CBSE Class 12 Chemistry The p Block Elements Assignment Set 02

Question. Why is Bi(v) a stronger oxidant than Sb(v)?
Answer: The stability of the \( +5 \) oxidation state reduces and that of the \( +3 \) state rises down the group because of the inert pair effect. Consequently, Bi(V) readily gains two electrons to get reduced to Bi(III):
\( \text{Bi}^{5+} + 2e^- \rightarrow \text{Bi}^{3+} \)
In simple words: Bi(V) is less stable than Bi(III) because of the inert pair effect, so it easily takes two electrons from other substances to become stable.

Exam Tip: Inert pair effect is a key reason for the stability of lower oxidation states in heavier p-block elements. Always mention this term and write the chemical equation.

Question. Which is a stronger oxidizing agent Bi(v) or Sb(v)?
Answer: Bi(V) serves as a stronger oxidizing agent because of the inert pair effect.
In simple words: Bi(V) wants to gain electrons more than Sb(V) due to the inert pair effect, making it a stronger oxidizer.

Exam Tip: Clearly state that Bi(V) is stronger and attribute this directly to the inert pair effect for full marks.

Question. Why is red phosphorus less reactive than white phosphorus?
Answer: This is because white phosphorus experiences high angular strain within its \( \text{P}_4 \) molecules, where the bond angle is merely \( 60^\circ \).
In simple words: White phosphorus has a lot of strain in its shape because of its tight 60-degree angles, which makes it very unstable and highly reactive compared to red phosphorus.

Exam Tip: Mention 'angular strain' and the specific bond angle of \( 60^\circ \) in white phosphorus to get full marks.

Question. Why does NO₂ dimerise?
Answer: The \( \text{NO}_2 \) molecule possesses an odd number of electrons, specifically 23 valence electrons. Nitrogen has only seven valence electrons in this state, making it unstable. In order to achieve stability, it dimerizes to yield \( \text{N}_2\text{O}_4 \).
In simple words: \( \text{NO}_2 \) has an odd number of electrons, which makes it unstable. It pairs up with another \( \text{NO}_2 \) molecule to share electrons and become stable.

Exam Tip: Be sure to state the total number of valence electrons (23 odd electrons) and how dimerization results in a stable octet.

Question. What is the oxidation number of phosphorus in H3PO2 molecule?
Answer: In the \( \text{H}_3\text{PO}_2 \) molecule, let the oxidation state of \( \text{P} \) be \( x \). Since hydrogen is \( +1 \) and oxygen is \( -2 \), we have:
\( 3(+1) + x + 2(-2) = 0 \)
\( \implies 3 + x - 4 = 0 \)
\( \implies x - 1 = 0 \)
\( \implies x = +1 \)
Therefore, the oxidation state of phosphorus in \( \text{H}_3\text{PO}_2 \) is \( +1 \).
In simple words: To find the oxidation state of phosphorus, we use the known charges of hydrogen (+1) and oxygen (-2). Setting the sum of all charges to zero gives a charge of +1 for phosphorus.

Exam Tip: Always show the step-by-step calculation. Remember that the overall charge of the neutral molecule is zero.

Question. Fluorine does not exhibit any positive oxidation state. Why?
Answer: Because fluorine is the most electronegative element and lacks d-orbitals in its valence shell, it is unable to expand its octet. Consequently, it never displays a positive oxidation state, whereas other halogens have vacant d-orbitals and can show various positive oxidation states.
In simple words: Fluorine is extremely greedy for electrons and has no empty space (d-orbitals) to hold extra electrons, so it only gets negative oxidation states.

Exam Tip: Key phrases to include: 'most electronegative element' and 'absence of d-orbitals'.

