Ncert Solutions Chemistry Class 12th

Allylic halide have halogen bonded to sp³ carbon which is adjacent to > C = C <

Initially, after protonation followed by loss of water, secondary carbocation is formed, further Hydride shift leads to 3° carbocation.

The greater stability of Cu²? (aq) rather than Cu? (aq) is due to the much more negative ΔhydH of Cu²? (aq) than Cu? , which more than compensates for the second ionisation enthalpy of Cu.

When the reactants and the catalyst are in different phases, then the catalysis is known as heterogenous catalysis.

1. N? (g) + 3H? (g) - (Fe (s)-> 2NH? (g) (Heterogenous catalysis)

2. 2SO? (g) + O? (g) - (NO (g)-> 2SO? (g) (Homogenous catalysis)

3. C? H? O? (aq) + H? O (l) → C? H? O? (aq) + C? H? O? (aq) (Homogenous catalysis)

4. NO (g) + O? (g) → NO? (g) + O? (g)

Since, atom B forms CCP structure. Therefore, there will be 4-B atoms.

Now, atom A occupies 1/3 of tetrahedral voids.

Hence, number of A atoms = 1/3 * 8 = 8/3

The correct formula of the compound = A? /? B?

= A? B?

x + y = 2 + 3 = 5

H? O < H? S < H? Se < H? Te
Down the group acidic strength increases
So pK? value decreases

n (O? ) = 4/32 = 1/8 mol
n (H? ) = 2/2 = 1 mol
n (Total) = n (O? ) + n (H? ) = 1/8 + 1 = 9/8 mol
PV = nRT
P (Total) * 1 = (9/8) * 0.082 * 273
P (Total) = 25.18 atm

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

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