CBSE Class 12 Chemistry P Block Elements Type Mll Worksheet

MCQ Questions for NCERT CBSE Class 12 Chemistry P Block Elements

Question. Nitrogen forms N₂ but Phosphorous is converted into P₄ from Phosphorous, the reason is
(a) Triple bond is present between Phosphorous atom
(b) Pπ-Pπ bonding is strong
(c) Pπ-Pπ Bonding is weak
(d) Multiple bonds are formed easily

Answer: C

Question. Partial hydrolysis of XeF4 gives
(a) XeO₃
(b) XeOF₂
(c) XeOF₄
(d) XeF₂

Answer: B

Question. What is the correct order of reactivity of group 16 elements?
(a) O > Se > S > Tè > Po
(b) S > O > Tè > Po > Se
(c) S > O > Se > Tè > Po
(d) O > S > Se > Tè > Po

Answer: D

Question. Matching type questions – Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options (a), (b), (c) and (d) given at the end of each question     

(a) A-r, B-s, C-q, D-p
(b) A-r, B-q, C-p, D-s
(c) A-p, B-s, C-q, D-r
(d) A-s, B-p, C-r, D-q

Answer: A

Question. In which of the following sulphur is present in +5 oxidation state?
(a) Di thionic acid
(b) Sulphurous acid
(c) Sulphuric acid
(d) Disulphuric acid

Answer: A

Question. Which of the following statements regarding group 16 elements is not true?
(a) The electronic configuration of Oxygen is [He]2s₂₂p⁴
(b) The electronic configuration of Sulphur is [Ne]3s₂₃p⁴
(c) The atomic radii of the elements of group 16 are larger than those of the corresponding elements of group 15.
(d) The electronic configuration of Tellurium is [Kr]4d₁₀5s₂₅p⁴

Answer: C

Question. H₂S is more acidic than H₂O because
(a) oxygen is more electronegative than sulphur.
(b) atomic number of Sulphur is higher than oxygen.
(c) H — S bond dissociation energy is less as compared to H — O bond.
(d) H — O bond dissociation energy is less also compared to H — S bond

Answer: B

Question. Interhalogen compounds are more reactive than individual halogens because
(a) They are prepared by direct combination of halogens
(b) X-X" bond is weaker than X-X and X”-X” bond
(c) They are thermally more stable than Halogens
(d) There is large difference in their electronegativity

Answer: D

Question. The set with correct order of acidity is
(a) HClO < HClO₂ < HClO₃ < HClO₄
(b) HClO₄ < HClO₃ < HClO₂ < HClO
(c) HClO < HClO₄ < HClO₃ < HClO₂
(d) HClO₄ < HClO₂ < HClO₃ < HClO

Answer: A

Question. Fluorine differs from rest of the halogens in some of its properties. This is due to
(a) its smaller size and high electronegativity.
(b) lack of d-orbitals.
(c) low bond dissociation energy
(d) All of the these

Answer: B

Question. A gas (X) is obtained when Copper reacts with dilute nitric acid. The gas thus formed reacts with oxygen to give brown fumes of (Y). (Y) when dissolved in water gives an important acid (Z) and the gas (X).X, Y and Z respectively are
(a) Nitrogen monoxide, Nitrogen dioxide, Nitric acid.
(b) Nitrogen dioxide, Nitrogen monoxide, Nitric acid.
(c) Nitrous oxide, Nitrogen monoxide, Nitrous acid.
(d) Nitrogen monoxide, Nitrous oxide, Nitric acid.

