NCERT Exemplar: p-Block Elements (Group 15-18) | Chemistry for JEE Main & Advanced PDF Download

MULTIPLE CHOICE QUESTIONS (TYPE - I)

Q.1. On addition of conc. H₂SO₄ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because
(i) H₂SO₄ reduces HI to I₂
(ii) HI is of violet colour
(iii) HI gets oxidised to I₂
(iv) HI changes to HIO₃
Ans. (iii)
Solution.
Hydrogen iodide (HI) is stronger oxidising agent than H₂SO₄. So, it reduces H₂SO₄ to SO₂ and itself oxidises to I₂. Colour of I₂ is violet hence on adding conc. H₂SO₄ to HI, it gets oxidised to I₂.


Q.2. In qualitative analysis when H₂S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO₃, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________.
(i) Deep blue precipitate of Cu (OH)₂
(ii) Deep blue solution of [Cu (NH₃)₄]²⁺
(iii) Deep blue solution of Cu(NO₃)₂
(iv) Deep blue solution of Cu(OH)₂.Cu(NO₃)₂
Ans. (ii)
Solution.
In qualitative analysis when H₂S is passed through an aqueous solution of salt acidified with dil. HCl a black ppt of CuS is obtained.

On boiling CuS with dil. HNO₃ it forms a blue coloured solution and the following reactions occur
3CuS+8HNO₃ → 3Cu(NO₃)₂ + 2NO+3S+4H₂O
S+2HNO₃ → H₂SO₄ + NO



Q.3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
(ii) 6 double bonds; 6 single bonds
(iii) 3 double bonds; 12 single bonds
(iv) Zero double bonds; 12 single bonds
Ans. (iii)
Solution.
Cyclotrimetaphosphoric acid contains three double bonds and 9 single bonds as shown below
a, b, c are three π bonds and numbers 1 to 12 represent sigma (σ) bonds.

Q.4. Which of the following elements can be involved in pπ–dπ bonding?
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
Ans. (iii)
Solution.
Among given four elements i.e., carbon, nitrogen, phosphorus and boron. As only phosphorus has vacant d-orbital so, only phosphorus can form pπ-dπ bond.

Q.5. Which of the following pairs of ions are isoelectronic and isostructural? 
(i)
(ii)
(iii)
(iv)
Ans. (i)
Solution.
Compounds having same value of total number of electrons are known as isoelectronic.
For
Total number of electrons
= 6 + 8 x 3 + 2
= 6 + 24 + 2
= 32
For
Total number of electrons
= 7 + 8 x 3 + 1
= 7 + 25
= 32
Hence,and are isoelectronic. These two ions have similar structure so they are isostructural also.

Both have triangular planar structure as in both the species carbon and nitrogen are sp² hybridised. Hence, (i) is the correct choice.

Q.6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? 
(i) HF
(ii) HCl
(iii) HBr
(iv) HI
Ans. (i)
Solution.
On moving down the group, atomic radii of halogens increases leading to increased H– X bond length due to which its bond dissociation enthalpy increases. So the highest bond dissociation enthalpy is of HF.

Q.7. Bond dissociation enthalpy of E—H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent?

(i) NH₃
(ii) PH₃
(iii) AsH₃
(iv) SbH₃
Ans. (iv)
Solution.
On moving top to bottom, size of central atom increases. Due to this, the bond length of X–H bond increases and bond dissociation energy decreases. Hence, reducing nature increases down the group. So, SbH₃ acts as the strongest reducing agent here.

Q.8. On heating with concentrated NaOH solution in an inert atmosphere of CO₂, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(i) It is highly poisonous and has smell like rotten fish.
(ii) It’s solution in water decomposes in the presence of light.
(iii) It is more basic than NH₃.
(iv) It is less basic than NH₃.
Ans. (iii)
Solution.
White phosphorous on reaction with NaOH solution in the presence of inert atmosphere of CO2 it produces phosphine gas which is less basic than NH₃.


