CBSE Class 12 Chemistry The p Block Elements Assignment Set 01

LEVEL-1 QUESTIONS

 

1. NH3 has higher boiling point than phosphine. Why?

Ans—Ammonia is a polar molecule it form in-termolecular hydrogen bonding.

 

2. Why BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements

Ans- Less B.E or longer bond length.

 

3. Why H₂S is acidic and H₂S is neutral ?

Ans-The S---H bond is weaker than O---H bond because the size of S atomis bigger than that of O atom . Hence H₂S can dissociate to give H+ Ions in aqueous solution .

 

4. Name two poisonous gases which can be prepared from chlorine gas ?

Ans- Phosgene (COCl₂) , tear gas (CCl₃NO₂)

 

5. Name the halogen which does not exhibit positive oxidation state .

Ans- Flourine being the most electronegative element does not show positive oxidation state .

 

6. Iodine forms I₃^- but F₂ does not form F₃- ions .why?

Ans- Due to the presence of vacant D-orbitals , I₂ accepts electrons from I-ions to form I₃-ions , but because of d-orbitals F2 does not accept electrons from F-ions to form F3 ions.

 

7. Draw the structure of peroxosulphuric acid .

 

Important Questions for NCERT Class 12 Chemistry The p Block Elements

Question. Which would quickly absorb oxygen?
(a) Alkaline solution of pyrogallol
(b) Conc. H₂SO₄
(c) Lime water
(d) Alkaline solution of CuSO₄

Answer     A

Question. Oxygen will directly react with each of the following elements except
(a) P
(b) Cl
(c) Na
(d) S 

Answer   B

Question. It is possible to obtain oxygen from air by fractional distillation because
(a) oxygen is in a different group of the periodic table from nitrogen
(b) oxygen is more reactive than nitrogen
(c) oxygen has higher b.pt. than nitrogen
(d) oxygen has a lower density than nitrogen.

Answer     C

Question. The angular shape of ozone molecule (O₃) consists of
(a) 1s and 1p bond
(b) 2s and 1p bond
(c) 1s and 2p bonds
(d) 2s and 2p bonds. 

Answer    B

Question. The gases respectively absorbed by alkaline pyrogallol and oil of cinnamon are
(a) O₃, CH₄
(b) O₂, O₃
(c) SO₂, CH₄
(d) N₂O, O₃

Answer     B

Question. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?
(a) Is soluble in water.
(b) Is used as a food preservative.
(c) Forms ‘acid-rain’.
(d) Is a reducing agent.

Answer     B

Question. Which of the following oxoacid of sulphur has — O — O — linkage
(a) H₂SO₃, sulphurous acid
(b) H₂SO₄, sulphuric acid
(c) H2S₂O₈, peroxodisulphuric acid
(d) H₂S₂O₇, pyrosulphuric acid 

Answer    C

Question. Identify the correct formula of oleum from the following :
(a) H₂S₂O₇
(b) H₂SO₃
(c) H₂SO₄
(d) H₂S₂O₈

Answer    A

Question. In which pair of ions both the species contain S — S bond?
(a) S₄O₆^2–, S₂O₃^2–
(b) S₂O₇^2–, S₂O₈^2–
(c) S₄O₆^2–, S₂O₇^2–
(d) S₂O₇^2–, S₂O₃^2–

Answer    A

Question. Oleum is
(a) castor oil
(b) oil of vitriol
(c) fuming H₂SO₄
(d) none of these.

Answer   C

Question. Which of the following statements is not true for halogens?
(a) All form monobasic oxyacids.
(b) All are oxidizing agents.
(c) All but fluorine show positive oxidation states.
(d) Chlorine has the highest electron-gain enthalpy. 

Answer   C

Question. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) Br₂ > I₂ > F₂ > Cl₂
(b) F₂ > Cl₂ > Br₂ > I₂
(c) I₂ > Br₂ > Cl₂ > F₂
(d) Cl₂ > Br₂ > F₂ > I₂

Answer    D

Question. The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl.
What explains the higher boiling point of hydrogen fluoride?
(a) There is strong hydrogen bonding between HF molecules.
(b) The bond energy of HF molecules is greater than in other hydrogen halides.
(c) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.
(d) The electronegativity of fluorine is much higher than for other elements in the group. 

