p-Block Elements — Page 20 | JEE Chemistry

Q191

Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y' . Compounds 'X' and 'Y' and the oxidation state of Xe are respectively :

A XeO2(+4) and XeO3(+6)

B XeOF4(+6) and XeO3(+6)

C XeO2F2(+6) and XeO2(+4)

D XeOF4(+6) and XeO2F2(+6)

Correct Answer

Option D

Solution

XeF6 on hydrolysis with water can produces 3 compounds XeOF4, XeO2F2 and XeO3.

Here XeOF4 and XeO2F2 are produced on partial hydrolysis and XeO3 is produced on complete hydrolysis.

XeF6 + H2O

\overset{{Partial}}\longrightarrow

XeOF4 + 2HF XeF6 + 2H2O

\overset{{Partial}}\longrightarrow

XeO2F2 + 4HF XeF6 + 3H2O

\overset{{Complete}}\longrightarrow

XeO3 + 6HF So, the compound X and Y are XeOF4(+6) or XeO2F2(+6).

Q192

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation state. Reason R : Tl metal has fourteen f electrons in its electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below :

A A is correct but R is not correct

B A is not correct but R is correct

C Both A and R are correct but R is NOT the correct explanation of A

D Both A and R are correct and R is the correct explanation of A

Correct Answer

Option C

Solution

A : Due to inert pair effect, Tl is more stable in +1 oxidation state Hence TlI3 and CSI3 are isomorphous R : Electronic configuration of Tl (81) = [Xe] 4f14 5d10 6s2 6p1 Both A and R are correct but R is not the correct explanation of A.

Q193

Match with . )_4}Si$$

List - I		List - II
(B)	$(C{H_3})Si{(OH)_3}$	(I)	Chain Silicone
(C)	${(C{H_3})_2}Si{(OH)_2}$	(II)	Dimeric Silicone
(D)	${(C{H_3})_3}Si(OH)$	(III)	Silane
()		(IV)	2D - Silicone

A (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

B (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

C (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

D (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Correct Answer

Option D

Solution

(CH3)4Si is a silane.

(CH3)Si(OH)3 polymerise to form 2D silicone.

(CH3)2Si(OH)2 polymerise to form chain silicone.

(CH3)3Si(OH) form dimer (CH3)3Si-O-Si(CH3)3

Q194

The pair in which phosphorous atoms have a formal oxidation state of +3 is :

A Orthophosphorous and pyrophosphorous acids

B Pyrophosphorous and hypophosphoric acids

C Orthophosphorus and hypophosphoric acids

D Pyrophosphorous and pyrophosphoric acids

Correct Answer

Option A

Solution

Phosphorous acids contain

P

in

+3

oxidation state. .tg .tg Acid Formula Oxidation state of Phosphorous Pyrophosphorous acid H4P2O5 +3 Pyrophophoric acid H4P2O7 +5 Orthophosphorous acid H3PO3 +3 Hypophosphoric acid H4P2O6 +4

Q195

Match the items in Column I with its main use listed in Column II : .tg .tg Column I Column II

List - I		List - II
(A)	Silica gel	(i)	Transistor
(B)	Silicon	(ii)	Ion-exchanger
(C)	Silicone	(iii)	Drying agent
(D)	Silicate	(iv)	Sealant

A (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

B (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)

C (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

D (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

Correct Answer

Option A

Solution

(A) Silica gel (SiO2) is a good dehydrating agent.

It is used good dehydrating agent.

It is used to absorb moisture.

SiO2 + 2H2O $\to$ SiO2.2H2O (B) Silicon is used in transistor as it is a semiconductor. (c) Silicon is a good resistant of heat, water.

So it used as sealant to seal some element. (d) Silicate including zeolites used in ion-exchange beds in domestic and commercial water purification, softening and other applications.

Q196

Match with : .tg .tg (compound) (effect/affected species)

List - I		List - II
(a)	Carbon monoxide	(i)	Carcinogenic
(b)	Sulphur dioxide	(ii)	Metabolized by pyrus plants
(c)	Polychlorinated biphenyls	(iii)	haemoglobin
(d)	Oxides of Nitrogen	(iv)	Stiffness of flower buds

A (a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)

B (a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)

C (a) - (i), (b) - (ii), (c) - (iii), (d) - (iv)

D (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

Correct Answer

Option A

Solution

A.

If haemoglobin binds CO, it prevents the binding of O2 to haemoglobin due to the competition for same binding sites.

B.

Higher conc.

SO2 in air can cause stiffness of flower buds.

C.

Polychlorinated biphenyls are carcinogenic (cancer causing) in humans as well as animals.

D.

Certain plants, e.g. pinus, Juniparus, quercus, pyrus and vitis can metabolise nitrogen oxides.

Hence, the correct match is (a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)

Q197

Match with . .tg .tg (Metal) (Application)

List - I		List - II
(A)	Cs	(I)	High temperature thermometer
(B)	Ga	(II)	Water repellent sprays
(C)	B	(III)	Photoelectric cells
(D)	Si	(IV)	Bullet proof vest Choose the most appropriate answer from the options given below :

A (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

B (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

C (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

D (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Correct Answer

Option A

Solution

Caesium is used in devising photoelectric cells.

Boron fibres are used in making bullet–proof vest.

Silicones being surrounded by non–polar alkyl groups are water repelling in nature.

Gallium is less toxic and has a very high boiling point, so it is used in high temperature thermometers.

Q198

The amorphous form of silica is :

A kieselguhr

B quartz

C tridymite

D cristobalite

Correct Answer

Option A

Solution

Quartz, tridymite and cristobalite are crystalline forms of silica. Kieselguhr is an amorphous form of silica.

Q199

Match List I with List II. .tg .tg List-I (Anion) List-II (gas evolved on reaction with dil.H

_2

SO

_2

) (A) CO

{_3^{2 - }}

I. Colourless gas which turns lead acetate paper black. (B) S

^{2 - }

II. Colourless gas which turns acidified potassium dichromate solution green. (C) SO

{_3^{2 - }}

III. Brown fumes which turns acidified KI solution containing starch blue. (D) NO

{_2^ - }

IV. Colourless gas evolved with brisk effervescence, which turns lime water milky. Choose the correct answer from the options given below :

A A-III, B-I, C-II, D-IV

B A-II, B-I, C-IV, D-III

C A-IV, B-I, C-III, D-II

D A-IV, B-I, C-II, D-III

Correct Answer

Option D

Solution

$\mathrm{CO}_{3}^{2-}$ : On action of dil sulphuric acid, $\mathrm{CO}_{2}$ gas is released which turns lime water milky.

$\mathrm{S}^{2-}$ : On action of dil sulphuric acid, $\mathrm{H}_{2} \mathrm{~S}$ gas is released which turns lead acetate paper black.

$\mathrm{SO}_{3}^{2-}$ : On action of dil $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}$ gas is evolved which turns acidified potassium dichromate solution green.

$\mathrm{NO}_{2}^{-}$ : On action of dil $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{NO}_{2}$ gas is evolved which turns $\mathrm{KI}$ solution containg starch blue.

Q200

C60 an allotrope of carbon contains :

A 12 hexagons and 20 pentagons.

B 20 hexagons and 12 pentagons.

C 16 hexagons and 16 pentagons

D 18 hexagons and 14 pentagons.

Correct Answer

Option B

Solution

Fullerene (C60) contains 20 hexagons and 12 pentagons.