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D and F - Block Elements Test - 7

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D and F - Block Elements Test - 7
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  • Question 1
    4 / -1

     

    Which is most basic oxide of chromium?

     

    Solution

     

    Oxide having lower oxidation number is basic and as oxidation number increase, acidic nature increases.

     

  • Question 2
    4 / -1

    The basic character of the transition metal monoxides follows the order

    (Atomic Number, Ti = 22, V = 23, Cr = 24, Fe = 26)

    Solution

    TiO > VO > CrO > FeO with the decrease in size of metal atom from Ti to Fe, the basic character of their monoxides decreases.

  • Question 3
    4 / -1

    MnO4-  + S2- + H+ → Mn2+ +S + H2O When this equation is balanced, the total number of coefficients on the left hand side are

    Solution

  • Question 4
    4 / -1

    when this equation is balanced, the num ber of OF- ions added

    Solution

  • Question 5
    4 / -1

    One mole of each KMnO4 and K2Cr2O7 can oxidise ........... moles of ferrous ion.

    Solution

    One equivalent of KMnO4 = One equivalent of K2Cr2O7 = One equivalent of Fe2+ ion or 1/5 moles of KMnO4 = 1/6 moles of K2Cr2O7 = One mole of Fe2+ ion.
    Hence, one mole of KMnO4 reacts with 5 moles of Fe2+ ion and one mole of K2Cr2O7 reacts with 6 moles of Fe2+ ion.

  • Question 6
    4 / -1

    is of intense pink colour, though Mn is in (+7) oxidation state. It is due to

    Solution

    This is called ligand to metal charge transfer phenomenon.

  • Question 7
    4 / -1

    Which of the following is incorrect statement?

    Solution

    The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

  • Question 8
    4 / -1

    Pyrolusite in MnO2 is used to prepare KMnO4. Steps are

    Here, I and II are

    Solution


  • Question 9
    4 / -1

    Which is the correct statem ent about  structure?

    Solution

    The correct answer is Option A.
    In a dichromate dianion structure, the negative charge can be on any O attached to Cr, therefore all Cr - O are equivalent. Also have Cr - O - Cr bond.

  • Question 10
    4 / -1

    Out of Cr (VI) as  , which is better oxidising agent?

    Solution

     + 4H2O + 3e- → Cr(OH)3 + 5OH-, E° = - 0.13 V 
    + 14H+ + 6e- → 2Cr3+ + 7H2O , E° = 1.33 V

  • Question 11
    4 / -1

     

    The number of moles of acidified KMnO4 required to convert one mole of sulphite ion into sulphate ion is-

     

    Solution

     

     

  • Question 12
    4 / -1

     

    CrO3 dissolves in aqueous NaOH to give-

     

    Solution


     

  • Question 13
    4 / -1

     

    Solution of MnO4- is purple-coloured due to-

     

    Solution

    MnO4– has purple colour due to change transfer

     

  • Question 14
    4 / -1

    Which of the following statements is not true?

    Solution

    Potassium dichromate is used as a primary standard in volumetric analysis.

  • Question 15
    4 / -1

     

    Transition elements having more tendency to form complex than representative elements (s and p-block elements) due to-

     

    Solution

     

    Complex pormation α (Zeff = Z –σ) tendency

  • Question 16
    4 / -1

    Comprehension Type

    Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

    Passage

    Oxides are generally formed by the reaction of metals with oxygen at high temperatures. All the metals except scandium form MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond 7th group, higher oxides coinciding with the group number do not form. As the oxidation number of metal increases covalent character and acidity increases. Mn2O7, CrO3 are acidic and their lower oxides are basic and amphoteric. Stability of higher oxides increases down the group.

    Q. 

    Which of the following reacts with both acids and bases?

    Solution

    ZnO and V2O5 are am photeric oxides.

  • Question 17
    4 / -1

    Oxides are generally formed by the reaction of metals with oxygen at high temperatures. All the metals except scandium form MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond 7th group, higher oxides coinciding with the group number do not form. As the oxidation number of metal increases covalent character and acidity increases. Mn2O7, CrO3 are acidic and their lower oxides are basic and amphoteric. Stability of higher oxides increases down the group.

    Q. 

    Which of the following undergo disproportionation?

    Solution

  • Question 18
    4 / -1

    Matching List Type

    Direction (Q. Nos. 18 and 19) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

    Q.

    Match the Column I with Column II and mark the correct option from the codes given below.

    Solution

    (i) → (q,t) (ii) → (p.r) (iii) → (q.r.t) (iv) → (q,s)

  • Question 19
    4 / -1

    Match the Column I with Column II and mark the correct option from the codes given below.

    Solution

    (i) → (q) (ii) → (s) (iii) → (r) (iv) → (p)

    (i) In acidic medium,

    Hence, = M/5
    (ii) In strongly alkaline medium

    Hence, = M/1
    (iii) In neutral or weakly basic medium

    Hence, = M/3

    Hence, equivalent weight is M/6.

  • Question 20
    4 / -1

     

    During estimation of oxalic acid Vs KMnO4, self indicator is-

     

    Solution

     

    KMnO4 acts as self indicator.

  • Question 21
    4 / -1

     

    50 cc of 0.04 M K2Cr2O7 in acidic medium oxidises a sample of H2S gas to sulphur. Volume of 0.03 M KMnO4 required to oxidise the same amount of H2S gas to sulphur in acidic medium is 10 x x. Here, the value of x is

     

    Solution

     

    Milliequivalents of K2Cr2O7 reacted with H2S (N1V1) = miliiequivalents of KMnO4 reacted with H2S(N2V2), i.e. milliequivalent of KMnO4 = milliequivalent of H2S Therefore, 50 x 0.04 x 6 = V2 x 0.03 x 5
    or V2 = 80 mL

     

  • Question 22
    4 / -1

     

    The number of sigma bonds in ion are

     

    Solution

     

     

  • Question 23
    4 / -1

     

    If molarity of K2Cr2O7 solution is 0.5 M, its normality value is

     

    Solution

     

    Normality = molarity x number of electrons gained per mole of dichromate in a redox reaction. Hence, Normality = Molarity x 6

  • Question 24
    4 / -1

    Statement Type

    Direction (Q. Nos. 24 and 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Q. 

    Statement I ; When a solution of potassium chromate is treated with an excess of dilute nitric acid chromate undergoes oxidation.

    Statement II : Dichromate ions are produced in above process.

    Solution

  • Question 25
    4 / -1

    Statement I : The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution has green colour.

    Statement II : The product has Cr3+ ions.

    Solution

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