Answer    A

Question. Among the following which is the strongest oxidising agent?
(a) Br₂
(b) I₂
(c) Cl₂
(d) F₂ 

Answer    D

Question. Which one of the following arrangements does not give the correct picture of the trends indicated against it?
(a) F₂ > Cl₂ > Br₂ > I₂ : Bond dissociation energy
(b) F₂ > Cl₂ > Br₂ > I₂ : Electronegativity
(c) F₂ > Cl₂ > Br₂ > I₂ : Oxidizing power
(d) F₂ > Cl₂ > Br₂ > I₂ : Electron gain enthalpy

Answer     A,D

Question. Which one of the following orders is not in accordance with the property stated against it?
(a) F₂ > Cl₂ > Br₂ > I₂ : Bond dissociation energy
(b) F₂ > Cl₂ > Br₂ > I₂ : Oxidising power
(c) HI > HBr > HCl > HF : Acidic property in water
(d) F₂ > Cl₂ > Br₂ > I₂ : Electronegativity 

Answer    A

Question. Which statement is wrong?
(a) Bond energy of F₂ > Cl₂
(b) Electronegativity of F > Cl
(c) F is more oxidising than Cl
(d) Electron affinity of Cl > F 

Answer    A

Question. Which of the following has the greatest electron affinity?
(a) I
(b) Br
(c) F
(d) Cl 

Answer   D

Question. Which of the following displaces Br2 from an aqueous solution containing bromide ions?
(a) I₂
(b) I₃–
(c) Cl₂
(d) Cl– 

Answer    C

Question. Which of the following species has four lone pairs of electrons?
(a) I
(b) O
(c) Cl–
(d) He

Answer   C

Chapter 7 - p-Block Elements

LEVEL-1 QUESTIONS

Question 1. \(\text{NH}_3\) has higher boiling point than phosphine. Why?
Answer: Ammonia forms intermolecular hydrogen bonding due to the high electronegativity and small size of nitrogen. Phosphine does not form hydrogen bonding because of lower electronegativity of phosphorus. Hence, more energy is required to break hydrogen bonds in liquid ammonia, resulting in a higher boiling point.
In simple words: Molecules of ammonia stick together more strongly using hydrogen bonds, needing more heat to boil.

Exam Tip: Mention intermolecular hydrogen bonding explicitly as the key reason.

Question 2. Why is \(\text{BiH}_3\) the strongest reducing agent amongst all the hydrides of group 15 elements?
Answer: As we move down group 15 from nitrogen to bismuth, the atomic size increases and the metal-hydrogen (\(\text{E-H}\)) bond length increases, leading to a decrease in bond dissociation enthalpy. Since the \(\text{Bi-H}\) bond is the weakest and easily breaks to release hydrogen gas, \(\text{BiH}_3\) acts as the strongest reducing agent.
In simple words: The bismuth-hydrogen bond is very weak and breaks easily, making it give away hydrogen readily.

Exam Tip: Relate reducing power directly to bond dissociation enthalpy and bond length down the group.

Question 3. Why is \(\text{H}_2\text{S}\) acidic and \(\text{H}_2\text{O}\) is neutral?
Answer: The \(\text{S-H}\) bond is weaker than the \(\text{O-H}\) bond because the size of the sulfur atom is much larger than that of oxygen, resulting in poor orbital overlap. Hence, \(\text{H}_2\text{S}\) can easily dissociate to give \(\text{H}^+\) ions in aqueous solution, making it acidic.
In simple words: Sulfur is bigger than oxygen, so its bond with hydrogen is weaker and releases protons more easily.

Exam Tip: Use atomic size and bond strength to explain acidic character.

Question 4. Name two poisonous gases which can be prepared from chlorine gas.
Answer: Phosgene (\(\text{COCl}_2\)) and tear gas (\(\text{CCl}_3\text{NO}_2\)).
In simple words: Phosgene and chloropicrin (tear gas) are toxic gases prepared using chlorine.

Exam Tip: Memorize both common names and chemical formulas for these hazardous gases.