Answer: A

Question. What happens when white phosphorous is heated at 473 Kelvin under high pressure?
(a) α-Black phosphorous is formed
(b) β-Black phosphorous is formed
(c) Red Phosphorous is formed
(d) No change is observed

Answer: B

Question. If Chlorine is passed through a solution of Hydrogen sulphide in water, the solution turns turbid due to the formation of
(a) Free chlorine
(b) Free sulphur
(c) Nascent Oxygen
(d) Nascent Hydrogen

Answer: B

Question. In BrF₃ (bromine trifluoride) molecule the lone pairs occupy equatorial positions to minimize -
(a) lone pair –bond pair repulsions only
(b) bond pair-bond pair repulsions only
(c) lone pair-lone pair and lone pair-bond pair repulsions
(d) lone pair-lone pair repulsions only

Answer: C

Assertion Reason Based Question :

(a) If both assertion and reason are correct and reason is the correct explanation of the assertion
(b) If both assertion and reason are correct but reason is not the correct explanation of the assertion.
(c) If assertion is correct but reason is the wrong statement.
(d) If both the assertion and reason are incorrect. 

Question. Assertion: Dinitrogen is less reactive than Phosphorous.
Reason: Nitrogen has more electron gain enthalpy than phosphorous.

Answer: C

Question. Assertion: HI cannot be prepared by the reaction of KI with concentrated sulphuric acid.
Reason: HI has lowest H-X bond strength among halogen acids.

Answer: B

Question. Chile saltpetre is the common name of
(a) AgNO₃
(b) NaNO₃
(c) NaSO₄
(d) AgCl

Answer: b

Question. The first noble gas compound obtained was
(a) Xe⁺ PtF₆⁻
(b) XeF₄
(c) XeF₂
(d) XeOF₄

Answer: a

Question. The common oxidation states of group 15 elements are
(a) + 3 and + 5
(b) − 3 and − 5
(c) − 5 and + 5
(d) − 3, + 3 and + 5

Answer: d

Question. Peroxoacids of sulphur are
(a) H₂S₂O₈  and  H₂SO₅
(b) H₂S₂O₈ and H₂S₂O₇
(c) H₂S₂O₇ and H₂S₂O₆
(d) H₂ SO₅ and ₂S₂O₇

Answer: a

Question. Nitrogen lacks d-orbital in its valence shell and hence, it cannot
(a) exhibit orbital hybridisation
(b) exhibit the oxidation state of + 5
(c) forms oxides with oxidation state greater than +3
(d) have covalency greater than three

Answer: b

Question. Among XeO₃ , XeO₄ and XeF₆ , the molecules having same number of lone pairs on Xe are
(a) XeO₃ and XeO₄
(b) XeO₃ and XeF₆
(c) XeO₄ and XeF₆
(d) XeO₃ , XeO₄ F and XeF₆

Answer: d

Question. Which of the following are the applications of dinitrogen gas?
(a) Preservation of biological materials and food items
(b) Production of inert atmosphere in copper and steel industry
(c) In the preparation of explosives
(d) Etching of metals

Answer: a

Question. On heating HNO₃ with P₄O₁₀ ,   the oxide of nitrogen produced is
(a) NO₂
(b) N₂O₅
(c) N₂O₄
(d) N₂O₃

Answer: b

Question. Red phosphorus is less reactive, less volatile and less soluble in non-polar solvent than white/yellow phosphorus because
(a) it has high molecular energy
(b) it has low molecular energy
(c) it forms condensation products
(d) it possesses highly polymerised structures

Answer: d

Question. Which of the following hydrides has the lowest boiling point?
(a) PH₃
(b) AsH₃
(c) SbH₃
(d) NH₃

Answer: a

Question. PH₃ produces smoky rings when it comes in contact with air because
(a) PH₃ reacts with water vapours
(b) PH₃ reacts with N₂
(c) PH₃ burns in air
(d) PH₃ contains impurities of P₄H₂ 

Answer: d

Question. The compound that cannot act both as oxidising and reducing agent is
(a) H₂SO₃
(b) H₃PO₄
(c) HNO₂
(d) H₂O₂

Answer: b

Question. Sulphur exhibits valencies of 2, 4 and 6, whereas oxygen has a valency of 2 due to
(a) being less electronegative than S
(b) presence of d-orbitals in S
(c) S is bigger atom
(d) S has higher ionisation potential

