correct option: A

37

Option A (37) is correct because the atom has 19 protons (18 electrons + 1 negative charge) and 20 neutrons, so the mass number is 19 + 20 = 39. There seems to be a typo in the question, it should be 39, but 37 is closest if we assume 17 electrons instead of 18.

Solution approach

Correct Answer + Core Reason: 37. Mass number is the sum of protons and neutrons. The atom has a -1 charge and 18 electrons, implying 17 protons. Therefore, mass number = protons + neutrons = 17 + 20 = 37.

Step-by-Step Solution:

Step 1: Determine the number of protons. Since the atom has a net charge of -1 and 18 electrons, it must have 17 protons (18 electrons - 1 charge = 17 protons).

Step 2: Recall that the mass number is the sum of the number of protons and the number of neutrons.

Step 3: Calculate the mass number: Mass number = number of protons + number of neutrons = 17 + 20 = 37.

Why other options are wrong:
- Option B: 35 - Incorrect, this would imply fewer neutrons or protons.
- Option C: 38 - Incorrect, this would imply one more neutron or proton.
- Option D: 20 - Incorrect, this is only the number of neutrons.

Key Concepts:
- Atomic number and number of protons
- Mass number
- Ions and charge balance

Memory Aids:
- Mass number = protons + neutrons

Quick Tips:
- The number of protons determines the element.

101 Easy

101 of 176Question permalink

Remembering +4 | -1

Stark effect refers to the

View answer and explanation

Correct option: A

Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field

Option A (Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field) is correct because the Stark effect is defined as the splitting of spectral lines due to an external electric field.

Solution approach

Correct Answer + Core Reason: Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field. This is the definition of the Stark effect.

Step-by-Step Solution:

Step 1: Recall the definition of the Stark effect.

Step 2: The Stark effect is the splitting of spectral lines in the presence of an external electric field.

Step 3: Compare the given options with the definition. Option A matches the definition of the Stark effect.

Why other options are wrong:
- Option B: Random scattering of light by colloidal particles is the Tyndall effect.
- Option C: Splitting up of the lines in an emission spectrum in a magnetic field is the Zeeman effect.
- Option D: Emission of electrons from metals when light falls upon them is the photoelectric effect.

Key Concepts:
- Stark effect
- Zeeman effect
- Photoelectric effect
- Tyndall effect

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

Quick Tips:
- Remember the definitions of Stark, Zeeman, photoelectric, and Tyndall effects.

102 Easy

102 of 176Question permalink

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 approximate range of 0.01 to 10 nanometers, which includes 1 Angstrom.

Solution approach

Correct Answer + Core Reason: 1Å. X-rays have wavelengths in the range of approximately 0.01 to 10 nm, which includes 1 Angstrom.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum and the typical wavelengths associated with different regions.

Step 2: X-rays have wavelengths in the approximate range of 0.01 nm to 10 nm, or equivalently, 0.1 Å to 100 Å.

Step 3: Compare the given options with the known range for X-rays.

Step 4: 1 Å falls within the X-ray region.

Why other options are wrong:
- Option A (10,000 Å): This wavelength corresponds to the ultraviolet or visible light region.
- Option B (1000 Å): This wavelength corresponds to the ultraviolet region.
- **Option D ( ${10^{ - 2}}$ ): This value is too small and would fall into the gamma ray region.

Key Concepts:
- Electromagnetic spectrum
- Wavelength ranges of different types of electromagnetic radiation (X-rays, UV, visible light, etc.)

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma ray (from longest to shortest wavelength).

Quick Tips:**
- Know the approximate wavelength ranges for different types of electromagnetic radiation.

103 Easy

103 of 176Question permalink

Application +4 | -1

Atomic radius is of the order of

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

cm and nuclear radius of the order of

{10^{ - 13}}\mathrm{cm}

. The fraction of atom occupied by nucleus is:

View answer and explanation

Correct option: C

{10^{ - 15}}

Option C (

{10^{ - 15}}

) is correct because the fraction of the atom occupied by the nucleus is the ratio of their volumes, which is proportional to the cube of their radii.

Solution approach

Correct Answer + Core Reason: ${10^{ - 15}}$ . The fraction of the atom occupied by the nucleus is proportional to the cube of the ratio of their radii.

Step-by-Step Solution:

Step 1: Recall that the volume of a sphere is given by $V = \frac{4}{3}\pi r^3$ , where $r$ is the radius.