Question 5. Name the halogen which does not exhibit positive oxidation state.
Answer: Fluorine. Being the most electronegative element in the periodic table, it exhibits only \(-1\) oxidation state in all its compounds.
In simple words: Fluorine is the greediest element for electrons, so it never shows positive charges.

Exam Tip: Fluorine lacks vacant d-orbitals and is the most electronegative element, so it never has positive oxidation states.

Question 6. Iodine forms \(\text{I}_3^-\) but \(\text{F}_2\) does not form \(\text{F}_3^-\) ions. Why?
Answer: Due to the presence of vacant \(d\)-orbitals, iodine (\(\text{I}_2\)) can accept electrons from iodide ions (\(\text{I}^-\)) to form \(\text{I}_3^-\) ions. Fluorine lacks vacant \(d\)-orbitals in its valence shell and cannot expand its octet, hence it does not form \(\text{F}_3^-\) ions.
In simple words: Iodine has empty rooms (d-orbitals) to accommodate extra electrons, but fluorine does not.

Exam Tip: Always mention the availability of vacant \(d\)-orbitals for larger halogens.

Question 7. Draw the structure of peroxosulphuric acid.
Answer:
Peroxosulphuric acid (\(\text{H}_2\text{SO}_5\), also known as Caro's acid) has a peroxy linkage (\(-O-O-\)).
Structure:
\(\text{HO}-\text{S}(=\text{O})_2-\text{O}-\text{O}-\text{H}\)
Visually represented with a central sulfur atom double-bonded to two oxygens, single-bonded to an \(-\text{OH}\) group, and single-bonded to a peroxy oxygen atom linked to another hydrogen.
In simple words: Caro's acid contains a special oxygen-oxygen peroxide bridge in its structure.

Exam Tip: Clearly show the peroxy linkage (\(-O-O-\)) when drawing structures containing 'peroxo' in their name.

Question 8. Phosphorous forms \(\text{PCl}_5\) but nitrogen cannot form \(\text{NCl}_5\). Why?
Answer: Due to the availability of vacant \(d\)-orbitals in the valence shell of phosphorus, it can expand its octet and form five covalent bonds (\(\text{PCl}_5\)). Nitrogen belongs to the second period and lacks vacant \(d\)-orbitals, restricting its maximum covalency to 4.
In simple words: Phosphorus has empty d-orbitals to expand its bonding capacity, but nitrogen is too small and lacks them.

Exam Tip: Reference the absence of \(d\)-orbitals in period 2 elements.

Question 9. Give reasons for the following:
(i) Conc. \(\text{HNO}_3\) turns yellow on exposure to sunlight.
(ii) \(\text{PCl}_5\) behaves as an ionic species in solid state.
Answer:
(i) Concentrated nitric acid decomposes in the presence of sunlight to form brown colored nitrogen dioxide gas (\(\text{NO}_2\)), which dissolves in the remaining liquid \(\text{HNO}_3\), turning it yellow.
\(4\text{HNO}_3 \xrightarrow{\text{Sunlight}} 4\text{NO}_2 + 2\text{H}_2\text{O} + \text{O}_2\)
(ii) In the solid state, \(\text{PCl}_5\) exists as an ionic compound composed of tetrahedral [\(\text{PCl}_4\)]\(^+\) cation and octahedral [\(\text{PCl}_6\)]\(^-\) anion.
In simple words: Sunlight breaks down nitric acid into brown gas that colors the liquid, and solid phosphorus pentachloride splits into ions.

Exam Tip: Write both the decomposition reaction and the solid-state ionic formula [\(\text{PCl}_4\)]\(^+\)[\(\text{PCl}_6\)]\(^-\).

Question 10. How is ozone estimated quantitatively?
Answer: When ozone reacts with an excess of potassium iodide (\(\text{KI}\)) solution buffered with a borate buffer (\(\text{pH} = 9.2\)), iodine (\(\text{I}_2\)) is liberated. The liberated iodine is then titrated against a standard solution of sodium thiosulphate (\(\text{Na}_2\text{S}_2\text{O}_3\)) using starch as an indicator. This stoichiometric reaction provides a quantitative estimation of ozone gas.
In simple words: Ozone frees iodine from potassium iodide, and we measure the iodine using sodium thiosulphate.