Q.9. Which of the following acids forms three series of salts?
(i) H₃PO₂
(ii) H₃BO₃
(iii) H₃PO₄
(iv) H₃PO₃
Ans. (iii)
Solution.
Structure of H₃PO₄ is
H₃PO₄ has three –OH groups i.e., has three ionisable H-atoms and hence forms three series of salts. These three possible series of salts for H₃PO₄ are as follows:
NaH₂PO₄, Na₂HPO₄ and Na₃PO₄

Q.10. Strong reducing behaviour of H₃PO₂ is due to
(i) Low oxidation state of phosphorus
(ii) Presence of two –OH groups and one P–H bond
(iii) Presence of one –OH group and two P–H bonds
(iv) High electron gain enthalpy of phosphorus
Ans. (iii)
Solution.
Strong reducing behaviour of H₃PO₂ is due to presence of two P–H bonds and one P– OH bond

Q.11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(i) N₂O, PbO
(ii) NO₂, PbO
(iii) NO, PbO
(iv) NO, PbO₂
Ans. (ii)
Solution.
On heating lead nitrate, it produces brown coloured nitrogen dioxide (NO₂) and lead (II) oxide.


Q.12. Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
Ans. (i)
Solution.
The single N-N bond is weak because of high inter-electronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen that is why it does not show allotropy.

Q.13. Maximum covalency of nitrogen is __________.
(i) 3
(ii) 5
(iii) 4
(iv) 6
Ans. (iii)
Solution.
Maximum covalency of nitrogen is 4 in which one electron is made available by sorbital and 3 electrons are made available by p-orbitals. Hence, total four electrons are available for bonding.

Q.14. Which of the following statements is wrong?
(i) Single N–N bond is stronger than the single P–P bond.
(ii) PH₃ can act as a ligand in the formation of coordination compound with transition elements.
(iii) NO₂ is paramagnetic in nature.
(iv) Covalency of nitrogen in N₂O₅ is four.
Ans. (i)
Solution.
Single N–N bond is stronger than the single P–P bond.
True statement is that single N–N bond is weaker than the single P–P bond. This is because of high interelectronic repulsion of the non-bonding electrons in nitrogen which leads to small bond length.
PH₃ acts as a ligand in the formation of coordination compound due to presence of lone pair of electrons.
NO₂ is paramagnetic in nature due to presence of one unpaired electron. Structure of NO₂ is
Covalency of nitrogen in N₂O₅ is 4.

Q.15. A brown ring is formed in the ring test for NO^–₃ ion. It is due to the formation of
(i) [Fe(H₂O)₅ (NO)]²⁺
(ii) FeSO₄.NO₂
(iii) [Fe(H₂O)₄(NO)2]²⁺
(iv) FeSO₄.HNO₃
Ans. (i)
Solution.
When freshly prepared solution of FeSO₄ is added in a solution containing NO₃ ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.


Q.16. Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(i) Bi₂O₅
(ii) BiF₅
(iii) BiCl₅
(iv) Bi₂S₅
Ans. (ii)
Solution.
Stability of +5 oxidation state decreases top to bottom and +3 oxidation state increases top to bottom due to inert pair effect. Meanwhile compound having +5 oxidation state of Bi is BiF₅. It is due to smaller size and high electronegativity of fluorine.

Q.17. On heating ammonium dichromate and barium azide separately we get 
(i) N₂ in both cases
(ii) N₂ with ammonium dichromate and NO with barium azide
(iii) N₂O  with ammonium dichromate and N₂ with barium azide
(iv) N₂O with ammonium dichromate and NO₂ with barium azide
Ans. (i)
Solution.
On heating ammonium dichromate and barium azide, it produces N₂ gas separately.


Q.18. In the preparation of HNO₃, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH₃ will be ______.
(i) 2
(ii) 3
(iii) 4
(iv) 6
Ans. (i)
Solution.

