In Kipp’s apparatus, ${{\rm{H}}_2}{\rm{S}}$ is prepared:

${\rm{By\;FeS}} + {\rm{dil}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{By\;FeS}} + {\rm{conc}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{By\;Fe}} + {\rm{dil}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{\;and\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_3}$ can be distinguished by the addition of:

${\rm{FeC}}{{\rm{l}}_3}{\rm{\;solution}}$

${\rm{NaHS}}{{\rm{O}}_4}{\rm{\;solution}}$

Addition of ${H_2}S{O_4}$ to $Ba{O_2}$ produces

$BaO,S{O_2}\,\,\,{\text{and}}\,\,\,\,{H_2}O$

$BaHS{O_4}\,\,{\text{and}}\,\,\,{O_2}$

$BaS{O_4},\,\,{H_2}\,\,{\text{and}}\,\,\,{O_2}$

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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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91 Easy

91 of 248

Application +4 | -1

The geometry of

XeO{F_4}

molecule is

View answer and explanation

Correct option: B

Square pyramidal

Option B (Square pyramidal) is correct because

XeOF_4

has one lone pair and five bonding pairs, leading to a square pyramidal geometry.

92 Easy

92 of 248

Remembering +4 | -1

In Kipp’s apparatus,

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

is prepared:

View answer and explanation

Correct option: C

\mathrm{By\;FeS} + \mathrm{dil}.\mathrm{\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

Option C (

\mathrm{By\;FeS} + \mathrm{dil}.\mathrm{\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

) is correct because Kipp's apparatus is designed to produce H2S gas by reacting ferrous sulfide (FeS) with dilute sulfuric acid.

93 Hard

93 of 248

Remembering +4 | -1

Phosphate + conc.

HN{O_3} + {\left( {N{H_4}} \right)_2}\;Mo{O_4}

solution

\to

Yellow precipitate.
The composition of yellow precipitate is:

View answer and explanation

Correct option: B

{\left( {N{H_4}} \right)_3}P{O_4}.12Mo{O_3}

Option B (

{\left( {N{H_4}} \right)_3}P{O_4}.12Mo{O_3}

) is correct because the yellow precipitate formed in the reaction of phosphate with concentrated nitric acid and ammonium molybdate is ammonium phosphomolybdate, which has the formula

(NH_4)_3[P(Mo_{12}O_{40})]

94 Medium

94 of 248

Application +4 | -1

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}\mathrm{\;and\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}

can be distinguished by the addition of:

View answer and explanation

Correct option: B

\mathrm{FeC}{\mathrm{l}_3}\mathrm{\;solution}

Option B (

\mathrm{FeC}{\mathrm{l}_3}\mathrm{\;solution}

) is correct because

FeCl_3

can oxidize

H_2SO_3

H_2SO_4

, while

H_2SO_4

cannot reduce

Fe^{3+}

Fe^{2+}

95 Medium

95 of 248

Understanding +4 | -1

Stability of hydrides generally increases with:

View answer and explanation

Correct option: A

Increase in bond angle

Option A (Increase in bond angle) is correct because as the bond angle increases, the s-character in the bonding orbitals increases, leading to greater stability.

96 Medium

96 of 248

Analyzing +4 | -1

A salt X gives white precipitates with lead acetate solution, insoluble in hot water and nitric acid. The salt X most probably contains:

View answer and explanation

Correct option: C

SO_4^{2 - }

Option C (

SO_4^{2 - }

) is correct because lead sulfate (

PbSO_4

) is a white precipitate insoluble in hot water and nitric acid.

97 Medium

97 of 248

Application +4 | -1

The oxidation states of

{P}

in the following compounds

{H}_3 {PO}_3, {PCl}_3, {Ca}_3 {P}_2, {Na}_3 {PO}_4, {POF}_3

are respectively be

View answer and explanation

Correct option: A

+3, +3, -3, +5, +5

Option A (+3, +3, -3, +5, +5) is correct because it accurately represents the oxidation states of P in each compound, calculated based on electronegativity differences and charge neutrality.

98 Easy

98 of 248

Analyzing +4 | -1

In which case, the order of acidic strength is not correct?

View answer and explanation

Correct option: B

HI{O_4} > HBr{O_4} > HCI{O_4}

Option B (

HIO_4 > HBrO_4 > HClO_4

) is not correct because the acidic strength of perhalic acids decreases down the group due to decreasing electronegativity of the central atom.

