RBSE Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements

Rajasthan Board RBSE Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements Important Questions and Answers.

RBSE Class 12 Chemistry Chapter 7 Important Questions The p-Block Elements

Multi Choice Questions (MCQ):
 
Question 1. 
Which of the following elements occur in free state: 
(a) Nitrogen 
(b) Arsenic 
(c) Phosphorus 
(d) Antimony
 Answer:
(a) Nitrogen 

 
Question 2. 
PCl₅ exist but NCl₅ does not because: 
(a) Nitrogen does not have free d-orbitals 
(b) NCl₅ is stable 
(c) Nitrogen atom is very small 
(d) Nitrogen is very inert in nature. 
Answer:
(a) Nitrogen does not have free d-orbitals  

Question 3. 
In solid state PCl₅ exist as : 
(a) In the form of PCl₅ 
(b) In the form of PCl₆^- 
(c) In the form of PCl₄⁺ 
(d) In the form of [PCl₄⁺] [PCl^-] 
 Answer:
(d) In the form of [PCl₄⁺] [PCl^-] 
 
Question 4. 
The mixture used in holme signals: 
(a) CaC₂ + CaCl₂ 
(b) CaC₂ + Ca₃P₂ 
(c) CaCl₂ + Ca₃P₂ 
(d) CaC₂ + CaN₂ 
 Answer:
(b) CaC₂ + Ca₃P₂ 
 
Question 5. 
Nitrogen show anomalous behaviour because of: 
(a) small size and high electro negativity 
(b) Absence of d-orbitals in valence shell 
(c) Tendency to from multiple bonds easily 
(d) All of the above 
 Answer:
(d) All of the above 
 
Question 6. 
The H-N-H bond angle in anmonia is : 
(a) 109°28' 
(b) 108° 
(c) 105° 
(d) 106°25' 
 Answer:
(d) 106°25' 
 
Question 7. 
Which of the following element is metalloid : 
(a) Phosphorus 
(b) Antimony 
(c) Nitrogen 
(d) Bismuth 
 Answer:
(b) Antimony 
 
Question 8. 
Nitrogen is inert in nature because: 
(a) It is a mutiple bonded molecule 
(b) Bond polarity is absence 
(c) Inter nuclear distance is very less 
(d) Bond energy is very high 
Answer:
(d) Bond energy is very high 


 
Question 9. 
Which of the following has minimum melting point: 
(a) NH₃ 
(b) PH₃ 
(c) AsH₃ 
(d) SbH₃ 
Answer:
(b) PH₃ 

Question 10. 
Which of the following is most basic: 
(a) NH₃ 
(b) PH₃ 
(c) AsH₃ 
(d) SbH₃ 
Answer:
(a) NH₃ 

Question 11. 
The shape of ammonia molecule is : 
(a) See-saw shape
(b) Triangular planer  
(c) Octahedial 
(d) Pyramidal 
Answer:
(d) Pyramidal 

Question 12. 
When ammonia reacts with the axcess of chlorine then which fo the following compound is obtained: 
(a) N₂ and NCl₃ 
(b) N₂ and HCl
(c) N₂ and NH₄Cl 
(d) HCl, and HCl 
Answer:
(d) HCl, and HCl 

Question 13. 
What is the outer most electronic configiuation of nitrogen atom: 
(a) ns² np⁵ 
(b) ns² np⁴ 
(c) ns² np³ 
(d) ns² np¹ 
Answer:
(c) ns² np³ 

Question 14. 
Which of the following is used in drying the ammonia gas: 
(a) conc. H₂SO₄
(b) P₂O₅  
(c) only CaCl₂ 
(d) CaO 
Answer:
(d) CaO 

Question 15. 
The correct order of first ionisation potential is: 
(a) C < N < O < F 
(b) C > N > O > F 
(c) C < N > O < F 
(d) C > N < O < F 
Answer:
(c) C < N > O < F 

Question 16. 
The basicity of H₃PO₃ is: 
(a) one 
(b) two 
(c) three 
(d) zero 
Answer:
(b) two 

Question 17. 
Nitrogen is adi-atomic molecule. Which of the following is while phosphorus molecule: 
(a) P₂ 
(b) P₄ 
(c) P₆ 
(d) P₈ 
Answer:
(b) P₄ 

Question 18. 
Which of the following element has the highest electro negativity: 
(a) P 
(b) C 
(c) As 
(d) N 
Answer:
(d) N 

Question 19. 
White phosphorus is strored in which of the following liquid: 
(a) Chloroform 
(b) Carbon tetra chloride 
(c) Benzene 
(d) Water 
Answer:
(d) Water 

Question 20. 
Nitrolim is: 
(a) Ca(CN)₂ + C 
(b) CaC₂ 
(c) CaCN₂ + C 
(d) Ca(CN)₂
Answer:
(c) CaCN₂ + C 

Very Short Answer Questions:
 
Question 1. 
Give reason for the following: Ozone is thermodynamically less stable than oxygen. 
Answer:
Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heart (∆H is negative) and increase in entropy (∆S is positive). These two effects reinforce each other that results in large negative Gibbs energy change (∆G) for its conversion into oxygen. Hence, high concentration of ozone can be dangerously explosive. 

Question 2. 
Give reason for the following Above 1000 K sulphur shows paramagnetism. 
Answer:
Above 1000 K, sulphur show paramagnetism due to its existence in vapour state as S2 (i.e. diatomic state). 

