CBSE Class 12 Chemistry Chapter 7 The p Block Elements Competency Based Questions Set 01

Multiple Choice Questions (MCQs)

Question 1. \( \text{H}_2\text{S} \) is more acidic than \( \text{H}_2\text{O} \) because
(a) oxygen is more electronegative than sulphur.
(b) atomic number of sulphur is higher than oxygen.
(c) \( \text{H}-\text{S} \) bond dissociation energy is less as compared to \( \text{H}-\text{O} \) bond.
(d) \( \text{H}-\text{O} \) bond dissociation energy is less also compared to \( \text{H}-\text{S} \) bond.
Answer: (c) \( \text{H}-\text{S} \) bond dissociation energy is less as compared to \( \text{H}-\text{O} \) bond.
In simple words: The bond between hydrogen and sulfur is weaker and splits more easily than the bond between hydrogen and oxygen, which releases acid ions much faster.

Exam Tip: Remember that acidic strength in hydrides down a group depends primarily on bond dissociation enthalpy rather than electronegativity.

Question 2. The boiling points of hydrides of group 16 are in the order
(a) \( \text{H}_2\text{O} > \text{H}_2\text{Te} > \text{H}_2\text{S} > \text{H}_2\text{Se} \)
(b) \( \text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te} \)
(c) \( \text{H}_2\text{O} > \text{H}_2\text{Te} > \text{H}_2\text{Se} > \text{H}_2\text{S} \)
(d) None of the options
Answer: (c) \( \text{H}_2\text{O} > \text{H}_2\text{Te} > \text{H}_2\text{Se} > \text{H}_2\text{S} \)
In simple words: Water has a very high boiling point because of strong hydrogen bonding. For the rest of the hydrides, boiling point increases as they get heavier and have stronger attractions.

Exam Tip: Always highlight the role of intermolecular hydrogen bonding in water when explaining its abnormally high physical constants compared to other group hydrides.

Question 3. In the manufacture of sulphuric acid by contact process Tyndall box is used to
(a) convert \( \text{SO}_2 \) and \( \text{SO}_3 \)
(b) test the presence of dust particles
(c) filter dust particles
(d) remove impurities
Answer: (b) test the presence of dust particles
In simple words: The Tyndall box shines light through the gas to detect any dust particles, ensuring the mixture is clean before reaching the catalyst.

Exam Tip: Remember that any residual dust can poison the catalyst (divanadium pentoxide), which is why the Tyndall box is vital for quality control in the contact process.

Question 4. Fluorine differs from rest of the halogens in some of its properties. This is due to
(a) its smaller size and high electronegativity.
(b) lack of d-orbitals.
(c) low bond dissociation energy.
(d) All of the options
Answer: (d) All of the options
In simple words: Fluorine behaves differently because it is tiny, has strong pull for electrons, lacks empty d-orbitals, and has a very weak single bond.

Exam Tip: Anomalous behavior of second-period elements like fluorine is always a popular exam topic; memorize all three main reasons (small size/high electronegativity, low F-F bond energy, and lack of d-orbitals).

Question 5. The set with correct order of acidity is
(a) \( \text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4 \)
(b) \( \text{HClO}_4 < \text{HClO}_3 < \text{HClO}_2 < \text{HClO} \)
(c) \( \text{HClO} < \text{HClO}_4 < \text{HClO}_3 < \text{HClO}_2 \)
(d) \( \text{HClO}_4 < \text{HClO}_2 < \text{HClO}_3 < \text{HClO} \)
Answer: (a) \( \text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4 \)
In simple words: Acidity increases as we add more oxygen atoms because they stabilize the negative charge left behind after the hydrogen ion departs.

Exam Tip: For oxoacids of the same halogen, acidity increases directly with the oxidation state of the halogen atom (from +1 in \( \text{HClO} \) to +7 in \( \text{HClO}_4 \)).

Question 6. When chlorine reacts with cold and dilute solution of sodium hydroxide, it forms
(a) \( \text{Cl}^- \) and \( \text{ClO}^- \)
(b) \( \text{Cl}^- \) and \( \text{ClO}_2^- \)
(c) \( \text{Cl}^- \) and \( \text{ClO}_3^- \)
(d) \( \text{Cl}^- \) and \( \text{ClO}_4^- \)
Answer: (a) \( \text{Cl}^- \) and \( \text{ClO}^- \)
In simple words: Cold and dilute alkali reacts with chlorine to produce a mixture of simple chloride and hypochlorite ions.

