- Messages
- 120
- Reaction score
- 17
- Points
- 28
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For 5, A n B dnt qualify, as nitrogen is bonded 2 only 2 atoms, so bond angles hv 2 b gr8r thn 109 or 104,pls do tell me
ok, lets go thru thm 1 by 1.no i still posted the questions earlier
No I didintok, lets go thru thm 1 by 1.
Did u gt ques 11??
For 2, luk at the electronic configuration...any one pls q 2 aand 5 pls be quick
if u dnt get d ans, tell me...any one pls q 2 aand 5 pls be quick
2 be honest, i hvnt cmpltly understood it as well...No I didint
ok2 be honest, i hvnt cmpltly understood it as well...
the eqn is perfectly balanced..
i read tht Ba(NO3)2 is an oxidising agent... maybe this is related...
Personally, i thnk tht Al reacts more readily wth Oxygen thn Nitrate, n so tht is the correct ans..
Hope i helped...
Let's move 2 d nxt ques, if thts ok wth u
any one pls q 2 aand 5 pls be quick
did u get a valid conclusion.....for it is not a matter of belief or dis belief!!lol......I can not believe it... question 11 took me half a page to do. -_____-
Yeah, I did get the answer. If you want, I'll put up a picture of how I did it (if you're THAT fond of getting shocks )did u get a valid conclusion.....for it is not a matter of belief or dis belief!!lol......
Cant it be B as trigonal planar(120) means 3 bonding pair hence there are 2 bonding and a lone pair which dec the anlgle a bit so it shld be around 109??/?For 5, A n B dnt qualify, as nitrogen is bonded 2 only 2 atoms, so bond angles hv 2 b gr8r thn 109 or 104,
also since thr is a single lone pair (N) d bond angle cn nt b 180.
Which leaves only C as the ans...
Q11) Notice the words in the question '[...]reaction between powdered aluminium and powdered barium nitrate in which heat is evolved[...]'. The formation of Al2O3 is a highly exothermic reaction.http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q11 Why cant answer be D
Q13 What to do in this?
Q31 How can radius be same?
Q34,Q35,Q37
Got it i miss the trick of 2 p thanks alotFor 2, luk at the electronic configuration...
For Be, 1S2, 2S2... so, 1 pair of electrons in s orbital of 2nd shell, n no electrons in p orbital... thus unequal... 1 pair n 0 unpaired
For C, 1S2, 2S2, 2P2...so, 1 pair of electrons in s orbital, n 2 unpaired electrons in p orbital...thus unequal... 1 pair n 2 unpaired.
For N, 1S2, 2S2, 2P3.... so, 1 pair of e in s orbital, n 3 unpaired electrons in p... thus unequal... 1 pair n 3 unpaired.
For O, 1S2, 2S2, 2P4..... so, 1 pair of e in s orbital, 1 pair of e in p orbital, n 2 unpaired in p.... thus....2 pairs n 2 unpaired....thus equal
so, D is correct..
dont get ur point i am AS student do not involve phenol pls Cant it be B as trigonal planar(120) means 3 bonding pair hence there are 2 bonding and a lone pair which dec the anlgle a bit so it shld be around 109??/?for question 2 ) View attachment 7509
for question 5) the answer is 120 because you have two most electronegative element of the periodic table,which means when the electron are attracted by oxygen,then the lone pair of nitrogen takes the place of bonded electron(it is same like the resonance principle of phenol)...so as there are no lone pair in nitrogen which means the bond angle wouldn't decrease instead it would be 120 !!
MS says B :SA. All of them.
Ooops, sorry. I misread the question.MS says B :S
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