Question 3. 
Complete the following reaction. 
PbS(s) + O₃ → 
Answer:
PbS(s) + 4O₃(g) → PbSO₄(s) + 4O₂(s) 

Question 4. 
Assign suitable reason for the following SF₆ is inert towards hydrolysis. 
Answer:
SF₆ is sterically protected by six F-atoms and hence does not allow H₂O molecule to attack the S-atom. Thus, SF is inert towards hydrolysis. 

Question 5. 
Draw the structure of H₂S₂O₈. 
Answer:
Structure of H₂S₂O₈ is given below: 

Question 6. 
Sulphur in vapour state exhibits paramagnetic behaviour. Give reason. 
Answer:
In vapour state, sulphur partly exists as S₂ molecule which has two unpaired electron in the antibonding p-orbitals like O₂ and hence, exhibits paramagnetism. 

Question 7. 
Complete the following chemical equation: 
Fe³⁺ + SO₂ + H₂O → 
Answer:
Fe³⁺ + SO₂ + H₂O → 2Fe²⁺ + SO₄ + 4H⁺ 

Question 8. 
Oxygen is a gas but sulphur is a solid at room temperature. 
Answer:
Oxygen atoms, owing to small size, form рл-рë ьоnd between atoms and exist as diatomic (O₂) molecules. As a result van der Waals forces acting on these mol- ecules are very less and the molecules exist in gaseous state. On the other hand, sulphur atoms, unable to form л- bonds due to large size, form single covalent bond between atoms which results in formation of cyclic molecules comprising of 8 atoms (Sg). Hence, van der Waals' forces act on these molecules to larger extent and a result sulphur exists in solid state. 

Question 9. 
Write the formulae of any two oxoacids of sulphur. 
Answer: 

1. H₂SO₄ (Sulphuric acid) 
2. H₂SO₃ (Sulphurous acid) 

Question 10. 
Which allotrope of sulphur is thermally stable at room temperature? 
Answer: 
Rhombic sulphur is themally stable at room temperature. 

Question 11. 
Draw the structure of the following H₂SO₄ 
Answer:
(i) The structure of H₂SO₄ is 

Question 12. 
Complete the following chemical equation: 
Cu + H₂SO₄ (conc.) → 
Answer:
When conc. H₂SO₄ is added to Cu, then CuSO₄ and SO₂ are formed. In this reaction, hot conc. H₂SO₄ acts as a moderately strong oxidising agent. Here, metal is oxidised by conc. H₂SO₄ and itself it is reduced to SO₂. 
Cu + 2H₂SO₄(conc) → CUSO₄ + SO₂ + 2H₂O

Question 13. 
Draw the structure of the following compound: 
Answer:

Question 14. 
Give reason for the following: 
Oxygen has less electrons gain enthalpy with negative sign than sulphur. 
Answer:
Because of the compact nature (small size) of oxygen atom, it has less negative electron gain enthalpy than sulphur. 

Question 15. 
Account for the following: 
Oxygen shows catenation behaviour less than sulphur. 
Or 
Sulphur has a greater tendency for catenation than oxygen. Why? 
Or 
Sulphur exhibits tendency for catenatio but oxygen does not do so. Give reason. 
Answer:
Bond energy of S - S bond (213 kJ mol^-1) is greater than O-O bond (138 kJ mol^-1). Due to small size of oxygen atom, there is greater lp-bp repulsion in O - O, resulting in weakening of O - O bond more than in S - S bond. Therefore, the tendency of catenation in oxygen is lower than sulphur. 

Question 16. 
Predict the shape and the asked angle (90° or more or less) in the following case SO₃ and the angle in O - S - O. 
Answer:
Structure of SO₃ is pyramidal as shown below: 

Shape: Pyramidal 
Angle in O - S - O: More than 90°. 

Question 17. 
Of PH₃ and H₂S, which is more acidic and why? 
Answer:
H₂S is more acidic than PH₃ because S is more electronegative than P. So, S-H bond is more polar than P-H bond resulting in easier removal of H⁺ ion from H₂S. 

Question 18. 
All the bonds in SF₄ are not equivalent. Explain, why? 
Answer:
In SF₄, sulphur is sp³ d-hybridised. It has trigonal bipyramidal structure in which one of the equatorial positions is occupied by a lone pair of electrons (see- saw geometry). That's why, all the bonds in SF₄ are not equal.
 
Question 19. 
Give reason : H₂S is more acidic than H₂O. 
Answer:
Due to decrease in (E - H) bond dissociation enthalpy down the group, acidic character increases. Thus, H₂S is more acidic than H₂O. 

Question 20. 
Draw the structure of H₂S₂O₇. 
Answer: 

Question 21. 
SF₆ is kinetically inert substance. Explain. 
Answer:
In SF molecule the six atoms of fluorine sterically protect the sulphur atom from attack by a reagent. Hence, SF₆ is called kinetically inert substance. 