Question. Nitrogen is relatively inert as compared to phosphorus. Why?
Answer: This is because the \( \text{P}-\text{P} \) single bond is far weaker than the \( \text{N}\equiv\text{N} \) triple bond. Nitrogen has a very short bond length and an exceptionally high bond dissociation enthalpy, rendering it chemically inert, whereas phosphorus is much more reactive.
In simple words: Nitrogen molecules are held together by a very strong triple bond that is extremely hard to break, while phosphorus has weaker single bonds, making phosphorus much more active.

Exam Tip: Highlight the presence of the triple bond in \( \text{N}_2 \) and its high bond dissociation energy as the primary reasons for its inertness.

Question. Which one of PCl₄⁺ and PCl₄^- is not likely to exist and why?
Answer: The species \( \text{PCl}_4^- \) is unlikely to exist. This is because phosphorus in \( \text{PCl}_4^- \) would have 10 valence shell electrons, which cannot be accommodated within standard \( sp^3 \) hybrid orbitals without involving higher d-orbitals, leading to less stable geometry.
In simple words: PCl4- is unstable and unlikely to exist because phosphorus would have too many electrons to fit comfortably into its standard outer shell orbitals.

Exam Tip: Clearly specify that \( \text{PCl}_4^- \) does not exist, explaining that \( \text{P} \) has 10 valence electrons which exceed the capacity of \( sp^3 \) hybridization.

Question. Of PH₃ and H₂S which is more acidic and why?
Answer: Between \( \text{PH}_3 \) and \( \text{H}_2\text{S} \), \( \text{H}_2\text{S} \) is more acidic. This is due to the greater electronegativity of sulfur compared to phosphorus, which polarizes the \( \text{S}-\text{H} \) bond and makes the release of \( \text{H}^+ \) ions easier.
In simple words: H2S is more acidic because sulfur is more electronegative than phosphorus, which makes it easier for the hydrogen to break away as an acid.

Exam Tip: To explain trends in acidity across a period, focus on the electronegativity of the central atom.

Question. Which is a stronger reducing agent, SbH₃ or BiH₃, and why?
Answer: \( \text{BiH}_3 \) is the stronger reducing agent. This is because the \( \text{Bi}-\text{H} \) bond is very weak due to the large size of bismuth, which maximizes its tendency to liberate hydrogen.
In simple words: BiH3 is a stronger reducing agent because the bond between bismuth and hydrogen is very weak, so it easily lets go of hydrogen.

Exam Tip: Correlate reducing character directly with the decreasing bond dissociation energy of the \( \text{M}-\text{H} \) bond down the group.

Question. What is the basicity of H₃PO₂ acid and why?
Answer: The basicity of \( \text{H}_3\text{PO}_2 \) is 1 (it is monobasic). This is because it contains only one replaceable hydrogen atom directly attached to oxygen as an \( -\text{OH} \) group.
In simple words: H3PO2 is monobasic because only the one hydrogen atom connected to oxygen can be released as an acid ion.

Exam Tip: Always clarify that only the hydrogen atoms bonded directly to oxygen (in \( -\text{OH} \) groups) are ionizable, while those bonded to phosphorus are not.

Question. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why?
Answer: Although nitrogen shows a \( +5 \) oxidation state, it cannot form pentahalides. This is due to the complete absence of vacant d-orbitals in its valence electronic shell, preventing it from expanding its octet beyond four.
In simple words: Nitrogen does not have any d-orbitals in its outer shell, so it cannot hold more than eight shared electrons to form five bonds.

Exam Tip: The non-availability of d-orbitals is the primary reason for nitrogen's maximum covalency of 4. Make sure to use the term 'covalency' or 'octet expansion'.

Question. Noble gases have low boiling points. Why?
Answer: Since noble gases exist as monoatomic molecules, they experience no interatomic attractions other than extremely weak London dispersion forces. Consequently, they can only be liquefied at extremely low temperatures, resulting in very low boiling points.
In simple words: Noble gas atoms do not bond with each other and are held together only by very weak forces, so they boil and turn into gas at very low temperatures.

Exam Tip: Mention that noble gases are monoatomic and that the only intermolecular forces present are weak dispersion (London) forces.