Exam Tip: Mention the use of potassium iodide and titration with sodium thiosulphate.

Question 11. Are all the five bonds in \(\text{PCl}_5\) molecule equivalent? Justify your answer.
Answer: No, all five bonds are not equivalent. \(\text{PCl}_5\) has a trigonal bipyramidal geometry. The three equatorial \(\text{P-Cl}\) bonds lie in one plane and are equivalent, whereas the two axial \(\text{P-Cl}\) bonds make a \(90^\circ\) angle with the equatorial plane and experience greater repulsion, making them longer and weaker than equatorial bonds.
In simple words: Axial bonds are pushed apart more strongly, making them longer and weaker than the three equatorial bonds.

Exam Tip: State that axial bonds are longer due to greater repulsion from equatorial bond pairs.

Question 12. \(\text{NO}_2\) is colored and readily dimerises. Why?
Answer: \(\text{NO}_2\) contains an odd number of valence electrons (\(17\) electrons) and behaves as an odd-electron molecule, which imparts a brown color to it. On dimerization, two \(\text{NO}_2\) molecules combine to form a stable \(\text{N}_2\text{O}_4\) molecule with an even number of electrons, eliminating unpaired electrons and becoming colorless.
In simple words: Having an odd electron makes it colored and unstable, so it pairs up by joining into a dimer (\(\text{N}_2\text{O}_4\)).

Exam Tip: Highlight the presence of an odd number of valence electrons as the root cause.

Question 13. Account for the following:
(i) \(\text{SF}_6\) is less reactive than \(\text{SF}_4$.
(ii) Of the noble gases, only xenon forms chemical compounds.
Answer:
(i) In \(\text{SF}_6\), the sulfur atom is sterically hindered and completely surrounded by six fluorine atoms, resulting in less repulsion between fluorine atoms and preventing attack by reagents compared to \(\text{SF}_4\).
(ii) Xenon has a relatively large atomic size, low ionization enthalpy, and high polarising power compared to lighter noble gases, enabling it to form compounds with highly electronegative elements like fluorine and oxygen.
In simple words: Sulfur hexafluoride is tightly packed and shielded by fluorines, while xenon is large enough and has low enough ionization energy to form chemical bonds.

Exam Tip: Use steric crowding for \(\text{SF}_6\) and low ionization enthalpy for xenon.

Question 14. With what neutral molecule is \(\text{ClO}^-\) isoelectronic? Is that molecule a Lewis base?
Answer: The hypochlorite ion \(\text{ClO}^-\) has \(7 (\text{Cl}) + 6 (\text{O}) + 1 (\text{negative charge}) = 14\) valence electrons. The neutral molecule is chlorine monofluoride (\(\text{ClF}\)), which also has \(7 (\text{Cl}) + 7 (\text{F}) = 14\) valence electrons. Yes, \(\text{ClF}\) is a Lewis base because it contains lone pairs of electrons on the chlorine and fluorine atoms that can be donated.
In simple words: ClF has the same total number of electrons as \(\text{ClO}^-\) and acts as a Lewis base due to its lone pairs.

Exam Tip: Count total valence electrons carefully when checking for isoelectronic species.

Question 15. (i) Why is helium used in diving apparatus?
(ii) Noble gases have very low boiling points. Why?
(iii) Why is \(\text{ICl}\) more reactive than \(\text{I}_2\)?
Answer:
(i) Helium has very low solubility in blood even under high pressure, preventing the painful condition known as bends.
(ii) Noble gases are monoatomic elements held together only by weak London dispersion forces, requiring very little thermal energy to separate them.
(iii) The \(\text{I-Cl}\) bond is a polar covalent bond and is weaker than the non-polar \(\text{I-I}\) bond, allowing it to cleave easily during reactions.
In simple words: Helium doesn't dissolve in blood under pressure, noble gases only have weak forces holding their single atoms together, and interhalogen bonds are weaker than pure halogen bonds.

Exam Tip: Remember solubility in blood for helium, weak dispersion forces for noble gases, and bond polarity for interhalogens.