Exam Tip: Be careful not to confuse this with hot, concentrated NaOH, which yields chloride (\( \text{Cl}^- \)) and chlorate (\( \text{ClO}_3^- \)) ions instead.

Question 7. The formation of \( \text{O}_2^+ [\text{PtF}_6]^- \) is the basis for the formation of first xenon compound. This is because
(a) \( \text{O}_2 \) and Xe have different sizes.
(b) both \( \text{O}_2 \) and Xe are gases.
(c) \( \text{O}_2 \) and Xe have comparable electro-negativities.
(d) \( \text{O}_2 \) and Xe have comparable ionisation enthalpies.
Answer: (d) \( \text{O}_2 \) and Xe have comparable ionisation enthalpies.
In simple words: Since molecular oxygen and xenon take nearly the same amount of energy to release an electron, Bartlett realized xenon could form a similar compound.

Exam Tip: Always remember that the ionization enthalpy of Xe (\( 1170 \text{ kJ/mol} \)) is very close to that of \( \text{O}_2 \) (\( 1175 \text{ kJ/mol} \)). This value is a crucial point for exams.

Question 8. Partial hydrolysis of \( \text{XeF}_4 \) gives
(a) \( \text{XeO}_3 \)
(b) \( \text{XeOF}_2 \)
(c) \( \text{XeOF}_4 \)
(d) \( \text{XeF}_2 \)
Answer: (b) \( \text{XeOF}_2 \)
In simple words: When xenon tetrafluoride reacts with a limited amount of water, it partially splits to form xenon oxydifluoride.

Exam Tip: Keep the stoichiometry of water in mind; partial hydrolysis of \( \text{XeF}_4 \) gives \( \text{XeOF}_2 \), whereas complete hydrolysis gives \( \text{XeO}_3 \) and Xe.

Question 9. Helium is preferred to be used in balloons instead of hydrogen because it is
(a) incombustible
(b) lighter than hydrogen
(c) more abundant than hydrogen
(d) non polarizable
Answer: (a) incombustible
In simple words: Helium is completely safe because it is an inert gas that cannot catch fire or explode, unlike hydrogen.

Exam Tip: Although hydrogen has a higher lifting power because it is lighter, safety concerns make non-flammable helium the preferred choice for meteorological balloons.

Question 10. The increasing order of reducing power of the halogen acids is
(a) \( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
(b) \( \text{HI} < \text{HBr} < \text{HCl} < \text{HF} \)
(c) \( \text{HBr} < \text{HCl} < \text{HF} < \text{HI} \)
(d) \( \text{HCl} < \text{HBr} < \text{HF} < \text{HI} \)
Answer: (a) \( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
In simple words: As we go down the halogen group, the bond with hydrogen becomes weaker, making it easier to release hydrogen and reduce other substances.

Exam Tip: Reducing power is inversely proportional to bond dissociation enthalpy. Since the H-I bond is the weakest, HI is the strongest reducing agent among hydrides.

Assertion-Reason Questions

Question. Assertion: \( \text{N}_2 \) is less reactive than \( \text{P}_4 \).
Reason: Nitrogen has more electron gain enthalpy than Phosphorous.
Answer: (c) Assertion is correct statement but reason is wrong statement.
Because of the great stability of the \( \text{N}_2 \) molecule, the triple covalent bond between its atoms is hard to break down due to its tiny size and strong \( p\pi-p\pi \) bonding.
In simple words: Nitrogen is highly unreactive because its two atoms are locked together by a very strong triple bond that is difficult to split.

Exam Tip: Remember that nitrogen has less negative electron gain enthalpy than phosphorus due to its small size and high inter-electronic repulsion.

Question. Assertion: \( \text{HI} \) cannot be prepared by the reaction of \( \text{KI} \) with Concentrated \( \text{H}_2\text{SO}_4 \).
Reason: \( \text{HI} \) has lowest \( \text{H}-\text{X} \) bond strength among halogen acids.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Since concentrated sulfuric acid acts as a strong oxidizing agent, it readily oxidizes the formed hydrogen iodide to iodine while being reduced to sulfur dioxide. Consequently, a non-oxidizing acid like concentrated phosphoric acid is preferred for the preparation.
In simple words: Concentrated sulfuric acid is a strong oxidizer that turns the newly formed hydrogen iodide back into iodine, so we must use phosphoric acid instead.