Question 22. 
Complete the following chemical equation: 
SO₃ + H₂SO₄(conc.) → 
Answer:

Question 23. 
What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt? 
Answer:
Fe³⁺ + SO₂ + H₂O → 2Fe²⁺ + SO₄ + 4H⁺ 

Question 24. 
Complete the following equation: 
C + H₂SO₄(conc.) → 
Answer:
C + 2H₂SO₄ (conc.) → CO₂ + 2SO₂ + 2H₂O 

Question 25. 
Explain : The two oxygen-oxygen bond lengths in ozone (O₃) molecule are same. 
Answer:
The two oxygen-oxygen bond lengths in the ozone molecule are identical (128 pm). Because it is a resonance hybrid of the following two main forms: 

Question 26. 
Elements of group 16 generally show lover value of first ionisation enthalpy as compared to the corresponding elements in the period of group 15. Explain  why? 
Answer:
Due to the presence of extra stable half-filled np - orbitals in the electronic configurations of group 15 elements (ns²np³ configuration), greater amount of energy is required to remove electrons as compared to group 16 elements (ns² np¹ configuration). Thus, ionisation energy of group 16 elements is lower than that of the corresponding elements of group 15. 

Question 27. 
O₃ acts as a powerful oxidising agent. Give reason. 
Answer:
Ozone acts as a powerful oxidising agent because it liberates nascent oxygen. 
O₃ → O₂ + [O] 
e.g. it oxidises lead sulphide to lead sulphate. 
PbS(s) + 4O₃(g) → PbSO₄(s) + 4O₂(g) 

Question 28. 
The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen. Give reason. 
Answer:
Because of the compact nature (small size) of oxygen atom, it has less negative electron gain enthalpy than sulphur. 

Question 29. 
Draw the structure of O₃ molecule. 
Answer:
Structure of O₃ molecule is given below. 

Question 30. 
H₂S is less acidic than H₂Te. Why? 
Answer:
Due to decrease in (E - H) bond dissociation enthalpy down the group, acidic character increases. Thus, H₂S is less acidic than H₂Te. 

Question 31. 
OF₆ compound is not known. Why? 
Answer:
Due to the absence of d-orbitals in O, it limits its covalency to four, therefore OF₆ is not formed. 

Question 32. 
Write the balanced chemical equation for the following reaction: 
Excess of SO₂ reacts with sodium hydroxide solution.
Answer:
The asked reaction results in formation of sodium hydrogen sulphite. 

Question 33. 
Sulphur hexafluoride is less reactive than sulphur tetrafluoride. Why? 
Answer:
SF₆ is sterically protected so, it is less reactive than SF₄.

Question 34. 
Why are the two S-O bonds in SO₂ molecule of equal strength? 
Answer:
Due to resonance, the two S-O bonds in SO₂ are of equal strength. 

Question 35. 
What happen when, chlorine gas reacts with cold and dilute solution of NaOH? 
Answer:
When chlorine reacts with cold and dilute solution of NaOH, sodium hypochlorite is formed. 

Question 36. 
Draw the structure of the following HCIO₃ 
Answer:
The structure of HCIO₃ is given below 

Question 37. 
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only. 
Answer:

HCI can be prepared from Cl₂ on treating it with water. 

Question 38. 
Arrange the following in increasing order of property indicated, giving reason: 
Hydrides of group 17-acidic strength. 
Answer:
Increasing order of acidic strength of hydrides of group- 17 follows the order 
H - F < H - Br < H - C < H - I 
This is because the bond dissociation enthalpy of H - X bond decrease from H - F to H - I as the size of the atom increases from F to I. 

Question 39. 
Despite lower value of its electron gain enthalpy with negative sign, fluorine, F₂ is a stronger oxidising agent than Cl₂. 
Or 
Although electron gain enthalpy of fluorine is less negative than that of chlorine, yet flourine is a better oxidising agent than chlorine. 
Answer:
Fluorine is the stronger oxidising agent as it is more electronegative than chlorine, it accepts electron most easily. It oxidises other halide ions in solution or even in solid phase. 
F₂ + 2X^- → 2F + X₂  (X = Cl, Br or I) 

Question 40. 
Give reason: 
When Cl₂ reacts with excess of F₂, CIF₃ is formed and not FCl₃. 
Answer:
CIF₃ is formed because it is stable. FCl₃ does not exist actually because 

1. Cl has vacant d-orbitals and hence, can show an oxidation state of +3, but F has no d-orbitals, there- fore, it cannot show positive oxidation states. 
2. Further, since F can show only -1 oxidation state due to its highest electronegativity. Therefore, it forms only CIF or CIF₃. 

Question 41. 
Arrange the following in the decreasing order of their reducing character. 
HF, HCl, HBr, HI 
Answer:
Decreasing order of depends on the ease with which it decomposes to give H₂ and X₂, which in turn de- pends on bond dissociation energy. Bond dissociation energy is least for Hl, hence it has maximum reducing character. 

Question 42. 
Give reason, electron gain enthalpies of halogens are largely negative. 
Answer:
Halogens have the smallest size in their repective periods and therefore, have high effective nuclear charge. As a result, they readily accept one electron to acquire the stable electronic configuration of the nearest noble gas. In other words, large amount of energy is released when a halogen atom accepts an electron to form the corresponding halide ion and thus, halogens have maximum negative electron gain enthalpies. 

Question 43. 
F₂ is a stronger oxidising agent than Cl₂. Why? 
Answer:
F2 is better oxidizing agent in water than chlorine due to greater hydration energy.

Question 44. 
Name two poisonous gases which can be prepared from chlorine gas. 
Answer:
Phosgene (COCl₂), tear gas (CCl₃NO₂), mustard gas (ClCH₂CH₂SCH₂CH₂Cl) all are obtained from chlorine gas (you can write any two gases). 

Question 45. 
Give reason for the following: 
Fluorine does not exhibit any positive oxidation state. 
Answer:
Fluorine is the most electronegative element and hence, it cannot exhibit any positive oxidation state. 

