115)why the dipole moment for CH3O-Ph-OCH3 is not zero? Shouldn't it be zero since the same groups are attached at the opposite side of benzene ring. (I havn't drawn the figure, but in the figure in my book both the CH3O- groups are attached at the para position of benzene ring, at the opposite side of course)
6)Elements cannot have fractional oxidation state in any compound. This statement is true or false. I think its false because in superoxides O has -0.5 which is a fraction. (in one of my frnd's book this statement is given true, so am confused, plz help me)
7)Statement 1:- Flourine acts as a stronger oxidising agent than Oxygen
Staement 2:-Flourine is more electronegative than oxygen
I know both statements are correct and according to me statement 2 is correct explanation for statement 1, but me answer is not matching with my book, my book says its not the correct explanation. So who's right?
Please provide explanation for your answers friends.
Thanks [1]
1well right now i can't type answers for all so just attempting last wala
7) i think both statements are correct but statement -2 is not a correct explanation of statement-1 because oxidising nature is related to ELECTRON AFFINTIY not ELECTRONEGATIVITY
1
kunl
·2011-01-22 18:07:39
6)Well to be very true it is both TRUE and FALSE[3] ...because the question is ambiguous as it does not specify which type of oxidation state it is referring to-specific or overall oxidation state...sounds cofusing??
well here is an example
Fe3O4=FeO.Fe2O3
1
kunl
·2011-01-22 18:09:28
5.i think u can answer for urself now after this hint: TWO LONE PAIR OF ELECTRONS ON OXYGEN
1
kunl
·2011-01-22 18:12:09
4.(b) and (c) are correct as per rules it is exclusively n that determines energy for hydrogen
baki question baad mein dekhte hein
11
Khyati
·2011-01-24 07:15:13
1st, 2nd and 3rd are remaining
11 . ans is b - 8
acc to ques uncertainity in position is equal to de broglie wavelenght , therefore dx = h/mv
now , acc to uncertainity principle , dx . dp ≥ h /4∩ . as min is required .so mdv.dx = h/4∩.put the value of dx from above . u get dv/v =1/4∩ . now find out the percentage
11
Khyati
·2011-01-25 08:19:13
2)The number of regions of maximum probability is one for and why ?(more than 1 option may be correct)
(a) 1s (b) 3p (c)3d (d) 4f
Answer are option (a),(b) and ?(d)
http://books.google.co.in/books?id=h-gDAAAAMBAJ&pg=PA1050&lpg=PA1050&dq=The+number+of+regions+of+maximum+probability+is+one+for&source=bl&ots=NfYSeX2atR&sig=1hVhqZxdYdd68DRpdpBl4ReBX-4&hl=en&ei=0fY-TYzFEYnQrQe5pqjDCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q=The%20number%20of%20regions%20of%20maximum%20probability%20is%20one%20for&f=false
11
Khyati
·2011-01-25 08:30:32
3) The correct answer is option(a), (b) and (d) because X and Y both represents K(potassium) atom's electronic configuration