P Block Elements

P Block Elements Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

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This is a short answer type question as classified in NCERT Exemplar

SO₃ generates a dense fog of sulphuric acid that does not condense quickly, it is not absorbed directly in water to form H₂SO_4.

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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) also give necessary equations of all the steps involved.

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P Block Elements Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

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This is a long answer type question as classified in NCERT Exemplar

A = NH₄NO₂

B = N₂

C = NH₃

D = HNO₃

(i) NH₄NO₂→N₂ + 2H₂O

(ii) N₂ + 3H₂→2NH₃

(iii) 4NH₃ + 5O₂→4NO + 6H₂O

4NO + O₂→4NO₂

3NO₂ + H₂O→2HNO₃ + NO

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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 write reactions involved and draw the structures of 'B' and 'C'.

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P Block Elements Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Here, 'A' is NO₂ (Nitrogen dioxide)

'B' is N₂O₄ (dinitrogen tetraoxide)

'C' is N₂O₃ (dinitrogen trioxide)

A brown gas is produced when lead nitrate (II) is heated

2Pb (NO₃)₂ 2PbO + 4NO₂ + O₂

2NO₂ N₂O₄

2NO + N₂O₄ 2N₂O₃

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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.

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This is a long answer type question as classified in NCERT Exemplar

'A' is S₈ 'B' is SO₂ gas

S₈ + 8O₂ 8SO₂

2MnO₄ ^– + 5SO₂ + 2H₂O → 5 SO₄^2– + 4H⁺ + 2Mn²⁺

(violet) (colourless)

2Fe³⁺+ SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2– + 4H⁺

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7.70 Arrange the following in the order of property indicated for each set: (i) F2 , Cl2 , Br2 , I2 - increasing bond dissociation enthalpy. (ii) HF, HCl, HBr, HI - increasing acid strength. (iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.

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7.70

As Bond dissociation energy generally decreases on moving down the group as the atomic size of the element increases. However, among halogens, the bond dissociation energy of F₂ is lower than that of Cl₂ and F₂ due to the small atomic size of

Thus increasing order for bond dissociation energy among halogens is as follows:I₂2

As Bond dissociation energy of H-X molecules where X is the halogen decreases with increase in the atomic HI is the strongest acid as it loses H atom easily due to weak bonding between H and I.

So Increasing acid strength is as follows: HF

Basic strength decreases as we move from Nitrogen to Bismuth down the group

...more

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7.69 How are XeO₃ and XeOF₄ prepared?

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7.69

(i) XeO₃ can be produced by hydrolysis of XeF₄ and XeF₆ under controlled pH of the medium in which reaction is taking place as shown below:

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

XeF₆ + 3H₂O → XeO₃ + 6HF

(ii) XeOF₄ can be obtained on partial hydrolysis of XeF₆ as shown below:

XeF₆ + H₂O → XeOF₄ + 2HF

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7.68 With what neutral molecule is ClO– isoelectronic? Is that molecule a Lewis base?

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7.68

ClO^- isisoelectronic to ClF as both the compounds contain 26 electrons in all. ClO^- : 17+8+1 = 26

ClF : 17+9 = 26

Yes, ClF Molecule is a Lewis base as it accepts electrons from F to form ClF₃.

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7.67 How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

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7.33

Xe and F₂ combine under different conditions to produce XeF₂, XeF₄, XeF₆ as follows:

Ratio

Temperature & Pressure Condition

Reaction

Excess	at {673K,1bar}	Xe (g) + F2 (g) → XeF2 (s)
1:5 ratio	at {873K,7bar}	Xe (g) + 2F2 (g) → XeF4 (s)
1:20 ratio	at {573K,60-70bar}	Xe (g) + 3F2 (g) → XeF6 (s)

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7.66 Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in the presence of MnO₂. (ii) Chlorine gas is passed into a solution of NaI in water.

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7.66

4NaCl + MnO₂ + 4H₂SO₄ MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂

Manganese (IV) oxide reacts with sodium chloride and sulfuric acid to produce manganese (II) chloride, chlorine, sodium bisulfate and water.

This reaction takes place at a temperature near 100°C.

Cl₂ + NaI 2NaCl + I₂

Chlorine reacts with sodium iodide to produce sodium chloride and iodine. Chlorine - diluted solution.

Sodium iodide - cold solution.

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7.65 What are the oxidation states of phosphorus in the following: (i) H₃PO₃ (ii) PCl₃ (iii) Ca₃P₂ (iv) Na₃PO₄ (v) POF₃?

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