Question. Write a reaction to show the reducing behaviour of H₃PO₂.
Answer: The reducing nature of \( \text{H}_3\text{PO}_2 \) is demonstrated by its ability to reduce silver nitrate (\( \text{AgNO}_3 \)) to metallic silver (\( \text{Ag} \)):
\( 4\text{AgNO}_3 + 2\text{H}_2\text{O} + \text{H}_3\text{PO}_2 \rightarrow 4\text{Ag} + 4\text{HNO}_3 + \text{H}_3\text{PO}_4 \)
In simple words: H3PO2 can turn silver nitrate into solid silver metal by donating electrons to it.

Exam Tip: Write the balanced chemical equation clearly, marking the reduction of \( \text{Ag}^+ \) to \( \text{Ag} \).

Question. Why does PCl₃ fume in moisture?
Answer: \( \text{PCl}_3 \) fumes in moist air because it undergoes rapid hydrolysis to release gaseous hydrogen chloride (\( \text{HCl} \)):
\( \text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3\text{HCl} \)
In simple words: PCl3 reacts quickly with water in the air to produce hydrogen chloride gas, which looks like white fumes.

Exam Tip: Include the balanced chemical reaction of hydrolysis to support your qualitative explanation.

Question. Why does NO₂ dimerise ?
Answer: \( \text{NO}_2 \) contains an odd count of valence electrons, acting as a classic odd-electron molecule. Upon dimerization, it converts into a stable \( \text{N}_2\text{O}_4 \) molecule featuring an even number of paired electrons.
In simple words: Because NO2 has an unpaired electron, it joins with another NO2 molecule to pair up that electron and form a stable molecule.

Exam Tip: Clearly link dimerization to the pairing of the odd electron to achieve stable electronic configuration.

Question. Why is ICl more reactive than I2?
Answer: \( \text{ICl} \) is more reactive than \( \text{I}_2 \) because the polar \( \text{I}-\text{Cl} \) bond is significantly weaker and easier to break than the non-polar \( \text{I}-\text{I} \) bond.
In simple words: The bond between iodine and chlorine is weaker than the bond between two iodine atoms, making ICl break apart and react more easily.

Exam Tip: Explain that interhalogen bonds (like I-Cl) are polar and weaker than homonuclear halogen bonds (like I-I).

Question. Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements?
Answer: The reducing capacity of hydrides depends directly on the stability of the \( \text{M}-\text{H} \) bond. Since the bond stability decreases from nitrogen to bismuth hydrides due to increasing bond length, \( \text{BiH}_3 \) becomes the strongest reducing agent.
In simple words: As you go down the group, the central atom gets bigger, making the bond with hydrogen weaker and easier to break to release hydrogen.

Exam Tip: Explain that larger atomic size down the group leads to weaker orbital overlap and lower bond dissociation enthalpy.

Question. What is the covalency of nitrogen in N₂O₅?
Answer: The covalency of nitrogen in \( \text{N}_2\text{O}_5 \) is 4. This is because each nitrogen atom shares exactly four pairs of electrons, as it cannot expand its octet beyond four.
In simple words: Even though nitrogen is in its +5 oxidation state, it can only form a maximum of four bonds because it has no d-orbitals to hold more electrons.

Exam Tip: Do not confuse oxidation state (+5) with covalency (4). Nitrogen can never have a covalency of 5.

Question. What inspired N.Bartlett for carrying out reaction between Xe and PtF6?
Answer: Bartlett observed that \( \text{PtF}_6 \) could oxidize molecular oxygen to form \( \text{O}_2^+[\text{PtF}_6]^- \). He reasoned that \( \text{PtF}_6 \) should similarly oxidize xenon to \( \text{Xe}^+ \) because the first ionization enthalpy of xenon (\( 1170 \text{ kJ mol}^{-1} \)) is very close to that of oxygen (\( 1175 \text{ kJ mol}^{-1} \)).
In simple words: Neil Bartlett realized that oxygen and xenon require almost the same amount of energy to lose an electron, so if a compound could react with oxygen, it should also react with xenon.