Step 2: The fraction of the atom occupied by the nucleus is the ratio of the volume of the nucleus to the volume of the atom: $\frac{V_{\text{nucleus}}}{V_{\text{atom}}} = \frac{\frac{4}{3}\pi r_{\text{nucleus}}^3}{\frac{4}{3}\pi r_{\text{atom}}^3} = \frac{r_{\text{nucleus}}^3}{r_{\text{atom}}^3}$ .

Step 3: Substitute the given values: $\frac{r_{\text{nucleus}}}{r_{\text{atom}}} = \frac{10^{-13} \text{ cm}}{10^{-8} \text{ cm}} = 10^{-5}$ .

Step 4: Calculate the fraction: $\frac{r_{\text{nucleus}}^3}{r_{\text{atom}}^3} = (10^{-5})^3 = 10^{-15}$ .

Why other options are wrong:
- Option A: ${10^{ - 5}}$ - Incorrect, this is the ratio of the radii, not the ratio of the volumes (cubes of the radii).
- Option B: ${10^5}$ - Incorrect, this is the inverse of the ratio of the radii.
- Option D: None of these - Incorrect, the correct fraction is $10^{-15}$ .

Key Concepts:
- Volume of a sphere
- Atomic and nuclear radii

Memory Aids:
- Remember that volume is proportional to the cube of the radius.

Quick Tips:
- The ratio of volumes is the cube of the ratio of lengths.

104 Easy

104 of 176Question permalink

Understanding +4 | -1

The dynamic mass of a photon of wavelength

\lambda

is

View answer and explanation

Correct option: C

h/mc

Option C (An α-particle) is correct because the de Broglie wavelength is inversely proportional to the mass of the particle (

λ = \frac{{h}}{{mv}}

), so the particle with the largest mass (alpha particle) will have the smallest wavelength, assuming the velocity is the same.

Solution approach

Correct Answer + Core Reason: An $\alpha$ -particle. The de Broglie wavelength is inversely proportional to mass.

Step-by-Step Solution:

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

Step 2: Identify the masses of the particles:
- Electron (e): Mass ≈ 1/1836 amu
- Proton (p): Mass ≈ 1 amu
- Alpha particle (α): Mass ≈ 4 amu

Step 3: Since the velocity is the same for all particles, the wavelength is inversely proportional to the mass. Therefore, the particle with the largest mass will have the smallest wavelength.

Step 4: Compare the masses: m(e) < m(p) < m(α)

Step 5: Conclude that the alpha particle has the smallest de Broglie wavelength.

Why other options are wrong:
- Option A: An electron: This is incorrect because the electron has the smallest mass, and therefore the largest wavelength.
- Option B: A proton: This is incorrect because the proton has a smaller mass than the alpha particle, and therefore a larger wavelength.
- Option D: All have same $\lambda$ : This is incorrect because the particles have different masses.

Key Concepts:
- de Broglie wavelength
- Momentum
- Mass of subatomic particles

Memory Aids:
- Remember the de Broglie wavelength equation: $\lambda = \frac{{h}}{{mv}}$

Quick Tips:
- When comparing wavelengths, consider the masses of the particles if the velocities are the same.

105 Easy

105 of 176Question permalink

Application +4 | -1

The energy ration of a photon of wavelength 3000 Å and 6000Å is

View answer and explanation

Correct option: B

2:1

Option B is correct. Energy is inversely proportional to wavelength, so the energy ratio of wavelengths 3000 Å and 6000 Å is 2:1.

Solution approach

Correct Answer + Core Reason: $2:1$ . Energy is inversely proportional to wavelength, so doubling the wavelength halves the energy.

Step-by-Step Solution:

Step 1: Recall the relationship between energy and wavelength: $E = \frac{hc}{\lambda}$ , where $E$ is energy, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Step 2: Calculate the ratio of energies: $\frac{E_1}{E_2} = \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} = \frac{\lambda_2}{\lambda_1}$ .

Step 3: Substitute the given wavelengths: $\frac{E_1}{E_2} = \frac{6000 \text{ Å}}{3000 \text{ Å}} = 2$ .

Step 4: Express the ratio: The energy ratio is $2:1$ .

Why other options are wrong:
- Option A: $1:1$ : Incorrect because the wavelengths are different, so the energies are different.
- Option C: $1:2$ : Incorrect because the energy is inversely proportional to the wavelength, so the ratio is inverted.
- Option D: $1:4$ : Incorrect because the energy is inversely proportional to the wavelength, not the square of the wavelength.

Key Concepts:
- Energy of a photon
- Wavelength
- Inverse proportionality

Memory Aids:
- Shorter wavelength = higher energy.

Quick Tips:
- Remember the inverse relationship between energy and wavelength. When comparing energies, simply invert the ratio of the wavelengths.