Question 46. 
Draw the structure of BrF₂ molecule. 
Answer:
Structure of BrF₃ is given below: 

Question 47. 
Give reason for the following: 
F₂ is more reactive than CIF₃ but CIF₃ is more reac- tive than Cl₂. 
Answer:
Fluorine due to its small size, high electronegativity and low bond energy is more reactive than CIF₃ but bond energy of Cl - Cl bond is higher than Cl - F bond, therefore CIF₃ is more reactive than Cl₂. 

Question 48. 
Bond enthalpy of F₂ is less than that of Cl₂. Why? 
Answer:
This is due to relatively large electronic repulsion among the lone pairs in F₂ molecule where they are much closer to each other than in case of Cl₂. 

Question 49. 
Give reason for the following: 
PbCl is more covalent than PbCl₂. 
Answer:
If the metal exhibits more than one oxidation states, the halide in the higher oxidation state will be more covalent than the one in lower oxidation state. Therefore, PbCl is more covalent than PbCl₂ due to more than one oxidation states of Pb. 

Question 50. 
Account for the following: HF is not stored in glass bottles but is kept in wax-coated bottles? 
Answer:
HF forms fluorosilicate ion on reaction with glass. Hence, it is stored in wax-coated bottles. 

Question 51. 
Predict the shape and the asked angle (90° or more or less) in the following case: 
CIF₃ and the angle: F - Cl - F. 
Answer:
Structure of CIF₃ is given below: 

Shape Bent T-shaped 
Angle F-Cl-F: Less than 90° 

Question 52. 
Why is ICI more reactive than I₂? 
Answer:
Interhalogen compounds are more reactive than halogens (except fluorine) because X - X bond (I - Cl bond in question) in interhalogens is weaker than X-X bond (I-Ibond) in halogens, except F-F bond. 

Question 53. 
The halogens are coloured, why? 
Answer: 
All halogens are coloured due to the absorption of radiations in visible region which results in the exci- tation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours. 

Question 54. 
Draw the structure of CIF₃ molecule. 
Answer:
Structure of CIF₃ is given below: 

Shape Bent T-shaped 
Angle F-Cl-F: Less than 90° 

Question 55. 
Draw the structure of HCIO₄. 
Answer:
Structure of HCIO₄ is given below: 

Question 56. 
Complete the following reaction: 
NaOH (hot and conc.) + Cl₂ → 
Answer:

Question 57. 
Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water. 
Answer:
Higher boiling point of H₂O is due to the extensive H-bonding than HF. 

Question 58. 
Complete the following chemical equation: 
Cl₂ + F₂ (excess) → 
Answer:

Question 59. 
Account for the following: 
Chlorine water losses its yellow colour on standing. 
Answer:
Due to formation of HCl and HClO, chlorine water loses its yellow colour on standing. 

Question 60. 
F₂ is most reactive of all the four common halogens. Explain. 
Answer:
It is due to low bond dissociation energy and high hydration energy and high electron affinity. 

Question 61. 
What happens when chlorine gas is passed through a hot conc. solution of NaOH? 
Answer:
When chlorine gas reacts with hot and concentrated NaOH solution, it disproportionates into Chloride (Cl^-) and Chlorate (ClO₃^-) ions. When chlorine gas reacts with hot and concentrated NaOH solution, it disproportionates into Chloride (Cl^-) and Chlorate (ClO₃^-) ions.

Question 62. 
Why does fluorine not play the role of a central atom in interhalogen compounds? 
Answer:
Fluorine does not have d-orbitals and it cannot show higher oxidation state. Therefore, it does not play the role of a central atom in interhalogen compounds. 

Question 63. 
How would you account for the following? The oxidising power of oxoacids of chlorine follows the order HClO, < HClO₃ < HClO, < HClO. 
Answer:
As the oxidation number of halogen atom in oxoacid increases, its oxidising power decreases. Therefore, HClO is least stable and gives [O] most easily, so its 
oxidising power is greater than HCIO₄. Thus, the oxidising power of oxoacids of chlorine is 
 HClO₄ < HClO₃ < HClO₂ < HClO 

 
Question 64. 
Complete the following chemical equation: 
Br₂ + F₂ (excess) → 
Answer:
Br₂ + 5F₂ (excess) → 2BrF₅ 

Question 65. 
Explain in aqueous medium HCl is stronger acid than HF. 
Answer:
H-Cl is a storage acid than HF in aqueous medium because with increase in H - X bond length, bond dissociation energy decreases and H⁺ ion is easily produced. 

Question 66. 
Draw the structure of HOCIO₂ molecule. 
Answer:
Structure of HOCIO₂ is given below: 

 
Question 67. 
Complete the following reaction: 
I₂ + H₂O + Cl₂ → 
Answer:
I₂ + 6H₂O + 5Cl₂ → 2HIO₃ + 10HCl 

Question 68. 
Electron gain enthalpy with negative sign for fluorine is less than that for chlorine. 
Answer:
Due to small size of fluorine atom there are strong interelectronic repulsions in the relativity smaller 2p- orbitals of fluorine. Thus, incoming electron does not experience much attraction. 