Exam Tip: For preparing volatile hydrides like \( \text{HI} \) or \( \text{HBr} \), always use non-oxidizing phosphoric acid (\( \text{H}_3\text{PO}_4 \)) instead of sulfuric acid.

Question. Assertion: Ozone is a powerful oxidising agent in comparison to \( \text{O}_2 \).
Reason: Ozone is diamagnetic but oxygen is paramagnetic.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Because ozone easily breaks down to release highly reactive nascent oxygen, it behaves as an exceptionally strong oxidizing agent.
In simple words: Ozone is a strong oxidizer because it easily decomposes to release a single, highly reactive oxygen atom.

Exam Tip: Be sure to write the decomposition equation \( \text{O}_3 \rightarrow \text{O}_2 + \text{O} \) when explaining the strong oxidizing nature of ozone.

Question. Assertion: Noble gases have highest ionization energies in their respective periods.
Reason: Noble gases have a stable closed shell electronic configuration.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Noble gases have completely filled valence shells, making them extremely stable. As a result, a massive amount of energy is required to remove an electron from their stable shells.
In simple words: Since noble gases have completely full outer electron shells, they are highly stable and require a lot of energy to lose any electrons.

Exam Tip: Remember to state the electronic configuration \( ns^2 np^6 \) to explain the high chemical stability and maximum ionization energy of noble gases.

Question. Assertion: Fluorine has lower reactivity.
Reason: \( \text{F}-\text{F} \) bond has low bond dissociation energy.
Answer: (d) Assertion is wrong statement but reason is correct statement.
Fluorine is actually highly reactive due to its low \( \text{F}-\text{F} \) bond dissociation energy and high hydration energy. Thus, the assertion is false, while the reason is a true statement.
In simple words: Fluorine is actually the most reactive halogen, not the least, even though its single bond is very easy to break.

Exam Tip: Don't forget that fluorine's high reactivity is primarily due to its low bond dissociation energy and high hydration enthalpy, which outweigh its slightly lower electron gain enthalpy.

Question. Assertion: \( \text{HClO}_4 \) is a stronger acid than \( \text{HClO}_3 \).
Reason: Oxidation state of \( \text{Cl} \) in \( \text{HClO}_4 \) is +7 and in \( \text{HClO}_3 \), it is +5.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
As the oxidation state of the central chlorine atom increases, the stabilization of the conjugate base through charge dispersion over more oxygen atoms increases, resulting in a stronger acid.
In simple words: Since chlorine has a higher positive charge (+7) in perchloric acid, it pulls electrons more strongly, stabilizing the remaining ion and making the acid stronger.

Exam Tip: For oxoacids of the same halogen, remember that the greater the number of oxygen atoms, the more stable the conjugate base becomes due to resonance delocalization.

Question. Assertion: At room temperature, oxygen exists as a diatomic gas whereas sulphur exists as a solid.
Reason: The catenated \( -\text{O}-\text{O}-\text{O}- \) chains are less stable as compared to \( \text{O}=\text{O} \) molecule.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Oxygen forms stable diatomic molecules with double bonds due to its small size and effective orbital overlap, which exist as a gas. In contrast, sulfur forms single-bonded crown-shaped \( \text{S}_8 \) rings due to its tendency for catenation, making it a solid.
In simple words: Oxygen easily forms small double-bonded gas molecules, while sulfur prefers to connect in large, heavy rings of eight atoms, making it a solid at room temperature.

Exam Tip: Always associate the gas/solid difference of oxygen and sulfur with their ability to form \( p\pi-p\pi \) multiple bonds versus single-bond catenation.

Question. Assertion: \( \text{PCl}_5 \) is covalent in gaseous state but ionic in solid state.
Reason: \( \text{PCl}_5 \) exists as tetrahedral \( [\text{PCl}_4]^+ \) cation and octahedral \( [\text{PCl}_6]^- \) anion.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
While phosphorus pentachloride consists of neutral trigonal bipyramidal molecules in the gaseous phase, it achieves greater stability in the solid state by organizing into a crystalline lattice of tetrahedral and octahedral complex ions.
In simple words: In gaseous form, phosphorus pentachloride is made of neutral molecules, but in the solid state, it splits into positive and negative ions to form a stable crystal.

Exam Tip: Remember the specific geometries: \( [\text{PCl}_4]^+ \) is tetrahedral (\( sp^3 \)) and \( [\text{PCl}_6]^- \) is octahedral (\( sp^3d^2 \)). This is a common question on hybridizations.