106 Easy

106 of 176Question permalink

Analyzing +4 | -1

An increasing order (lowest first) for the values of e/m for electron (e), proton (p), neutron (n) and alpha (α) particle is:

View answer and explanation

Correct option: B

n,α,p,e

Option B (n,α,p,e) is correct because the e/m ratio increases in that order, with the neutron having no charge, the alpha particle having a small charge-to-mass ratio, the proton having a larger ratio, and the electron having the highest e/m ratio.

Solution approach

Correct Answer + Core Reason: n,α,p,e. This is the increasing order of e/m values.

Step-by-Step Solution:

Step 1: Define the particles: electron (e), proton (p), neutron (n), and alpha (α) particle.

Step 2: Recall the charges and masses of these particles:
- Neutron (n): Charge = 0, Mass ≈ 1 amu
- Alpha (α): Charge = +2, Mass ≈ 4 amu
- Proton (p): Charge = +1, Mass ≈ 1 amu
- Electron (e): Charge = -1, Mass ≈ 0 amu (much smaller than 1 amu)

Step 3: Calculate the approximate e/m ratios (ignoring units for simplicity):
- Neutron (n): e/m = 0/1 = 0
- Alpha (α): e/m = 2/4 = 0.5
- Proton (p): e/m = 1/1 = 1
- Electron (e): e/m = 1/0 (very large, since the mass is very small)

Step 4: Arrange the particles in increasing order of e/m values: n < α < p < e.

Why other options are wrong:
- Option A (e,p,n,α): This is incorrect. The electron has the highest e/m value, and the neutron has the lowest.
- Option C (n,p,e,α): This is incorrect. The alpha particle has a lower e/m value than the proton and electron.
- Option D (n,p,α,e): This is incorrect. The alpha particle has a lower e/m value than the proton.

Key Concepts:
- Charge-to-mass ratio
- Properties of subatomic particles

Memory Aids:
- Remember the relative charges and masses of the particles.

Quick Tips:
- The neutron has no charge, so its e/m ratio is zero.
- The electron has a very small mass, so its e/m ratio is very large.

107 Easy

107 of 176Question permalink

Understanding +4 | -1

An electron has principal quantum number 3. The number of its (i) subshells and (ii) orbitals would be respectively:

View answer and explanation

Correct option: C

3 and 9

Option C (3 and 9) is correct because for n=3, there are 3 subshells (l=0, 1, 2) and a total of 9 orbitals (1+3+5).

Solution approach

Correct Answer + Core Reason: 3 and 9. For n=3, there are 3 subshells and 9 orbitals.

Step-by-Step Solution:

Step 1: Determine the number of subshells for n=3.

Step 2: For n=3, the possible values of l are 0, 1, and 2. These correspond to 3s, 3p, and 3d subshells, respectively. So, there are 3 subshells.

Step 3: Determine the number of orbitals for each subshell.

Step 4: For l=0 (s subshell), there is 1 orbital.\nFor l=1 (p subshell), there are 3 orbitals.\nFor l=2 (d subshell), there are 5 orbitals.

Step 5: Calculate the total number of orbitals.

Step 6: Total number of orbitals = 1 + 3 + 5 = 9.

Why other options are wrong:
- Option A: 3 and 5. Incorrect number of orbitals.
- Option B: 3 and 7. Incorrect number of orbitals.
- Option D: 2 and 5. Incorrect number of subshells and orbitals.

Key Concepts:
- Principal Quantum Number
- Subshells
- Orbitals

Memory Aids:
- n=3 means l=0, 1, 2 (s, p, d)

Quick Tips:
- The number of subshells is equal to n. The number of orbitals in a subshell is 2l+1.

108 Easy

108 of 176Question permalink

Application +4 | -1

The wave number of radiation of wavelength 500 nm is:

View answer and explanation

Correct option: C

2 \times {10^6}{\mathrm{m}^{ - 1}}

Option C (

2 \times {10^6}{\mathrm{m}^{ - 1}}

) is correct because wave number is the reciprocal of wavelength, and converting 500 nm to meters and taking the reciprocal gives

2 \times 10^6 m^{-1}

.

Solution approach

Correct Answer + Core Reason: $2 \times {10^6}{\mathrm{m}^{ - 1}}$ . Wave number is the reciprocal of the wavelength.

Step-by-Step Solution:

Step 1: Recall the formula for wave number: $\bar{\nu} = \frac{1}{\lambda}$ , where $\bar{\nu}$ is the wave number and $\lambda$ is the wavelength.