Question 69. 
Arrange F₂, Cl₂, Br₂ and I₂ in the order of increasing bond dissociation enthalpy. 
Answer:
The order of increasing bond dissociation enthalpy is as follows: 
I₂ < F₂ < Br₂ < Cl₂ 

Short Answer Questions:
 
Question 1. 
Complete the following reaction. 
XeF₆ + NaF → 
Answer:
XeF₆ + NaF → Na + [XeF₇]^-

Question 2. 
What happens when XeF₂ undergoes hydrolysis? 
Answer:
XeF₂ readily hydrolysed even by traces of water. 
2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g) 

Question 3. 
Out of noble gases only xenon is known to form established chemical compounds. 
Answer:
Only xenon is well known to form chemical com pounds because xenon is large in size and having higher atomic mass. Due to large atomic radius the force of attraction be tween the outer electron and the protons is weak hence they are easily available to form compounds. 

Question 4. 
Complete the following equation : 

Answer:

Question 5. 
Draw the structure of the following: 
XeF₄ 
Answer:
Structure of XeF₄ is as follow 

Question 6. 
Complete the following reaction: 
XeF₄ + SbF₅  → 
Answer:
XeF₄ + SbF₅  → [XeF₃] + [SbF₆]^- 

Question 7. 
Draw the structure of the following XeOF₄ 
Answer:
Structure of XeOF₄ is as follows 

Question 8. 
Complete the following reaction: 
XeF₆ + 2H₂O → 
Answer:
XeF₆ + 2H₂O → XeO₂ F₂ + 4HF 

Question 9. 
(i) Draw the structures of the following molecules. 
(a) XeOF₄ 
(b) H₂SO₄ 
Answer:
(i) (a) Structure of XeOF₄ is as follows 

(b) Helium is used as a diluent for oxygen in modern div- ing apparatus because of its very low solubility in blood.

Question 10. 
Draw the structure of XeF₂ molecule. 
Answer:
Structure of XeF₂ molecule is given below: 

Question 11. 
Helium is used in diving equipments. Why? 
Answer:
Helium is used as a diluent for oxygen in modern div- ing apparatus because of its very low solubility in blood. 

Question 12. 
What inspired N Bartlett for carrying out reaction between Xe and PtF₆? 
Answer:
Bartlett found the first ionisation enthalpy of molecular oxygen in O₂ [PtF₆] to be almost similar with that of xenon. Thus, he got inspired for carrying out reaction between Xe and PtF₆ and prepared Xe+ [PtF₆]^- by mixing Xe and PtF₆. 

Question 13. 
Draw the molecular structure of XeF₆. 
Answer:
Structure of XeF₆ is given below: 

Question 14. 
Predict the shape and the asked angle (90° or more or less) in the following case: 
XeF₂ and the angle F-Xe-F 
Answer:
Structure of XeF₂ is given below: 

Shape: Linear 
Angle: F - Xe - F: 180° 

Question 15. 
Answer:
According to the valence bond approach, covalent bonds are formed by the overlapping of half-filled atomic orbitals. But xenon has fully-filled electronic configuration. Hence, the structure of xenon fluorides cannot be explained by VBT. 

Question 16. 
Helium forms no real chemical compound. Why? 
Answer:
Due to its small and high ionisation energy helium is chemically unreactive. That's why, it forms no real chemical compound. 

Question 17. 
Complete the following chemical equation: 
XeF₄ + H₂O → 
Answer:
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂ 

Question 18. 
Why do noble gases have very low boiling points? 
Answer:
Due to weak van der waals forces of attraction between atoms of noble gases (weak dispersion forces), these have very low boiling points. 

Question 19. 
Complete the following chemical equation: 
XeF₆ + H₂O (Excess) → 
Answer:
XeF₆ + 3H₂O → XeO₃ +6HF 

Note: Partial hydrolysis of XeF₆ gives oxyfluorides, 
XeOF₄ and XeO₂F₂ 
XeF₆ + H₂O → XeOF₄ + 2HF 
XeF₆ + 2H₂O → XeO₂F₂ + 4HF 

Question 20. 
Complete the following reaction equation: 
XeF₂ + PF₅ → 
Answer:
XeF₂ + PF₅ → [XeF]⁺ [PF6]^-

Question 21. 
XeF₂ is linear molecule without a bent. Explain. 
Or 
XeF₂ has a straight linear structure and not a bent angular structure. Explain. 
Answer:
XeF₂ is linear molecule. According to VSEPR theory, the three lone pairs will occupy the equatorial positions and two bond pairs will occupy axial positions to minimise lp-lp and Ip-bp repulsions. 

Question 22. 
Noble gases are least reactive elements. Give reason. 
Answer:
Least reactivity of the noble gases is due to the following reasons: 

1. The noble gases except helium 1s² have completely filled ns²np electronic configuration in their valence shell. 
2. They have high ionisation enthalpy and more positive electron gain enthalpy. 

Question 23. 
What happens when XeF₆ is is hydrolysed? 
Answer:
The complete hydrolysis of XeF₆ produces XeO₃ which is xenon trioxide. This xenon trioxide is highly explosive and acts as a powerful oxidising agent in solution.

Question 24. 
Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. 
Answer:
SO₂ is a pungent smelling gas. Two tests are used to detect the presence of SO₂ that are as follows: 
(i) SO₂ turns the pink-violet colour of KMnO₄ solution to colourless due to the reduction of MnO^-₄ to Mn²⁺. 

(ii) SO₂ turns organge coloured acidified K₂Cr₂O₇ solution to green due to the reduction of Cr₂O₂^- to Cr³⁺ ion. 

