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 Consider the following reaction, $xMnO_{4}^{-}+yC_{2}O_{4}^{2-}+zH^{+}\rightarrow x Mn^{2+}+2yCO_{2}+\frac{z}{2} H_{2}O$

 The values of x, y, and z in the reaction are respectively

Answer:

Explanation:

The half equations of the reaction are

   $MnO_{4}^{-}\rightarrow Mn^{2+}$

  $C_{2}O_{4}^{2-}\rightarrow CO_{2}$

  The balanced half equation are

             $MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$

                   $C_{2}O_{4}^{2-}\rightarrow2CO_{2}+2e^{-}$

   On equating number of electrons, we get

 $2MnO_{4}^{-}+16H^{+}+10e^{-}\rightarrow2Mn^{2+}+8H_{2}O$

  $5C_{2}O_{4}^{2-}\rightarrow10CO_{2}+10e^{-}$

 On adding both equations , we get

  $2MnO_{4}^{-}+5C_{2}O_{4}^{-}+16H^{+}\rightarrow 2Mn^{2+}+2\times 5CO_{2}+\frac{16}{2}H_{2}O$

  Thus, x,y, and z are 2,5 and 16 respectively.

The order of stability of the following carbocations

                     +                                     +  

$CH_{2}=CH-CH_{2}(I); CH_{3}-CH_{2}-CH_{2}$ (II);

Answer:

Explanation:

 The order of stability  pf carbocation will be

Which of the following arrangements does not represent the correct order of the property stated against it"

Answer:

Explanation:

(a)   $V^{2+}$  = 3 unpaired elecrons

  $Cr^{2+}$  = 4 unpaired electrons

 $Mn^{2+}$  = unpaired electrons

  $Fe^{2+}$= 4 unpaired electrons

 Hence, the order of paramagnetic behaviour should be

 $V^{2+}$  < $Cr^{2+}$=   $Fe^{2+}$ <   $Mn^{2+}$

(b) ionic size decreases from left to right in the same period

 (c) $Co^{3+}$/   $Co^{2+}$= 1.97

  $Fe^{3+}$/  $Fe^{2+}$ = 0.77

  $Cr^{3+}$/  $Cr^{2+}$= -.041

   $Sc^{3+}$ is highly stable ( it does not show +2)

(d) The oxidation  states increases as we go from group 3 to group 7 in the same period

 A compound with molecular mass 180 is acylated with CH₃ COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is 

Answer:

Explanation:

since each -COCH₃ group displace one H atom in the reaction of one mole of 

  with one -NH₂  group the molecular mass increases with 42 unit. Since the mass increases by (390-180)=210

hence the number of -NH₂   group is 210/42=5

The rate of reaction  doubles when its temperature changes from 300K  to 310 K. Activation energy of such a reaction  will be (R=8.314 JK^-1 mol^-1  and log 2= 0.301)

Answer:

Explanation:

From Arrhenius equation,

  $\log \frac{{k_{2}}}{k_{1}}=\frac{-E_{a}}{2.303 R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$

  Given,   $\frac{{k_{2}}}{k_{1}}=2;T_{2}=310 K$

          T₁=300 K

 On putting values, 

   $\Rightarrow\log 2=\frac{-E_{a}}{2.303\times8.314}\left(\frac{1}{310}-\frac{1}{300}\right)$

   $  \Rightarrow$         $ E_{a}=53598.6 J/mol$

   =53.6 kJ/mol

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