Exam Tip: Always mention the specific values of the first ionization enthalpies of \( \text{O}_2 \) and \( \text{Xe} \) to show the close comparison.

Question. What happens when ethyl chloride is treated with aqueous KOH?
Answer: When ethyl chloride is heated with aqueous potassium hydroxide (\( \text{KOH} \)), it undergoes nucleophilic substitution to produce ethanol and potassium chloride:
\( \text{C}_2\text{H}_5\text{Cl} + \text{aq. KOH} \xrightarrow{\Delta} \text{C}_2\text{H}_5\text{OH} + \text{KCl} \)
In simple words: Ethyl chloride reacts with water-based KOH to replace the chlorine atom with an alcohol group, forming ethanol.

Exam Tip: Ensure you write 'aq. KOH' rather than alcoholic KOH, as alcoholic KOH would lead to elimination (formation of ethene) instead of substitution.

Question. Name two poisonous gases which can be prepared from chlorine gas.
Answer: Two highly toxic gases that can be synthesized from chlorine gas are:
(i) Phosgene (\( \text{COCl}_2 \))
(ii) Chloropicrin or tear gas (\( \text{CCl}_3\text{NO}_2 \))
In simple words: Chlorine can be used to make phosgene and chloropicrin, both of which are highly poisonous gases.

Exam Tip: Memorize both the common names and the molecular formulas for full credit.

Question. Which aerosol depletes ozone layer?
Answer: Aerosols such as sprays and foams contain chlorofluorocarbons (CFCs), commonly referred to as freons, which cause the depletion of the ozone layer.
In simple words: Aerosol sprays often release freons (CFCs) into the air, which break down the protective ozone layer.

Exam Tip: Mention 'freons' or 'chlorofluorocarbons (CFCs)' as the active agents responsible for ozone depletion.

Question. What is the basicity of H₃PO₃ and why?
Answer: The basicity of phosphorus acid (\( \text{H}_3\text{PO}_3 \)) is 2. This is because only two hydrogen atoms are bonded to highly electronegative oxygen atoms to form ionizable \( -\text{OH} \) groups, while the third hydrogen is bonded directly to phosphorus.
In simple words: H₃PO₃ is dibasic because only the two hydrogens attached to oxygens can separate to act as acids, while the one attached directly to phosphorus cannot.

Exam Tip: Always draw the structure of \( \text{H}_3\text{PO}_3 \) to show the two \( -\text{OH} \) groups and one \( \text{P}-\text{H} \) bond to clearly justify its basicity of 2.

Question. Why does PCl₃ fume in moisture?
Answer: Phosphorus trichloride reacts rapidly with atmospheric moisture, yielding phosphorus acid and hydrogen chloride gas through a highly exothermic hydrolysis reaction:
\( \text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3\text{HCl} \)
In simple words: PCl3 has a highly reactive, exothermic reaction with water in the air, releasing hydrochloric acid gas which appears as white fumes.

Exam Tip: Make sure to write the balanced reaction. Note that the HCl gas formed is what actually produces the visible fumes.

Question. Draw the structure of H₃PO₂ molecule.
Answer: The structure of the \( \text{H}_3\text{PO}_2 \) (hypophosphorous acid) molecule is tetrahedral. It features one \( \text{P}=\text{O} \) bond, one \( \text{P}-\text{OH} \) bond, and two \( \text{P}-\text{H} \) bonds directly connected to the central phosphorus atom.
In simple words: H₃PO₂ has a phosphorus atom in the middle bonded to one double-bonded oxygen, one OH group, and two hydrogen atoms.

Exam Tip: Drawing this structure helps show why the basicity is 1 (only one \( \text{P}-\text{OH} \) bond is present).

