Name of gas that can readily decolourises acidified ${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ solution:

${{\rm{P}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$

The oxoacid of ${\rm{S}}{{\rm{O}}_2}$ is:

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_3}$

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${{\rm{H}}_2}{{\rm{S}}_2}{{\rm{O}}_8}$

Statement- I: Acid strength increases in the order given as $HF < < HCl < < HBr < < HI.$ Statement- II: As the size of the elements $F,\,\,Cl,\,\,Br,\,\,I$ increases down the group, the bond strength of $HF,HCl,HBr{\rm{ and }}HI$ decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below.

Statement I is correct but Statement II is false

Statement I is incorrect but Statement II is true

Both Statement I and Statement II are false

Aqueous solution of ${\rm{NaOCl}}$

In nitroprusside ion, the iron and NO exist as ${\rm{F}}{{\rm{e}}^{{\rm{II}}}}$ and $N{O^ + }$ rather than $F{e^{III}}$ and NO. These forms can be differentiated by:

Estimating the concentration of iron

Measuring the concentration of ${\rm{C}}{{\rm{N}}^ - }$

Thermally decomposing the compound

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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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Showing 1-30 of 93 questions

1 Medium

1 of 93

Remembering +4 | -1

Which gas is filled in electric bulbs/tubes?

View answer and explanation

Correct option: C

Ar

Option C (Ar) is correct because argon is an inert gas that prevents the filament from oxidizing and prolongs the life of the bulb.

2 Medium

2 of 93

Application +4 | -1

The correct order of the oxidation states of nitrogen in

NO,\,N{O_2},\,N{O_2}

and

{N_2}{O_3}

View answer and explanation

Correct option: B

{N_2}O < NO < {N_2}{O_3} < N{O_2}

Option B (

{N_2}O < NO < {N_2}{O_3} < N{O_2}

) is correct because the oxidation states of nitrogen in these compounds are +1, +2, +3, and +4, respectively.

3 Medium

3 of 93

Remembering +4 | -1

Weldon mud is:

View answer and explanation

Correct option: C

2CaO.Mn{O_2}

Option C (

2CaO.Mn{O_2}

) is correct because Weldon mud is a mixture of calcium oxide and manganese dioxide, chemically represented as

2CaO.Mn{O_2}

4 Medium

4 of 93

Analyzing +4 | -1

Which of the following is least likely to exist?

View answer and explanation

Correct option: D

HBr{O_2}

Option D (

HBrO_2

) is least likely to exist because bromine has a lower tendency to form multiple bonds with oxygen compared to chlorine, making

HBrO_2

less stable.

5 Medium

5 of 93

Understanding +4 | -1

Name of gas that can readily decolourises acidified

\mathrm{KMn}{\mathrm{O}_\mathrm{4}}

solution:

View answer and explanation

Correct option: C

\mathrm{S}{\mathrm{O}_\mathrm{2}}

Option C (

\mathrm{S}{\mathrm{O}_\mathrm{2}}

) is correct because sulfur dioxide readily decolorizes acidified

\mathrm{KMn}{\mathrm{O}_\mathrm{4}}

solution due to its reducing properties.

6 Medium

6 of 93

Remembering +4 | -1

The oxoacid of

\mathrm{S}{\mathrm{O}_2}

is:

View answer and explanation

Correct option: A

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}

Option A (

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}

) is correct because sulfur dioxide (

\mathrm{S}{\mathrm{O}_2}

) reacts with water to form sulfurous acid (

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}

7 Medium

7 of 93

Understanding +4 | -1

Statement- I: Acid strength increases in the order given as

HF < < HCl < < HBr < < HI.

Statement- II: As the size of the elements

F,\,\,Cl,\,\,Br,\,\,I

increases down the group, the bond strength of

HF,HCl,HBr\mathrm{ and }HI

decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: C

Both Statement I and Statement II are true

Option C (Both Statement I and Statement II are true) is correct because acid strength increases down the halogen group due to decreasing bond strength.

8 Medium

8 of 93

Analyzing +4 | -1

Which does not react with

KMn{O_4}

solution?

