p-Block Elements — Page 7 | JEE Chemistry

Q61

In KO2, the nature of oxygen species and the oxydation state of oxygen atom are, respectively :

A Oxide and

-

2

B Superoxide and

-

1/2

C Peroxide and

-

1/2

D Superoxide and

-

1

Correct Answer

Option B

Solution

KO2 is called potassium superoxide. Here O

_2^ -

is the superoxide ion. Oxidation state of K is +1 and let oxidation state of oxygent atom is = x

\therefore\,\,\,

1 + 2(x) = 0 $\Rightarrow$

\,\,\,

x = $-$

{1 \over 2}

Q62

In XeO3F2, the number of bond pair(s),

\pi

-bond(s) and lone pair(s) on Xe atom respectively are :

A 5, 2, 0

B 4, 2, 2

C 5, 3, 0

D 4, 4, 0

Correct Answer

Option C

Solution

\therefore\,\,\,

Number of Bond pairs = 5 Number of $\pi$ bonds = 3 Number of lone pairs = 0

Q63

A group 13 element 'X' reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form Cl3X

\leftarrow

NH3 adduct ; however, XCl3 does not dimerize X is :

A B

B Al

C Ga

D In

Correct Answer

Option A

Solution

Here BCl3 is electron deficient compound as B has 6 electrons.

That is why it accept electron pair from NH3 to form an adduct.

BCl3 does not form dimer like Al, Ga or In, because its electron deficiency is complemented by the formation of co-ordinate bond between lone pair of electron of chlorine and empty unhybridized P-orbital of boron forming P $\pi$ $-$ P $\pi$ bonding.

Q64

The number of pentagons in C60 and trigons (triangles) in white phosphorus, respectively, are :

A 12 and 3

B 20 and 3

C 20 and 4

D 12 and 4

Correct Answer

Option D

Solution

Pentagons in C60 = 12 Triangles in P4 = 4

Q65

The C–C bond length is maximum in :

A C60

B diamond

C graphite

D C70

Correct Answer

Option B

Solution

Structure of diamond is tetrahedral and diamond has sp3 hybridisation.

C60 , C70 , and graphite has sp2 hybridisation .

As % s Character is less in sp3 so C - C bond length is maximum in diamond.

Q66

The noble gas that does not occur in the atmosphere is :

A Ne

B He

C Kr

D Ra

Correct Answer

Option D

Solution

Inert gas Radon(Ra) is not present in atmosphere. Remaining all the inert gas can be found in the atmosphere.

Q67

Given below are two statements : Statement I :

\alpha

and

\beta

forms of sulphur can change reversibly between themselves with slow heating or slow cooling. Statement II : At room temperature the stable crystalline form of sulphur is monoclinic sulphur. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false.

B Statement I is false but Statement II is true.

C Both Statement I and Statement II are true.

D Both Statement I and Statement II are false.

Correct Answer

Option A

Solution

\alpha - sulphur\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{ < 369K}}^{{ > 369K}}} \beta - sulphur

at room temperature $\alpha$ -sulphur (Rhombic) is most stable form.

Q68

On treating a compound with warm dil. H2SO4, gas X is evolved which turns K2Cr2O7 paper acidified with dil. H2SO4 to a green compound Y. X and Y respectively are :

A X = SO2, Y = Cr2O3

B X = SO3, Y = Cr2(SO4)3

C X = SO3, Y = Cr2O3

D X = SO2, Y = Cr2(SO4)3

Correct Answer

Option D

Solution

SO2 + K2Cr2O7 + dil. H2SO4 $\to$ SO3 + Cr2(SO4)3 (green)

Q69

Match with : Industrial process Application

List - I		List - II
(a)	Haber's process	(i)	$HN{O_3}$ synthesis
(b)	Ostwald's process	(ii)	Aluminium extraction
(c)	Contact process	(iii)	$N{H_3}$ synthesis
(d)	Hal-Heroult process	(iv)	${H_2}S{O_4}$ synthesis

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

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

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

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

Correct Answer

Option B

Solution

(a) Haber's process is used for NH3 synthesis. (b) Ostwald's process is used for HNO3 synthesis. (c) Contact process is used for H2SO4 synthesis. (d) In Hall-Heroult process, electrolytic reduction of impure alumina can be done.

(Aluminium extraction)

Q70

A group of 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is :

A As

B Sb

C Bi

D P

Correct Answer

Option C

Solution

BiH3 is the strongest reducing agent among the hydrides of 15 group elements as Bi – H bond dissociation energy is very less.