Question. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Why is it so?
Answer: This occurs because fluorine is the most electronegative element, and it completely lacks d-orbitals in its outer shell, making it impossible to expand its octet to show positive states.
In simple words: Fluorine is extremely electronegative, so it always pulls electrons toward itself to get a -1 state, and it cannot expand its shell due to having no d-orbitals.

Exam Tip: Make sure to state both reasons: fluorine's high electronegativity and the absence of d-orbitals.

Question. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why?
Answer: This is because the nitrogen atom has no vacant d-subshell in its valence layer, which restricts its maximum covalency to four.
In simple words: Nitrogen lacks d-orbitals to expand its bonding capacity, so it cannot form five bonds to make pentahalides.

Exam Tip: Keep your response focused on the lack of d-orbitals, which limits nitrogen's maximum covalency to 4.

Question. Bond enthalpy of fluorine is lower than that of chlorine. Why?
Answer: This is because the \( \text{F}_2 \) molecule is very small, which results in exceptionally strong interelectronic repulsions between the non-bonding lone pairs on the two adjacent fluorine atoms.
In simple words: Fluorine atoms are so small that their lone pair electrons are squeezed very close together, creating a strong repulsion that weakens the bond.

Exam Tip: Focus on the high interelectronic repulsion between the lone pairs due to the extremely small size of the fluorine atom.

Question. HF is a weaker acid than HCl. Why?
Answer: This is because the \( \text{H}-\text{F} \) bond has a much higher bond dissociation enthalpy and there is strong intermolecular hydrogen bonding in \( \text{HF} \), which resists the release of protons compared to \( \text{HCl} \).
In simple words: HF has a stronger bond and strong hydrogen bonding holding its molecules together, making it harder to release hydrogen ions compared to HCl.

Exam Tip: Attribute the weaker acidity of HF to both its exceptionally high bond dissociation enthalpy and strong hydrogen bonding.

Question. What is the basicity of H₃PO₃?
Answer: The basicity of \( \text{H}_3\text{PO}_3 \) is 2. Basicity is defined as the number of ionizable hydrogen ions in an acid, and \( \text{H}_3\text{PO}_3 \) acts as a dibasic acid since it contains only two replaceable hydrogen atoms as \( -\text{OH} \) groups.
In simple words: H3PO3 has a basicity of 2 because it has only two acidic hydrogen atoms that can be released when dissolved in water.

Exam Tip: Clearly state that \( \text{H}_3\text{PO}_3 \) is a dibasic acid due to its two ionizable \( -\text{OH} \) bonds.

Question. Why does NO₂ dimerise?
Answer: The \( \text{NO}_2 \) molecule contains an odd count of valence electrons (23 electrons), which leaves nitrogen with only seven valence electrons, making it less stable. To attain a stable octet, it readily dimerizes to produce \( \text{N}_2\text{O}_4 \).
In simple words: NO₂ dimerizes to share its unpaired electron with another molecule, allowing both to complete their stable octets.

Exam Tip: Always write the total number of valence electrons (23) and point out that dimerization solves the instability of the unpaired electron.

Question. Why does NH₃ act as a Lewis base?
Answer: This is because nitrogen in \( \text{NH}_3 \) possesses a non-bonding lone pair of electrons that can be easily donated to electron-deficient species.
In simple words: NH₃ can act as a Lewis base because its nitrogen atom has an extra pair of electrons that it can share with other atoms.

Exam Tip: Explicitly mention the 'presence of a lone pair of electrons on the nitrogen atom' as the reason for its donor ability.

Question. Why is F₂ a stronger oxidising agent than Cl₂?
Answer: This is because fluorine has a very low bond dissociation enthalpy and a highly negative electron gain enthalpy combined with high hydration energy. Consequently, it has a strong tendency to accept electrons:
\( \text{F} + e^- \rightarrow \text{F}^- \). Thus, \( \text{F}_2 \) serves as a potent oxidizing agent compared to \( \text{Cl}_2 \).
In simple words: Fluorine molecules break apart very easily and fluorine atoms love to gain electrons, making it a very powerful oxidizing agent.