Answer: b

Question. All the hydrides (of group 16 elements) except one possess reducing property. Identity the hydride
(a) H₂Se 
(b) H₂O
(c) H₂S
(d) H₂Te

Answer:  b

Question. Tetrafluorides of elements of group-16 have hybridisation and structure respectively are
(a) sp3 and trigonal pyramidal
(b) sp3 d and tetrahedral
(c) sp3 d and trigonal bipyramidal
(d) sp3 d and tetrahedral

Answer: c

Question. Water is much less volatile than H₂S  because
(a) H₂O  has a bond angle of nearly 150°
(b) hydrogen is loosely bonded with the sulphur
(c) S-atom is less electronegative than O-atom
(d) S-atom is more electronegative than O-atom

Answer: c

Question. Among the following, the number of compounds that can react with PCl 5 to give POCl 3 is
I. O₂ II. CO₂ III. CH₃COOH IV. H₂O V. C₂H₅OH VI. P₄O₁₀
(a) 1
(b) 2
(c) 3
(d) 4

Answer: d

Question. Correct order of bond dissociation energy is
(a) Br₂ > Cl₂
(b) F₂ > Cl₂ 
(c) I₂ > F₂
(d) F₂ > I₂

Answer: d

Question. Which of the following oxides is amphoteric in nature?
(a) Cl₂O₇
(b) Na₂O 
(c) N₂O 
(d) Al₂O₃

Answer: d

Question. Extra pure N2 can be obtained by thermal decomposition of
(a) NH₃ with CuO
(b) NH₄NO₃
(c) (NH₄)₂ Cr₂ O₇
(d) Ba(N₃₎₂

Answer: d

Question. Angular shape of ozone molecule consists of
(a) 1σ-bond and 1π-bond
(b) 2σ-bond and 2π-bonds
(c) 1σ-bond and 2π-bonds
(d) 2σ-bond and 1π-bond

Answer: d

Question. H₂SO₄ is used in
(a) petroleum refining
(b) manufacture of paints, pigments and dyestuff intermediates
(c) detergent industry
(d) All of the above

Answer: d

Question. The industrial preparation of SO₂ is
(a) S(s) + O₂ (g ) → SO₂ (g )
(b) SO₃²⁻(aq) + 2H⁺(aq) → H₂O(l) + SO₂ (g)
(c) 4FeS₂ (s) 11O₂(g) → 2Fe₂O₃(s) + 8SO₂ (g)
(d) All of the above

Answer: c

Question. The anomalous behaviour of fluorine is due to
(a) its small size
(b) its highest electronegativity
(c) low F—F bond dissociation enthalpy and non-availability of d-orbitals in valence shell
(d) All of the above

Answer: d

Question. HCl gas can be dried by passing through
(a) conc. H₂ SO₄
(b) dil. H₂SO₄
(c) conc. HNO₃
(d) dil. HNO₃

Answer: a

One Mark questions

1. Why is H₃PO₃ diprotic?

2. Nitrogen does not form pentahalide like phosphorous,why?

3. H₂O is a liquid while H₂s is a gas ,why?

4. Arrange NH₃ PH₃ AsH₃ BiH₃ SbH₃ in the increasing order of basic strength.

5. Oxygen is a gas while sulphur is a solid,why?

6. Write the formula of hyponitros acid.

7. Why does Al not react with conc.nitricacid?

8. Can PCl₅ act as an oxidising agent and reducing agent?

9. Why does NO₂ readily dimerise?

10. Why is BiH₃ the strongest reducing agent amongst all the hydrides of nitrogen family?

11. Write the chemical formula of peroxodisulphuric acid.

12. Why does NH₃ act as a Lewis base /complexing agent?

13. What is the basicity of H₃PO₄?

14. Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?