From above equation, it is clear that two moles of NH₃ will produce 2 moles of NO on catalytic oxidation of ammonia in preparation of nitric acid.

Q.19. The oxidation state of central atom in the anion of compound NaH₂PO₂ will be ______.
(i) +3
(ii) +5
(iii) +1
(iv) –3
Ans. (iii)
Solution.
Let oxidation state of P in NaH₂PO₂ is x.
1 + 2 x 1 + x + 2x – 2 = 0
1 + 2 + x – 4 = 0
x – 1 = 0
x = + 1

Q.20. Which of the following is not tetrahedral in shape?
(i)
(ii) SiCl₄
(iii) SF₄
(iv)
Ans. (iii)
Solution.
SF₄ has sea-saw shaped structure as shown below:
It has trigonal bipyramidal geometry having sp³d hybridisation.

Q.21. Which of the following are peroxoacids of sulphur?
(i) H₂SO₅ and H₂S₂O₈
(ii) H₂SO₅ and H₂S₂O₇
(iii) H₂S₂O₇ and H₂S₂O₈
(iv) H₂S₂O₆ and H₂S₂O₇
Ans. (i)
Solution.
Peroxoacids of sulphur must contain one —O—O— bond as shown below:


Q.22. Hot conc. H₂SO₄ acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc.
H₂SO₄ into two gaseous products?
(i) Cu
(ii) S
(iii) C
(iv) Zn
Ans. (iii)
Solution.
H₂SO₄ is a moderately strong oxidising agent which oxidises both metals and nonmetals as shown below:
Cu+2H₂SO₄ (conc) → CuSO₄ + SO₂ + 2H₂O
S+2H₂SO₄ (conc) → 3SO₂ + 2H₂O
On the other hand, carbon on oxidation with H₂SO₄ produces two types of oxides CO₂ and SO₂.
C+2H₂SO₄ (conc)  → CO₂ + 2SO₂ + 2H₂O

Q.23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH₃ an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.
(i) –3 to +3
(ii) –3 to 0
(iii) –3 to +5 
(iv) 0 to –3
Ans. (i)
Solution.
Black coloured compound MnO₂ reacts with HCl to produce greenish yellow coloured gas of Cl₂

Cl₂ on further treatment with NH₃ produces NCl₃.

NH₃ (–3) changes to NCl₃ (+3) in the above reaction. Hence, (i) is the correct choice.

Q.24. In the preparation of compounds of Xe, Bartlett had taken  as a base compound. This is because
(i) Both O₂ and Xe have same size.
(ii) Both O₂ and Xe have same electron gain enthalpy.
(iii) Both O₂ and Xe have almost same ionisation enthalpy.
(iv) Both Xe and O₂ are gases.
Ans. (iii)
Solution.
Bertlett had taken   as a base compound because O₂ and Xe both have almost same ionisation enthalpy. The ionisation enthalpies of noble gases are the highest in their respective periods due to their stable electronic configurations.

Q.25. In solid state PCl₅ is a _________.
(i) Covalent solid
(ii) Octahedral structure
(iii) Ionic solid with [PCl₆]⁺ octahedral and [PCl₄]^– tetrahedra
(iv) Ionic solid with [PCl₄]⁺ tetrahedral and [PCl₆]^– octahedra
Ans. (iv)
Solution.
In solid state PCl₅ exists as an ionic solid with [PCl₄]⁺ tetrahedral and [PCl₆]^– octahedral.

Q.26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

Ion
Reduction potential EΘ/V	EΘ=1.19V	EΘ=1.65V	EΘ=1.74V

(i) 
(ii) 
(iii)
(iv)
Ans. (iii)
Solution.
Higher the standard reduction potential of species higher will be its oxidising power.

Q.27. Which of the following is isoelectronic pair?
(i) ICl₂, ClO₂
(ii)
(iii) ClO₂, BrF
(iv) CN^–, O₃
Ans. (ii)
Solution.
Isoelectronic pair involves species with same number of electrons.