Exam Tip: Attribute the high oxidizing power of \( \text{F}_2 \) to its low bond dissociation energy and very high hydration energy of the fluoride ion.

Question. What is the basicity of H₃PO₄?
Answer: The basicity of \( \text{H}_3\text{PO}_4 \) is 3. This is because there are three ionizable \( -\text{OH} \) groups present in the phosphoric acid molecule.
In simple words: H₃PO₄ has a basicity of 3 because it has three hydrogen-oxygen bonds that can release hydrogen ions.

Exam Tip: State that all three hydrogen atoms are part of \( -\text{OH} \) groups and are therefore replaceable, making \( \text{H}_3\text{PO}_4 \) tribasic.

Question. Write the formulae of any two oxoacids of sulphur.
Answer: Two common oxoacids of sulfur are:
(i) Sulfurous acid (\( \text{H}_2\text{SO}_3 \))
(ii) Sulfuric acid (\( \text{H}_2\text{SO}_4 \))
In simple words: Two acid compounds containing sulfur and oxygen are sulfurous acid (H2SO3) and sulfuric acid (H2SO4).

Exam Tip: Be sure to write both the names and molecular formulas accurately.

Question. On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu²⁺ ion. Identify the gas.
Answer: The pungent, colorless gas is ammonia (\( \text{NH}_3 \)). The deep blue complex formed with copper ions is tetraamminecopper(II) ion, \( [\text{Cu}(\text{NH}_3)_4]^{2+} \).
In simple words: The gas is ammonia. When it reacts with copper ions, it creates a bright blue substance called a tetraamminecopper complex.

Exam Tip: Identify the gas as \( \text{NH}_3 \) and write the formula of the blue complex, \( [\text{Cu}(\text{NH}_3)_4]^{2+} \), to get full credit.

Question. Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerization on cooling. Identify the gas.
Answer: The brown gas is nitrogen dioxide (\( \text{NO}_2 \)), which dimerizes to form colorless dinitrogen tetroxide (\( \text{N}_2\text{O}_4 \)) upon cooling. The decomposition reaction is:
\( 2\text{Pb}(\text{NO}_3)_2 \xrightarrow{\Delta} 2\text{PbO} + 4\text{NO}_2 + \text{O}_2 \)
In simple words: Heating lead nitrate produces nitrogen dioxide gas, which is brown and turns into N2O4 when it gets cold.

Exam Tip: Specify the identity of the gas as \( \text{NO}_2 \) and provide the balanced thermal decomposition equation of lead nitrate.

Question. Write the formula of the compound of phosphorus which is obtained when conc. HNO₃ oxidises P4.
Answer: When concentrated nitric acid (\( \text{HNO}_3 \)) oxidizes white phosphorus (\( \text{P}_4 \)), orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)) is produced:
\( \text{P}_4 + 20\text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + 20\text{NO}_2 + 4\text{H}_2\text{O} \)
In simple words: Reacting white phosphorus with strong nitric acid produces phosphoric acid (H3PO4) along with nitrogen dioxide and water.

Exam Tip: Write the balanced chemical reaction to show the formation of orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)).

Question. Write the formula of the compound of sulphur which is obtained when conc. HNO₃ oxidises S8.
Answer: The oxidation of sulfur (\( \text{S}_8 \)) by concentrated nitric acid yields sulfuric acid (\( \text{H}_2\text{SO}_4 \)):
\( \text{S}_8 + 48\text{HNO}_3 \rightarrow 8\text{H}_2\text{SO}_4 + 48\text{NO}_2 + 16\text{H}_2\text{O} \)
In simple words: Concentrated nitric acid oxidizes solid sulfur (S8) to create sulfuric acid (H2SO4).

Exam Tip: Ensure you balance the equation correctly, showing that \( \text{H}_2\text{SO}_4 \) is the sulfur-containing product.