View answer and explanation

Correct option: A

{O_3}

Option A (

{O_3}

) is correct because ozone does not react with

KMnO_4

solution.

9 Medium

9 of 93

Remembering +4 | -1

Anhydrone is:

View answer and explanation

Correct option: C

Anhydrous magnesium perchlorate

Option C is correct. Anhydrone is anhydrous magnesium perchlorate [Mg(ClO4)2], used as a drying agent.

10 Medium

10 of 93

Understanding +4 | -1

Strongest reducing agent among the following is

View answer and explanation

Correct option: B

Bi{H_3}

Option B (

Bi{H_3}

) is correct because reducing character increases down the group due to the decreasing bond strength of the E-H bond (where E is the group 15 element), making it easier to release hydrogen.

11 Medium

11 of 93

Remembering +4 | -1

Which of the following is used to provide smoke screens :

View answer and explanation

Correct option: A

Calcium phosphide

Option A (Calcium phosphide) is correct because calcium phosphide reacts with water to produce phosphine, which spontaneously ignites in air, creating smoke.

12 Medium

12 of 93

Analyzing +4 | -1

Which of the following does not exist?

View answer and explanation

Correct option: B

{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }

Option B (

{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }

) is correct because the cation

{\left[ {Sb{F_4}} \right]^ + }

is not stable and does not exist.

13 Medium

13 of 93

Understanding +4 | -1

The element which evolves two gases on reacting with conc.

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

is:

View answer and explanation

Correct option: B

\mathrm{C}

Option B (

\mathrm{C}

) is correct because carbon reacts with concentrated

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

to produce

CO_2

and

SO_2

gases.

14 Medium

14 of 93

Remembering +4 | -1

Javelle water is:

View answer and explanation

Correct option: C

Both (1) and (2)

Option C is correct. Javelle water is an aqueous solution of sodium hypochlorite (NaOCl) and is used as a bleaching agent.

15 Medium

15 of 93

Application +4 | -1

\text{XeF}_{4}

reacts with

\mathrm{SbF}_{5}

to form

\left[\mathrm{XeF}_{\text{m}}\right]^{\text{n}+}\left[\mathrm{SbF}_{\text{y}}\right]^{2-}

\mathrm{m}+\mathrm{n}+\mathrm{y}+\mathrm{z}=\text {? }

View answer and explanation

Correct option: A

Option A (11) is correct because the reaction forms

[XeF_3^+][SbF_6^-]

, so m=3, n=1, y=6, and z=1, thus m+n+y+z = 3+1+6+1 = 11.

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16 Medium

16 of 93

Understanding +4 | -1

On heating silver nitrate strongly ……. is obtained finally:

View answer and explanation

Correct option: D

All

Option D (All) is correct because on heating silver nitrate strongly, it decomposes to form silver metal, nitrogen dioxide, and oxygen gas.

17 Medium

17 of 93

Remembering +4 | -1

Atomicity of sulphur in rhombic sulphur is

View answer and explanation

Correct option: A

Option A (8) is correct because rhombic sulfur exists as S8 molecules, meaning each molecule contains 8 sulfur atoms.

18 Medium

18 of 93

Analyzing +4 | -1

The perhalate ion with maximum oxidizing power is:

View answer and explanation

Correct option: B

BrO_4^ -

Option B (

BrO_4^ -

) is correct because among the perhalate ions,

BrO_4^-

has the highest reduction potential and thus the maximum oxidizing power.

19 Medium

19 of 93

Understanding +4 | -1

In ramsay and rayleigh’s isolation of noble gases from air, the nitrogen of the air is finally converted into

View answer and explanation

Correct option: D

NaN{O_2}

and

NaN{O_3}

Option D (

NaNO_2

and

NaNO_3

) is correct because in Ramsay and Rayleigh's method, nitrogen is converted into sodium nitrite and sodium nitrate.

20 Medium

20 of 93

Application +4 | -1

In the reaction,

HN{O_3} + {P_4}{O_{10}} \to 4HP{O_3} + X

the product X is

View answer and explanation

Correct option: B

{N_2}{O_5}

Option B (

{N_2}{O_5}

) is correct because the reaction of nitric acid with phosphorus pentoxide produces metaphosphoric acid and dinitrogen pentoxide.