From the above table, it is clear that only (ii) is the correct choice.

MULTIPLE CHOICE QUESTIONS (TYPE - II)

Note : In the following questions two or more options may be correct.
Q.28. If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________.
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to –1
(iv) 0 to +1
Ans. (i, iii)
Solution.
When chlorine gas is passed through hot NaOH solution it produces NaCl and NaClO₃.

Oxidation state varies from 0 to –1 and 0 to +5.
Hence, (i) and (iii) are correct choices

Q.29. Which of the following options are not in accordance with the property mentioned against them?

Ans. (ii, iii)
Solution.
(ii) As ability to gain electron increases oxidising property increases. Here, F is the most electronegative element having highest value of SRP hence it has highest oxidising power.
Correct order of ionic character of metal halide is:
MI < MBr < MCl < MF
As electronegativity difference between metal and halogen increases, ionic character increases.
(iii) Due to electronic repulsion among lone pair in F₂ molecule, the correct order of bond dissociation enthalpy is as follows:
Cl₂ > Br₂ > F₂>I₂

Q.30. Which of the following is correct for  P₄ molecule of white phosphorus?
(i) It has 6 lone pairs of electrons.
(ii) It has six P–P single bonds.
(iii) It has three P–P single bonds.
(iv) It has four lone pairs of electrons.
Ans. (ii, iv)
Solution.
Structure of P₄ molecule can be represented as

It has total four lone pairs of electrons with one pair situated at each P-atom.
It has six P–P single bonds.

Q.31. Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(ii) Leaving F—F bond, all halogens have weaker X—X bond than X—X' bond in interhalogens.
(iii) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Ans. (i, iii, iv)
Solution.
(i) Among halogens, radius ratio between iodine and fluorine is maximum because iodine has largest radius and fluorine has the smallest radius out of all halogens.
(ii) It can be correctly stated as in general interhalogen compounds are more reactive than halogen compounds. This is because X–X' bond in interhalogen is weaker than X—X bond in halogens except F–F bond.
(iii) Among interhalogen compounds, maximum number of atoms are present in iodine fluoride because radius ratio of iodine and fluorine has maximum value.
(iv) Interhalogen compounds are more reactive than halogens due to weaker X–X' bond as compared to X–X of halogen compounds.

Q.32. Which of the following statements are correct for SO₂ gas?
(i) It acts as bleaching agent in moist conditions.
(ii) It’s molecule has linear geometry.
(iii) It’s dilute solution is used as disinfectant.
(iv) It can be prepared by the reaction of dilute H₂SO₄ with metal sulphide.
Ans. (i, iii)
Solution.
(i) In moist condition SO₂ gas acts as a bleaching agent.
e.g., it converts Fe (III) to Fe (II) ion and decolourises acidified KMnO₄ (VII).

(ii) SO₂ molecule has bent structure.
(iii) Its dilute solution is used as a disinfectant.
(iv) It can be prepared by the reaction of O₂ with sulphide ore.
4FeS₂ + 11O₂ →  2Fe₂O₃ + 8 SO₂
However, metal on treatment with H₂SO₄ produces H₂S.
Hence, options (i) and (iii) are correct choices.

Q.33. Which of the following statements are correct?
(i) All the three N—O bond lengths in HNO₃ are equal.
(ii) All P—Cl bond lengths in PCl₅ molecule in gaseous state are equal.
(iii) P₄ molecule in white phohsphorus have angular strain therefore white phosphorus is very reactive.
(iv) PCl₅ is ionic in solid state in which cation is tetrahedral and anion is octahedral.
Ans. (iii, iv)
Solution.
(i) All the three N–O bond lengths in HNO₃ are not equal.
(ii) All P–Cl bond lengths in PCl₅ molecule in gaseous state are not equal. Axial bond is longer than equatorial bond.
(iii) P₄ molecule in white phosphorous has angular strain therefore white phosphorous is very reactive.
(iv) PCl₅ is ionic in solid state in which cation is tetrahedral and anion is octahedral.
Cation → [PCl₄]⁺
Anion → [PCl₆]^–

Q.34. Which of the following orders are correct as per the properties mentioned against each?