Question. Write the formula of the compound of iodine which is obtained when conc. HNO₃ oxidises I2.
Answer: When iodine (\( \text{I}_2 \)) is oxidized by concentrated nitric acid, iodic acid (\( \text{HIO}_3 \)) is obtained:
\( \text{I}_2 + 10\text{HNO}_3 \rightarrow 2\text{HIO}_3 + 10\text{NO}_2 + 4\text{H}_2\text{O} \)
In simple words: Reacting iodine with strong nitric acid produces iodic acid, which has the chemical formula HIO3.

Exam Tip: Remember that iodic acid (\( \text{HIO}_3 \)) is the oxidation product of iodine, not hydrogen iodide.

Question. Complete the following chemical reaction equations:
(i) \( \text{P}_4(\text{s}) + \text{NaOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \)
(ii) \( \text{I}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{O}_3(\text{g}) \rightarrow \)
Answer:
(i) White phosphorus reacts with hot, concentrated aqueous sodium hydroxide to undergo disproportionation, forming phosphine gas and sodium hypophosphite:
\( \text{P}_4(\text{s}) + 3\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{PH}_3(\text{g}) + 3\text{NaH}_2\text{PO}_2(\text{aq}) \)
(ii) Iodide ions are oxidized by ozone in the presence of water to liberate molecular iodine:
\( 2\text{I}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{O}_3(\text{g}) \rightarrow \text{I}_2(\text{s}) + 2\text{OH}^-(\text{aq}) + \text{O}_2(\text{g}) \)
In simple words: First, phosphorus reacts with basic water to make phosphine and sodium hypophosphite. Second, ozone reacts with iodide ions to create iodine gas and oxygen.

Exam Tip: These are highly repeating equations in CBSE boards. Practice balancing them carefully.

Question. Complete the following chemical reaction equations:
(i) \( \text{XeF}_2(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \)
(ii) \( \text{PH}_3 + \text{HgCl}_2 \rightarrow \)
Answer:
(i) \( 2\text{XeF}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Xe}(\text{g}) + 4\text{HF}(\text{aq}) + \text{O}_2(\text{g}) \)
(ii) \( 2\text{PH}_3 + 3\text{HgCl}_2 \rightarrow \text{Hg}_3\text{P}_2 + 6\text{HCl} \)
In simple words: First, XeF2 reacts with water to produce xenon gas, hydrofluoric acid, and oxygen. Second, phosphine gas reacts with mercuric chloride to form mercuric phosphide and hydrochloric acid.

Exam Tip: Both of these reactions represent important chemical tests and properties. Make sure all stoichiometry numbers are fully balanced.

Question. Complete the following chemical reaction equations:
(i) \( \text{I}_2 + \text{HNO}_3\text{(Conc.)} \rightarrow \)
(ii) \( \text{HgCl}_2 + \text{PH}_3 \rightarrow \)
Answer:
(i) \( \text{I}_2 + 10\text{HNO}_3\text{(conc.)} \rightarrow 2\text{HIO}_3 + 10\text{NO}_2 + 4\text{H}_2\text{O} \)
(ii) \( 2\text{PH}_3 + 3\text{HgCl}_2 \rightarrow \text{Hg}_3\text{P}_2 + 6\text{HCl} \)
In simple words: First, hot concentrated nitric acid converts iodine into iodic acid. Second, phosphine gas precipitates mercuric phosphide from a mercuric chloride solution.

Exam Tip: Take extra care with balancing these specific red-ox and precipitation reactions as they are frequently tested.

Question. Complete the following chemical reaction equations:
(i) \( \text{NaOH} + \text{Cl}_2 \rightarrow \text{ (cold and dilute)} \)
(ii) \( \text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{ (excess)} \)
Answer:
(i) \( 2\text{NaOH}\text{(cold, dilute)} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O} \)
(ii) \( \text{XeF}_6 + 3\text{H}_2\text{O}\text{(excess)} \rightarrow \text{XeO}_3 + 6\text{HF} \)
In simple words: First, cold and dilute NaOH reacts with chlorine to make salt, sodium hypochlorite, and water. Second, excess water completely hydrolyzes XeF6 to form xenon trioxide and hydrofluoric acid.