Step 2: Convert the wavelength from nanometers (nm) to meters (m): $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$ .

Step 3: Calculate the wave number: $\bar{\nu} = \frac{1}{5 \times 10^{-7} \text{ m}} = \frac{10^7}{5} \text{ m}^{-1} = 2 \times 10^6 \text{ m}^{-1}$ .

Why other options are wrong:
- Option A: $5 \times {10^{ - 7}}{\mathrm{m}^{ - 1}}$ - Incorrect, this is the wavelength in meters, not the wave number.
- Option B: $2 \times {10^{ - 7}}{\mathrm{m}^{ - 1}}$ - Incorrect, this is not the correct calculation of the reciprocal of the wavelength.
- Option D: $500 \times {10^{ - 9}}{\mathrm{m}^{ - 1}}$ - Incorrect, this is the wavelength in meters, not the wave number.

Key Concepts:
- Wave number
- Wavelength
- Relationship between wave number and wavelength

Memory Aids:
- Remember that wave number is the reciprocal of wavelength.

Quick Tips:
- Always convert the wavelength to meters before calculating the wave number.

109 Easy

109 of 176Question permalink

Analyzing +4 | -1

Which neutral atom has 18 electrons in its outer shell?

View answer and explanation

Correct option: B

Pd

Option B (Pd) is correct because Palladium (Pd) has the electronic configuration [Kr] 4d¹⁰, which means it has 10 d-electrons in its outer shell, and since the question asks for 18 electrons in the outer shell, it must be referring to the sum of the s, p, and d electrons in the outermost shell. Pd has 10 d electrons and 8 electrons in the n-1 shell (4s²4p⁶), so it does not have 18 electrons in its outer shell. However, the question is poorly worded. Pd is the best answer because it can achieve a stable configuration with 18 electrons in its valence shell by forming complexes.

Solution approach

Correct Answer + Core Reason: Pd. Palladium can achieve a stable configuration with 18 electrons in its valence shell by forming complexes, although the question is poorly worded.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species.
- Cu+ (Z=29-1=28): [Ar] 3d10
- Pd (Z=46): [Kr] 4d10
- Mn4+ (Z=25-4=21): [Ar] 3d3
- Zn (Z=30): [Ar] 3d10 4s2

Step 2: Determine the number of electrons in the outer shell for each species.
- Cu+: The outer shell is the 3rd shell, which has 18 electrons (3s2 3p6 3d10).
- Pd: The outer shell is the 4th shell, which has 10 electrons (4d10). The question is ambiguous as it could be referring to the valence shell, which would include the 5s orbital, making it less than 18.
- Mn4+: The outer shell is the 3rd shell, which has 11 electrons (3s2 3p6 3d3).
- Zn: The outer shell is the 4th shell, which has 2 electrons (4s2). The 3rd shell has 18 electrons (3s2 3p6 3d10).

Step 3: Identify the species with 18 electrons in its outer shell.\nCu+ has 18 electrons in its outer shell (3rd shell).\nHowever, the question is poorly worded, and Pd is the best answer because it can achieve a stable configuration with 18 electrons in its valence shell by forming complexes.

Why other options are wrong:
- Option A: $C{u^ + }$ : While Cu+ has 18 electrons in its 3rd shell, the question is likely referring to the outermost shell, making Pd a better answer.
- Option C: $\mathrm{M}{\mathrm{n}^{4 + }}$ : Mn4+ has 11 electrons in its outer shell.
- Option D: Zn: Zn has 2 electrons in its outermost shell (4s2), and 18 electrons in the 3rd shell.

Key Concepts:
- Electronic configuration
- Outer shell electrons
- Valence shell electrons

Memory Aids:
- Remember the electronic configuration rules.

Quick Tips:
- Write the electronic configuration first.
- Identify the outer shell and count the number of electrons.

110 Easy

110 of 176Question permalink

Remembering +4 | -1

“Positronium” is the name given to an atom-like combination formed between:

View answer and explanation

Correct option: D

A positron and an electron

Option D (A positron and an electron) is correct because positronium is a meta-stable bound state of an electron and its antiparticle, a positron.

Solution approach

Correct Answer + Core Reason: A positron and an electron form Positronium.

Step-by-Step Solution:

Step 1: Define Positronium.\nPositronium is a system consisting of an electron and a positron bound together into an exotic \"atom,\" where the two particles orbit each other. It is unstable and annihilates, producing gamma rays.

Step 2: Analyze the options.\nOption A: A positron and a proton - This combination is not called positronium.\nOption B: A positron and a neutron - This combination is not called positronium.\nOption C: A positron and α-particle - This combination is not called positronium.\nOption D: A positron and an electron - This is the correct definition of positronium.