Ans. (i, iv)
Solution.
(i)
(ii) Correct order is
(iii) Correct order for more negative electron gain enthalpy is
S < O < F< Cl
(iv)
Thermal stability decreases on moving top to bottom due to increase in its bond length.

Q.35. Which of the following statements are correct?
(i) S–S bond is present in H₂S₂O₆.
(ii) In peroxosulphuric acid (H₂SO₅) sulphur is in +6 oxidation state.
(iii) Iron powder along with Al₂O₃ and K₂O is used as a catalyst in the preparation of NH₃ by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of SO₃ by catalytic oxidation of SO₂.
Ans. (i, ii)
Solution.
(i) Structure of H₂S₂O₆ is as shown below
It contains one S–S bond.
(ii) In peroxosulphuric acidd (H₂SO₅) sulphur is in +6 oxidation state.
Structure of H₂SO₅ is

Let oxidation state of S = x
2 x (+1) + x + 3 x (–2) + 2 x (–1) = 0
x – 6 = 0
x = 6
(iii) During preparation of ammonia, iron oxide with small amount of K₂O and Al₂O₃ is used to increase the rate of attainment of equilibrium.
(iv) Change in enthalpy is negative for preparation of SO₃ by catalytic oxidation of SO₂.

Q.36. In which of the following reactions conc. H₂SO₄ is used as an oxidising reagent?
(i) CaF₂ + H₂SO₄ → CaSO₄ + 2HF
(ii) 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
(iii) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
(iv) NaCl + H₂SO₄ → NaHSO₄ + HCl
Ans. (ii, iii)
Solution.
In the above given four reactions, (ii) and (iii) represent oxidising behaviour of H₂SO₄. In reaction (ii), it oxidizes HI and itself reduces to SO₂. Oxidation state of central atom Sulphur decreases from +6 to +4.

In reaction (iii), it oxidizes copper and itself get reduced to SO₂.


37. Which of the following statements are true?
(i) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of XeF6 is a redox reaction.
(iv) Xenon fluorides are not reactive.
Ans. (i, ii)
Solution.
(i) Only one type of interactions between particles of noble gases are due to weak dispersion forces.
(ii) lonisation enthalpy of molecular oxygen is very close to that of xenon. This is the reason for the formation of xenon oxides.
(iii) Hydrolysis of XeF₆is not a redox reaction.
(iv) Xenon fluorides are highly reactive hydrolysis readily even by traces of water.

SHORT ANSWER TYPE QUESTIONS

Q.38. In the preparation of H₂SO₄ by Contact Process, why is SO₃ not absorbed directly in water to form H₂SO₄?
Ans. In Contact process SO₃ is not absorbed directly in water to from H₂SO₄ because the reaction is highly exothermic. Acid mist is formed which is difficult to condense.

Q.39. Write a balanced chemical equation for the reaction showing catalytic oxidation of NH₃ by atmospheric oxygen.
Ans. Ammonia (NH₃) on catalytic oxidation by atmospheric oxygen in presence of Rh/Pt gauge at 500K under pressure of 9 bar produces nitrous oxide.
Balanced chemical reaction can be written as

Q.40. Write the structure of pyrophosphoric acid.
Ans. Molecular formula of pyrophosphoric acid is H₄P₂O₇ and its structure is as follows:

Q.41. PH₃ forms bubbles when passed slowly in water but NH₃ dissolves. Explain why?
Ans. Dissolution of NH₃ and PH₃ in water can be explained on the basis of H-bonding. NH₃ forms H-bond with water so, it is soluble but PH₃ does not form H-bond with water so it remains as gas and forms bubble in water.