Exam Tip: Differentiate clearly between the reactions of chlorine with cold/dilute NaOH vs. hot/concentrated NaOH, and partial vs. complete hydrolysis of \( \text{XeF}_6 \).

Question. Explain the following giving an appropriate reason in each case.
(i) O₂ and F2 both stabilize higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of Xenon fluorides cannot be explained by Valence Bond approach.
Answer:
(i) This occurs because oxygen is capable of forming multiple bonds (double bonds) with metal atoms, which helps in stabilizing higher oxidation states more effectively than fluorine, which can only form single bonds.
(ii) Valence Bond theory cannot explain the structures of xenon fluorides because the energy required to promote paired valence electrons from xenon's closed shell \( 5s \) and \( 5p \) orbitals to empty \( 5d \) orbitals is exceptionally high, which is thermodynamically unfavorable under standard VB assumptions.
In simple words: First, oxygen can form strong double bonds to metals, allowing it to stabilize high oxidation states better than fluorine. Second, promoting xenon's electrons to make bonds requires way too much energy for simple valence bond theory to explain.

Exam Tip: Emphasize 'oxygen's ability to form multiple bonds' for the first part and 'high electronic promotion energy' of xenon for the second.

Question. Explain the following:
(a) Xenon does not form such fluorides as XeF3 and XeF5.
(b) Out of noble gases, only Xenon is known to form real chemical compounds.
Answer:
(a) When paired electrons in xenon's filled valence shell unpair, they are promoted to higher energy orbitals, which always creates an even number of half-filled orbitals (2, 4, or 6). As a result, xenon only forms stable fluorides with even coordination numbers like \( \text{XeF}_2 \), \( \text{XeF}_4 \), and \( \text{XeF}_6 \), rather than odd ones like \( \text{XeF}_3 \) or \( \text{XeF}_5 \).
(b) Xenon has a large atomic radius and a relatively low first ionization enthalpy compared to lighter noble gases. This weaker nuclear hold on its valence electrons allows highly electronegative elements like fluorine and oxygen to easily share or polarize these electrons, making compound formation feasible.
In simple words: First, unpairing electrons always creates even numbers of bonding sites, so xenon cannot form odd compounds like XeF3 or XeF5. Second, xenon is so large that its outer electrons are far from the nucleus, making it easier for strong elements like fluorine to react with them.

Exam Tip: To explain the lack of odd-valent fluorides, describe how promotion of paired electrons always yields an even number of half-filled valence orbitals. For xenon's reactivity, mention its 'large atomic size' and 'low ionization enthalpy'.

Question. (a) Which form of sulphur shows paramagnetic behaviour and why ?
(b) Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5 or +7 oxidation states also. Explain as to why.
Answer:
(a) In the vapor phase, sulfur exists partially as \( \text{S}_2 \) molecules. Similar to oxygen (\( \text{O}_2 \)), \( \text{S}_2 \) has two unpaired electrons located in its antibonding \( \pi^* \) molecular orbitals, resulting in paramagnetic behavior.
(b) Since fluorine is the most electronegative element, it always displays a negative oxidation state (\( -1 \)). Furthermore, because it does not possess vacant d-orbitals in its valence shell, it is unable to expand its valence shell to exhibit positive oxidation states.
In simple words: First, in vapor form, sulfur exists as S2, which has two unpaired electrons that make it magnetic. Second, fluorine is too electronegative to ever lose electrons and lacks d-orbitals to expand its shell.

Exam Tip: Identify the paramagnetic form as \( \text{S}_2 \) in vapor state, explaining it using molecular orbital theory. For fluorine's oxidation state, highlight its electronegativity and absence of d-orbitals.