Question 16. Account for the following:
(i) Noble gases form compounds with \(\text{F}_2\) and \(\text{O}_2\) only.
(ii) Sulphur shows paramagnetic behavior.
(iii) \(\text{HF}\) is much less volatile than \(\text{HCl}\).
(iv) White phosphorous is kept under water.
(v) Ammonia is a stronger base than phosphine.
Answer:
(i) Fluorine and oxygen are the two most powerful oxidizing agents capable of oxidizing noble gases.
(ii) In the vapor state, sulfur partly exists as \(\text{S}_2\) molecules, which contain two unpaired electrons in the antibonding \(\pi^*\) orbitals (similar to \(\text{O}_2\)), hence exhibiting paramagnetism.
(iii) \(\text{HF}\) molecules are associated with strong intermolecular hydrogen bonding, whereas \(\text{HCl}\) has only weaker dipole-dipole interactions.
(iv) White phosphorus has a very low ignition temperature (\(303\text{ K}\)) and spontaneously catches fire in air forming \(\text{P}_4\text{O}_{10}\). Storing it under water cuts off contact with air.
(v) Due to the smaller size of nitrogen, its lone pair of electrons is concentrated and readily available for donation compared to the larger phosphorus atom in phosphine.
In simple words: Fluorine and oxygen are super strong oxidizers, sulfur vapor has unpaired electrons, hydrogen fluoride molecules stick together with strong H-bonds, white phosphorus burns spontaneously in air, and nitrogen's small size makes its lone pair easier to share.

Exam Tip: Address each point using fundamental concepts: oxidizing strength, molecular orbitals, hydrogen bonding, ignition temperature, and charge density.

Question 17. Arrange the following in the increasing order of the property mentioned:
(i) \(\text{HOCl}, \text{HClO}_2, \text{HClO}_3, \text{HClO}_4\) (Acidic strength)
(ii) \(\text{As}_2\text{O}_3, \text{ClO}_2, \text{GeO}_3, \text{Ga}_2\text{O}_3\) (Acidity)
(iii) \(\text{NH}_3, \text{PH}_3, \text{AsH}_3, \text{SbH}_3\) (\(\text{H-E-H}\) bond angle)
(iv) \(\text{HF}, \text{HCl}, \text{HBr}, \text{HI}\) (Acidic strength)
(v) \(\text{MF}, \text{MCl}, \text{MBr}, \text{MI}\) (Ionic character)
Answer:
(i) \(\text{HOCl} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4$
(ii) \(\text{Ga}_2\text{O}_3 < \text{GeO}_3 < \text{As}_2\text{O}_3 < \text{ClO}_2$
(iii) \(\text{SbH}_3 < \text{AsH}_3 < \text{PH}_3 < \text{NH}_3$
(iv) \(\text{HF} < \text{HCl} < \text{HBr} < \text{HI}$
(v) \(\text{MI} < \text{MBr} < \text{MCl} < \text{MF}\)
In simple words: Acid strength increases with oxygen count or bond breaking ease, bond angles decrease down group 15, and ionic character grows when paired with more electronegative fluorine.

Exam Tip: Higher oxidation state of halogen in oxoacids means higher acidity.

LEVEL-2 QUESTIONS

Question 1. Draw the structure of following compounds:
a) \(\text{N}_2\text{O}, \text{NO}, \text{N}_2\text{O}_3, \text{NO}_2\) and \(\text{N}_2\text{O}_5\).
b) Oxoacids of Phosphorous, Sulphur and chlorine.
c) Compounds of Xenon.
Answer:
a) Nitrogen oxides structures:
- \(\text{N}_2\text{O}\): Linear, \(\text{N}\equiv\text{N}\rightarrow\text{O}\)
- \(\text{NO}\): Bent/Angular radical with 3-electron bond resonance.
- \(\text{N}_2\text{O}_3\): Planar asymmetric structure with \(\text{N-N}\) single bond.
- \(\text{NO}_2\): Angular radical with O-N-O bond angle of \(134^\circ\).
- \(\text{N}_2\text{O}_5\): Planar symmetric structure \(\text{O}_2\text{N}-\text{O}-\text{NO}_2\).
b) Oxoacids:
- Phosphoric acids: Orthophosphoric, Pyrophosphoric, Hypophosphorous, and Metaphosphoric acids containing characteristic \(\text{P=O}\), \(\text{P-OH}\), and \(\text{P-H}\) bonds.
- Sulphur acids: Sulphurous, Sulphuric, Peroxodisulphuric (Marshall's acid), and Pyrosulphuric acid (Oleum) containing \(\text{S=O}\), \(\text{S-OH}\), and peroxy/bridge linkages.
- Chlorine acids: Hypochlorous, Chlorous, Chloric, and Perchloric acids.
c) Xenon compounds:
- \(\text{XeF}_2\): Linear
- \(\text{XeF}_4\): Square planar
- \(\text{XeF}_6\): Distorted octahedral
- \(\text{XeOF}_4\): Square pyramidal
- \(\text{XeO}_3\): Pyramidal
In simple words: Refer to standard VSEPR geometries and chemical bonding principles for oxides, oxoacids, and xenon fluorides/oxides.