Q.42. In PCl₅, phosphorus  is in sp³d hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
Ans. It has trigonal bipyramidal geometry, in which two Cl atoms occupy axial position while three occupy equatorial positions. All five P–Cl bonds are not identical. There are two types of bond lengths (i) Axial bond lengths (ii) Equatorial bond lengths.
Thus, difference in bond length is due to fact that axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

Q.43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
Ans. In gaseous state, NO₂ exists as a monomer which has one unpaired electron but in solid state, it dimerises to N₂O₄ so no unpaired electron left. Therefore, NO₂ is paramagnetic in gaseous state but diamagnetic in solid state.

Q.44. Give reason to explain why ClF₃ exists but FCl₃ does not exist.
Ans. Existance of ClF₃ and FCl₃ can be explained on the basis of size of central atom. Because fluorine is more electronegative as compared to chlorine and has smaller size. Thus, one large Cl atom can accomodate three smaller F atoms but reverse is not true.

Q.45. Out of H₂O and H₂S, which one has higher bond angle and why?
Ans. Bond angle of H₂O ( H — O — H = 104.5°) is larger than that of H₂S (H — S — H = 92°) because oxygen is more electronegative than sulphur therefore, bond pair electron of O—H bond will be closer to oxygen and there will be more bond pair—bond pair repulsion between bond pairs of two O—H bonds.

Q.46. SF₆ is known but SCl₆ is not. Why?
Ans. Fluorine atom is smaller in size so, six F^- ions can surround a sulphur atom. The case is not so with chlorine atom due to its large size. So, SF₆ is known but SCl₆ is not known due to interionic repulsions between larger Cl^- ions.

Q.47. On reaction with Cl₂, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowish-white powder but halide ‘B’ is colourless oily liquid. Identify A and B and write the formulas of their  hydrolysis products.
Ans. Phosphorus on reaction with Cl₂ forms two types of halides A and B.
'A' is PCl₅ and 'B' is PC1₃.
P₄ +10Cl₂ → 4PCl₅
P₄ + 6Cl₂ → 4PCl₃
When 'A' and 'B' are hydrolysed
(a)
(b)

Q.48. In the ring test of  ion reduces nitrate ion to nitric oxide, which combines with Fe²⁺ (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.
Ans.

This test is known as brown ring test of nitrates generally used to identify the presence of nitrate ion in given solution.

Q.49. Explain why the stability of oxoacids of chlorine increases in the order given below: HClO < HClO₂ < HClO₃ < HClO₄
Ans. Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from  ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below
Due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the same order
HClO < HClO₂ < HClO₃ < HClO₄

Q.50. Explain why ozone is thermodynamically less stable than oxygen.
Ans. Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat (ΔH is negative) and an increase in entropy (ΔS is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change (ΔG) for its conversion into oxygen.

Q.51. P₄O₆ reacts with water according to equation P₄O₆ + 6H₂O → 4H₃PO₃. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P₄O₆ in H₂O.
Ans. Given reaction equation is:
P₄O₆ + 6H₂O → 4H₃PO₃      ... (i)
Neutralisation reaction is:
H₃PO₃ + 2NaOH → Na₂HPO₃ + 2H₂O] × 4    ...(ii)
Adding equations (i) and (ii):

Number of moles of P₄O₆,

Molar mass of P₄O₆ = (4 x 31) + (6 x 16) = 220
∵ Product formed by 1 mole of P₄O₆ is neutralised by 8 moles NaOH
∴ Product formed by 1/200 moles of P₄O₆ will be neutralised by NaOH

Given, Molarity of NaOH = 0.1 M = 0.1 mol/L


∴ 400 ml NaOH is required.