99 Medium

99 of 248

Analyzing +4 | -1

Which of the following reaction gives a gas which act as oxidising agent as well as reducing agent :

View answer and explanation

Correct option: A

N{a_2}S{O_3} + HCl

Option A (

N{a_2}S{O_3} + HCl

) is correct because the reaction produces

SO_2

, which can act as both an oxidizing and reducing agent.

100 Easy

100 of 248

Remembering +4 | -1

For blasting purpose TNT is mixed with:

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}

Option B (

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}

) is correct because ammonium nitrate is an oxidizer that enhances the explosive power of TNT.

101 Easy

101 of 248

Application +4 | -1

Normality of pure sulphuric acid is:

View answer and explanation

Correct option: D

36 N

Option D (36 N) is correct because the normality of pure sulfuric acid is approximately 36 N, calculated using its density and molar mass.

102 Easy

102 of 248

Analyzing +4 | -1

Which of the following statements is wrong

View answer and explanation

Correct option: C

The stability of hydrides increases from

N{H_3}\mathrm{ to }Bi{H_3}

in Group 15

Option C (The stability of hydrides increases from

N{H_3}\mathrm{ to }Bi{H_3}

in Group 15) is correct because the stability of hydrides decreases down the group due to decreasing bond strength.

103 Easy

103 of 248

Understanding +4 | -1

The anhydride of

\mathrm{HN}{\mathrm{O}_3}

is:

View answer and explanation

Correct option: D

{\mathrm{N}_2}{\mathrm{O}_5}

Option D (

{\mathrm{N}_2}{\mathrm{O}_5}

) is correct because dinitrogen pentoxide (

N_2O_5

) is formed by removing water from two molecules of nitric acid (

HNO_3

104 Easy

104 of 248

Understanding +4 | -1

Which of the following oxoacid of sulphur has

- O - O -

linkage

View answer and explanation

Correct option: C

{H_2}{S_2}{O_8}

Option C (

{H_2}{S_2}{O_8}

) is correct because peroxodisulfuric acid (

{H_2}{S_2}{O_8}

) contains a peroxide linkage (-O-O-).

105 Hard

105 of 248

Remembering +4 | -1

In the manufacture of

{H_2}S{O_4}

the nitrated acid from the Gay-Lussac’s tower is chemically:

View answer and explanation

Correct option: D

HS{O_4}.NO

Option D (

HS{O_4}.NO

) is correct because the nitrated acid from the Gay-Lussac's tower contains nitrosylsulfuric acid, represented as

HS{O_4}.NO

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106 Medium

106 of 248

Application +4 | -1

Iron sulphide is heated in air to form A. an oxide of sulphur. A is dissolved in water to give an acid. The basicity of this acid is….

View answer and explanation

Correct option: A

Option A (2) is correct because when iron sulphide is heated in air, it forms sulphur dioxide, which dissolves in water to form sulphuric acid (

{H_2}S{O_4}

), a dibasic acid.

107 Hard

107 of 248

Understanding +4 | -1

A greenish yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fireworks and safety matches. The gas and halate respectively are:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{l}_2},\mathrm{\;KCl}{\mathrm{O}_3}

Option B (

\mathrm{C}{\mathrm{l}_2},\mathrm{\;KCl}{\mathrm{O}_3}

) is correct because chlorine gas reacts with alkali metal hydroxide to form potassium chlorate, which is used in fireworks and safety matches.

108 Medium

108 of 248

Understanding +4 | -1

Sulphuric acid has great affinity for water because

View answer and explanation

Correct option: C

Acid forms hydrates with water

Option C (Acid forms hydrates with water) is correct because the dissolution of sulphuric acid in water is highly exothermic due to the formation of hydrates, releasing a large amount of energy.

109 Easy

109 of 248

Analyzing +4 | -1

Arrange the acids (I)

{H_2}S{O_3}

, (II)

{H_3}P{O_3}

, and (III)

HCl{O_3}

in the decreasing order of acidic nature.

View answer and explanation

Correct option: C

\mathrm{III} > I > II

Option C (

\mathrm{III} > I > II

) is correct because the acidity depends on the number of oxygen atoms bonded to the central atom; more oxygen atoms lead to higher acidity.

