CBSE Class 11 Chemistry Structure Of Atom Worksheet Set 03

Structure Of Atom MCQ Questions with Answers Class 11 Chemistry

Question- The correct order of increasing energy of atomic orbitals is

(a) 5 p < 4 f < 6 s < 5 d

(b) 5 p < 6 s < 4 f < 5 d

(c) 5 p < 5 d < 4 f < 6 s

(d) None of these

Answer-(b)

Question- The number of orbitals present in the fifth shell will be

(a) 25

(b) 10

(c) 50

(d) 20

Answer-(a)

Question-The total number of electrons that can be accommodated in all orbitals having principal quantum number 2 and azimuthal quantum number 1 is

(a) 2

(b) 4

(c) 6

(d) 8

Answer-(c)

Question-The order of filling of electrons in the orbitals of an atom will be

(a) 3d, 4s, 4p, 4d, 5s

(b) 4s, 3d, 4p, 5s, 4d

(c) 5s, 4p, 3d, 4d, 5s

(d) 3d, 4p, 4s, 4d, 5s

Answer-(b)

Question- The following quantum numbers are possible for how many orbital(s) n = 3, l = 2, m = +2 ?

(a) 1

(b) 3

(c) 2

(d) 4

Answer-(a)

Question- Number of unpaired electrons in N²⁺ is

(a) 2

(b) 0

(c) 1

(d) 3

Answer-(c)

Question-Which of the following is not permissible arrangement of electrons in an atom?

(a) n = 5, l = 3, m = 0, s = + 1/2

(b) n = 3, l = 2, m = – 3, s = – 1/2

(c) n = 3, l = 2, m = – 2, s = – 1/2

(d) n = 4, l = 0, m = 0, s = – 1/2

Answer-(b)

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

(a) Cu⁺

(b) Th⁴⁺

(c) Cs⁺

(d) K⁺

Answer-(a)

Question-Which of the following sets of quantum numbers represents the highest energy of an atom?

(a) n = 3, l = 0, m = 0, s = +1/2

(b) n = 3, l = 1, m = 1, s = +1/2

(c) n = 3, l = 2, m = 1, s = +1/2

(d) n = 4, l = 0, m = 0, s = +1/2

Answer-(c)

Question-Which of the following represents correct set of the four quantum numbers for an electron in a 4d subshell ?

(a) 4, 2, 1, 0

(b) 4, 2, 1, – 1/2

(c) 4, 3, 2, + 1/2

(d) 4, 3, – 2, – 1/2

Answer-(b)

Question-What will be the sum of all possible values of l and m for n = 5 ?

(a) 12

(b) 13

(c) 4

(d) 9

Answer-(b)

Question-The orbitals are called degenerate when

(a) they have the same wave functions

(b) they have the same wave functions but different energies

(c) they have different wave functions but same energy

(d) they have the same energy

Answer-(d)

Question- The number of spherical nodes in 3p orbitals are

(a) one

(b) three

(c) two

(d) None of these

Answer-(a)

Question-If there are five radial nodes, then what can be the correct representation of the orbital for n = 8.

(a) 8d

(b) 8s

(c) 8p

(d) 8f

Answer-(a)

Question-What is the correct orbital designation of an electron with the quantum number, n = 4, l = 3, m = – 2, s = 1/2 ?

(a) 3s

(b) 4 f

(c) 5p

(d) 6s

Answer-(b)

Question-For azimuthal quantum number l = 3, the maximum number of electrons will be

(a) 2

(b) 6

(c) 0

(d) 14

Answer-(d)

Question-What can be the representation of the orbital having 3 angular nodes and n = 5.

(a) 5d

(b) 5f

(c) 5p

(d) 5s

Answer-(b)

Question-Arrange the orbital of same shell in the increasing order of shielding strength of the outer shell of electrons. s, f, d, p

(a) s < p < d < f

(b) s > p < d < f

(c) s > p > d < f

(d) s > p > d > f

Answer-(d)

Question-The number of d-electrons retained in Fe²⁺ (At. no. of Fe = 26) ion is

(a) 4

(b) 5

(c) 6

(d) 3

Answer-(c)

Question-In a given atom no two electrons can have the same values for all the four quantum numbers. This is called

(a) Hund’s Rule

(b) Aufbau principle

(c) Uncertainty principle

(d) Pauli’s exclusion principle

Answer-(d)

Question- Maximum number of electrons in a subshell of an atom is determined by the following:

(a) 2 l + 1

(b) 4 l – 2

(c) 2 n2

(d) 4 l + 2

Answer-(d)