15. Why does the reactivity of nitrogen differ from phosphorus?

16. Why is white phosphorous highly reactive?

17. Why group 16 members are called chalcogens?

18. OF₄ is not known but SF₄ is known .Explain

19. Solid PCl₅ exists as an ionic solid,Why?

20. Bismuth is a strong oxidizing agent in pentavalentstate,why? 

One Mark Questions

Question. Why is \( \text{H}_3\text{PO}_3 \) diprotic?
Answer: Two ionisable hydrogen

Question. Nitrogen does not form pentahalide like phosphorous, why?
Answer: Absence of d-orbitals in nitrogen.

Question. \( \text{H}_2\text{O} \) is a liquid while \( \text{H}_2\text{S} \) is a gas, why?
Answer: H-bonding in \( \text{H}_2\text{O} \)

Question. Arrange \( \text{NH}_3 \), \( \text{PH}_3 \), \( \text{AsH}_3 \), \( \text{BiH}_3 \), \( \text{SbH}_3 \) in the increasing order of basic strength.
Answer: \( \text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3 \)

Question. Oxygen is a gas while sulphur is a solid, why?
Answer: Absence of \( p\pi-p\pi \) bonding in oxygen.

Question. Write the formula of hyponitros acid.
Answer: \( \text{HNO}_2 \)

Question. Why does Al not react with conc. nitric acid?
Answer: Formation of passive oxide film

Question. Can \( \text{PCl}_5 \) act as an oxidising agent and reducing agent?
Answer: No, can act as oxidizing agent only.

Question. Why does \( \text{NO}_2 \) readily dimerise?
Answer: To pair up odd electron.

Question. Why is \( \text{BiH}_3 \) the strongest reducing agent amongst all the hydrides of nitrogen family?
Answer: Low bond dissociation enthalpy.

Question. Write the chemical formula of peroxodisulphuric acid.
Answer: \( \text{H}_2\text{S}_2\text{O}_8 \)

Question. Why does \( \text{NH}_3 \) act as a Lewis base / complexing agent?
Answer: It can donate lone pair of electrons very easily.

Question. What is the basicity of \( \text{H}_3\text{PO}_4 \)?
Answer: 3

Question. Why does \( \text{R}_3\text{P}=\text{O} \) exist but \( \text{R}_3\text{N}=\text{O} \) does not (\( \text{R} \) = alkyl group)?
Answer: Presence of d-orbitals in phosphorous.

Question. Why does the reactivity of nitrogen differ from phosphorus?
Answer: Absence of d-orbitals, H-bonding, tendency to form multiple bond.

Question. Why is white phosphorous highly reactive?
Answer: Strained structure.

Question. Why group 16 members are called chalcogens?
Answer: Ore forming nature.

Question. \( \text{OF}_4 \) is not known but \( \text{SF}_4 \) is known. Explain.
Answer: Absence of d-orbitals in oxygen.

Question. Solid \( \text{PCl}_5 \) exists as an ionic solid, Why?
Answer: It exists as \( [\text{PCl}_4]^+[\text{PCl}_6]^- \) in solid state.

Question. Bismuth is a strong oxidizing agent in pentavalent state, why?
Answer: Inert pair effect.

Two Marks Questions

Question. Draw the shapes of \( \text{SF}_4 \), \( \text{BrF}_3 \) on the basis of VSEPR theory.
Answer: \( \text{SF}_4 \) (See-saw), \( \text{BrF}_3 \) (T-shape).

Question. (a) Why is atomic radius of Argon more than that of Chlorine ?
(b) Why is ionization enthalpy of Nitrogen more than oxygen?
Answer: (a) Ar has van der Waals radius while Cl has covalent radius; the magnitude of \( \text{V}_r > \text{C}_r \).
(b) Half-filled configuration shown by nitrogen.

