Match the catalysis(Column - I) with products (Column-II) Column-I (Catalyst): (a) $\mathrm{V}_{2} \mathrm{O}_{5}$ (b) $\mathrm{TiCl}_{4} / \mathrm{Al}(\mathrm{Me})_{3}$ (c) $\mathrm{PdCl}_{2}$ (d) Iron oxide Column-II (Product): (i) Polyethylene (ii) Ethanal (iii) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (iv) $\mathrm{NH}_{3}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Surface Chemistry

Match the catalysis(Column - I) with products (Column-II)

Column-I (Catalyst):
(a)

V_{2} O_{5}

(b)

TiCl_{4} / Al (Me)_{3}

(c)

PdCl_{2}

(d) Iron oxide

Column-II (Product):
(i) Polyethylene
(ii) Ethanal
(iii)

H_{2} SO_{4}

(iv)

NH_{3}

Answer: A → III, B → I, C → II, D → IV

Each catalyst goes with the process it drives.

V X_{2} O X_{5}

is the catalyst of the Contact process that makes sulphuric acid,

H X_{2} SO X_{4}

TiCl X_{4} / Al (Me) X_{3}

is the Ziegler-Natta catalyst that polymerises ethene to polyethylene.

PdCl X_{2}

runs the Wacker oxidation of ethene to ethanal. And iron oxide is the catalyst of the Haber process that fixes nitrogen as ammonia,

NH X_{3}

. So

V X_{2} O X_{5}

pairs with

H X_{2} SO X_{4}

TiCl X_{4} / Al (Me) X_{3}

with polyethylene,

PdCl X_{2}

with ethanal, and iron oxide with ammonia.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Match the columns