NEET Chemistry Solved Question Paper 2004

Only salts of (weak acid + strong base) and (strong acid + weak base) get hydrolysed (i.e., show alkalinity or acidity in water) . KClO₄ is a salt of strong acid and strong base ,therefore it does not get hydrolysed in water .

KClO₄ $\rightleftharpoons$ K⁺ + ClO₄^-

H₂O $\rightleftharpoons$ $\underset{\stackrel{⥮}{\stackrel{\mathrm{KOH}}{\left(\mathrm{strong}\right)}}}{{\mathrm{OH}}^{-}} + \underset{\stackrel{⥮}{\underset{\left(\mathrm{strong}\right)}{{\mathrm{HClO}}_4}}}{{\mathrm{H}}^{+}}$

2KI + CuSO₄ $\to$ $\underset{\mathrm{unstable}}{{\mathrm{CuI}}_2}$ + K₂SO₄

2CuI₂ $\to$ Cu₂I₂ + I₂

Hence , solution contains K₂SO₄ + Cu₂I₂ + I₂

The ionisation energy of Tin (Sn) is less than that of lead (Pb) .It is due to the poor sheilding of d - and f - electron in Pb due to which it feels greater attraction from nucleus .

Boron (B) , Si , Ge , As , Sb , Te and At are the metalloid elements .Bismuth (Bi) and tin (Sn) are metals while carbon (C) is non-metal .

0.05 M Ba(OH)₂ Solution

= 2 x 0.05 N Ba(OH)₂

= 0.10 N Ba(OH)₂

$\therefore$ pOH = -log [OH^-]

= - log (0.10) = 1

pH = 14 - pOH

= 14 - 1 = 13

K_c = $\frac{[\mathrm{HI}{]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}$ $\frac{{\left({\displaystyle \frac{0.50}{2}}\right)}^2}{\left({\displaystyle \frac{0.15}{2}}\right)\left({\displaystyle \frac{0.15}{2}}\right)}$

= $\frac{0.5 \times 0.5}{0.15 \times 0.15}$ = 11.11

Fluorine is the most electronegative element in the periodic table so it never shows positive oxidation state .

Ionisation depends upon concentration , temperature , nature of solute and on nature of solvent .Ionisation increases on increasing dilution .

Ionisation potential generally increase when we move in a period from left to right but IE₁ of N₂ is greater than that of O₂ .It is due to the more stable (half-filled orbitals) configurations of N .

$\underset{\mathrm{s} \mathrm{mol}/\mathrm{lit}}{{\mathrm{Sb}}_2{\mathrm{S}}_3}$ $\rightleftharpoons$ $\underset{2\mathrm{s}}{2{\mathrm{Sb}}^{3+}}$ + $\underset{3\mathrm{s}}{3{\mathrm{S}}^{2-}}$

Solubility product

(K_sp) = [Sb³⁺]² [S²]³

= (2s)² (3s)³

= 108 s⁵

= 109 x (1.0 x 10^-5)⁵

= 108 x 10^-25