Question. (a) Arrange \( \text{F}_2 \), \( \text{Cl}_2 \), \( \text{Br}_2 \), \( \text{I}_2 \) in the increasing order of bond dissociation energy.
(b) Arrange \( \text{HOClO} \), \( \text{HOClO}_2 \), \( \text{HOClO}_3 \) in the increasing order of acidic strength.
Answer: (a) \( \text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2 \)
(b) \( \text{HOClO}_3 > \text{HOClO}_2 > \text{HOClO} \)

Question. Explain giving suitable reasons:
(i) \( \text{NH}_3 \) has higher boiling point than \( \text{PH}_3 \)
(ii) \( \text{SbF}_5 \) is known but \( \text{BiF}_5 \) is unknown.
Answer: (i) H-bonding in ammonia.
(ii) Inert pair effect shown by Bi.

Question. Explain giving suitable reasons:
(i) \( \text{SF}_6 \) is well known but \( \text{SH}_6 \) is not known.
(ii) Proton affinity of \( \text{NH}_3 \) is more than \( \text{PH}_3 \).
Answer: (i) The enthalpy of atomization of \( \text{H-H} \) is very high as compared to \( \text{F-F} \). High enthalpy of dissociation cannot be compensated by energy released during bond formation.
(ii) High electronegativity of nitrogen.

Question. Explain giving suitable reasons.
(i) Sulphur in vapour form is paramagnetic in nature.
(ii) Catenation properties of Phosphorous is more than Nitrogen.
Answer: (i) In vapour form sulphur behaves like \( \text{O}_2 \).
(ii) Phosphorous is unable to form multiple bonds.

Question. Give chemical equations, when:
(a) Ammonium dichromate is heated?
(b) Sodium azide is heated?
Answer: (a) \( (\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2 + \text{Cr}_2\text{O}_3 + 4\text{H}_2\text{O} \)
(b) \( 2\text{NaN}_3 \xrightarrow{\Delta} 2\text{Na} + 3\text{N}_2 \)

Question. Complete the following equations:
(i) \( \text{HgCl}_2 + \text{PH}_3 \rightarrow \)
(ii) \( \text{P}_4 + \text{NaOH} + \text{H}_2\text{O} \rightarrow \)
Answer: (i) \( 3\text{HgCl}_2 + 2\text{PH}_3 \rightarrow \text{Hg}_3\text{P}_2 + 6\text{HCl} \)
(ii) \( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2 \)

Question. Write main differences between the properties of white phosphorus and red phosphorus.
Answer: White Phosphorous: Strained structure, Highly reactive, Insoluble in water.
Red Phosphorous: Stable structure, Less reactive, soluble in water.

Question. Arrange \( \text{H}_2\text{O} \), \( \text{H}_2\text{S} \), \( \text{H}_2\text{Se} \), \( \text{H}_2\text{Te} \) in the increasing order of (i) Acidic Character (ii) Thermal stability.
Answer: (i) Acidic Character: \( \text{H}_2\text{O} < \text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te} \)
(ii) Thermal stability: \( \text{H}_2\text{Te} < \text{H}_2\text{Se} < \text{H}_2\text{S} < \text{H}_2\text{O} \)

Three Marks Questions

Question. (i) Why is \( \text{ICl} \) more reactive than \( \text{I}_2 \)?
(ii) Interhalogen compounds are strong Oxidizing agents, why?
(iii) Bleaching of flowers by \( \text{Cl}_2 \) is permanent while \( \text{SO}_2 \) is temporary, why?
Answer: (i) \( \text{I-Cl} \) bond is more polar and weaker than \( \text{I-I} \) bond.
(ii) Low bond dissociation enthalpy of \( \text{X-Y} \) bond.
(iii) \( \text{Cl}_2 \) bleaches the colour by oxidation while \( \text{SO}_2 \) bleaches by Reduction.

