All India Engineering / Architecture Entrance Examination (AIEEE)

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AIEEE 2009 Chemistry

 Q. 15. The two functional groups present in a typical carbohydrate are:

1. –OH and –CHO
   
   
2. –OH and –COOH
   
   
3. –CHO and –COOH
   
   
4. 

Sol.:   Carbohydrate is a polyhydroxy aldehyde or polyhydroxy ketone or any compound which give these on hydrolysis. group includes aldehydic as well as ketonic group.

Answer : (4)

Q. 16. Calculate the wavelength (in nanometer) associated with a proton moving at

1. 14.0 nm
   
   
2. 0.032 nm
   
   
3. 0.40 nm
   
   
4. 2.5 nm

Sol:

Answer: (3)

Q. 17. The bond dissociation energy of B–F in BF₃ is 646 kJ mol–1 whereas that of C–F in CF₄ is 515 kJ mol–1. The correct reason for higher B–F bond dissociation energy as compared to that of C–F is

1. lower degree of  interaction between B and F in BF₃than that between C and F in CF₄.
   
   
2. smaller size of B atoms as compared to that of C atom.
   
   
3. stronger  bond between B and F in BF3 as compared to that between C and F in CF₄
   
   
4. significant  interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF₄ .

Sol: Boron in BF3 has a vacant p–orbital, allowing  back bonding while carbon in CF₄ has no vacant orbital, so no back bonding is feasible. Thus, B–F bond is stronger than CF₄.

Answer: (4)