Question. Assertion: Sulphuric Acid is more viscous than water.
Reason: Sulphuric Acid has a strong affinity for water.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Sulfuric acid is highly viscous because of the extensive network of hydrogen bonding among its molecules, which hinders their movement. Its strong affinity for water is a true statement but does not explain its internal viscosity.
In simple words: Sulfuric acid is thick and syrupy because its molecules form strong hydrogen bonds with each other, restricting their flow.

Exam Tip: Viscosity in liquids is directly related to the strength of intermolecular forces such as hydrogen bonding, rather than their chemical reactivity or affinity for other solvents.

Question. Assertion: Both \( \text{ClF}_3 \) and \( \text{SF}_4 \) show same hybridization but \( \text{ClF}_3 \) is T-Shaped.
Reason: In \( \text{ClF}_3 \), five electron pairs are present, out of which 2 are lone pairs and three are bond pairs.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Both molecules utilize \( sp^3d \) hybridization because of five electron pairs. In \( \text{ClF}_3 \), the placement of the two lone pairs in the equatorial positions to reduce electron repulsion results in its characteristic T-shaped geometry.
In simple words: Since both have five electron pairs, they share the same hybridization, but the two lone pairs on chlorine squeeze the other bonds into a T-shape.

Exam Tip: When explaining molecular shapes, always use VSEPR theory to show how lone pairs occupy equatorial positions in a trigonal bipyramidal arrangement to minimize repulsion.

Case Based Questions

Case 1
All the elements of group 16 have \( ns^2 np^4 \) configuration in their outermost shell. Therefore, the atoms of these elements try to gain or share two electrons to achieve noble gas configuration. Sulphur and other elements of group 16 are less electronegative than oxygen, so they cannot accept electrons easily. By sharing of two electrons with other elements, these element acquire \( ns^2 np^6 \) configuration and exhibit +2 oxidation state. Except oxygen, group 16 elements have vacant d-orbitals in their valence shell to which electrons can be promoted from p- and s-orbitals of the same shell. As a result, they can show +4 and +6 oxidation states also.

Question (i) Oxygen shows +2 oxidation state in
(a) \( \text{OF}_2 \)
(b) \( \text{H}_2\text{O} \)
(c) \( \text{Cl}_2\text{O} \)
(d) \( \text{H}_2\text{O}_2 \)
Answer: (a) \( \text{OF}_2 \)
Because fluorine is the only element more electronegative than oxygen, oxygen has a positive oxidation state of +2 when combined with it in oxygen difluoride.
In simple words: Fluorine attracts electrons more strongly than oxygen, which forces oxygen to have a positive (+2) oxidation state in this compound.

Exam Tip: Remember that oxygen normally has negative oxidation states (-2 or -1), but with fluorine, it is forced to exhibit positive oxidation states like +2 in \( \text{OF}_2 \).

Question (ii) Like sulphur, oxygen is not able to show +4 and +6 oxidation states because
(a) oxygen is a gas while sulphur is a solid.
(b) sulphur has high ionisation enthalpy as compared to oxygen.
(c) oxygen has no d-orbitals in its valence shell.
(d) oxygen has high electron affinity as compared to sulphur.
Answer: (c) oxygen has no d-orbitals in its valence shell.
Unlike other group members, oxygen does not have vacant d-orbitals in its outer shell, so it cannot promote its valence electrons to show +4 or +6 oxidation states.
In simple words: Oxygen has no extra space (d-orbitals) in its outer shell to spread out its electrons and form more than two bonds.

Exam Tip: The absence of vacant d-orbitals in second-period elements is the standard reason why they cannot expand their covalency beyond four.

Question  (iii) Compounds of sulphur with +4 oxidation state acts as a/an
(a) oxidising agent.
(b) reducing agent.
(c) both oxidising as well as reducing agent.
(d) Cannot be predicted.
Answer: (c) both oxidising as well as reducing agent.
With an intermediate +4 oxidation state, sulfur can either lose two electrons to reach +6 or gain electrons to arrive at lower oxidation states, enabling it to act as both an oxidant and a reductant.
In simple words: Since +4 is right in the middle of sulfur's possible oxidation states (-2 to +6), it can either gain or lose electrons, doing both jobs.

Exam Tip: If an element is in its maximum oxidation state, it can only act as an oxidizing agent. If it is in its minimum state, it can only act as a reducing agent. Intermediate states can do both.

