Test Index

JEE Advanced 2013 Paper 2

Paragraph for Next 2 Questions

The reactions of Cl

_2

gas with cold - dilute and hot - concentrated NaOH in water give sodium salts to two(different) oxoacids of chlorine, P and Q , respectively. The Cl

_2

gas reacts with SO

_2

gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus T

Section: Chemistry

Question : 36 of 60

Marks: +1, -0

R,S and T, respectively, are

[JEE Adv 2013 P2]

SO $_{2}$ Cl $_{2}$ , PCl $_{5}$ and H $_{3}$ PO $_{4}$
SO $_{2}$ Cl $_{2}$ , PCl $_{3}$ and H $_{3}$ PO $_{3}$
SOCl $_{2}$ , PCl $_{3}$ and H $_{3}$ PO $_{2}$
SOCl $_{2}$ , PCl $_{5}$ and H $_{3}$ PO $_{4}$

Validate

Solution:

(35 & 36)
Cl

_{2}

+ cold dil. NaOH → NaOCl + NaCl
Cl

_{2}

+ hot conc. NaOH → NaClO

_{3}

+ NaCl
NaOCl is salt of hypochlorus acid = P
NaOCl

_{3}

is salt of chloric acid = Q
Cl

_{2}

+ SO

_{2}

\xrightarrow[]{\text{Charcoal}}

_{2}

_{2}

(R)
SO

_{2}

_{2}

+ P

_{4}

→ PCl

_{5}

(S) + SO

_{2}

PCl

_{5}

+ H

_{2}

O → H

_{3}

_{4}

(T) + HCl

Go to Question:

Prev Question

Next Question