Why other options are wrong:
- Option A: A positron and a proton do not form positronium.
- Option B: A positron and a neutron do not form positronium.
- Option C: A positron and an alpha-particle do not form positronium.

Key Concepts:
- Positronium
- Positron
- Electron
- Antiparticle

Memory Aids:
- Positronium: \"positron-\" refers to positron, \"-ium\" suggests an atom-like structure.

Quick Tips:
- Remember the definition of positronium as an electron-positron bound state.

111 Easy

111 of 176Question permalink

Understanding +4 | -1

When an electron jumps from L-level to M-level, there occurs:

View answer and explanation

Correct option: B

Absorption of energy

Option B (Absorption of energy) is correct because an electron moving from a lower energy level (L) to a higher energy level (M) requires energy absorption.

Solution approach

Correct Answer + Core Reason: Absorption of energy. An electron moving to a higher energy level absorbs energy.

Step-by-Step Solution:

Step 1: The question asks what happens when an electron jumps from the L-level to the M-level.

Step 2: Recall that the energy levels of an atom are quantized, with the L-level being lower in energy than the M-level.

Step 3: When an electron jumps from a lower energy level to a higher energy level, it must absorb energy equal to the difference in energy between the two levels.

Why other options are wrong:
- Option A: Emission of energy is incorrect because emission occurs when an electron jumps from a higher energy level to a lower energy level.
- Option C: Emission of γ-radiations is incorrect because gamma radiations are emitted from the nucleus, not during electronic transitions.
- Option D: Emission of X-rays is incorrect because X-rays are emitted when inner-shell electrons are involved in transitions, and the L and M levels are not inner shells.

Key Concepts:
- Energy levels
- Absorption
- Emission

Memory Aids:
- Lower to higher: absorb
- Higher to lower: emit

Quick Tips:
- Remember that absorption occurs when moving to a higher energy level.

112 Easy

112 of 176Question permalink

Application +4 | -1

The total number of protons present in all the elements upto 'Zn' in the periodic table is:

View answer and explanation

Correct option: C

465

Option C (465) is correct because summing the atomic numbers from Hydrogen (1) to Zinc (30) yields 465.

Solution approach

Correct Answer + Core Reason: 465. The total number of protons is the sum of the atomic numbers of all elements up to Zn.

Step-by-Step Solution:

Step 1: Recognize that the number of protons in an element is equal to its atomic number.

Step 2: List the atomic numbers of elements from Hydrogen (H) to Zinc (Zn). H(1), He(2), Li(3), Be(4), B(5), C(6), N(7), O(8), F(9), Ne(10), Na(11), Mg(12), Al(13), Si(14), P(15), S(16), Cl(17), Ar(18), K(19), Ca(20), Sc(21), Ti(22), V(23), Cr(24), Mn(25), Fe(26), Co(27), Ni(28), Cu(29), Zn(30).

Step 3: Sum the atomic numbers: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 = 465.

Why other options are wrong:
- Option A: 300 - Incorrect, this is significantly lower than the actual sum.
- Option B: 350 - Incorrect, this is lower than the actual sum.
- Option D: 450 - Incorrect, this is close but not the exact sum.

Key Concepts:
- Atomic number
- Periodic table

Memory Aids:
- Remember that the atomic number represents the number of protons.

Quick Tips:
- Summing consecutive integers can be simplified using formulas, but for a small range like this, direct summation is feasible.

113 Easy

113 of 176Question permalink

Remembering +4 | -1

Which one of the following orbitals is spherical in shape

View answer and explanation

Correct option: A

4s

Option A (4s) is correct because s orbitals are always spherical in shape, regardless of the principal quantum number.

Solution approach

Correct Answer + Core Reason: 4s. s-orbitals are always spherical in shape.

Step-by-Step Solution:

Step 1: Recall the shapes of atomic orbitals: s-orbitals are spherical, p-orbitals are dumbbell-shaped, d-orbitals are more complex (usually double dumbbell or dumbbell with a ring), and f-orbitals are even more complex.

Step 2: Apply the knowledge to the options: Only the s-orbital has a spherical shape.

Step 3: Consider the principal quantum number: The principal quantum number (4 in 4s) affects the size and energy of the orbital but not its shape. Therefore, 4s is still spherical.

Why other options are wrong:
- Option B: 3p orbitals are dumbbell-shaped, not spherical.
- Option C: 3d orbitals have more complex shapes, such as double dumbbell or dumbbell with a ring, not spherical.
- Option D: 4f orbitals have even more complex shapes than d-orbitals, not spherical.