Question (iv) Oxidation State of sulphur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is
(a) 7/2
(b) 5/2
(c) 1/2
(d) 3/2
Answer: (b) 5/2
By setting up the oxidation number equation: \( 2(+1) + 4x + 6(-2) = 0 \implies 4x = 10 \implies x = \frac{5}{2} \). The average oxidation state of the four sulfur atoms is +2.5.
In simple words: When we solve for the unknown charge of sulfur in sodium tetrathionate, we get an average value of 2.5, which is 5/2.

Exam Tip: Be aware that this is an average oxidation state. In terms of structure, the two central sulfur atoms have an oxidation state of 0, while the two terminal ones have +5.

Question (v) The Oxidation State of sulphur in \( \text{S}_8 \), \( \text{SO}_3 \) and \( \text{H}_2\text{S} \) are respectively
(a) 0, +6 and –2
(b) +6, 0 and –2
(c) –2, 0 and +6
(d) +2, +6 and –2
Answer: (a) 0, +6 and –2
Elemental sulfur in \( \text{S}_8 \) has an oxidation state of 0. In trioxide form, it exhibits +6, and in the hydride form, it has a state of -2.
In simple words: Pure sulfur has a charge of 0, while in sulfur trioxide it is +6, and in hydrogen sulfide it is -2.

Exam Tip: Remember that any element in its free or allotropic state (like \( \text{S}_8 \), \( \text{O}_2 \), \( \text{P}_4 \)) always has an oxidation state of zero.

Case 2
Ozone is an unstable, dark blue diamagnetic gas. It absorbs the UV radiation strongly, thus protecting the people on earth from the harmful UV-radiation from the sun. The use of chlorofluorocarbon (CFC) in aerosol and refrigerators and their subsequent escape into the atmosphere, is blamed for making holes in the ozone layer over the Antarctica. Ozone acts as a strong oxidising agent in acidic and alkaline medium. For this property, ozone is used as a germicide and disinfectant for sterilizing water. It is also used in laboratory for the ozonolysis of organic compounds and in industry for the manufacture of potassium permanganate, artificial silk, etc.

Question (i) Which of the following statements is not correct for ozone?
(a) It oxidises lead sulphide.
(b) It oxidises potassium iodide.
(c) It oxidises mercury.
(d) It cannot act as bleaching agent in dry state.
Answer: (d) It cannot act as bleaching agent in dry state.
Unlike chlorine, ozone is a strong dry bleaching agent that works by direct oxidation, so the claim that it cannot bleach in a dry state is incorrect.
In simple words: Ozone does not need water to bleach things because it does so through direct oxidation, meaning it works perfectly fine even in a dry state.

Exam Tip: Remember that ozone's bleaching action is permanent and occurs via oxidation, unlike sulfur dioxide's temporary bleaching which occurs via reduction.

Question (ii) Ozone gives carbonyl compounds with
(a) alkyl chloride.
(b) alkanes.
(c) alkenes followed by decomposition with \( \text{Zn}/\text{H}_2\text{O} \).
(d) alcohols followed by decomposition with \( \text{Zn}/\text{H}_2\text{O} \).
Answer: (c) alkenes followed by decomposition with \( \text{Zn}/\text{H}_2\text{O} \).
Reacting alkenes with ozone creates unstable ozonide intermediates, which yield carbonyl compounds when cleaved under reducing conditions using zinc and water.
In simple words: Ozone breaks the double bonds in alkenes to form ozonides, which then split apart when treated with zinc and water to produce aldehydes or ketones.

Exam Tip: Ozonolysis is a key organic reaction used to locate the position of double bonds in unknown alkene chains.

Question (iii) Ozone reacts with moist iodine to give
(a) \( \text{HI} \)
(b) \( \text{HIO}_3 \)
(c) \( \text{I}_2\text{O}_5 \)
(d) \( \text{I}_2\text{O}_4 \)
Answer: (b) \( \text{HIO}_3 \)
Moist iodine is oxidized by ozone to produce iodic acid according to the reaction: \( \text{I}_2 + 5\text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{HIO}_3 + 5\text{O}_2 \).
In simple words: Ozone is a strong oxidizer that turns wet iodine into iodic acid while releasing oxygen gas.

Exam Tip: Be careful with the state of the reactant: dry iodine reacts with ozone to form yellow solid \( \text{I}_4\text{O}_9 \), while moist iodine yields \( \text{HIO}_3 \).

