CBSE Class 11 Chemistry HOTs Structure of Atom Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): Most of the α-particles passed straight through the gold foil in Rutherford's experiment.
Reason (R): Most of the volume of an atom is empty space.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
Both statements are true. The electrons are very light and the nucleus occupies only a tiny fraction of the atomic volume, so an α-particle usually meets nothing that can deflect it. R explains A.
In simple words: Undeflected particles are the result of an atom being mostly empty.

Teacher's Note:
1. Quote both observation and conclusion in your answer.
2. Large deflections need the rare close approach to a nucleus.

Question 2. Assertion (A): For light above the threshold frequency, the number of photoelectrons increases when the intensity of light is increased.
Reason (R): Each photon can eject only one electron, and brighter light has more photons per second.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
A brighter beam carries more photons every second, and each photon above the threshold frees one electron, so the number of electrons rises. The kinetic energy of each electron stays the same because the photon energy hν is unchanged.
In simple words: More photons means more ejected electrons.

Teacher's Note:
1. Intensity changes the number of electrons; frequency changes their energy.
2. This is the particle picture of light.

Question 3. Assertion (A): If the intensity of light of frequency below the threshold frequency is increased enough, electrons will eventually be emitted from the metal.
Reason (R): The energy of each photon is proportional to its frequency.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (d).
A is false: a photon with hν below the work function cannot free an electron, and one photon gives its energy to one electron, so intensity cannot help. R is true: E = hν.
In simple words: Below threshold, extra brightness is useless because each photon is too weak.

Teacher's Note:
1. The work function is the minimum energy for one photon to eject one electron.
2. Classical wave theory wrongly predicts emission at high intensity.

Question 4. Assertion (A): The line emission spectrum of an element is its fingerprint.
Reason (R): The energy levels in an atom are quantised and are different for the atoms of each element.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
The allowed energy levels, and hence the energy gaps and the wavelengths emitted, are unique to each element. This is why spectroscopy identifies elements.
In simple words: Different elements have different energy gaps, so they emit different colours.

Teacher's Note:
1. Quantised levels give lines, not a continuous band.
2. Helium was first found in the Sun from its spectrum.

Question 5. Assertion (A): Bohr's model cannot explain the spectrum of the helium atom.
Reason (R): Bohr's model correctly predicts that the radius of the first orbit of hydrogen is 52.9 pm.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (b).
A is true: Bohr's equations were derived for one electron, and in helium the repulsion between its two electrons is not included. R is also true, but the success with hydrogen's orbit radius is not the reason for the failure with helium.
In simple words: The model handles one electron only, so helium is out of reach.

Teacher's Note:
1. It works for H, He⁺ and Li²⁺ but not for He.
2. Wave mechanics handles many-electron atoms.

Question 6. Assertion (A): An electron cannot exist inside the nucleus.
Reason (R): According to the uncertainty principle, confining an electron within about 10⁻¹⁵ m would require a speed greater than that of light.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
Taking Δx = 10⁻¹⁵ m, Δv = h/(4π m Δx) = 5.8 × 10¹⁰ m s⁻¹, which is far more than c = 3 × 10⁸ m s⁻¹. Such a speed is impossible, so an electron cannot be confined in the nucleus.
In simple words: Squeezing an electron into the nucleus would make its speed absurdly large.

Teacher's Note:
1. The calculation uses Δx · m Δv ≥ h/4π.
2. Electrons come out of nuclei in β-decay only as new particles.

Question 7. Which of the following statements about a 3d orbital are correct? (more than one may be correct)
(a) It has n = 3 and l = 2
(b) It has two angular nodes
(c) It has one radial node
(d) There are five degenerate 3d orbitals in the absence of a field
Answer:
Answer: (a), (b) and (d).
3d means n = 3 and l = 2. Angular nodes = l = 2. Radial nodes = n − l − 1 = 0, so (c) is wrong. The 2l + 1 = 5 orbitals have equal energy without an external field.
In simple words: A 3d orbital has two nodal planes and no spherical node.

Teacher's Note:
1. Work out n − l − 1 for the radial nodes.
2. Total nodes = n − 1 = 2, all of them angular here.

Question 8. Which of the following ions have five unpaired electrons? (more than one may be correct; Mn Z = 25, Fe Z = 26, Cr Z = 24, Zn Z = 30)
(a) Mn²⁺
(b) Fe³⁺
(c) Cr³⁺
(d) Zn²⁺
Answer:
Answer: (a) and (b).
Mn²⁺ is 3d⁵ and Fe³⁺ is 3d⁵, each with five unpaired electrons. Cr³⁺ is 3d³ (three unpaired) and Zn²⁺ is 3d¹⁰ (none).
In simple words: Both ions have a half-filled 3d subshell.

Teacher's Note:
1. Write the neutral atom first, remove 4s electrons first, then 3d.
2. Spin-only moment for five unpaired electrons is 5.92 B.M.

Question 9. The 4s orbital is filled before 3d in a neutral atom, yet in forming Fe²⁺ the two electrons are lost from 4s. Explain.
Answer:
For filling, 4s has the lower (n + l) value (4 against 5), so the 4s orbital is filled first.
Once the 3d orbitals hold electrons, their energy drops below that of 4s, because 3d electrons are in a more compact region and the nuclear charge is felt more strongly.
Also the 4s electrons have the larger principal quantum number and are farther from the nucleus, so they are removed first. Fe²⁺ is therefore [Ar] 3d⁶, not [Ar] 3d⁴ 4s².
In simple words: 4s fills first but empties first, because it is the outermost shell.

Teacher's Note:
1. Cations lose electrons from the highest n.
2. The same happens in all the first-row transition metals.

Question 10. Explain, using exchange energy, why chromium is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².
Answer:
Exchange energy depends on the number of possible exchanges between electrons of parallel spin.
In 3d⁵ 4s¹ all six electrons can have the same spin, giving 6 × 5/2 = 15 possible exchanges.
In 3d⁴ 4s² only five electrons (four in 3d and one in 4s) can have parallel spin, giving 5 × 4/2 = 10 exchanges; the paired 4s electrons have opposite spins.
More exchanges release more energy, so 3d⁵ 4s¹ is more stable. The symmetrical half-filled 3d subshell adds to this.
In simple words: More parallel-spin electrons means more exchange energy and more stability.

Teacher's Note:
1. Number of exchanges for k parallel electrons = k(k − 1)/2.
2. Cu follows the same idea with a full d subshell.

Question 11. Why does the hydrogen atom, which has only one electron, give a spectrum with many lines?
Answer:
A hydrogen gas sample contains a very large number of atoms. When excited, the electrons of different atoms are raised to different levels.
As they fall back to lower levels by many different routes, they emit photons of many different energies. A fall from n = 5 alone can give 4 + 3 + 2 + 1 = 10 different lines.
Together these give all the lines of the Lyman, Balmer, Paschen and other series.
In simple words: Many atoms, many possible jumps, many lines.

Teacher's Note:
1. Lines are separate because the levels are quantised.
2. Maximum lines from level n = n(n − 1)/2.