P Block Elements

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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

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

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

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

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.

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

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

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NO₃⁻ + 3Fe₂⁺ + 4H⁺→ NO⁺ + 3Fe₃⁺ + 2H₂O

[Fe (H₂O)₆]²⁺ + NO → [Fe (H₂O)₅ (NO)]²⁺ + H₂O ( brown ring )

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A is PCl₅ (yellowish white powder)

P₄ + 10Cl₂→4PCl₅

B is PCl₃ (colourless oily liquid)

Hydrolysis of the following product

PCl₃+3H₂O→H3PO₃+3HCl

PCl₅+4H₂O→H3PO₄+5HCl

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

Six F atoms can be accommodated around sulphur due to fluorine's tiny size, however the chloride ion is comparably bigger, resulting in interatomic repulsion.

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Oxygen is more electronegative than sulphur, the bond angle of H₂O is greater, and the bod pair electron of an OH bond will be closer to oxygen, causing bond-pair repulsion between bond pairs of two OH bonds

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

Chlorine has vacant d-orbitals, which become excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.

Due to the lack of unoccupied d-orbitals in the second energy shell, fluorine cannot expand its octet. As a result, it can only have one covalency.