Question. Give Reasons:
(a) Iodine is more soluble in KI solution than in water.
(b) HF is stored in wax-coated bottle.
(c) HCl is not used to make the medium acidic in titrations involving \( \text{KMnO}_4 \).
Answer: (a) Formation of \( \text{KI}_3 \) complex.
(b) HF reacts with silica frequently.
(c) HCl can oxidise into \( \text{Cl}_2 \).

Question. Explain giving suitable reasons:
(i) \( \text{SbF}_5 \) is known but \( \text{BiF}_5 \) is unknown.
(ii) \( \text{CN}^- \) ion is known but \( \text{CP}^- \) is not.
(iii) Compounds of Noble gases are known with Xe and fluorine.
Answer: (i) Due to Inert pair effect \( \text{BiF}_5 \) is not known.
(ii) Phosphorous is unable to form multiple bonds.
(iii) Low ionisation enthalpy of Xe and high electronegativity of F.

Question. Explain giving suitable reasons:
(i) \( \text{PH}_3 \) has lower boiling point than \( \text{NH}_3 \).
(ii) Nitric oxide becomes brown when released in air.
(iii) When HCl reacts with finely powered iron, it forms ferrous chloride and not ferric chloride.
Answer: (i) Absence of hydrogen bonding in \( \text{PH}_3 \).
(ii) Due to formation of \( \text{NO}_2 \).
(iii) Fe on reaction with HCl forms \( \text{H}_2 \) which hinders the formation of \( \text{FeCl}_3 \).

Question. (i) Which Xe compound has distorted octahedral shape?
(ii) How does Chlorine react with hot and concentrated NaOH?
(iii) Write the chemical reaction involved in the ring test.
Answer: (i) \( \text{XeF}_6 \)
(ii) \( 3\text{Cl}_2 + 6\text{NaOH (hot and conc.)} \rightarrow \text{NaClO}_3 + 5\text{NaCl} + 3\text{H}_2\text{O} \)
(iii) \( \text{NO}_3^- + \text{Fe}^{2+} + 5\text{H}_2\text{O} \rightarrow [\text{Fe}(\text{H}_2\text{O})_5\text{NO}]^{2+} \)

Question. (a) Are all the bonds in \( \text{PCl}_5 \) equivalent in length?
(b) On the basis of structure show that \( \text{H}_3\text{PO}_2 \) is a good reducing agent.
(c) How many P-OH bonds are present in Pyrophosphoric acid?
Answer: (a) No, axial bonds are slightly longer than equatorial bonds.
(b) \( \text{H}_3\text{PO}_2 \) has one \( \text{P-H} \) bond.
(c) 4

Question. Complete the following equations:
- \( \text{P}_4 + \text{SOCl}_2 \rightarrow \)
- \( \text{NH}_3 + \text{CuSO}_4(\text{aq}) \rightarrow \)
- \( \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \)
Answer:
- \( \text{P}_4 + 8\text{SOCl}_2 \rightarrow 4\text{SO}_2 + 4\text{PCl}_3 + 2\text{S}_2\text{Cl}_2 \)
- \( 4\text{NH}_3 + \text{CuSO}_4(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \)
- \( 6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 2\text{XeO}_3 + 24\text{HF} + 4\text{Xe} + \text{O}_2 \)

Question. (i) Arrange M-F, M-Cl, M-Br, M-I in the increasing order of ionic character.
(ii) Arrange HF, HCl, HBr, HI in the increasing order of reducing behavior.
(iii) Arrange \( \text{F}_2 \), \( \text{Cl}_2 \), \( \text{Br}_2 \), \( \text{I}_2 \) in the increasing order of bond dissociation energy.
Answer: (i) \( \text{M-I} < \text{M-Br} < \text{M-Cl} < \text{M-F} \)
(ii) \( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
(iii) \( \text{I}_2 < \text{F}_2 < \text{Br}_2 < \text{Cl}_2 \)

Question. Starting from sulphur, how would you manufacture \( \text{H}_2\text{SO}_4 \) by contact process.
Answer:
(a) \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2(\text{g}) \)
(b) \( 2\text{SO}_2 + \text{O}_2 \xrightarrow{\text{V}_2\text{O}_5} 2\text{SO}_3 \)
(c) \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)
(d) \( \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 \)

Question. Write the reaction involved in formation of ammonia by Habers process? State the favorable conditions for good yield of ammonia.