Question (iv) Ozone acts as an oxidising agent due to
(a) liberation of nascent oxygen.
(b) liberation of oxygen gas.
(c) both (a) and (b).
(d) none of the options
Answer: (a) liberation of nascent oxygen.
Because of its thermodynamic instability, ozone easily decomposes to yield a molecule of oxygen and a highly reactive nascent oxygen atom.
In simple words: Ozone breaks down very easily to set free a single, highly active oxygen atom which instantly oxidizes other elements.

Exam Tip: Nascent oxygen (\( \text{O} \)) is a single unbonded oxygen atom that is highly unstable and reacts instantly, making ozone a powerful oxidizing agent even at room temperature.

Question (v) The colour of ozone molecule is
(a) white
(b) blue
(c) pale green
(d) pale yellow
Answer: (b) blue
Depending on its physical state, ozone appears as a pale blue gas, a deep blue liquid, or a dark violet solid.
In simple words: Ozone has a light blue color in its gas form, which darkens into a deep blue as a liquid.

Exam Tip: Memorize the colors of ozone in all three states: pale blue gas, dark blue liquid, and violet-black solid.

Case 3
The halogen elements show great resemblances to one another in their chemical behaviour and properties of their compounds with other elements. There is, however, a progressive change in properties from F through Cl, Br and I to At. F is most reactive among the halogens and infact, from all other elements and it has certain other properties that set it apart from the other halogens.

Question (i) Assertion: \( \text{F}_2 \) has high reactivity.
Reason: \( \text{F}_2 \) has low bond dissociation enthalpy.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
The high reactivity of fluorine is driven by its strong electronegativity coupled with the low energy required to break the \( \text{F}-\text{F} \) bond.
In simple words: Fluorine reacts incredibly fast because its molecules split easily and it is extremely eager to attract electrons.

Exam Tip: Remember that despite fluorine having a lower electron gain enthalpy than chlorine, its low bond dissociation energy is what makes it far more reactive.

Question (ii) Assertion: The bond between \( \text{F}-\text{F} \) is weaker than between \( \text{Cl}-\text{Cl} \).
Reason: Atomic size of \( \text{F} \) is smaller than that of \( \text{Cl} \).
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Because fluorine has a very small atomic size, the non-bonding valence electron pairs on adjacent fluorine atoms repel each other strongly, weakening the \( \text{F}-\text{F} \) bond.
In simple words: Since fluorine atoms are tiny, their outer electrons are packed closely together, causing them to push each other away and weaken the bond.

Exam Tip: The unexpected weakness of the \( \text{F}-\text{F} \) bond compared to \( \text{Cl}-\text{Cl} \) is due to high inter-electronic repulsion of lone pairs on the small fluorine atoms.

Question (iii) Assertion: Fluoride does not show oxidation number greater than zero.
Reason: The halogens chlorine, bromine and iodine can show positive oxidation states of +1, +3 and +7.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
As fluorine is the most electronegative element, it always exhibits a negative oxidation state of -1 and can never show positive oxidation numbers. The ability of other halogens to show positive states is true but does not explain this.
In simple words: Fluorine is the most electronegative element on the periodic table, meaning it always pulls electrons toward itself and never gets a positive charge.

Exam Tip: Always state that fluorine lacks d-orbitals and has the highest electronegativity, which prevents it from showing any positive oxidation states.

Question (iv) Assertion: \( \text{F} \) atom has less negative electron affinity than \( \text{Cl} \) atom.
Reason: Additional electrons are repelled more effectively by 3p- electrons in \( \text{Cl} \) than by 2p- electrons in \( \text{F} \) atom.
Answer: (c) Assertion is correct statement but reason is wrong statement.
Due to the extremely compact size of the fluorine atom, its 2p cloud is highly crowded, causing incoming electrons to experience strong inter-electronic repulsion. Consequently, the electron gain enthalpy of fluorine is less negative than that of chlorine, making the reason statement false.
In simple words: The outer shell of fluorine is so tiny and crowded that incoming electrons face a lot of pushing back, which is why chlorine actually has a more negative electron affinity.

Exam Tip: Be ready to explain why chlorine has a more negative electron gain enthalpy than fluorine, focusing on the high inter-electronic repulsion in fluorine's small 2p subshell.

Question (v) Assertion: Fluorine is strongest oxidising agent in halogens.
Reason: It displaces other halogens from its aqueous solution.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
The exceptionally high oxidizing power of fluorine is due to its low \( \text{F}-\text{F} \) bond dissociation energy and very high hydration enthalpy, which allow it to easily oxidize other halide ions in aqueous solution and displace them.
In simple words: Fluorine is a superb oxidizer because it is extremely eager to take electrons from other halogens in water, easily pushing them out of solution.