Question 1. The uncertainty in momentum of an electron is \(1.0 \times 10^{-5} \text{ Kg ms}^{-1}\). What is the uncertainty in its position?
Answer: According to Heisenberg's Uncertainty Principle:
\(\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\)
Where:
\(\Delta p = 1.0 \times 10^{-5} \text{ Kg ms}^{-1}\)
Planck's constant \(h = 6.626 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}\)
Substitute the known values to find uncertainty in position \((\Delta x)\):
\(\Delta x = \frac{h}{4 \pi \cdot \Delta p}\)
\(\Delta x = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 1.0 \times 10^{-5}}\)
\(\Delta x = \frac{6.626 \times 10^{-34}}{12.5664 \times 10^{-5}}\)
\(\Delta x = 5.27 \times 10^{-30} \text{ m}\)
In simple words: Heisenberg's uncertainty principle formula is used to calculate position uncertainty when momentum uncertainty is known.

Exam Tip: Always verify that SI units are used consistently before substituting values into quantum mechanics formulas.

Question 2. Find the de Broglie wavelength in \(\text{Å}\) of a particle with mass \(1 \text{ g}\) and velocity \(100 \text{ m/s}\).
Answer: Given data:
Mass \(m = 1 \text{ g} = 10^{-3} \text{ kg}\)
Velocity \(v = 100 \text{ m/s} = 10^2 \text{ m/s}\)
Planck's constant \(h = 6.626 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}\)
Using de Broglie's equation:
\(\lambda = \frac{h}{mv}\)
\(\lambda = \frac{6.626 \times 10^{-34}}{10^{-3} \times 10^2}\)
\(\lambda = \frac{6.626 \times 10^{-34}}{10^{-1}}\)
\(\lambda = 6.626 \times 10^{-33} \text{ m}\)
Converting meters to Angstroms (\(1 \text{ m} = 10^{10} \text{ Å}\)):
\(\lambda = 6.626 \times 10^{-33} \times 10^{10} \text{ Å}\)
\(\lambda = 6.626 \times 10^{-23} \text{ Å}\)
In simple words: Substitute mass in kg and velocity in m/s into the de Broglie wavelength formula and then convert the resulting meters into Angstroms.

Exam Tip: Remember to convert grams into kilograms to maintain consistency with SI units in Planck's constant.

Question 3. Find wavelength of photon having energy \(3.03 \times 10^{-19} \text{ J}\).
Answer: Given data:
Energy of photon \(E = 3.03 \times 10^{-19} \text{ J}\)
Planck's constant \(h = 6.626 \times 10^{-34} \text{ J s}\)
Speed of light \(c = 3 \times 10^8 \text{ m/s}\)
Using the Planck energy-wavelength relation:
\(E = \frac{hc}{\lambda}\)
Rearranging for wavelength \((\lambda)\):
\(\lambda = \frac{hc}{E}\)
\(\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{3.03 \times 10^{-19} \text{ J}}\)
\(\lambda = \frac{19.878 \times 10^{-26}}{3.03 \times 10^{-19}}\)
\(\lambda = 6.56 \times 10^{-7} \text{ m}\) or \(656 \text{ nm}\)
In simple words: Use the energy formula \(E = hc/\lambda\) and solve for wavelength by dividing \(hc\) by the given energy.

Exam Tip: Keep track of exponent arithmetic carefully when dividing numbers in scientific notation.

Question 4. What will be the uncertainty in velocity of an electron (mass of electron = \(9.1 \times 10^{-28} \text{ g}\)) moving with a velocity of \(3.0 \times 10^4 \text{ m s}^{-1}\) accurate up to \(0.011\%\)?
Answer: Given data:
Mass of electron \(m = 9.1 \times 10^{-28} \text{ g} = 9.1 \times 10^{-31} \text{ kg}\)
Actual velocity \(v = 3.0 \times 10^4 \text{ m s}^{-1}\)
The uncertainty in velocity \((\Delta v)\) is given as \(0.011\%\) of the actual velocity:
\(\Delta v = \frac{0.011}{100} \times 3.0 \times 10^4 \text{ m s}^{-1}\)
\(\Delta v = 1.1 \times 10^{-4} \times 3.0 \times 10^4\)
\(\Delta v = 3.3 \times 10^0 = 3.3 \text{ m s}^{-1}\)
In simple words: The uncertainty in velocity is calculated directly by finding the specified percentage of the given velocity value.

Exam Tip: Pay attention to whether the question asks for uncertainty in velocity directly from percentage or via Heisenberg's formula.