Q.52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Ans. Equations for the reactions

Phosphorus used = 31 × 4 = 124g  HCl
produced = 12 × 36.5= 438.0g
∵ 124 g of white phosphorus produces HCl = 438 g
∴ 62 g of white phosphorus will produces HCl =

Q.53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Ans. Three oxoacids of nitrogen having oxidation state + 3 are
(a) HNO₂, nitrous acid
(b) HNO₃, nitric acid
(c) Hyponitrous acid, H₂N₂O₂
In HNO₂, N is in + 3 oxidation state
Disproportionation reaction


Q.54. Nitric acid forms an oxide of nitrogen on reaction with P₄O₁₀. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
Ans. P₄O₁₀ being a dehydrating agent, on reaction with HNO₃ removes a molecule of water and forms anhydride of HNO₃.
4HNO₃ + P₄O₁₀ → 4HPO₃ + 2N₂O₅
Resonating structures of N₂O₅ are

Q.55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Ans. Main points of difference between three allotropic forms of phosphorus are tabulated below:

Q.56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Ans. Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal.
3Cu + 8HNO₃ (Dil.) → 3Cu (NO₃)₂ + 2NO + 4H₂O
Cu + 4HNO₃ (Conc.) → Cu (NO₃)₂ + 2NO₂ + 2H₂O

Q.57. PCl₅ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH₃ solution. Write the reactions involved to explain what happens.
Ans. PCl₅ on reaction with finely divided silver produces silver halide.
PCl₅ + 2Ag → 2AgCl + PCl₃
AgCl on further reaction with aqueous ammonia solution produces a soluble complex of [Ag(NH₃)₂]⁺Cl^-


Q.58. Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Ans. Among various forms of oxoacids, phosphinic acid has stronger reducing property

Reaction showing reducing behaviour of phosphinic acid is as follows
4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag ↓ + 4HNO₃ + H₃PO₄

MATCHING TYPE

Note : Match the items of Column I and Column II in the following questions. 
Q.59. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.

Code :
(i) A → (1) B → (3) C → (4) D → (2)
(ii) A → (1) B → (2) C → (4) D → (3)
(iii) A → (4) B → (3) C → (1) D → (2)
(iv) A → (4) B → (1) C → (2) D → (3)
Ans. (i)
Solution.



Q.60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

Code :
(i) A → (1) B → (2) C → (3) D → (4)
(ii) A → (4) B → (1) C → (2) D → (3)
(iii) A → (3) B → (2) C → (4) D → (1)
(iv) A → (4) B → (3) C → (1) D → (2)
Ans. (ii)
Solution.

Mn₂O₇ on dissolution in water produces acidic solution.
Bi₂O₃ on dissolution in water produces basic solution.

Q.61. Match the items of Columns I and II and mark the correct option.

Code :
(i) A → (4) B → (3) C → (1) D → (2)
(ii) A → (3) B → (4) C → (1) D → (2)
(iii) A → (4) B → (1) C → (2) D → (3)
(iv) A → (2) B → (1) C → (3) D → (4)
Ans. (i)
Solution.
A. H₂ SO₄ is used in storage batteries.
B. CCl₃NO₂ is known as tear gas.
C. Cl₂ has highest electron gain enthalpy.
D. Sulphur is a member of chalcogen i.e., ore producing elements.

Q.62. Match the species given in Column I with the shape given in Column II and mark the correct option.

Column I	Column II
(A) SF4	(1) Tetrahedral
(B) BrF3	(2) Pyramidal
(C)	(3) Sea-saw shaped
(D)	(4) Bent T -shaped

Code :
(i) A → (3) B → (2) C → (1) D → (4)
(ii) A → (3) B → (4) C → (2) D → (1)
(iii) A → (1) B → (2) C → (3) D → (4)
(iv) A → (1) B → (4) C → (3) D → (2)
Ans. (ii)
Solution.
Explanation of this answer is given below:


Q.63. Match the items of Columns I and II and mark the correct option.

Code :
(i) A → (1) B → (4) C → (2) D → (3)
(ii) A → (1) B → (2) C → (3) D → (4)
(iii) A → (2) B → (1) C → (4) D → (3)
(iv) A → (1) B → (3) C → (2) D → (4)
Ans. (iii)
Solution.
(A) Partial hydrolysis of XeF₆ does not change oxidation state of central atom.