Question 25. 
What happens when 
(i) conc. H₂SO₄ is added to Cu? 
(ii) SO₃ is passed through water? Write the equations. 
Answer:
(i) Structure of BrF₃ is given below: 

(ii) SO₃ reacts vigorously with water, evolving a large amount of heat and forming H₂SO₄. 
SO₃ + H₂O → H₂SO₄ 

Question 26. 
Name the two most important allotropes of sulphur. Which one of the two is stable at room temperature? What happens when the stable form is heated above 370 K? 
Answer:
Two most important allotropes of sulphur are 

1. rhombic sulphur (a-sulphur) 
2. monoclinic sulphur 

The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur when heated above 370K. 

Question 27. 
1. Write the conditions to maximise the yield H₂SO₄ by contact process. 
2. Why is Ka₂ << Ka₁ for H₂SO₄ in water? 
Answer:

1. Low temperature (about 720 K), high pressure about 2 bar and presence of catalyst (V₂O₅) are the favourable conditions for the manufacture of H₂SO₄ by contact process. 
2. H₂SO₄ is very strong acid in water largely because of its ionisation to H₃O⁺ and HSO₄. Further, ionisation of HSO₄ to SO and H₃O⁺ is very small as losing H⁺  from a negatively charged HSO₄ is more difficult. That's why, Ka₂ << Ka₁.

Question 28. 
Account for the following: 
(i) Decomposition of O₃ molecule is a spontaneous process. 
(ii) SF is inert towards hydrolysis. 
Answer:
(i) For spontaneity of a reaction, AG must be negative. Decomposition of O₃ is an exothermic process (∆H = - ve) and occurs with increase in entropy (∆S = +ve). These two effects reinforce each other which results in large negative Gibbs energy change. It favours its decomposition into oxygen.
 
(ii) In SF₆,S atom is sterically protected by six F atoms and does not allow water molecules to attack the S atom. Further, F does not have d-orbitals to accept the electrons donated by H₂O molecules. Due to these reasons, SF₆ is kinetically an inert substance.

Question 29. 
Account for the following: 
1. H₂S is less acidic than H₂Te. 
2. SO₂ is an air pollutant. 
Answer:

1. In binary acids such as H₂S and H₂Se, the H - Se bonds is longer than the H - S bonds as Se is larger than S. The H - Se bond is therefore weaker than the H - S bond and H₂Se is thus a stronger acid than H₂S. 
2. SO₂ is water soluble, therefore it dissolves in rain-water causing acid rain. Moreover, when released in air, it mixes with it and leads to several diseases like eyes irritation, redness, asthma, bronchitis, etc. Thus, it is considered as an air pollutant. 

Question 30. 
Draw the structures of O₃ and S₈ molecules. 
Answer:

Question 31. 
(i) Assign reasons for the following: 
(a) H₂S is more acidic than H₂O. 
(b) Sulphur has a greater tendency for catenation than oxygen. 
Answer:
(a)  H₂S is more acidic than water because OH bonds are stronger than SH bonds. Sulfur is also larger than oxygen, so that helps stabilize the SH^- anion relative to OH^-.

(b) Sulphur has greater catenation property than oxygen because S-S bond energy is more than O-O bond energy. This is because, oxygen atom is small in size and the lone pair of electrons in oxygen repel bond pair electrons of O-O to a greater extent than sulphur, in turn making the O-O bond weak.

Question 32. 
Complete the following reactions: 
1. Cl₂ + H₂O → 
2. XeF₆ + 3H₂O → 
Answer:

1. Cl₂ + H₂O → HCl + HOCI 
2. XeF₆ + 3H₂O → XeO₃ + 6HF 

Question 33. 
Write the structure of the following: 
(i) BrF₃  
(ii) XeF₄ 
Answer:
The structure of 
(i) BrF₃: Bent T-shaped 


(ii) Structure of XeF₄ is as follow 

Question 34. 
What happens when 
(i) SO₂ gas is passed through an aqueous solution of Fe³⁺ salt? 
(ii) XeF₄ reacts with SbF₅? 
Answer:
(i) When SO₂ gas is passed through an aqueous solution of Fe³⁺ salt, then SO₂ acts as a reducing agent and reduces Fe³⁺ ions present in the salt to Fe²⁺ ions. 
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₂ ̄ + 4H⁺
(ii) XeF₄ reacts with SbF₅ to form a ionic species. XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^-

Question 35. 
Complete the following chemical reaction equation: 

Answer:

(ii) 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂ 

Question 36. 
Complete the following equations : 
(i) C + H₂SO₄ → 
(Conc.) 
(ii) XeF₂ + H₂O→ 
Answer:
(i) C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O 
(ii) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂ 

Question 37. 
Write the formula and the structures of noble gas species (one each) which are isostructural with 
(i) ICI 
(ii) BrO₃ 
Answer:
(i) XeF₄ (isostructural to ICI₄) 

(ii) XeO₃ (isostructural to BrO₃) 

Question 38. 
(i) Account for the following: 
(a) Ozone is thermodynamically unstable. 
(b) Fluorine forms only one oxoacid HOF. 
Answer:
(i) (a) Ozone is thermodynamically unstable because its decomposition into oxygen results in the liberation of heat (AH is negative) and increase in entropy (AS is positive). These two effects reinforce each other resulting in large negative Gibbs energy change (AG) for its conversion into oxygen. 
(b) Due to high electronegativity and small size, fluorine forms only one oxoacid, HOF known as fluoric (I) acid or hypofluorous acid. 

