Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

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The following reaction with silver nitrate demonstrates phosphinic acid reducing behaviour:

4AgNO₃ + 2H₂O + H₃PO₂→4Ag + 4HNO₃ + H₃PO₄

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PCl₅ + 2Ag→2AgCl + PCl₃

AgCl + 2NH₃ (aq)→ [Ag (NH₃)₂] + Cl^- 

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When nitric acid is mixed with copper metal, it produces distinct oxidation products.

3Cu + 8HNO₃ (dil.)→3Cu (NO₃)₂ + 2NO + 4H₂O

Cu + 4HNO₃ (Conc)→3Cu (NO₃)₂ + 2NO₂ + 2H₂O

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The crystal structure of red phosphorus features a sophisticated network of bonding, whereas white phosphorus is made up of P₄ molecules. To avoid natural combustion, white phosphorus must be stored in water, but red phosphorus is stable in air.

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On interaction with P₄O₁₀ forms an oxide of nitrogen,  N₂O₅, and metaphosphoric acid, 3, nitric acid is formed. HPO₃

HNO₃ + P₄O₁₀→4HPO₃ + 2N₂O₅

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Three oxoacids are

(a) HNO₂, Nitrous acid

(b) HNO₃, Nitric acid

(c) Hyponitrous acid, H₂N₂O₂

3HNO₂ + H₂O + 2NO

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P₄+6Cl₂→4PCl₃…  (i)

PCl₃+3H₂O→H₃PO₃+3HCl *4

On adding eq. (i) and (ii)

P₄+6Cl₂+12H₂O→4H₃PO₃+12HCl

1 mol of white phosphorus produces 12 mol of HCl

62 g of white phosphorus has been taken which is equivalent to $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{ }}$ = $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$ mol. Therefore 6 molHCl will be formed.

Mass of 6 molHCl=6*36.5=219.0g HCl

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P₄O₆ + 6H₂O→4H₃PO₃  (i)

Next is neutralization

4 times H₃PO₃ + 2NaOH→Na₂HPO₃ + 2H₂O  (ii)

On addition of 2 equations

P₄O₆+8NaOH→4Na₂HPO₃+2H₂O

P₄O₆ ( mol.mass )= (4*31+16*6)=220

Number of moles of P₄O₆ = $\mathit{?}/\mathit{V}$    = $\mathit{?}$

The product formed is neutralized 8 moles of NaOH

P₄O₆ = 8* $\mathit{?}$  = $\mathit{?}$  molNaOH

Molarity of NaOH in 1 litre  =  0.1M

 Molarity = $\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

 Volume = $\frac{1.1}{220}$  

= $\frac{1.1}{220}$ * $\frac{8.8}{220}$ = 0.4 L

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Due its disintegration into oxygen results in the release of heat ΔH is negative) and an increase in entropy (ΔS is positive), ozone is thermodynamically unstable in comparison to oxygen. These two reactions combine to produce a substantial negative Gibbs energy change (ΔG) for its conversion to oxygen.

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Oxygen is more electronegative than chlorine, the negative charge on chlorine disperses from ClO⁻ to ClO₄⁻ ion as the number of oxygen atoms linked to chlorine grows. As a result, ion stability will improve in the following order:

As the stability of the conjugate base rises, the acidic strength of the corresponding acid increases in the following order:

HClO < HClO₂ < HClO₃ < HClO₄