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2.Assertion Reason
I at room temp O exists as diatomic molecule whereas S is solid.
II the catenated -O-O- chains are less stable as compared to doubly bonded O2 molecule.
ii think both are right and II explains I correctly.Ans given as II doesnt explain I.plz clarify

13 Answers

1
archana anand ·

1)b??(coz five membered as well as deactivating group attahed)

1
pranav ·

but then it can be d also na......but i too felt (B) first!

1
archana anand ·

5 member ring is more stable than 7 membered and hence more acidic.

1
Arka Halder ·

ya answer gives is B archana.
plz clarify 2.

1
archana anand ·

for 2nd

the existence of oxygen as diatomic molecules is due to its ability to form p(pi)-p(pi) multiple bond.The overlapping between two oxygen atom leads to formation of O=O.whereas higher member are unable to form p(pi)-p(pi) bond due to larger size.

1
Manmay kumar Mohanty ·

HMM I GOT IT

Q1) answer is (b)
five memebered ring is more stable than seven membered and deactivating group is attached so we can compare between (b) and (d) according to their stability......
hence ( b ).............................[1]

Q2) the reason is true but doesn't explain correctly
since assertion is O2 is gas whereas S is solid at room temperature and this can be justified like
O=O is much stronger than O - O ( about 3 times )
whereas S-S is more stable than S=S

hence O2 is gas but S is solid. the reason given hence partially describes i.e only abt Oxygen but not S

1
Arka Halder ·

1.O2,O3,CH4 r all gases at standard temperature.what is the order of their molar entropies?
2.the eqn of line which is at max dist from (1,2) and belongs to the family of lines x+2y+1+λ(x+y-2)=0 is
a)4x+5y=25 b)4x-5y=25 c)4x-3y=35 d)4x-5y=35
is any of d options satisfying the ans?

1
1.618 ·

1) CH4<O3<O2 ?

1
Arka Halder ·

ans given CH4<O2<O3
cud u tell ur reasoning ray?

1
1.618 ·

Maths wala nahi ho raha...none matching.

1
ABHI ·

ya none matchin in d maths ques....it was in rtpf i think...i think d ans gvn was a.....

1
1.618 ·

Reason- I used standard enthalpy of formation...
then used (del)S=T(del)H

1
1.618 ·

Don't rely too much on my posts. I am a newbie[1]

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