Question 39. 
(i) Account for the following: 
(a) H₂S has lower boiling point than H₂O. 
(b) Reducing character decreases from SO₂ to TeO₂. 
Answer:
(i) (a) Because of the small size and high electronegativity of oxygen molecules, water molecules are highly associated through hydrogen bonding resulting in higher boiling point of H₂O than H₂S. 

(b) Since, the stability of +6 oxidation state decrease or +4 oxidation state increases from S to Te due to inert pair effect, therefore, the reducing character of diox- ides decreases from SO, to TeO₂. 

Question 40. 
(i) Account for the following: 
Acidic character increases from HF to HI. 
(ii) Draw the structure of the following: 
(a) CIF₃ 
(b) XeF₄ 
Answer:
(i) (a) The acidic strength of hydrogen halides increases from HF to HI. This is because the stability of these halides decreases down the group due to decrease in bond dissociation enthalpy of H-X bond from HF to HI. 

(b) (i) XeF₄ (isostructural to ICI₄) 

Long Answer Questions:
 
Question 1. 
(i) Elements of group 16 generally show lower value of first ionisation enthalpy compared to the corresponding periods of group 15. Why? 
(ii) What happens when 
(a) concentrated H₂SO₄ is added to CaF₂? 
(b) sulphur dioxide reacts with chlorine in the presence of charcoal? 
(c) ammonium chloride is treated with Ca(OH)₂? 
Answer:
(i) Two most important allotropes of sulphur are 

1. rhombic sulphur (a-sulphur) 
2. monoclinic sulphur 

The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur when heated above 370K. 

(ii) (a) It forms hydrogen fluoride (HF) 
CaF₂ + H₂SO₄ → CaSO₄ + 2HE. 

(b) It forms sulphuryl chloride (SO₂Cl₂). 
SO₂(g) + Cl₂(g) → SO₂Cl₂(1) 

(c) Ammonia is formed. 
2NH₄Cl + Ca(OH)₂ → 2NH₃ + 2H₂O + CaCl₂ 

Question 2. 
(i) Write the formula and describe the structure of a noble gas species which is isostructural with 
(a) IBr^-₂ 
(b) BrO^-₃ 
(ii) Assign reasons for the following: 
(a) SF₆ is kinetically inert 
(b) HCl is a stronger acid than HF though flourine is more electronegative than chlorine. 
Answer:
(i) (a) IBr₂ Number of valence electron in this species is 7 + (2 × 7) + 1 = 22 The noble gas compound having 22 electrons is XeF₂ = 8 + (2 × 7) = 22 IBr₂ is linear and so, is XeF₂. 

(b) BrO₃ Number of valence electron in BrO₃ = 7 + (3 × 6) + 1 = 26. The noble gas compound having 26 electrons is XeO₃ = 8 + (6 × 3) = 26. The structure of BrO₃ is pyramidal and the structure of XeO₃ is also pyramidal. 

(ii) (a) In SF molecule the six atoms of fluorine sterically protect the sulphur atom from attack by a reagent. Hence, SF₆ is called kinetically inert substance.
(b) H-Cl is a storage acid than HF in aqueous medium because with increase in H - X bond length, bond dissociation energy decreases and H⁺ ion is easily produced.

Question 3. 
(i) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers? 
(ii) F₂ has lower bond dissociation enthalypy than Cl₂. Why? 
(iii) Which noble gas is used in filling ballons for meteorological observations? 
(iv) Complete the equation: 
XeF₂ + PE₅ →
Answer:
(i) Nirtogen oxide emitted from the exhausts of supersonic jet aeroplanes readily combine with ozone to form nitrogen dioxide and diatomic oxygen. Thereby depleting the concentration of ozone in the upper atmosphere. 
NO(g) + O₃(g) → NO₂(g) + O₂(g) 
(ii) Due to smaller size, the lone pair of electrons on the F- atoms repel the bond pair of F - F bond. In constrast, due to the comparatively larger size of Cl atoms, the lone pairs, on the Cl-atoms do not repel the bond pair of Cl - Cl bond to such a extent. As a result, F - F bond energy is lower than that of Cl - Cl bond energy. 
(iii) Helium, being inert, non-inflammable and light gas, is used in filling balloons for meteorological observations. 
(iv) XeF₂ + PF₅ → [XeF]⁺ [PF6]^-

COMPETITION KIT:

Question 1. 
The pair of compounds showing formal (+3) oxidation state. 
(a) Pyrophosphorus and hypophosphoric acid 
(b) Orthophosphorus and hypophosphoric acid 
(c) Pyrophosphorus and pyrophosphoric acid 
(d) Orthophosphorus and pyrophosphorus acid 
Answer:
(d) Orthophosphorus and pyrophosphorus acid 

Question 2. 
The compounds formed by the reaction of zinc with dilute and concentrate nitric acid respectively are. 
(a) NO, and NO 
(b) NO and H₂O 
(c) NO₂ and N₂O
(d) N₂O and NO₂ 
Answer:
(d) N₂O and NO₂ 

Question 3. 
The compound formed by the reaction between CaC₂ and nitrogen is 
(a) CaCN₂ 
(b) CaCN 
(c) CaCN₃ 
(d) Ca₂CN 
Answer:
(a) CaCN₂ 

Question 4. 
Which of the following statement is correct with respect of acid 
(a) Phosphinic acid is a diprotic acid while phosphoric acid is a monoprotic acid 
(b) Phosphinic acid is a monoprotic acid while phosphoric acid is diprotic acid. 
(c) Both are triprotic acid. 
(d) None of two acid is diprotic. 
Answer:
(b) Phosphinic acid is a monoprotic acid while phosphoric acid is diprotic acid. 