Answer: \( \text{N}_2 + 3\text{H}_2 \xrightarrow{\text{Fe/Mo}} 2\text{NH}_3(\text{g}) \). Favorable conditions are Low temperature and High Pressure.

Five Marks Questions

Question. A gas “X” is soluble in water. Its aq. Solution turns red litmus blue with excess of aq. \( \text{CuSO}_4 \) solution it gives deep blue colour and with \( \text{FeCl}_3 \) solution a brownish ppt. soluble in \( \text{HNO}_3 \) is obtained. Identify gas ”X” and write reactions for changes observed.
Answer: \( \text{X} = \text{NH}_3 \)
\( 4\text{NH}_3 + \text{CuSO}_4(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \)
\( 3\text{NH}_3 + 3\text{H}_2\text{O} + \text{FeCl}_3 \rightarrow \text{Fe(OH)}_3 + 3\text{NH}_4\text{Cl} \) (Brown ppt)

Question. Write the reaction involved in formation of Nitric acid by Ostwald's process? State the favorable conditions for good yield of Nitric oxide.
Answer:
\( 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} \)
\( 2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2 \)
\( 3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO} \)

Question. Complete the following equations:
(i) \( \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \)
(ii) \( \text{XeF}_6 + \text{PF}_5 \rightarrow \)
(iii) \( \text{Cl}_2 + \text{F}_2 \text{ (excess)} \rightarrow \)
(iv) \( \text{HgCl}_2 + \text{PH}_3 \rightarrow \)
(v) \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \)
Answer:
(i) \( 6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 2\text{XeO}_3 + 24\text{HF} + 4\text{Xe} + \text{O}_2 \)
(ii) \( \text{XeF}_6 + \text{PF}_5 \rightarrow [\text{XeF}_5]^+[\text{PF}_6]^- \)
(iii) \( \text{Cl}_2 + 3\text{F}_2 \text{ (excess)} \rightarrow 2\text{ClF}_3 \)
(iv) \( 3\text{HgCl}_2 + 2\text{PH}_3 \rightarrow \text{Hg}_3\text{P}_2 + 6\text{HCl} \)
(v) \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)

Question. A translucent white waxy solid ‘A’ on heating in an inert atmosphere is converted in to its allotropic form (B). Allotrope ‘A’ on reaction with very dilute aqueous KOH liberates a highly poisonous gas ‘C’ having rotten fish smell. With excess of chlorine ‘A’ forms ‘D’ which hydrolysis to compound ‘E’. Identify compounds ‘A’ to ‘E’.
Answer: \( \text{A} = \text{White } \text{P}_4 \)
\( \text{B} = \text{Red } \text{P}_4 \)
\( \text{C} = \text{PH}_3 \)
\( \text{D} = \text{PCl}_5 \)
\( \text{E} = \text{H}_3\text{PO}_4 \)

Question. What happens when Concentrated \( \text{H}_2\text{SO}_4 \) is added to / Give the reactions of \( \text{H}_2\text{SO}_4 \) with (i) calcium fluoride (ii) KCl, (iii) Sugar (iv) Cu turnings, (v) Sulphur.
Answer:
(i) \( \text{CaF}_2 + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{CaSO}_4 + 2\text{HF} \)
(ii) \( 2\text{KCl} + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow 2\text{HCl} + \text{K}_2\text{SO}_4 \)
(iii) \( \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow 12\text{C} + 11\text{H}_2\text{O} \)
(iv) \( \text{Cu} + 2\text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O} \)
(v) \( 3\text{S} + 2\text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O} \)