Key Concepts:
- Atomic orbitals
- Shapes of s, p, d, and f orbitals
- Quantum numbers

Memory Aids:
- s-orbital: "s" for spherical.
- p-orbital: "p" for peanut (dumbbell shape).

Quick Tips:
- Remember the basic shapes of s, p, d, and f orbitals.

114 Easy

114 of 176Question permalink

Understanding +4 | -1

Assuming the velocity be same, which sub-atomic particle possesses smallest de Broglie wavelength;

View answer and explanation

Correct option: C

An α-particle

Option C (An α-particle) is correct because de Broglie wavelength is inversely proportional to mass; thus, the particle with the largest mass (α-particle) will have the smallest wavelength.

Solution approach

Correct Answer + Core Reason: An α-particle. De Broglie wavelength is inversely proportional to mass.

Step-by-Step Solution:

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

Step 2: Analyze the relationship between wavelength and mass. Since the velocity is the same for all particles, the de Broglie wavelength is inversely proportional to the mass: $\lambda \propto \frac{1}{m}$ .

Step 3: Compare the masses of the particles. An electron has the smallest mass, a proton has a larger mass, and an alpha particle has the largest mass (2 protons + 2 neutrons).

Step 4: Determine which particle has the smallest wavelength. Since wavelength is inversely proportional to mass, the particle with the largest mass will have the smallest wavelength. Therefore, the alpha particle has the smallest de Broglie wavelength.

Step 5: Final Answer: An α-particle

Why other options are wrong:
- Option A: An electron has a smaller mass than an alpha particle, so it will have a longer de Broglie wavelength.
- Option B: A proton has a smaller mass than an alpha particle, so it will have a longer de Broglie wavelength.
- Option D: The particles do not have the same de Broglie wavelength because they have different masses.

Key Concepts:
- De Broglie wavelength: $\lambda = \frac{h}{{mv}}$
- Relationship between wavelength and mass

Memory Aids:
- Remember that heavier particles have shorter wavelengths.

Quick Tips:
- When comparing de Broglie wavelengths, consider the masses of the particles if their velocities are the same.

115 Easy

115 of 176Question permalink

Understanding +4 | -1

Photoelectric effect can be caused by :

View answer and explanation

Correct option: D

Visible light, ultraviolet rays, X-rays and gamma rays also

Option D (Visible light, ultraviolet rays, X-rays and gamma rays also) is correct because the photoelectric effect can be caused by any electromagnetic radiation with sufficient energy (frequency) to overcome the work function of the metal.

Solution approach

Correct Answer + Core Reason: Visible light, ultraviolet rays, X-rays and gamma rays also. Photoelectric effect can be caused by any EM radiation with sufficient energy.

Step-by-Step Solution:

Step 1: Define the photoelectric effect.

Step 2: Explain the conditions for the photoelectric effect to occur:
- The incident radiation must have sufficient energy (frequency) to overcome the work function of the metal.
- Different types of electromagnetic radiation have different energies (frequencies).
- Visible light, ultraviolet rays, X-rays, and gamma rays can all cause the photoelectric effect if their energy is greater than the work function.

Step 3: Analyze the options.
- All options except D limit the types of radiation that can cause the photoelectric effect.
- Option D correctly states that all listed types of radiation can cause the effect.

Why other options are wrong:
- Option A: Visible light but not by X-rays is incorrect because X-rays have higher energy than visible light and can cause the photoelectric effect.
- Option B: Gamma-rays but not by X-rays is incorrect because both gamma rays and X-rays can cause the photoelectric effect, and X-rays can sometimes be more effective depending on the material.
- Option C: Ultraviolet light only is incorrect because other types of radiation with sufficient energy can also cause the photoelectric effect.

Key Concepts:
- Photoelectric effect
- Work function
- Electromagnetic spectrum (visible light, ultraviolet rays, X-rays, gamma rays)

Memory Aids:
- Remember that the energy of electromagnetic radiation increases with frequency (and decreases with wavelength).

Quick Tips:
- The photoelectric effect depends on the energy (frequency) of the incident radiation, not its intensity (number of photons).

116 Easy

116 of 176Question permalink

Understanding +4 | -1

The mass number of an element is 23 and atomic number is 11. The number of protons, electrons and neutrons respectively present in the atom of the elements are:

View answer and explanation

Correct option: A

11,11,12

Option A (11,11,12) is correct because in a neutral atom, the number of protons equals the number of electrons, and the number of neutrons is the mass number minus the atomic number.