Exam Tip: Practice drawing VSEPR-based shapes for all interhalogen and noble gas species.

Question 2. What happens when white P is heated with conc. \(\text{NaOH}\) solution in an atmosphere of \(\text{CO}_2\)? Give equation.
Answer: Phosphorus gas/phosphine gas will be formed along with sodium hypophosphite.
\(\text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2\)
In simple words: White phosphorus reacts with hot concentrated sodium hydroxide to produce phosphine gas and sodium hypophosphite.

Exam Tip: This is a classic disproportionation reaction of white phosphorus.

Question 3. Give example of a molecule having geometry T-shaped.
Answer: Bromine trifluoride (\(\text{BrCl}_3\) or \(\text{BrF}_3\)). Here, bromine has 2 lone pairs (\(\text{lp} = 2\)) and 3 bond pairs (\(\text{bp} = 3\)).
In simple words: Interhalogen compounds like \(\text{BrF}_3\) with two lone pairs take a T-shape.

Exam Tip: Count valence electrons and lone pairs to determine T-shape geometry under VSEPR theory.

Question 4. Write the balanced chemical equation for the reaction of \(\text{Cl}_2\) with hot and concentrated \(\text{NaOH}\). Is this reaction a disproportionation reaction? Justify.
Answer:
\(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\)
Yes, this is a disproportionation reaction because chlorine gas (in oxidation state \(0\)) is simultaneously oxidized to chlorate (\(+5\) in \(\text{NaClO}_3\)) and reduced to chloride (\(-1\) in \(\text{NaCl}\)).
In simple words: Chlorine is both oxidized and reduced at the same time, making it a disproportionation reaction.

Exam Tip: Always assign oxidation states to prove disproportionation.

Question 5. Why is \(\text{HF}\) acid stored in wax coated glass bottles?
Answer: This is because \(\text{HF}\) does not attack wax, but it readily reacts with glass. It dissolves silica (\(\text{SiO}_2\)) present in glass to form water-soluble hydrofluorosilicic acid.
\(\text{SiO}_2 + 6\text{HF} \rightarrow \text{H}_2\text{SiF}_6 + 2\text{H}_2\text{O}\)
In simple words: Hydrofluoric acid dissolves glass, so it is stored in wax-lined containers that it cannot corrode.

Exam Tip: Highlight the reaction of \(\text{HF}\) with glass silica (\(\text{SiO}_2\)).

Question 6. What is laughing gas? Why is it so called? How is it prepared?
Answer: Nitrous oxide (\(\text{N}_2\text{O}\)) is called laughing gas because when inhaled in small amounts, it produces hysterical laughter and euphoria. It is prepared by gently heating ammonium nitrate.
\(\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}\)
In simple words: Nitrous oxide makes people laugh uncontrollably when inhaled and is made by gently heating ammonium nitrate.

Exam Tip: Write both the common name, formula, and the thermal decomposition equation of ammonium nitrate.

