p-Block Elements — Page 15 | JEE Chemistry

Q141

Choose the correct statements from the following A. All group 16 elements form oxides of general formula

\mathrm{EO}_2

and

\mathrm{EO}_3

, where

\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}

and

\mathrm{Po}

. Both the types of oxides are acidic in nature. B.

\mathrm{TeO}_2

is an oxidising agent while

\mathrm{SO}_2

is reducing in nature. C. The reducing property decreases from

\mathrm{H}_2 \mathrm{~S}

to

\mathrm{H}_2

Te down the group. D. The ozone molecule contains five lone pairs of electrons. Choose the correct answer from the options given below:

A A and B only

B C and D only

C A and D only

D B and C only

Correct Answer

Option A

Solution

(A) All group 16 elements form oxides of the

\mathrm{EO}_2

and

\mathrm{EO}_3

type where

\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}

or

\mathrm{Po}

. (B)

\mathrm{SO}_2

is reducing while

\mathrm{TeO}_2

is an oxidising agent. (C) The reducing property increases from

\mathrm{H}_2 \mathrm{S}

to

\mathrm{H}_2

Te down the group. (D)

Q142

Consider the oxides of group 14 elements

\mathrm{SiO}_2, \mathrm{GeO}_2, \mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{CO}

and

\mathrm{GeO}

. The amphoteric oxides are

A

\mathrm{SnO}_2, \mathrm{CO}

B

\mathrm{SiO}_2, \mathrm{GeO}_2

C

\mathrm{SnO}_2, \mathrm{PbO}_2

D

\mathrm{GeO}, \mathrm{GeO}_2

Correct Answer

Option C

Solution

SnO

_2

and PbO

_2

are amphoteric.

Q143

Give below are two statements: Statement - I: Noble gases have very high boiling points. Statement - II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points. In the light of the above statements, choose the correct answer from the options given below:

A Both Statement I and Statement II are false.

B Statement I is true but Statement II is false.

C Statement I is false but Statement II is true.

D Both Statement I and Statement II are true.

Correct Answer

Option A

Solution

Option A is the correct answer.

Both Statement I and Statement II are false.

Noble gases actually have low boiling points, not high.

This is because noble gases are indeed monoatomic gases that exist as single atoms.

They have a completely filled valence shell and do not easily form chemical bonds with other atoms.

As a result, the intermolecular forces present in noble gases are only temporary dipole-induced dipole attractions, also known as dispersion forces (or London forces).

However, these dispersion forces are very weak compared to other types of intermolecular forces like hydrogen bonding or dipole-dipole interactions that occur in other substances.

The dispersion forces in noble gases are weak because these gases are non-polar and since they are monoatomic, the electron cloud around the atom is symmetrical, meaning that there are no permanent dipoles within the molecules.

As the atomic size of noble gases increases down the group, the dispersion forces do increase in strength due to the larger electron cloud that can be more easily distorted, resulting in higher boiling points for the heavier noble gases.

Still, compared to other substances, the overall strength of the dispersion forces in noble gases is low, which means they can be liquefied at low temperatures and hence their boiling points are low, not high.

Therefore, the correct statement would be: "Noble gases are monoatomic gases.

They are held together by weak dispersion forces.

Because of this they are liquefied at very low temperatures.

Hence, they have very low boiling points."

Q144

Identify the incorrect pair from the following :

A Fluoroapatite

-3 \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 \cdot \mathrm{CaF}_2

B Carnallite

-\mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}

C Cryolite

-\mathrm{Na}_3 \mathrm{AlF}_6

D Fluorspar

-\mathrm{BF}_3

Correct Answer

Option D

Solution

(1) Fluorspar is

\mathrm{CaF}_2

Q145

On passing a gas, '

\mathrm{X}

', through Nessler's regent, a brown precipitate is obtained. The gas '

\mathrm{X}

' is

A

\mathrm{Cl}_2

B

\mathrm{CO}_2

C

\mathrm{NH}_3

D

\mathrm{H}_2 \mathrm{S}

Correct Answer

Option C

Solution

Nessler's Reagent Reaction :

\underset{\text{ (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text{ (lodine of Millon's base } \\ \text{ (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}

Q146

A and B formed in the following reactions are:

\begin{aligned} & \mathrm{CrO}_2 \mathrm{Cl}_2+4 \mathrm{NaOH} \rightarrow \mathrm{A}+2 \mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O}, \\ & \mathrm{A}+2 \mathrm{HCl}+2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{B}+3 \mathrm{H}_2 \mathrm{O} \end{aligned}

A

\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_5

B

\mathrm{A}=\mathrm{Na}_2 \mathrm{CrO}_4, \mathrm{~B}=\mathrm{CrO}_5

C

\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_4, \mathrm{~B}=\mathrm{CrO}_4

D

\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_3

Correct Answer

Option B

Solution

\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}

\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}

Q147

Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order

\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3> \mathrm{SbH}_3>\mathrm{BiH}_3

. B. The reducing ability of the hydride increases in the order

\mathrm{NH}_3 C. Among the hydrides,

\mathrm{NH}_3

is strong reducing agent while

\mathrm{BiH}_3

is mild reducing agent. D. The basicity of the hydrides increases in the order

\mathrm{NH}_3 Choose the most appropriate from the options given below :

A C and D only

B A and D only

C A and B only

D B and C only

Correct Answer

Option C

Solution

On moving down the group, bond strength of

\mathrm{M}-\mathrm{H}

bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

Q148

Given below are two statements : Statement (I) : The gas liberated on warming a salt with dil

\mathrm{H}_2 \mathrm{SO}_4

, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion. Statement (II) : In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

\mathrm{Na}_2 \mathrm{S}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{S}

\mathrm{{(C{H_3}COO)_2}Pb + {H_2}S \to \mathop {PbS}\limits_{Black\,lead\,sulphide} + 2C{H_3}COOH}

Q149

The Lassiagne's extract is boiled with dil

\mathrm{HNO}_3

before testing for halogens because,

A Silver halides are soluble in

\mathrm{HNO}_3

.

B

\mathrm{AgCN}

is soluble in

\mathrm{HNO}_3

.

C

\mathrm{Na}_2 \mathrm{S}

and

\mathrm{NaCN}

are decomposed by

\mathrm{HNO}_3

.

D

\mathrm{Ag}_2 \mathrm{S}

is soluble in

\mathrm{HNO}_3

.

Correct Answer

Option C

Solution

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

Q150

Aluminium chloride exists as dimer, Al2Cl6 in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives :

A Al3+ + 3Cl-

B Al2O3 + 6HCl

C [Al(OH)6]3-

D [Al(H2O)6]3+ + 3Cl-

Correct Answer

Option D

Solution

A{l_2}C{l_6} + 12{H_2}O\,\,\rightleftharpoons\,\,2{\left[ {Al{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} + 6C{l^ - }