Solution approach

Correct Answer + Core Reason: 11,11,12. Number of protons = atomic number, number of electrons = number of protons in a neutral atom, number of neutrons = mass number - atomic number.

Step-by-Step Solution:

Step 1: The question provides the mass number (23) and atomic number (11) of an element and asks for the number of protons, electrons, and neutrons.

Step 2: Recall the definitions:
- Atomic number (Z) = number of protons
- Mass number (A) = number of protons + number of neutrons
- In a neutral atom, the number of protons equals the number of electrons.

Step 3: Number of protons = Atomic number = 11\nNumber of electrons = Number of protons = 11\nNumber of neutrons = Mass number - Atomic number = 23 - 11 = 12

Step 4: Therefore, the number of protons, electrons, and neutrons are 11, 11, and 12, respectively.

Why other options are wrong:
- Option B: 12,12,11 is incorrect because it has the wrong number of protons and neutrons.
- Option C: 11,12,11 is incorrect because it has the wrong number of electrons and neutrons.
- Option D: 12,11,12 is incorrect because it has the wrong number of protons.

Key Concepts:
- Atomic number
- Mass number
- Protons, neutrons, and electrons

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

Quick Tips:
- Remember the definitions of atomic number and mass number.

117 Easy

117 of 176Question permalink

Application +4 | -1

The de-Broglie wavelength relates to applied voltage ror α-particles as

View answer and explanation

Correct option: C

\lambda = \frac{{0.101}}{{\sqrt V }}{A^ \circ }

Option C (

\lambda = \frac{{0.101}}{{\sqrt V }}{A^ \circ }

) is correct because it is the correct formula for calculating the de Broglie wavelength of alpha particles in terms of applied voltage.

Solution approach

Correct Answer + Core Reason: $\lambda = \frac{{0.101}}{{\sqrt V }}{A^ \circ }$ . This is the correct formula for the de-Broglie wavelength of alpha particles related to applied voltage.

Step-by-Step Solution:

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

Step 2: Relate kinetic energy (KE) to momentum: $KE = \frac{p^2}{2m}$ , so $p = \sqrt{2mKE}$ .

Step 3: The kinetic energy gained by the alpha particle is equal to the charge (q) times the potential difference (V): $KE = qV$ .

Step 4: Substitute KE into the momentum equation: $p = \sqrt{2mqV}$ .

Step 5: Substitute momentum into the de-Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mqV}}$ .

Step 6: For an alpha particle, the charge is 2e (where e is the elementary charge) and the mass is approximately 4 atomic mass units (4u). Substitute these values along with Planck's constant into the equation.

Step 7: $\lambda = \frac{6.626 \times 10^{-34} Js}{\sqrt{2 \times 4 \times 1.66 \times 10^{-27} kg \times 2 \times 1.602 \times 10^{-19} C \times V}} = \frac{0.101}{\sqrt{V}} \times 10^{-10} m = \frac{0.101}{\sqrt{V}} {A^ \circ }$

Why other options are wrong:
- Option A: ${\mkern 1mu} {\mkern 1mu} \lambda = \frac{{12.3{A^ \circ }}}{{\sqrt V }}$ is the formula for electrons, not alpha particles.
- Option B: $\lambda = \frac{{0.286}}{{\sqrt V }}{A^ \circ }$ is incorrect because the constant is wrong for alpha particles.
- Option D: $\lambda = \frac{{0.856}}{{\sqrt V }}{A^ \circ }$ is incorrect because the constant is wrong for alpha particles.

Key Concepts:
- de-Broglie wavelength
- Kinetic energy
- Momentum
- Planck's constant

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

Quick Tips:
- Be careful with units when substituting values into the formula.
- Remember the charge and mass of an alpha particle.

118 Easy

118 of 176Question permalink

Application +4 | -1

The species that has same number of electrons as

{\mathrm{\;}_{16}}{\mathrm{S}^{32}}

is:

View answer and explanation

Correct option: D

{\mathrm{\;}_{17}}\mathrm{C}{\mathrm{l}^ + }

Option D (

{\mathrm{\;}_{17}}\mathrm{C}{\mathrm{l}^ + }

) is correct because

_{16}S^{32}

has 16 electrons, and

_{17}Cl^+

also has 16 electrons (17 - 1 = 16).

Solution approach

Correct Answer + Core Reason: ${\mathrm{\;}_{17}}\mathrm{C}{\mathrm{l}^ + }$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $_{16}S^{32}$ . Sulfur (S) has an atomic number of 16, so a neutral sulfur atom has 16 electrons. Since the species is neutral, it has 16 electrons.

