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JEE Advanced Chemistry Test 14

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JEE Advanced Chemistry Test 14
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Weekly Quiz Competition
  • Question 1
    4 / -1

    Cyclopentadiene NA Which one of the following in not true about D.

    Solution

    Co – ordination no of Fe+2 in D is 6

     

  • Question 2
    4 / -1

    Which one of the following reaction is not possible:-

    Solution


     

  • Question 3
    4 / -1

    Directions For Questions

    (I) Co(CO)6

    (II) Co(CO)5NH3

    (III) Co(CO)4(NH3)2

    (IV) Co(CO)3(NH3)3

    (V) Co(CO)2(NH3)4

    (VI) Co(CO)(NH3)5

    ...view full instructions

    Which statement is correct for these compounds?

    Solution

    Synergic bonding.

     

  • Question 4
    4 / -1

    Directions For Questions

    On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic. The increasing order of stability of these compounds are is under: Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromatic follow the following conditions (according to valence bond theory)

    (i) The compounds must be be cyclic in structure having (4n + 2)π e, where n = Hückel’s number = 0, 1, 2, 3 et.c

    (ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.

    (iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure must undergo resonance .

    Compounds to be anti-aromatic, it must have 4nπe where n = 1, 2… and it must be planar and undergo resonance. Non-aromatic compounds the name itself spells that compounds must be non-planar irrespective of number of π electrons. Either it has 4nπe– or (4n + 2) π electrons it does not matter. The rate of reaction of any aromatic compounds depends upon the following factors:

    (i) Electron density

    (ii) stability of carbocation produced

    Higher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy. Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.) otherwise there will be no effect of kinetic labelling.

    ...view full instructions

    Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towards

    sulphonation?

    Solution

    2nd step is r.d.s. in sulphonation reaction.

     

  • Question 5
    4 / -1

    Directions For Questions

    On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic. The increasing order of stability of these compounds are is under:

    Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromatic follow the following conditions (according to valence bond theory)

    (i) The compounds must be be cyclic in structure having (4n + 2)π e, where n = Hückel’s number = 0, 1,2, 3 et.c
    (ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.
    (iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure must undergo resonance .

    Compounds to be anti-aromatic, it must have 4nπe where n = 1, 2… and it must be planar and undergo resonance. Non-aromatic compounds the name itself spells that compounds must be non-planar irrespective of number of π electrons. Either it has 4nπe– or (4n + 2) π electrons it  does not matter.

    The rate of reaction of any aromatic compounds depends upon the following factors:

    (i) Electron density
    (ii) stability of carbocation produced

    Higher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy. Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.) otherwise there will be no effect of kinetic labelling.

    ...view full instructions

    Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towards

    nitrations?

    Solution

    2nd step is fast step in nitration.

     

  • Question 6
    4 / -1

    Directions For Questions

    The Bertholet’s equation of state for 1 mole real gas is given as under:

    Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas). One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

    where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

    ...view full instructions

    where Ti is the inversion temperature, then which of the following is the correct relation of the Boyle’s temperature TB obtained by the Bertholet’s equation

    Solution

     

  • Question 7
    4 / -1

    Directions For Questions

    The Bertholet’s equation of state for 1 mole real gas is given as under:

    Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas).  One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

    where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

    ...view full instructions

    If the value of a and b increases by 8 times and 2 times respecively, then new Boyle’s temperature will increases by how many times and what will be effect on the ease of liquification ?

    Solution

     

  • Question 8
    4 / -1

    Directions For Questions

    The Bertholet’s equation of state for 1 mole real gas is given as under:

    Where a and b are van der Waal’s constants. van der Waal’s constant “a” signifies, the force of attraction among the gas molecules. van der Waal’s constant “b” signifies incompressible volume i.e. volume having no effect of compression and expansion. It is also known as co-volume. Thereby the volume  having effect of compression and expansion i.e. compressible volume = V – b (for 1 mole real gas). One of the form of van der Waal’s equation of state is virial equation. The virial equation for 1 mole real gas is as under:

    where A, B, C are known as 2nd, 3rd and 4th virial coefficients respectively. The temperature at which real gases obey ideal gas equation (PV = RT), is known as Boyles temperature i.e. TB.

    ...view full instructions

    If the third virial coefficient of 4 moles of mono-atomic as is 0.04 (lit/mol)2 at NTP, then what will be the real gas at NTP?

    Solution

    b2 = 0.04

    b = 0.2 lit/mol

    Vreal = 4 x 22.4 + 4 x 0.2 = 90.4 lit

     

  • Question 9
    4 / -1

    Which among the following statements is/are correct?

    Solution

    conjugate base of  

     

  • Question 10
    4 / -1

    4.5 g of Al (atomic mass = 27) is deposited at cathode from Al+3 solution by certain quantity of charge. The correct statement/s is/are:

    Solution

     

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