21 Medium

21 of 93

Remembering +4 | -1

The gas which is absorbed by ferrous sulphate solution giving blackish brown colour is:

View answer and explanation

Correct option: D

Option D (NO) is correct because nitric oxide (NO) reacts with ferrous sulfate to form a complex, resulting in a blackish-brown color.

22 Medium

22 of 93

Application +4 | -1

Oxidation states of

{P}

{H}_4 {P}_2 {O}_5, {H}_4 {P}_2 {O}_6, {H}_4 {P}_2 {O}_7

are respectively

View answer and explanation

Correct option: A

+3,+4,+5

Option A (+3, +4, +5) is correct because it accurately represents the oxidation states of P in each compound, calculated based on electronegativity differences and charge neutrality.

23 Medium

23 of 93

Remembering +4 | -1

A gas, that relights glowing splinter, is

View answer and explanation

Correct option: B

{\mathrm{O}_\mathrm{2}}

Option B (O2) is correct because oxygen supports combustion and can relight a glowing splinter due to its oxidizing properties.

24 Medium

24 of 93

Application +4 | -1

In the reduction of

HN{O_3}

and

{N_2}O

, the number of mole of electrons involved per mole of

HN{O_3}

View answer and explanation

Correct option: B

Option B (4) is correct because in the reduction of

HN{O_3}

{N_2}O

, the oxidation state of nitrogen changes from +5 to +1, involving 4 electrons.

25 Medium

25 of 93

Application +4 | -1

At T (K), 100 L of dry oxygen is present in a sealed container. It is subjeceted to silent electric discharge, till the volumes of oxygen and ozone becomes equal .what is the volume of ozone formed at T (K)?

View answer and explanation

Correct option: D

Option D (40) is correct because the volume of ozone formed is calculated based on the stoichiometry of the reaction and the given condition that the volumes of oxygen and ozone become equal.

26 Medium

26 of 93

Application +4 | -1

The amount of arsenic penta sulphide that can be obtained when 35.5 g arsenic acid is treated with excess

{H_2}S

in the presence of conc. HCl (assuming 100 % conversion) is:

View answer and explanation

Correct option: D

0.125 mole

Option D (0.125 mole) is correct because the amount of arsenic pentasulphide formed is calculated based on the stoichiometry of the reaction between arsenic acid and hydrogen sulphide.

27 Medium

27 of 93

Analyzing +4 | -1

Which of the following equations is not correctly formulated?

View answer and explanation

Correct option: B

3Zn + 8HN{O_3}\left( {\mathrm{very}\,\,\mathrm{dil}.} \right) \to 3Zn{\left( {N{O_3}} \right)_2} + 2NO + 4{H_2}O

Option B is not correctly formulated. With very dilute

HN{O_3}

, Zinc (

Zn

) produces

N{H_4}N{O_3}

instead of

NO

28 Medium

28 of 93

Analyzing +4 | -1

Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?

View answer and explanation

Correct option: A

Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF

Option A (

Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF

) is correct because there is no change in oxidation state of any element in this reaction, hence no oxidation or reduction.

29 Medium

29 of 93

Analyzing +4 | -1

In nitroprusside ion, the iron and NO exist as

\mathrm{F}{\mathrm{e}^{\mathrm{II}}}

and

N{O^ + }

rather than

F{e^{III}}

and NO. These forms can be differentiated by:

View answer and explanation

Correct option: C

Measuring the solid state magnetic moment

Option C is correct because the magnetic moment depends on the number of unpaired electrons, which differs between

\mathrm{F}{\mathrm{e}^{\mathrm{II}}}

and

F{e^{III}}

30 Medium

30 of 93

Remembering +4 | -1

With cold and dilute sodium hydroxide fluorine reacts to give

View answer and explanation

Correct option: A

\mathrm{NaF}\,\,\mathrm{and}\,\,\mathrm{O}{\mathrm{F}_\mathrm{2}}

Option A is correct. Fluorine reacts with cold and dilute sodium hydroxide to produce sodium fluoride (NaF) and oxygen difluoride (OF2).

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