Question 5. 
Which is correct about the nitrogen? 
(a) less electronegative 
(b) less ionisation enthalpy 
(c) presence of d-orbitals 
(d) have ability to form рл-рл bonding with itself. 
Answer:
(d) have ability to form рл-рл bonding with itself. 

Question 6. 
Which of the following do not form nitrogen dioxide on heating. 
(a) KNO₃ 
(b) Pb(NO₃)₂ 
(c) Cu(NO₃)₂ 
(d) AgNO₃ 
Answer:
(a) KNO₃ 

Question 7. 
Ammonium dichromate on thermal decomposition will give 
(a) N₂, H₂O and Cr₂O₃ 
(b) N₂, NH₃ and CrO₃ 
(c) (NH₄)₂ CrO₄ and H₂O 
(d) N₂, H₂O and CrO₃ 
(e) N₂, H₂O and CrO
Answer:
(a) N₂, H₂O and Cr₂O₃ 

Question 8. 
One mole of hydrozine (N₂H₄) on reaction forms a new compound (X) after losing 10 mol of electron, Let all the nitrogen atom of hydrazine is obtained in the form of new compound then the oxidation number of nitrogen in (X) compound will be (NOTE: There is no change in the oxidation state of hydrogen in the reaction.) 
(a) - 1 
(b) -3 
(c) +3 
(d) +5 
(e) +1 
Answer:
(c) +3 

Question 9. 
Good reducing property of H₂PO₃ is due to 
(a) Presence of one -OH group and two P-H bond. 
(b) High electron gain enthalpy of phosphorus. 
(c) High oxidation state of phosphorus. 
(d) Presence of two -OH group and one P-H bond. 
Answer:
(a) Presence of one -OH group and two P-H bond. 

Question 10. 
Aqua regia is 
(a) 1 : 3 conc. HNO₃ and conc. HCl
(b) 1 : 2 conc. HNO₃ and conc. HCl 
(c) 3 : 1 conc. HNO₃ and conc. HCl 
(d) 2 : 1 conc. HNO₃ and conc. HCl 
Answer:
(a) 1 : 3 conc. HNO₃ and conc. HCl

Question 11. 
Which element of the group-15 form the most stable pentavalent compound. 
(a) Phosphorus 
(b) Antimony 
(c) Bismuth 
(d) Arsenic 
Answer:
(a) Phosphorus 

Question 12. 
What is the basicity of orthophosphoric acid 
(a) one 
(b) two 
(c) three 
(d) four 
Answer:
(c) three 

Question 13. 
Sulphuryl chloride (SO₂Cl₂) on reaction with white phosphorus gives 
(a) PCl₅, SO₂ 
(b) OPCl₃, SOCl₂ 
(c) PCl₅, SO₂, S₂Cl₂ 
(d) OPCl₃, SO₂, S₂ Cl₃ 
Answer:
(a) PCl₅, SO₂ 

Question 14. 
What is the product of following reaction 
P₄(s) 3NaOH(aq) + H₂O(l) → 
(a) PH₃(g) + 3 Na₂HPO₄ (aq) 
(b) PH₂(g) + 3NaH₂PO₂ (aq) 
(c) 2PH₃(g) + 3Na₂HPO₂ (aq) 
(d) 2PH₃(g) + 3 NaH₂PO₄ (aq) 
Answer:
(b) PH₂(g) + 3NaH₂PO₂ (aq) 

Question 15. 
Which of the following gas will adsorb highly at given temperature and pressure 
(a) dihydrogen 
(b) dioxygen 
(c) ammonia 
(d) dinitrogen 
Answer:
(c) ammonia 

Question 16. 
When white phosphorus is heated with concentrated NaOH solution an inert atmosphere of CO₂ produce a gas. Which of the following statement is not correct in respect of that gas. 
(a) It is highly poisonous and smell like rotten fish 
(b) It is less basic than NH₃ 
(c) Its aqueous solution decomposes in presence of light 
(d) It is more basic than NH₃ 
Answer:
(d) It is more basic than NH₃ 

Question 17. 
Which of the following show sublimation 
(a) NH₄Cl 
(b) CaCO₃ 
(c) BaSO₄ 
(d) CaHPO₃ 
Answer:
(a) NH₄Cl 

Question 18. 
Which of the following oxide of nitrogen is colourless and neutral. 
(a) N₂O 
(b) N₂O₃ 
(c) N₂O 
(d) N₂O₅ 
Answer:
(a) N₂O 

Question 19. 
The product formed by the reaction of white phosphorus with SO₂Cl₂ is 
(a) PCl₃ 
(b) SO₂Cl₂ 
(c) SCl₂ 
(d) POCl₃ 
Answer:
(a) PCl₃ 

Question 20. 
Which is the strongest reducing agent [MP-PMT] 
(a) NH₃ 
(b) PH₃ 
(c) AsH₃ 
(d) SbH₃ 
Answer:
(d) SbH₃ 

Question 21. 
After a long time conversion of concentrated nitric acid into yellow-brown colour indicates the formation of 
(a) NO 
(b) NO₂ 
(c) N₂O 
(d) N₂O₄ 
Answer:
(b) NO₂