JEE Advance - Chemistry (2013 - Paper 2 Offline - No. 19)
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. | $$Pb{O_2} + {H_2}S{O_4}\buildrel ? \over \longrightarrow PbS{O_4} + {O_2} + Other\,products$$ |
1. | NO |
| Q. | $$N{a_2}{S_2}{O_3} + {H_2}O\buildrel ? \over \longrightarrow NaHS{O_4} + Other\,products$$ |
2. | $${I_2}$$ |
| R. | $${N_2}{H_4}\buildrel ? \over \longrightarrow {N_2} + Other\,products$$ |
3. | Warm |
| S. | $$Xe{F_2}\buildrel ? \over \longrightarrow Xe + Other\,products$$ |
4. | $$C{l_2}$$ |
P-4, Q-2, R-3, S-1
P-3, Q-2, R-1, S-4
P-1, Q-4, R-2, S-3
P-3, Q-4, R-2, S-1
Explanation
The reactions involved are as follows:
(P) $$Pb{O_2} + {H_2}S{O_4}\buildrel {Warm} \over \longrightarrow PbS{O_4} + {H_2}O + {1 \over 2}{O_2}$$
(Q) $$2N{a_2}{S_2}{O_3} + 2{H_2}O + C{l_2} \to 2NaCl + 2NaHS{O_4} + 2S$$
(R) $${N_2}{H_4} + 2{I_2} \to {N_2} + 4HI$$
(S) $$Xe{F_2} + 2NO \to Xe + 2NOF$$
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