110 Medium

110 of 248

Analyzing +4 | -1

Oxide of a non-metal possesses the following characteristics: (i) It is both a proton donor and proton acceptor. (ii) It is poor conductor of electricity. (iii) It reacts readily with basic and acidic oxides. (iv) It oxidses Fe at boiling point. The oxide is:

View answer and explanation

Correct option: A

{H_2}O

Option A (

{H_2}O

) is correct because water is amphoteric (proton donor and acceptor), a poor conductor of electricity, reacts with both acidic and basic oxides, and oxidizes Fe at boiling point.

111 Hard

111 of 248

Remembering +4 | -1

The compound is used in the estimation of carbon monoxide. Here,

A

refers to

View answer and explanation

Correct option: A

\mathrm{I}_2 \mathrm{O}_5

Option A (

\mathrm{I}_2 \mathrm{O}_5

) is correct because iodine pentoxide (

I_2O_5

) is used in the quantitative estimation of carbon monoxide.

112 Easy

112 of 248

Understanding +4 | -1

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

exhibits:

View answer and explanation

Correct option: B

Reducing properties

Option B (Reducing properties) is correct because

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

can donate electrons and get oxidized, thus acting as a reducing agent.

113 Hard

113 of 248

Understanding +4 | -1

Phosphine is not evolved when:

View answer and explanation

Correct option: D

Metaphosphoric acid is heated

Option D (Metaphosphoric acid is heated) is correct because metaphosphoric acid does not decompose to form phosphine upon heating.

114 Hard

114 of 248

Understanding +4 | -1

Addition of

{H_2}S{O_4}

Ba{O_2}

produces

View answer and explanation

Correct option: D

BaS{O_4}\,{\text{and}}\,\,{H_2}{O_2}

Option D (

BaSO_4 \text{ and } H_2O_2

) is correct because barium peroxide reacts with sulfuric acid to form barium sulfate and hydrogen peroxide.

115 Easy

115 of 248

Remembering +4 | -1

Oxide of nitrogen which is blue solid and is acidic in nature is

View answer and explanation

Correct option: D

{N_2}{O_3}

Option D is correct because

{N_2}{O_3}

is a blue solid and is acidic in nature.

116 Hard

116 of 248

Understanding +4 | -1

Ammonia on reaction with hypochlorite anion, can form

View answer and explanation

Correct option: B

{N_2}{H_4}

Option B (

{N_2}{H_4}

) is correct because ammonia reacts with hypochlorite to produce hydrazine (

N_2H_4

) via the Raschig process.

117 Easy

117 of 248

Application +4 | -1

A hydride of nitrogen having lowest oxidation number of

\mathrm{N}

View answer and explanation

Correct option: A

\mathrm{N}{\mathrm{H}_3}

Option A (

\mathrm{N}{\mathrm{H}_3}

) is correct because in ammonia (

NH_3

), the oxidation state of nitrogen is -3, which is the lowest among the given options.

118 Medium

118 of 248

Application +4 | -1

One mole of fluorine is reacted with two moles of hot and concentrated KOH. The products formed are KF

{H_2}O

and

{O_2}

. The molar ratio of KF,

{H_2}O

and

{O_2}

respectively is:

View answer and explanation

Correct option: B

0.084033565

Option B (0.084033565) is correct because the molar ratio of KF,

H_2O

, and

O_2

is 2:1:1, and the balanced equation is

F_2 + 2KOH \rightarrow 2KF + H_2O + \frac{1}{2}O_2

119 Medium

119 of 248

Understanding +4 | -1

Minimum melting point among the following is of

View answer and explanation

Correct option: B

P{H_3}

Option B (

P{H_3}

) is correct because melting point generally increases down the group for these hydrides, but

N{H_3}

has higher melting point due to hydrogen bonding, and

P{H_3}

has the lowest melting point among the remaining options.

120 Hard

120 of 248

Analyzing +4 | -1

Assertion (A):

{F}

atom has less negative electron affinity than

{Cl}

atom.

Reason (R): Additional electron is repelled more effectively by

3 {p}

-electrons in

{Cl}

than by

2 {p}

-electrons in

{F}

atom.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C is correct because fluorine has less negative electron affinity than chlorine, but the reason is due to smaller size and greater electron-electron repulsion in F.

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