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Chemistry Test 154
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Chemistry Test 154

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  • Question 1/10
    4 / -1

    Select the incorrect statement among following:

    Solutions

    (A) Because of E° = +1.74 V for Ce+4 to Ce+3    

    (B) Due to stability of half filled d–orbitals, Mn2+ is more stable than Mn3+ and thus its reduction is more spontaneous. Similarly Fe3+ is more stable than Fe2+ and thus its reduction is less spontaneous.

    (D) Co2+ is stable in aqueous solution because it get surrounded and weakly bonded to water molecules. In presence of strong ligands and air it gets oxidised to Co(III) as strong ligands get co–ordinated more strongly with Co(III). The electronic configuration of Co(II) and Co(III).

     

  • Question 2/10
    4 / -1

    Li +   → (P)

    (Q) + 2H2O → Ca(OH)2 + 2H2

    2NaBH4 + I2 → (R) + 2NaI + H2

    If equal masses of (P), (Q) and (R) treated with water, which will give more amount of H2 ?

    Solutions

     

  • Question 3/10
    4 / -1

    Directions For Questions

    The element Sulphur exists in many allotropes and each allotrope often exists in polymorphs having different crystal parameters i.e., axial lengths and angles (a, b, c, α, β, γ). Then consider the reaction for different crystal of Sulphur  :

    (i) S(a ≠ b ≠ c, α = β = γ = 90°)  +  → SO3(g),    ΔH1
    (ii) S(a ≠ b ≠ c, α = γ, β ≠ 90°) +   → SO3(g),  ΔH2

    (iii) S(a ≠ b ≠ c, α = β = γ = 90°) + O3(g)  → SO3(g),    ΔH3

    (iv) S(a ≠ b ≠ c, α = γ, β ≠ 90°) + O3(g)  → SO3(g),   ΔH4

    ...view full instructions

     

    Select incorrect option (only magnitude comparision).

    Solutions

    Rhombic sulphur is more thermodynamically stable than monoclinic sulphur and also oxygen is more stable than ozone.

        ∴  ΔH1 < ΔH2 > ΔH3 < ΔH4

     

  • Question 4/10
    4 / -1

    Which of the following is INCORRECT order for boiling points for unknowns (P), (Q), (R), (S), (T), (U) & (V). In the reaction sequence:

    Solutions

    (P) = H2O, (Q) = CCl4, (R) = SiCl4, (S) = H2, 

    (T) = NH3, (U) = AsH3, (V) = HF, 

    (X) = CH4 & (Y) = Cl2

     

  • Question 5/10
    4 / -1

    Directions For Questions

    CF3I + NaOH → A + B

    ...view full instructions

    Find the correct statement.

    Solutions

    A = CHF3, B = NaOI, C = CHI3, 

    D = HCOONa, P = CH3OH, Q = NaI

     

  • Question 6/10
    4 / -1

    Moles of  CO2 evolved during given reaction per mol?

  • Question 7/10
    4 / -1

    Which of the following compound is most harmful component of phtochemical smog?

    Solutions

    PAN– peroxy acetyl nitrate is most harmful.

     

  • Question 8/10
    4 / -1

    Some aqueous solutions are given below. Find the solution whose pH changes maximum by adding 10  millimoles amount of pure H2SO4. 

    Solutions

    (A) Initially, mms of OH-  =  100 ,
          mms of H+ = 10, [OH-] = 90/200
          ⇒ pOH=0.346
          Finally, mms of OH- = 100,
          ms of H+ =  30,  [OH-] = 70/200
          ⇒ pOH=0.456
          So, change in pH = 0.111 (small change)

    (B) After initial reaction,
          mms of HS- =  20, mms of H2S = 20  
          (its a buffer of HS–/H2S)
          So, change in pH will be very small.

    (C) Initially, mms of OH- = 10,
          mms of H+ =  0,  [OH–] = 10/200
          ⇒ pH=12.7
          Finally, mms of OH- = 10,
          mms of H+ =  20, [H+] = 10/200
          ⇒  pH=1.3
          So, change in pH = 11.4 (very large)

    (D) Initially, mms of AcOH =  30,
          mms of AcO- =  20,
          mms of H2S = 20 (its a buffer)
          So, change in pH will be very small.

     

  • Question 9/10
    4 / -1

    For a 3s–orbital. Find the maximum distance for radial node.

    Solutions

    Radial node occurs where probability of finding e– is zero.

     

  • Question 10/10
    4 / -1

    Packing fraction of NaCl unit cell [assuming ideal] if ions along one of its body diagonal are absent (Cl– forms FCC, r+ & r– are radius of cation and anion)

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