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AIPMT 2007 Chemistry Mains

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Q. 1.

  1. Two silver rods are dipped in 1M HCl and 1M HNO3 . In which of the two acids will the silver rods dissolve under standard conditions? Given:
  2. A 0.1M acetic acid solution ionizes to 1.2%. What is its Ka ?

Sol. Silver rod will dissolve more readily in HNO3 than HCl because HCl is a non-oxidizing acid. Moreso if we establish for the reaction

As the is positive the reaction would take place spontaneously under standard conditions

Q. 2.

  1. dissociates with a degree of dissociation as 0.4. Establish . Relation between and Kp and the value of Kp . Given total pressure = 1 atm and T = 315 K
  2. 1 Mole of nitrogen and 4 mole of Hydrogen react to form ammonia in a 20 ltr vessel. 10 litres of water are added and the vessel properly shaken. What will be the pressure of the residual gases?


The formed NH3 shall dissolve in water on shaking
When 10 litre H2O is added, the effective volume of the vessel occupied by H2
gas will be 20 – 10 = 10 ltr

Q. 3.

  1. An electron in which orbit of lithium will have same energy as an electron in the second orbit of hydrogen?
  2. Also find the value and unit of the rate constant from the data given above.
  3. For a photoelectron, the frequency is given by the expression . If the wavelength of
    the photoelectron is , what will be the value of ‘n’?


  1. Energy of electron in second orbit of hydrogen

    For lithium ion energy should be
    The electron must be in the sixth Bohr orbit of .
  2. As does not feature in the rate law expression so, the order of the reaction w.r.t will be zero.The total order of the reaction = 2 Rate

Q. 4.

  1. Complete the reaction given below

  2. Identify which of the following given compounds is optically active
    (i)2-chloro 3-methyl pent -1, 4-diene
    (ii) 3-methyl 3-hydroxy pentanol
    (iii) 2-chloro 2-methyl butane
  3. Convert:



  4. An alkene reacts withHBr both in the presence and absence of peroxides to give the same product. Identify the alkene.


  1. (i)
    (Optically active)

    (Optically active)

    (Optically inactive)
  2. (i)

  3. As 2-butene is symmetric so, its reaction with HBr with or without a peroxide would yield the same product 2-Bromo butane.

Q. 5.

  1. (i) is produced on reaction of an alkane with , which is not resolvable into optical isomers. Identify the compound?
  2. (ii) Make two possible dipeptides from the amino acids given below:
  3. The amino acid alanine when kept in a solution with pH less than its isoelectric point it coagulates at the cathode and if pH is greater than isoelectric point it coagulates at anode. Explain this phenomenon.
  4. Which out of 1-Butene and 2-Butene react easily with and why?


  1. The Alkanes must be
  2. The two dipeptides would be

  3. (iii) In an acidic solution alanine exists as

    As the colloid is now positively charged it coagulates at the negative electrode (cathode) when (Iso electric pt) In basic media

    This colloid is now negatively charged, so it coagulates at the anode (positive electrode)
  4. 1-butene is less crowded than 2-butene so will react more readly with

Q. 6.

  1. Why 1-Butyne gives sodium salt with but 2-butyne does not?
  2. Draw the structures for DNA purines?


  1. The terminal hydrogen in 1-Butyne is acidic. There is no such available hydrogen is 2-Butyne.

Q. 7.

  1. Why is more soluble in water than ?
  2. Why dimerizes but does not?
  3. The complex has 3 chlorine atoms bonded to platinum. Why is the chlorine atom lying opposite to ethene have higher bond length?


  1. Nitrogen in is electron dense and electronegative, it has the ability to hydrogen bond with water. Phosphorous does not show hydrogen bonding.
  2. is stabilized by back bonding due to which the electron deficiency of the central boron is satisfied. Backbonding is not seen in BH3 due to which it involves in bridge bonding.
  3. Ethene is a donor ligand when the cloud of ethane is polarized for donation into the d-orbitals of Pt, the chlorine lying opposite to it experiences repulsions. This causes the bond length to increase.

Q. 8.

  1. Why ions exist but ions don’t?
  2. Why is paramagnetic but is not?
  3. For a octahedral fields splitting when the pairing energy is less and when pairing energy is higher. Explain the spin magnetic moments acquired by and configurations of metal ions in this field.


  1. In order to form , nitrogen will have to exceed its octet and utilize d-orbitals to form multiple bonds with oxygen. As nitrogen does not have d-orbitals this is not possible.
  2. has a total of S with a M.O configuration

    has 12 electrons due to which the bonding molecular orbitals fill up completely making it diamagnetic.
  3. In a strong field
    Complex will be paramagnetic.
    Spin magnetic moment
    Complex will be diamagnetic.
    Spin magnetic moment = Zero In a weak field

    Spin magnetic moment

    Spin magnetic moment

Q. 9.

  1. The empirical formula of an insoluble compound is . On churning this compound with . The molecular formula of the compound will be?
  2. Out of trimethyl amine and triethyl phosphine, which one has higher dipole moment?


  1. From the products we understand that the complex must be a higher empirical derivative of
    The complex is tetraammineplatinum (II) tetrachloroplatinate (II)

  2. Both molecules shown above are pyramidal. Due to repulsions that arise between three Ethyl groups in triethyl phosphine the bond angle is larger which results in a lower Dipole moment.

Q. 10.

  1. Why is more reactive than ?
  2. Why is more oxidizing than
  3. Out of which is more basic and why?


  1. Due to high charge density of the Fluorine atom, inner electrons repel each other in the F – F bond making it weak and increasing the reactivity of .
  2. Most oxo ions of chromium are very strongly oxidizing. The most stable oxidation state of chromium is (+III), this results in spontaneous reduction of . For molybdenum (+III) state is uncommon, in fact 2-
    has Mo in (+VI) stat e which is the most stable.

  3. bonding due to the presence of vacant d-orbitals with Si. This however is not possible with carbon in due to absence of d-orbitals making it more basic.

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