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Chemistry: Post your doubts here!

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pls do tell me
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...
 
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any one pls q 2 aand 5 pls be quick
For 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..
 
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No I didint
2 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
 
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2 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
ok
 
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any one pls q 2 aand 5 pls be quick



for question 2 ) electron.JPG


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 !!
 

Jaf

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did u get a valid conclusion.....for it is not a matter of belief or dis belief!!lol...:p...
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 ;) )
 
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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,
also since thr is a single lone pair (N) d bond angle cn nt b 180.
Which leaves only C as the ans...
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??/?
 

Jaf

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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
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.

Q13) I'm pretty sure that they're testing our knowledge of solubility. We know all nitrates are soluble, so C and D can't possible be the answer. Likewise for halides. Except for a chosen few (eg: AgCl), all halides are soluble. Hence, the only salt that CAN precipitate is CaCO3. Also, CO3^(2-) ion is more reactive than the Cl- ion (recall the order of displacement in electrolysis).

31) The Mr is 72. This means one of the Cl atoms has nucleon number 35 and the other one has nucleon number 37.
Since the number of electrons and protons determine the radius and they are not changing, the radius remains the same. The nucleon number changes. The isotopic mass also changes.

34) Bronsted Lowry acids are proton donors and B-L bases are proton acceptors. Salts may or may not be formed. Hence answer is D.

35) This is purely a question on the properties of MgO. You need to know them in order to answer this.

36) Thermal stability DECREASES. Recall how introducing a hot rod into a jar with HI decomposes it.
Bond length increases. This can also be seen from the Data Booklet. (Bond strength decreases; stronger bonds are shorter)
Ease of oxidation increases. Concentrated H2SO4 oxidizes HI to I2 but can not oxidize HCl.
 
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For 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..
Got it i miss the trick of 2 p :p thanks alot
 
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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 !!
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??/?
 
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