Exam Tip: Remember that oxidizing power depends on three thermodynamic factors: sublimation enthalpy, dissociation enthalpy, and hydration enthalpy.

Case 4
In the last 10 years much has been learned about the molecular structure of elemental sulphur. It is now known that many different types of rings are sufficiently metastable to exist at room temperature for several days. It is known that at high temperature, the equilibrium composition allows for a variety of rings and chains to exist in comparable concentration, and it is known that at the boiling point and above, the vapor as well as the liquid contains small species with three, four, and five atoms. The sulfur atom has the same number of valence electrons as oxygen. Thus, sulphur atoms \( \text{S}_2 \) and \( \text{S}_3 \) have physical and chemical properties analogous to those of oxygen and ozone. \( \text{S}_2 \) has a ground state of 38 \( \sigma 3s^2 \sigma^* 3s^2 \sigma 3p_z^2 p 3p_x^2 = p 3p_y^2 p^* 3p_x^1 = p^* 3p_y^1 \). \( \text{S}_3 \), thiozone has a well known UV spectrum, and has a bent structure, analogous to its isovalent molecules \( \text{O}_3 \), \( \text{SO}_2 \) and \( \text{S}_2\text{O} \). The chemistry of the two elements, sulphur and oxygen, differs because sulfur has a pronounced tendency for catenation. The most frequently quoted explanation is based on the electron structure of the atom. Sulfur has low-lying unoccupied 3d orbitals, and it is widely believed that the 4s and 3d orbitals of sulfur participate in bonding in a manner similar to the participation of 2s and 2p orbitals in carbon.

Question (i) Assertion: Sulphur belongs to same group in the periodic table as oxygen.
Reason: \( \text{S}_2 \) has properties analogous to \( \text{O}_2 \).
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Sulfur belongs to the same oxygen family (Group 16) due to its outer shell configuration of \( ns^2 np^4 \). While \( \text{S}_2 \) shares analogous electronic and physical properties with \( \text{O}_2 \), this similarity is a consequence of their group relationship, not the cause.
In simple words: Sulfur sits in the same periodic group as oxygen because they share the same number of outer electrons, not because of their diatomic properties.

Exam Tip: The primary basis for placing elements in the same periodic group is their outer shell (valence) electronic configuration.

Question (ii) Assertion: Thiozone has bent structure like ozone.
Reason: Ozone has a lone pair which makes the molecule bent.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Both ozone (\( \text{O}_3 \)) and thiozone (\( \text{S}_3 \)) adopt bent geometries due to the presence of lone pairs on the central atom. The reason correctly describes ozone's shape but does not explicitly explain why thiozone also has a bent structure.
In simple words: Both molecules have a bent shape because their central atoms have lone pairs of electrons that push the side bonds down.

Exam Tip: Remember that isoelectronic species like \( \text{O}_3 \), \( \text{SO}_2 \), and \( \text{S}_3 \) often share similar bent geometries due to comparable lone-pair repulsions.

Question (iii) Assertion: \( \text{S}_2 \) is paramagnetic in nature.
Reason: The electrons in \( \pi^* 3p_x \) and \( \pi^* 3p_y \) orbitals in \( \text{S}_2 \) are unpaired.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Similar to oxygen, the vapor phase of sulfur contains \( \text{S}_2 \) molecules which are paramagnetic because they have two unpaired electrons in their antibonding \( \pi^* \) orbitals.
In simple words: In its gas form, sulfur exists as diatomic molecules with unpaired electrons in its outer orbits, making it paramagnetic just like oxygen gas.

Exam Tip: This is a highly high-scoring concept: remember that both \( \text{O}_2 \) and \( \text{S}_2 \) have two unpaired electrons in their antibonding \( \pi^* \) molecular orbitals, rendering them paramagnetic.

Question (iv) Assertion: Sulphur has a greater tendency for catenation than oxygen.
Reason: 3d and 4s orbitals of Sulphur have same energy.
Answer: (c) Assertion is correct statement but reason is wrong statement.
Sulfur has a much stronger tendency for catenation than oxygen because the \( \text{S}-\text{S} \) single bond is far stronger than the \( \text{O}-\text{O} \) single bond, which suffers from lone pair repulsion. The claim that the 3d and 4s orbitals have identical energies is incorrect.
In simple words: Sulfur atoms can link into long chains because their single bonds are strong and stable, whereas the reason is false because 3d and 4s orbitals do not have the same energy.

