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P Block Elements question

2013 · Shift 2 · Q19
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P Block Elements question

2013 · Shift 2 · Q19

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1

The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists :

List I List II
P. PbO2+H2SO4→?PbSO4+O2+Other productsPb{O_2} + {H_2}S{O_4}\xrightarrow{?} PbS{O_4} + {O_2} + Other\,productsPbO2​+H2​SO4​?​PbSO4​+O2​+Otherproducts 1. NO
Q. Na2S2O3+H2O→?NaHSO4+Other productsN{a_2}{S_2}{O_3} + {H_2}O\xrightarrow{?} NaHS{O_4} + Other\,productsNa2​S2​O3​+H2​O?​NaHSO4​+Otherproducts 2. I2{I_2}I2​
R. N2H4→?N2+Other products{N_2}{H_4}\xrightarrow{?} {N_2} + Other\,productsN2​H4​?​N2​+Otherproducts 3. Warm
S. XeF2→?Xe+Other productsXe{F_2}\xrightarrow{?} Xe + Other\,productsXeF2​?​Xe+Otherproducts 4. Cl2C{l_2}Cl2​

  1. A
    P-4, Q-2, R-3, S-1
  2. B
    P-3, Q-2, R-1, S-4
  3. C
    P-1, Q-4, R-2, S-3
  4. D
    P-3, Q-4, R-2, S-1
View written solutionFree

Correct answer: D

  1. Match reaction P

    PbO2+H2SO4→?PbSO4+O2+other productsPbO_2 + H_2SO_4 \xrightarrow{?} PbSO_4 + O_2 + \text{other products}PbO2​+H2​SO4​?​PbSO4​+O2​+other products

    Lead dioxide is a strong oxidizing agent and reacts with concentrated sulfuric acid on warming:

    2PbO2+2H2SO4→warm2PbSO4+2H2O+O22PbO_2 + 2H_2SO_4 \xrightarrow{\text{warm}} 2PbSO_4 + 2H_2O + O_22PbO2​+2H2​SO4​warm​2PbSO4​+2H2​O+O2​

    So, P→3  (Warm)P \to 3\; (\text{Warm})P→3(Warm)

  2. Match reaction Q

    Na2S2O3+H2O→?NaHSO4+other productsNa_2S_2O_3 + H_2O \xrightarrow{?} NaHSO_4 + \text{other products}Na2​S2​O3​+H2​O?​NaHSO4​+other products

    Thiosulfate is oxidized by chlorine in water. A known reaction is:

    Na2S2O3+4Cl2+5H2O→2NaHSO4+8HClNa_2S_2O_3 + 4Cl_2 + 5H_2O \to 2NaHSO_4 + 8HClNa2​S2​O3​+4Cl2​+5H2​O→2NaHSO4​+8HCl

    Hence the missing reagent is:

    Q→4  (Cl2)Q \to 4\; (Cl_2)Q→4(Cl2​)

  3. Match reaction R

    N2H4→?N2+other productsN_2H_4 \xrightarrow{?} N_2 + \text{other products}N2​H4​?​N2​+other products

    Hydrazine is oxidized by iodine:

    N2H4+2I2→N2+4HIN_2H_4 + 2I_2 \to N_2 + 4HIN2​H4​+2I2​→N2​+4HI

    Therefore, R→2  (I2)R \to 2\; (I_2)R→2(I2​)

  4. Match reaction S

    XeF2→?Xe+other productsXeF_2 \xrightarrow{?} Xe + \text{other products}XeF2​?​Xe+other products

    Xenon difluoride is reduced by nitric oxide:

    2XeF2+4NO→2Xe+4NOF2XeF_2 + 4NO \to 2Xe + 4NOF2XeF2​+4NO→2Xe+4NOF

    So, S→1  (NO)S \to 1\; (NO)S→1(NO)

  5. Final matching

    P−3,  Q−4,  R−2,  S−1P-3,\; Q-4,\; R-2,\; S-1P−3,Q−4,R−2,S−1

    This corresponds to Option D.

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