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JEE Advanced
Exam
JEE Advanced
Subject
Chemistry
Chapter
Practical Chemistry

JEE Advanced · Chemistry · Practical Chemistry

X(~s) dil HCl Y [ Na _3(Fe(CNO)_3(NO)]. Purple solution Gas Y has...

X( s)→ dil HClY↑→[ Na 3(Fe(CNO)3(NO)]X(\mathrm{~s}) \xrightarrow{\text { dil } \mathrm{HCl}} \mathrm{Y} \uparrow \xrightarrow{\left[\mathrm{\text{ Na }}_3\left(\mathrm{Fe}(\mathrm{CNO})_3(\mathrm{NO})\right]\right.} Purple solution Gas YY has been allowed to react with following Biack species in neutralacidic medium (A) FeCl3\mathrm{FeCl}_3 (B) CuSO4\mathrm{CuSO}_4 (C) BaCl2\mathrm{BaCl}_2 (D) SO2\mathrm{SO}_2 (E) Cr2O72−\mathrm{Cr}_2 \mathrm{O}_7^{2-} (F) CH3COONa−\mathrm{CH}_3 \mathrm{COONa}^{-} (G) Hg2+\mathrm{Hg}^{2+} Then calculate value of ( P+Q−R\mathrm{P}+\mathrm{Q}-\mathrm{R} ) PP : Number of species which undergoes redox reaction with gas YY Q : Number of species which give black ppt with gas YY P : Number of species with which gas Y produce no observable changeX(~s) dil HCl Y [ Na _3(Fe(CNO)_3(NO)]. Purple solution Gas Y has been allowed to react with following Biack species in

Correct answer

3

Step-by-step solution

X(~s) dil HCl Y [ Na _3(Fe(CNO)_3(NO)]. Purple solution Gas Y has been allowed to react with following Biack species in
(a) 2FeCl3+2H2 S↑→ Nevicil  Acitic 2Fe2++S↓+4H++6Cl−2 \mathrm{FeCl}_3+2 \mathrm{H}_2 \mathrm{~S} \uparrow \xrightarrow[\text { Nevicil }]{\text { Acitic }} 2 \mathrm{Fe}^{2+}+\mathrm{S} \downarrow+4 \mathrm{H}^{+}+6 \mathrm{Cl}^{-} (b) CuSO4+H2 S↑→ Nentral  Acidicl CuS1 Black +2H++SO42−\mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{~S} \uparrow \xrightarrow[\text { Nentral }]{\text { Acidicl }} \underset{\text { Black }}{\mathrm{CuS}_1}+2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-} (c) BaCl2+H2 S↑→ Neatral  Acidc Ba2+ No observable charge +2Cl−+2H++S2−\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{~S} \uparrow \xrightarrow[\text { Neatral }]{\text { Acidc }} \underset{\text { No observable charge }}{\mathrm{Ba}^{2+}}+2 \mathrm{Cl}^{-}+2 \mathrm{H}^{+}+\mathrm{S}^{2-} (d) SO2↑+2H2 S↑⟶2 S↓+H2O\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{~S} \uparrow \longrightarrow 2 \mathrm{~S} \downarrow+\mathrm{H}_2 \mathrm{O} or SO2↑+H2O⟶H2SO3→H2 s S↓+H2O\mathrm{SO}_2 \uparrow+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2 \mathrm{SO}_3 \xrightarrow{\mathrm{H}_2 \mathrm{~s}} \mathrm{~S} \downarrow+\mathrm{H}_2 \mathrm{O} (e) Cr2O72−+8H++3H2 S↑→ Nentril  Andol 2Cr3++S↓+7H2O\mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{+}+3 \mathrm{H}_2 \mathrm{~S} \uparrow \xrightarrow[\text { Nentril }]{\text { Andol }} 2 \mathrm{Cr}^{3+}+\mathrm{S} \downarrow+7 \mathrm{H}_2 \mathrm{O} (g) Hg2++H2 S→ Neveltal  Acede HgS Back ↓+2H+\mathrm{Hg}^{2+}+\mathrm{H}_2 \mathrm{~S} \xrightarrow[\text { Neveltal }]{\text { Acede }} \underset{\text { Back }}{\mathrm{HgS}} \downarrow+2 \mathrm{H}^{+} ∴P=3;Q=2;R=2\therefore P=3 ; \quad Q=2 ; \quad R=2 Hence value of (P+Q−R)=3+2−2=3(P+Q-R)=3+2-2=3
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