Exam Tip: Catenation capability depends on the strength of the element-element single bond. The \( \text{S}-\text{S} \) bond is stronger than the \( \text{O}-\text{O} \) bond due to less lone-pair repulsion in sulfur.

Case 5
Under normal conditions, noble gases are monoatomic and have closed shell electronic configuration. Lighter noble gases have low boiling points due to weak dispersion forces between the atoms and the absence of other interatomic interactions. Xenon, one of the important noble gas, forms a series of compounds with fluorine with oxidation number +2, +4 and +6. All xenon fluorides are strong oxidising agents. XeF4 reacts violently with water to give XeO3. The geometry of xenon compounds can be deduced by considering the total number of electron pairs in their valence shell.

Question (i) Among the noble gases (from He to Xe) only xenon reacts with fluorine to form stable xenon fluorides because xenon
(a) as the largest size.
(b) has the lowest ionisation enthalpy.
(c) has the highest heat of vapourisation.
(d) is the most readily available noble gas.
Answer: (b) has the lowest ionisation enthalpy.
As the atomic radius increases down the noble gas group, the ionization energy drops, allowing xenon to react with highly electronegative fluorine to form stable fluorides.
In simple words: Xenon's outer electrons are held loosely because of its large atomic size, which lets fluorine pull them into bonds.

Exam Tip: Remember that ionization energy decreases down the group, making xenon the most reactive of the non-radioactive noble gases.

Question (ii) The structure of \( \text{XeO}_3 \) is
(a) square planar
(b) Pyramidal
(c) linear
(d) T-shaped
Answer: (b) Pyramidal
In \( \text{XeO}_3 \), xenon has four electron pairs (three bonding pairs and one lone pair) with \( sp^3 \) hybridization, creating a pyramidal shape.
In simple words: The central xenon atom is bonded to three oxygen atoms and has one lone pair, pushing the bonds down into a pyramid shape.

Exam Tip: Make sure you do not confuse the pyramidal geometry of \( \text{XeO}_3 \) with the trigonal planar geometry of other trioxides, due to the presence of the active lone pair on xenon.

Question (iii) \( \text{XeF}_6 \) is expected to be
(a) oxidising agent.
(b) reducing agent.
(c) unreactive.
(d) strongly basic.
Answer: (a) oxidising agent.
Since xenon fluorides easily release fluorine or accept electrons to return to elemental xenon, \( \text{XeF}_6 \) acts as a very potent oxidizing agent.
In simple words: Xenon fluorides are very keen to give away their fluorine atoms, which makes them excellent at oxidizing other chemicals.

Exam Tip: Always remember that xenon compounds readily reduce back to elemental Xe, which is why all xenon fluorides are exceptionally strong oxidizing agents.

Question (iv) In the preparation of compound of xenon, Bartlett had taken \( \text{O}_2^+ [\text{PtF}_6]^- \) as a base compound. This is because
(a) both \( \text{O}_2 \) and Xe have same size.
(b) both Xe and \( \text{O}_2 \) have same electron gain enthalpy.
(c) both have almost same ionisation enthalpy.
(d) both Xe and \( \text{O}_2 \) are gases.
Answer: (c) both have almost same ionisation enthalpy.
Neil Bartlett realized that molecular oxygen and atomic xenon have nearly identical first ionization energies, prompting him to try synthesizing a corresponding xenon compound.
In simple words: Both oxygen molecules and xenon atoms require nearly the same amount of energy to lose an electron, which inspired the first xenon reaction.

Exam Tip: Memorize this historical fact: the ionization energy of Xe is \( 1170 \text{ kJ/mol} \) while that of oxygen is \( 1175 \text{ kJ/mol} \), which led to the synthesis of \( \text{Xe}^+ [\text{PtF}_6]^- \).

Question (v) The oxidation state of xenon in \( \text{XeO}_3 \) is
(a) +4
(b) +2
(c) +8
(d) +6
Answer: (d) +6
Since oxygen exhibits a -2 oxidation state, solving the charge equation for neutral xenon trioxide yields an oxidation state of +6 for xenon: \( x + 3(-2) = 0 \implies x = +6 \).
In simple words: Since each of the three oxygens has a -2 charge, the central xenon atom must have a +6 charge to balance them out.

Exam Tip: Always double check your simple algebra when calculating oxidation states, as this is an easy place to lose quick marks.