Question 5. Write the electronic configuration of following atoms/ions - \(F^-\), \(\text{Cr}\), \(\text{Mg}^{2+}\), \(\text{O}^-\), \(\text{Ca}\) (Atomic number of \(\text{F} = 9\), \(\text{Cr} = 24\), \(\text{Mg} = 12\), \(\text{O} = 8\), \(\text{Ca} = 20\))
Answer:
1. Fluoride ion (\(F^-\), Atomic Number = \(9 + 1 = 10\) electrons):
\(1s^2 2s^2 2p^6\) (or \([\text{Ne}]\))
2. Chromium atom (\(\text{Cr}\), Atomic Number = 24, exceptional configuration due to extra stability of half-filled subshells):
\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\) (or \([\text{Ar}] 3d^5 4s^1\))
3. Magnesium ion (\(\text{Mg}^{2+}\), Atomic Number = \(12 - 2 = 10\) electrons):
\(1s^2 2s^2 2p^6\) (or \([\text{Ne}]\))
4. Oxide radical/ion (\(\text{O}^-\), Atomic Number = \(8 + 1 = 9\) electrons):
\(1s^2 2s^2 2p^5\)
5. Calcium atom (\(\text{Ca}\), Atomic Number = 20):
\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) (or \([\text{Ar}] 4s^2\))
In simple words: Add electrons for negatively charged ions, subtract for positively charged ions, and account for Aufbau exceptions like chromium.

Exam Tip: Chromium (24) and Copper (29) are frequent exam favorites because they violate the standard Aufbau filling order to achieve stability.

Question 6. Which of the following species are isoelectronic? \(\text{Na}^+\), \(\text{O}^{2-}\), \(\text{F}^-\), \(\text{Ca}^{2+}\), \(\text{K}^+\)
Answer: Let us find the total number of electrons in each species:
- \(\text{Na}^+\): Atomic number = 11, Electrons = \(11 - 1 = 10\)
- \(\text{O}^{2-}\): Atomic number = 8, Electrons = \(8 + 2 = 10\)
- \(\text{F}^-\): Atomic number = 9, Electrons = \(9 - 1 = 10\)
- \(\text{Ca}^{2+}\): Atomic number = 20, Electrons = \(20 - 2 = 18\)
- \(\text{K}^+\): Atomic number = 19, Electrons = \(19 - 1 = 18\)
Therefore, the isoelectronic groups are:
Group 1: \(\text{Na}^+\), \(\text{O}^{2-}\), and \(\text{F}^-\) (each has 10 electrons)
Group 2: \(\text{Ca}^{2+}\) and \(\text{K}^+\) (each has 18 electrons)
In simple words: Isoelectronic species are atoms or ions that have the exact same number of total electrons.

Exam Tip: Always calculate electron counts by subtracting positive charges and adding negative charges to the atomic number.

Question 7. (i) Write values of \(n\) and \(l\) for \(4f\) orbital. (ii) Write all possible values of \(l\) and \(m\) for \(n = 2\).
Answer:
(i) For the \(4f\) orbital:
Principal quantum number \(n = 4\)
Azimuthal quantum number \(l = 3\) (since \(f\)-subshell corresponds to \(l = 3\))
(ii) For \(n = 2\):
Possible values of azimuthal quantum number \(l\) range from \(0\) to \((n - 1)\):
\(l = 0, 1\)
For each value of \(l\), the magnetic quantum number \(m\) ranges from \(-l\) to \(+l\):
- When \(l = 0\): \(m = 0\)
- When \(l = 1\): \(m = -1, 0, +1\)
In simple words: Quantum numbers define the energy level (\(n\)), subshell shape (\(l\)), and orbital orientation (\(m\)).

Exam Tip: Remember the subshell codes: \(s \rightarrow l=0\), \(p \rightarrow l=1\), \(d \rightarrow l=2\), \(f \rightarrow l=3\).

Question 8. Write one isobar of \(_18^{40}\text{Ar}\)
Answer: Isobars are atoms of different chemical elements that have the same mass number but different atomic numbers. An isobar of Argon (\(_18^{40}\text{Ar}\)) is Calcium (\(_20^{40}\text{Ca}\)) or Potassium (\(_19^{40}\text{K}\)).
Example: \(_20^{40}\text{Ca}\)
In simple words: Isobars have identical mass numbers (superscript) but different atomic numbers (subscript).

Exam Tip: Do not confuse isobars (same mass number) with isotopes (same atomic number) or isotones (same number of neutrons).

Question 10. What do you mean that energy of the electron is quantized?
Answer: Quantization of energy means that an electron can only possess certain specific, discrete energy values rather than a continuous range of energies. When an electron transitions between these fixed energy levels (orbits), it absorbs or emits a fixed quantum of energy corresponding to the energy difference between those levels.
In simple words: Energy quantization means an electron can only exist in specific permitted energy states, like steps on a ladder rather than a smooth ramp.

Exam Tip: Bohr's model introduced quantization as a foundational postulate to explain the stability of atoms and line spectra.