JEE 2015 :: Advanced 2015 :: Chemistry 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

 Copper is purified by electrolytic refining of blister copper. The correct statement(s0 about this process is/are 

Answer:

Explanation:

(a) is wrong statement, Impure copper is set as anode where copper is oxidised to Cu²⁺ and goes into electrolyte solutions.

(b) CuSO₄ is used as an electrolyte in the purification process

(c)  Pure copper is deposited at cathode as :

    $Cu^{2+}+2\overline{e}\rightarrow Cu $   :  (At  cathode )

 (d) Less active metals like Ag, Au etc. settle down as anode mud.

The correct statement (s) about Cr²⁺  and Mn³⁺  is/are  [atomic  number of Cr=24 and Mn=25]

Answer:

Explanation:

(a) In an aqueous solution  Cr²⁺ (3d⁴)  acts as a reducing agent, oxidising itself to  Cr³⁺ (3d³)  that gives a completely half-field t_2g level in the octahedral ligand field of H₂O.

(b)   Mn³⁺ (3d⁴) is an oxidising agent as it is reduced to Mn²⁺  (3d⁵) , a completely half-filled stable configuration.

(c) Both Cr²⁺  and Mn³⁺ have d⁴  configuration.

(d) $3d^{4}Cr^{2+}(aq)$         $\underrightarrow{R.A}$       $Cr^{3+}(aq)+\overline{e}$

 hence (d) is wrong statement.

 If the freezing point of a 0.01 molal  aqueous solution of a cobalt (III) chloride -ammonia  complex (which behaves as a strong electrolyte) is -0.0558° C, the number of chloride (s)  in the coordination  sphere of the complex is 

  [K_f   of water =1.86 K kg mol^-1]

Answer:

Explanation:

 $+\triangle T_{f}= iK_{f}m$

  $\triangle T_{f}=0-(-0.0588^{0}C)=0.0558^{0}C$

  $\Rightarrow$       i(vant Hoff's factor )=  $\frac{0.0558}{1.86\times 0.01}=3$

 This indicates that complex upon ionisation  produces  three ions as:

  $[Co(NH_{3})_{5}Cl]Cl_{2}\rightarrow\left[ Co\left(NH_{3}\right)_{5}Cl\right]^{2+}(aq)+2Cl^{-}(aq)$

 Thus, only  me Cl is inside the coordination sphere.

 All the energy released  from the reaction

   $X\rightarrow Y,\triangle_{r}G^{0}=-193kJmol^{-1}$

  is used for oxidising  M⁺ as   $M^{+}\rightarrow M^{3+}+2\overline{e}$, E⁰  =-0.25 V.

 Under standard conditions , the number of moles  of M⁺ oxidised  when one mole of X is converted to Y  is [F=96500 C mol^-]

Answer:

Explanation:

Energy obtained as one mole X is converted into Y is 193 KJ

Energy consumed in converting  one mole of M⁺ to M³⁺

      $= -nE^{0}F=2\times96500\times0.25 J=\frac{96500}{2}J$

   $\Rightarrow193\times10^{3}=n\left(\frac{96500}{2}\right)$

  $\Rightarrow$               n=4 mol

 Not considering  the electronic  spin, the degeneracy  of the second excited state (n=3)  of H-atom is 9,  while the degeneracy of the second excited state of H^-   is

Answer:

Explanation:

 In one electron  (hydrogenic) system, all orbitals of a shell remains degenerate, hence in second excited state, the degeneracy of H-atom is nine

 In case of many electrons system, different orbitals of a shell are non-degenerate, Hence,

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015