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The p-Block Elements Test - 16

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The p-Block Elements Test - 16
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

    Ion \(ClO_4^-\) \(IO_4^-\) \(BrO_4^-\)
    Reduction potential \(E^\ominus\)/V \(E^\ominus\) = 1.19V \(E^\ominus\) = 1.65V \(E^\ominus\) =1 .74V
    Solution

    \(BrO_4 ^- (1.74)> IO_4 ^- (1.65)> ClO_4^-(1.19)\)

    Higher is the reduction potential greater will be the oxidizing power.

  • Question 2
    1 / -0

    Which of the following is isoelectronic pair?

    Solution

    Isoelectronic species means no. of electron is same.

    \(BrO_2^-\) (no. of electron) = 35 + 16 + 1 = 52

    \(BrF_2^+\) (no. of electron) = 35 + 17 = 52

  • Question 3
    1 / -0

    Which of the following statements are correct for \(SO_2\) gas?

    (i) It acts as bleaching agent in moist conditions.

    (ii) It’s molecule has linear geometry.

    (iii) It’s dilute solution is used as disinfectant.

    (iv) It can be prepared by the reaction of dilute \(H_2SO_4\) with metal sulphide.

    Solution

    \(SO_2\) is used in bleaching of wool and silk and as an anti-chlor, disinfectant and preservative. By reaction of dil. \(H_2SO_4\) with metal sulphide results in the formation of \(H_2S\) gas not \(SO_2\).

    \(ZnS+H_2SO_4\longrightarrow ZnSO_4+H_2S\)

  • Question 4
    1 / -0

    Which of the following statements are correct?

    (i) All the three N—O bond lengths in \(HNO_3\) are equal.

    (ii) All P—Cl bond lengths in \(PCl_5\) molecule in gaseous state are equal.

    (iii) \(P_4\) molecule in white phohsphorus have angular strain therefore white phosphorus is very reactive.

    (iv) \(PCl_5\) is ionic in solid state in which cation is tetrahedral and anion is octahedral.

    Solution

    (i) All the three N-O bond length in \(HNO_3\) are not equal.

    (ii) In gaseous phase all P-Cl bond lengths in \(PCl_5\) molecule are not equal.

    (iii) White phosphorus is more reactive than the other solid phases under normal conditions because of angular strain in the \(P_4\) molecule.

    (iv) Solid state it exists as an ionic solid, \([PCl_4]^+[PCl_6]^-\) in which the cation, \([PCl_4]^+\) is tetrahedral and the anion, \([PCl_6]^-\) octahedral.

  • Question 5
    1 / -0

    Hot conc. \(H_2SO_4\) acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. \(H_2SO_4\) into two gaseous products?

    Solution

    Hot concentrated sulphuric acid is a moderately strong oxidising agent. In this respect, it is intermediate between phosphoric and nitric acids. Both metals and non-metals are oxidised by concentrated sulphuric acid, which is reduced to \(SO_2\). C is oxidised into two gaseous products.

    \(C + 2H_2SO_4(conc.) \rightarrow\) \(CO_2 + 2SO_2 + 2H_2O\)

  • Question 6
    1 / -0

    Which of the following orders are correct as per the properties mentioned against each?

    (i) \(As_2O_3 < SiO_2 < P_2O_3 < SO_2\)  Acid strength.

    (ii) \(AsH_3 < PH_3 < NH_3\)  Enthalpy of vapourisation.

    (iii) S < O < Cl < F  More negative electron gain enthalpy.

    (iv) \(H_2O > H_2S > H_2Se > H_2 Te\)  Thermal stability.

    Solution

    (i) \(As_2O_3 < SiO_2 < P_2O_3 < SO_2\)

    Order of acid strength

    (ii) Correct order of enthalpy of vaporization is \( NH_3>AsH_3 > PH_3 \)

    (iii) Correct order of more negative electron gain enthalpy S < O < F < Cl

    (iv) Order of thermal stability  \(H_2O > H_2Se > H_2Te\)

  • Question 7
    1 / -0

    In which of the following reactions conc. \(H_2SO_4\) is used as an oxidising reagent?

    (i) \(CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF \)

    (ii) \(2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O \)

    (iii) \(Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O \)

    (iv) \(NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\)

    Solution

    Among the above four (ii) and (iii) represent the oxidizing behavior of \(H_2SO_4\). In (ii) reaction it oxidizes HI and itself reduces to  \(SO_2 \)oxidation state of central atom Sulphur decreases from +6 to +4. In (iii) it oxidizes copper and itself get reduced to \(SO_2 \).

  • Question 8
    1 / -0

    Assertion : NaCl reacts with concentrated \(H_2 SO_4\) to give colourless fumes with pungent smell. But on adding \(MnO_2\) the fumes become greenish yellow.

    Reason : \(MnO_2\) oxidises HCl to chlorine gas which is greenish yellow.

    Solution

    NaCl reacts with concentrated \(H_2SO_4\) to give colourless fumes with pungent smell. But on adding \(MnO_2\) the fumes become greenish yellow. \(MnO_2\) oxidises HCl to chlorine gas which is greenish yellow.

    \(NaCl+H_2SO_4 \rightarrow NaHSO_4 + HCl\) (fumes of HCl is colourless)

    By heating manganese dioxide with concentrated hydrochloric acid.

    \(MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O\)

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