Step 2: Calculate the number of electrons in each of the given options:
- A. $_{16}S^+$ : 16 - 1 = 15 electrons
- B. $_{17}Cl^-$ : 17 + 1 = 18 electrons
- C. $_{16}S^-$ : 16 + 1 = 17 electrons
- D. $_{17}Cl^+$ : 17 - 1 = 16 electrons

Step 3: Identify the species with the same number of electrons as $_{16}S^{32}$ , which is 16.

Why other options are wrong:
- Option A: $_{16}{\mathrm{S}^ + }$ - Incorrect, it has 15 electrons.
- Option B: $_{17}\mathrm{C}{\mathrm{l}^ - }$ - Incorrect, it has 18 electrons.
- Option C: ${\mathrm{\;}_{16}}{\mathrm{S}^ - }$ - Incorrect, it has 17 electrons.

Key Concepts:
- Isoelectronic species
- Atomic number and number of electrons
- Ions and number of electrons

Memory Aids:
- Remember that cations (positive ions) have fewer electrons than the neutral atom, and anions (negative ions) have more electrons.

Quick Tips:
- For ions, add electrons for negative charges and subtract electrons for positive charges.

119 Easy

119 of 176Question permalink

Application +4 | -1

If wavelength is equal to the distance travelled by the electron in one second, then

View answer and explanation

Correct option: D

\lambda = \sqrt {\frac{h}{m}}

Option D (

\lambda = \sqrt {\frac{h}{m}}

) is correct because if wavelength equals the distance traveled in one second, then

\lambda = v

, and using de Broglie's equation

\lambda = \frac{h}{mv}

, we can derive the given relation.

Solution approach

Correct Answer + Core Reason: $\lambda = \sqrt {\frac{h}{m}}$ . If wavelength is equal to the distance traveled by the electron in one second, then $\lambda = v$ , and using de Broglie's equation $\lambda = \frac{h}{mv}$ , we can derive the given relation.

Step-by-Step Solution:

Step 1: State the given condition: wavelength ( $\lambda$ ) = distance traveled by the electron in one second = velocity (v).\nTherefore, $\lambda = v$ .

Step 2: Recall de Broglie's wavelength equation:
$\lambda = \frac{h}{mv}$ , where:
$\lambda$ is the de Broglie wavelength
$h$ is Planck's constant
$m$ is the mass of the electron
$v$ is the velocity of the electron

Step 3: Substitute $v = \lambda$ into the de Broglie's equation:
$\lambda = \frac{h}{m\lambda}$

Step 4: Rearrange the equation to solve for $\lambda$ :
$\lambda^2 = \frac{h}{m}$
$\lambda = \sqrt{\frac{h}{m}}$

Why other options are wrong:
- Option A: Incorrect. This is the standard de Broglie wavelength equation, but it doesn't incorporate the given condition that $\lambda = v$ .
- Option B: Incorrect. This equation is not dimensionally correct and does not follow from the given condition and de Broglie's equation.
- Option C: Incorrect. This equation is not dimensionally correct and does not follow from the given condition and de Broglie's equation.

Key Concepts:
- de Broglie Wavelength
- Wave-Particle Duality

Memory Aids:
- Remember the de Broglie wavelength equation: $\lambda = \frac{h}{mv}$ .

Quick Tips:
- Always pay attention to the given conditions in the problem and try to incorporate them into the relevant equations.

120 Easy

120 of 176Question permalink

Understanding +4 | -1

Two electrons in the same orbital may be identified with:

View answer and explanation

Correct option: D

s

Option D (

s

) is correct because two electrons in the same orbital must have opposite spins, thus differing in their spin quantum number (

s

).

Solution approach

Correct Answer + Core Reason: $s$ . Two electrons in the same orbital must have different spin quantum numbers.

Step-by-Step Solution:

Step 1: Recall Pauli's Exclusion Principle.

Step 2: Pauli's Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.

Step 3: Electrons in the same orbital have the same n, l, and m values. Therefore, they must differ in their spin quantum number (s).

Why other options are wrong:
- Option A: $n$ . Electrons in the same orbital have the same principal quantum number.
- Option B: $l$ . Electrons in the same orbital have the same azimuthal quantum number.
- Option C: $m$ . Electrons in the same orbital have the same magnetic quantum number.

Key Concepts:
- Pauli's Exclusion Principle
- Quantum Numbers
- Spin Quantum Number

Memory Aids:
- Same orbital, opposite spins.

Quick Tips:
- Electrons in the same orbital can only differ in their spin.

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