Chem

5.73

No2 has ON = 3

hydroxylamine has -1

hence nf=4

C involves 4f electeons not 5f

D Solution is yellow orange not colourless

4s2, 4p6, 4d10, 4f14

s2 is filled so total remaining orbitals are 15

it makes fecl2

6+1

a

ans A

ans CD

Ans BD

No and O2 are para

N2O is DIA

ACD

This is easy dude cmon

ans 5.6

benzoin condensation

Even tho we only have 1 mol of Br2/Naoh, when we add 1Br, that product becomes more reactive than the no Br ketone

So now the 1 Br waala ketone gets reacted on more and forms the 2Br ketone and so on and so forth

Thus, even tho we have 1 mol of each, haloform rxn undergoes completion

449

1 and 3 are true

2 and 4 are false

1. the rate is equal as the only condition here is that Oh and Br should be anti

1. Wittig

2. acetal turns to aldehyde

3. Aldol forms carboxylic and alcohol

4. carboxylic is removed

5. alcohol converts to bromide

6. AGAIN wittig

7. Final product

The main thing to note here is that when we form product 6, the mols of the reactants get halved, aka

2T->1S

This is on top of the fact ki the yield of rxn is 50%

Ans is 87.5

c

c

dG = VdP

delta G = V*DeltaP

c,d

a

415

9300

10

At 400K, volume becomes 400/3 ml

as 100ml stays in, 100/3 ml leaks out

so this is 100/3ml as measures at 400K and 1 atm

find the volume of gas at 600k and 5 atm

bc

Ad

4-3 = 1

54 and 42

b

8

abc

bcd

1. 6

2. 5

bcd

3 ka 16

4 ka 6