LEVEL-3 QUESTIONS

Question 1. A translucent white waxy solid (A) on heating in inert atmosphere is converted to its allotropic form (B). Allotrope (A) on reaction with very dilute aqueous \(\text{KOH}\) liberates a highly poisonous gas (C) having rotten fish smell. With excess of chlorine it forms (D) which hydrolyses to compound (E). Identify compounds (A) to (E).
Answer:
- (A) = White phosphorus
- (B) = Red phosphorus
- (C) = Phosphine (\(\text{PH}_3\))
- (D) = Phosphorus pentachloride (\(\text{PCl}_5\))
- (E) = Phosphoric acid (\(\text{H}_3\text{PO}_4\))
In simple words: White phosphorus converts to red phosphorus on heating, produces phosphine with alkali, and forms phosphorus pentachloride with excess chlorine which hydrolyzes to phosphoric acid.

Exam Tip: Use characteristic properties like "rotten fish smell" to identify phosphine immediately.

Question 2. When conc. \(\text{H}_2\text{SO}_4\) was added into an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were also added into this test tube. On cooling, the gas (A) changed into colorless gas (B).
a. Identify the gases A and B.
b. Write the equations for the reactions involved.
Answer:
a) Gas (A) is \(\text{NO}_2\) (nitrogen dioxide, brown gas) and Gas (B) is \(\text{N}_2\text{O}_4\) (dinitrogen tetroxide, colorless gas on cooling).
b) Equations involved:
\(\text{NaNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HNO}_3\)
\(4\text{HNO}_3 \rightarrow 4\text{NO}_2 + 2\text{H}_2\text{O} + \text{O}_2\)
\(\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{H}_2\text{O} + 2\text{NO}_2\)
\(2\text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4 \text{ (on cooling)}\)
In simple words: Nitrate salts with sulfuric acid give brown nitrogen dioxide gas, which turns into colorless dimer gas upon cooling.

Exam Tip: Note the intensification of brown fumes upon adding copper turnings due to reduction of nitric acid to \(\text{NO}_2\).

Question 3. A colorless inorganic salt A decomposes at about \(250^\circ\text{C}\) to give only two products B and C leaving no residue. The oxide C is a liquid at room temperature and is neutral to litmus paper while B is neutral oxide. White phosphorus burns in excess of B to produce Strong dehydrating agent. Give balanced equations for above processes.
Answer:
- (A) = \(\text{NH}_4\text{NO}_3$
- (B) = \(\text{N}_2\text{O}\) (Nitrous oxide, neutral oxide)
- (C) = \(\text{H}_2\text{O}\) (Water, liquid and neutral to litmus)
Equations involved:
\(\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}\)
\(10\text{N}_2\text{O} + \text{P}_4 \rightarrow 10\text{N}_2 + \text{P}_4\text{O}_{10}\)
In simple words: Ammonium nitrate decomposes into nitrous oxide and water, and white phosphorus burns in nitrous oxide to form phosphorus pentoxide (a strong drying agent).

Exam Tip: Identify neutral oxide liquid at room temperature as water and neutral gas as \(\text{N}_2\text{O}\).

Question 4. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colorless solid “B”. Solid “B” on heating with \(\text{NO}\) changes to a blue solid “C”. Identify “A”, “B” and “C”.
Answer:
\(2\text{Pb}(\text{NO}_3)_2 \rightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2\)
\(2\text{NO}_2 \rightarrow \text{N}_2\text{O}_4 \text{ (on cooling, colorless solid/liquid)}\)
\(\text{N}_2\text{O}_4 + 2\text{NO} \rightarrow 2\text{N}_2\text{O}_3 \text{ (blue solid)}\)
- (A) = \(\text{NO}_2$
- (B) = \(\text{N}_2\text{O}_4$
- (C) = \(\text{N}_2\text{O}_3$
In simple words: Lead nitrate decomposition gives brown \(\text{NO}_2\), which dimerizes to \(\text{N}_2\text{O}_4\), and reacting with NO forms blue dinitrogen trioxide.

Question 5. An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolorizes acidified aqueous \(\text{KMnO}_4\) solution and reduces \(\text{Fe}^{3+}\) to \(\text{Fe}^{2+}\). Identify the solid “A” and the gas “B”.
Answer:
- (A) = \(\text{S}_8\) (Sulphur)
- (B) = \(\text{SO}_2\) (Sulphur dioxide)
Equations involved:
\(\text{S}_8 + 8\text{O}_2 \rightarrow 8\text{SO}_2\)
\(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)
In simple words: Sulfur burns to form sulfur dioxide, which turns lime water milky and reduces permanganate solutions.