(B) He is used in modern diving apparatus.
(C) Ar is used to provide inert atmosphere for filling electrical bulbs.
(D) Central atom (Xe) of XeF₄ is in sp³d² hybridisation.

ASSERTION AND REASON TYPE

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Q.64. Assertion : N₂ is less reactive than P₄.
Reason : Nitrogen has more electron gain enthalpy than phosphorus.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (iii)
Solution.
N₂ is less reactive than P₄ due to high value of bond dissociation energy which is due to presence of triple bond between two N-atoms of N₂ molecule

Q.65. Assertion : HNO₃ makes iron passive.
Reason : HNO₃ forms a protective layer of ferric nitrate on the surface of iron.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (iii)
Solution.
HNO₃ makes iron passive due to formation of passive form of oxide on the surface. Hence, Fe does not dissolve in conc. HNO₃ solution.

Q.66. Assertion : HI cannot be prepared by the reaction of KI with concentrated H₂SO₄
Reason : HI has lowest H–X bond strength among halogen acids.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (ii)
Solution.
HI cannot be prepared by the reaction of KI with concentrated H₂SO₄ because HI is converted into I2 on reaction with H₂SO₄.

Q.67. Assertion : Both rhombic and monoclinic sulphur exist as S₈ but oxygen exists as O₂.
Reason : Oxygen forms pπ – pπ multiple bond due to small size and small bond length but pπ – pπ bonding is not possible in sulphur.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (i)
Solution.
Both rhombic and monoclinic sulphur exist as S₈ but oxygen exists as O₂, because oxygen forms pπ – pπ multiple bond due to its small size and small bond length. But pπ – pπ bonding is not possible in sulphur due to its bigger size as compared to oxygen.

Q.68. Assertion : NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell. But on adding MnO₂ the fumes become greenish yellow.
Reason : MnO₂ oxidises HCl to chlorine gas which is greenish yellow.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (i)
Solution.
NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell. Pungent smell is due to formation of HCl.
NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
But on adding MnO₂ the fumes become greenish yellow due to formation of chlorine gas.

Q.69. Assertion : SF6 cannot be hydrolysed but SF₄ can be.
Reason : Six F atoms in SF₆  prevent the attack of H₂O on sulphur atom of SF₆.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
Ans. (i)
Solution.
SF₄ can be hydrolysed but SF₆ cannot because six F-atoms in SF₆ prevent the attack of H₂O on sulphur atoms of SF₆.

LONG ANSWER TYPE QUESTIONS

Q.70. 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 decolourises acidified aqueous KMnO₄ solution and reduces Fe³⁺ to Fe²⁺. Identify the solid “A” and the gas “B” and write the reactions involved.
Ans. Since, the by-product of roasting of sulphide ore is SO₂, so A is S₈ 'A' = S₈; 'B' = SO₂
 (i) S₈ + 8O₂  8SO₂
(ii) Ca(OH)₂ + SO₂ → CaSO₃ + H₂O
(iii)
(iv) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe^2- + SO^2-₄ + 4H⁺

Q.71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.
Ans. Pb(NO₃)₂ on heating produces a brown coloured gas which may be NO₂. Since, on reaction with N₂O₄ and on heating it produces N₂O₃ and N₂O₄ respectively.
Structures
(i) N₂O₄

(ii) N₂O₃


Q.72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H₂) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
Ans. The main constituents of air are nitrogen (78%) and oxygen (21%). Only N₂ reacts with three moles of H₂ in the presence of a catalyst to give NH3 (ammonia) which is a gas having basic nature. On oxidation, NH₃ gives NO₂ which is a part of acid rain. So, the compounds A to D are as
A = NH₄NO₂; B = N2; C = NH₃; D = HNO₃ 
Reactions involved can be given as:
(i)
(ii)
(iii)
(iv) 2NO + O₂ → 2NO₂
(v)