Class 12th

Λ°? (CH? COOH) = Λ°? (H? ) + Λ°? (CH? COO? ) = 350 + 50 = 400Scm² mol? ¹
α = Λ? / Λ°?
α = 20/400 = 5 * 10? ²
K? (CH? COOH) = Cα²
= 0.007 * (5 * 10? ²)²
= 1.75 * 10? molL? ¹

NaBH₄ reduces aldehyde/ketone but does not reduce ester

R: CH? OH
Certain mild reducing agents like hypophosphorus acid or ethanol reduce diazonium salts to arene and themselves get oxidised to phosphorous acid and ethanal respectively.

[Fe (CN)? ]? ³ Fe? ³ = 3d?
Unpaired electron = 1, μ = 1.7BM
[Fe (H? O)? ]? ³ Fe? ³ = 3d?
Unpaired electrons = 5, μ = 5.9BM

[Fe (CN)? ]? Fe? ² = 3d?
Unpaired electron = 0, μ = 0BM
[Fe (H? O)? ]? ² Fe? ² = 3d?
Unpaired electrons = 4, μ = 4.9BM

(a) List-I List-II
CO, HCl Anhydrous AlCl? /CuCl → (ii) Gattermann-Koch reaction
(b) R – C=O .
(c) R – CH? OH + R'COOH - (conc. H? SO? )-> (iv) Esterification
(d) R – CH? COOH - ( (i) X? /RedP (ii) H? O )-> (i) Hell-Volhard Zelinsky reaction

CH? – CH? – COO? Na? - (NaOH, CaO, Heat)-> CH? – CH? + Na? CO?
Decarboxylation takes place by soda-lime (NaOH + CaO)

Most of the trivalent lanthanoid ions are coloured in the solid state.

Teflon are prepared by addition polymerization from tetrafluroethene.
CF? = CF? - (catalyst, High pressure)-> [–CF? – CF? –]? Teflon
Nylon-66, Novolac, Dacron are prepared by condensation polymerisation.

(4) C? H? O will have different alkyl group attached with polyvalent functional group that's why show metamerism

Only one arrangement possible so can not show metamerism.
(2) C? H? O → CH? – O – CH? – CH? Only one arrangement possible so can not show metamerism.
(1) No polyvalent functional group in C? H? , so can not show metamerism.

According to Boyle's law

P ∝ 1/V

At a given pressure,

V ∝ T