Exam Tip: The turning of lime water milky and decolorization of acidified \(\text{KMnO}_4\) are hallmark tests for \(\text{SO}_2\).

Question 6. Why does \(\text{PCl}_3\) fume in moisture?
Answer: In the presence of moisture (\(\text{H}_2\text{O}\)), \(\text{PCl}_3\) undergoes hydrolysis to give phosphorous acid and dense white fumes of \(\text{HCl}\) gas.
\(\text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3\text{HCl}\)
In simple words: Phosphorus trichloride reacts with water vapor in the air to produce steamy fumes of hydrochloric acid.

Exam Tip: Mention the formation of \(\text{HCl}\) gas as the cause of fuming in moisture.

Question 7. What happens when \(\text{H}_3\text{PO}_3\) is heated?
Answer: It undergoes a disproportionation reaction to give orthophosphoric acid and phosphine.
\(4\text{H}_3\text{PO}_3 \rightarrow 3\text{H}_3\text{PO}_4 + \text{PH}_3\)
In simple words: Heating phosphorous acid disproportionates it into phosphoric acid and phosphine gas.

Exam Tip: This is a standard disproportionation reaction of phosphorus oxoacids.

Question 8. When Conc. \(\text{H}_2\text{SO}_4\) was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were added into the test tube. On cooling, gas (A) changed into a colorless gas (B).
(a) Identify the gases 'A' and 'B'.
(b) Write the equations for the reactions involved.
Answer:
(a) Gas 'A' is \(\text{NO}_2\) (nitrogen dioxide) and Gas 'B' is \(\text{N}_2\text{O}_4\) (dinitrogen tetroxide).
(b) Equations:
\(\text{XNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{XHSO}_4 + \text{HNO}_3\)
\(\text{Cu} + 4\text{HNO}_3 \text{ (conc.)} \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}\)
\(2\text{NO}_2 \text{ (on cooling)} \rightarrow \text{N}_2\text{O}_4\)
In simple words: Nitrate salts with concentrated sulfuric acid produce brown \(\text{NO}_2\) gas, which dimerizes on cooling into colorless \(\text{N}_2\text{O}_4\).

Exam Tip: Include both salt action and copper turning reaction for complete credit.

Question 9. Complete the following equations:
(i) \(\text{XeF}_4 + \text{H}_2\text{O} \rightarrow$
(ii) \(\text{Ca}_3\text{P}_2 + \text{H}_2\text{O} \rightarrow$
(iii) \(\text{AgCl}_{(s)} + \text{NH}_3 \text{ (aq)} \rightarrow$
Answer:
(i) \(6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 4\text{Xe} + 2\text{XeO}_3 + 24\text{HF} + 3\text{O}_2$
(ii) \(\text{Ca}_3\text{P}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Ca}(\text{OH})_2 + 2\text{PH}_3$
(iii) \(\text{AgCl}_{(s)} + 2\text{NH}_3 \text{ (aq)} \rightarrow [\text{Ag}(\text{NH}_3)_2]\text{Cl}_{(aq)}\)
In simple words: Complete these hydrolysis and complex-formation reactions for xenon tetrafluoride, calcium phosphide, and silver chloride.

Question 10. (i) How is \(\text{XeOF}_4\) prepared? Draw its structure.
(ii) When \(\text{HCl}\) reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?
Answer:
(i) Prepared by partial hydrolysis of xenon hexafluoride:
\(\text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{XeOF}_4 + 2\text{HF}\)
Structure: Square pyramidal geometry with one lone pair on xenon.
(ii) The reaction of iron with hydrochloric acid produces hydrogen gas:
\(\text{Fe} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2\)
The liberation of hydrogen (\(\text{H}_2\)) creates a reducing atmosphere that prevents the oxidation of ferrous chloride to ferric chloride.
In simple words: Partial hydrolysis of \(\text{XeF}_6\) gives \(\text{XeOF}_4\), and hydrogen gas generated during iron corrosion prevents formation of ferric chloride by acting as a shield against oxidation.

Exam Tip: Mention the role of liberated hydrogen gas in preventing oxidation to ferric chloride.