Practice Test 11
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The compound YBa2Cu3O7 which shows super conductivity has copper in oxidation state ………. Assume that the rare earth element Yttrium is in its usual +3 oxidation state

  • Solution

    YBa2Cu3O7
    3 + 2 × 2 + 3x– (2 ×7) = 0
    3 + 4 +3x– 14 = 0
    3x= 7
    x = \(\frac{7}{3}\)

  • Solution

  • Solution

    Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.

When Cl2 gas reacts with hot and concentrated sodium hydroxide solution,the oxidation number of chlorine changes from :

  • Solution

  • Solution

    CrO2Cl2, \(MnO_{4}^{-}\)
    ; O.N. of Cr and Mn are +6 and +7 respectively.

In the equation
\(CrO_{4}^{2-}\) + \(SO_{3}^{2-}\) \(Cr(OH)_{4}^{-}\) + \(SO_{4}^{2-}\) the oxidation number of Cr changes from

  • Solution

    The oxidation number of Cr in \(CrO_{4}^{2-}\) is x– 8 = – 2; x = + 6
    The oxidation number of Cr is \(Cr(OH)_{4}^{-}\) is x– 4 = – 1 or x= + 3
    Thus the oxidation number of Cr in the given reaction changes from +6 to + 3

When K2Cr2O7 is converted into K2CrO4, the change in oxidation number of chromium is :

  • Solution

    O.N. of Cr in K2Cr2O7: 2 + 2x– 14 = 0
    2x= 12, x = 6
    O.N. of Cr in K2CrO4: 2 + x– 8 = 0
    x= 6
    ∴ Change in O.N. of Cr = 6 – 6 =0

Point out the correct statement of the following about Na2S4O6.

  • Solution

Fe shows an oxidation state of +1 in

  • Solution

    (a) x + 5 × 0 = 0 ⇒ x = 0
    (b) (x + 5 × 0 + 1) - 2 = 0 ⇒ x = +1
    (c) 4 × (+3) + [3x + 18(-1)] = 0 ⇒ x = +2
    (d) x + 4(-1) = -1 ⇒ x = +3

Oxidation number of ‘N’ in N3H (hydrazoic acid) is:

  • Solution

    Let oxidation number of N in N3H =x
    3x+ 1= 0 or 3x= – 1
    ∴ x= –\(\frac{1}{3}\)

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