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

${\rm{HP}}{{\rm{O}}_3} + {{\rm{H}}_2}{\rm{O\;}}\mathop \to \limits^{{\rm{Heat}}} ?$ The product is:

${{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}$

${{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_7}$

${{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3}$

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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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Showing 1-30 of 248 questions

1 Hard

1 of 248

Remembering +4 | -1

The source of most of the noble gases is:

View answer and explanation

Correct option: B

The atmospheric air

Option B (The atmospheric air) is correct because noble gases are extracted from the atmosphere through fractional distillation of liquefied air.

2 Easy

2 of 248

Analyzing +4 | -1

Which possesses least stable covalent P-H bond?

View answer and explanation

Correct option: D

PH_6^ +

Option D (

PH_6^ +

) is correct because the

PH_6^+

ion is highly unstable due to the expansion of the octet and high positive charge density on the phosphorus atom, making the P-H bonds very weak.

3 Medium

3 of 248

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.

4 Medium

4 of 248

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.

5 Easy

5 of 248

Understanding +4 | -1

Which of the following is not an acidic oxide?

View answer and explanation

Correct option: A

Option A (NO) is correct because nitric oxide (NO) is a neutral oxide, while the other options are acidic oxides that react with water to form acids.

6 Hard

6 of 248

Analyzing +4 | -1

On heating with

\mathrm{NaOH}

solution and zinc,

\mathrm{NaNO}_3

liberates a gas which can also be obtained by:

View answer and explanation

Correct option: B

Heating

\mathrm{NH}_4 \mathrm{NO}_2

with sodium hydroxide solution

Option B (Heating

\mathrm{NH}_4 \mathrm{NO}_2

with sodium hydroxide solution) is correct because the gas liberated is ammonia (

NH_3

), which can be produced by reducing

NaNO_3

with Zn and NaOH, or by heating

NH_4NO_2

with NaOH.

7 Medium

7 of 248

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}

8 Easy

8 of 248

Remembering +4 | -1

Fluorine is usually obtained from:

View answer and explanation

Correct option: A

Fluorspar

Option A is correct. Fluorspar (CaF2) is the primary source from which fluorine is usually obtained.

9 Easy

9 of 248

Understanding +4 | -1

The basic character of hydrides of V group is

View answer and explanation

Correct option: A

N{H_3} > P{H_3} > AS{H_3} > Sb{H_3}

Option A (

N{H_3} > P{H_3} > AS{H_3} > Sb{H_3}

) is correct because basicity decreases down the group due to the decrease in electronegativity and increase in bond length.

10 Medium

10 of 248

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.

11 Medium

11 of 248

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.

12 Easy

12 of 248

Analyzing +4 | -1

Which is planar molecule?

View answer and explanation

Correct option: B

Xe{F_4}

Option B (

Xe{F_4}

) is correct because it has a square planar geometry, which is planar.

13 Hard

13 of 248

Application +4 | -1

When one mole of bleaching powder is completely decomposed in presence of

\mathrm{C}{\mathrm{O}_2}

then the mass of chlorine gas that is liberated will be:

View answer and explanation

Correct option: B

70.90 g

Option B (70.90 g) is correct because one mole of bleaching powder (

CaOCl_2

) upon decomposition with

CO_2

liberates one mole of chlorine gas (

Cl_2

), which has a molar mass of 70.90 g/mol.

14 Easy

14 of 248

Analyzing +4 | -1

Mark the wrong statement. Halogens are all coloured.

View answer and explanation

Correct option: D

The excitation energy required by the small fluorine atoms is smaller than required by the large iodine atom

Option D (The excitation energy required by the small fluorine atoms is smaller than required by the large iodine atom) is the wrong statement because smaller atoms require higher excitation energy.

15 Medium

15 of 248

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}

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16 Medium

16 of 248

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.

17 Hard

17 of 248

Understanding +4 | -1

Calcium cyanamide on treatment with steam produces

View answer and explanation

Correct option: C

N{H_3} + CaC{O_3}

Option C (

N{H_3} + CaC{O_3}

) is correct because calcium cyanamide (

CaCN_2

) reacts with steam to produce ammonia (

NH_3

) and calcium carbonate (

CaCO_3

18 Medium

18 of 248

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.

19 Hard

19 of 248

Understanding +4 | -1

In the electrothermal process, the compound displaced by silica from calcium phosphate is

View answer and explanation

Correct option: D

Phosphorus pentoxide

Option D (Phosphorus pentoxide) is correct because in the electrothermal process, silica displaces phosphorus pentoxide from calcium phosphate.

20 Medium

20 of 248

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.

21 Medium

21 of 248

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.

22 Hard

22 of 248

Understanding +4 | -1

\mathrm{HP}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O\;}\mathop \to \limits^{\mathrm{Heat}} ?

The product is:

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because heating metaphosphoric acid with water leads to the formation of orthophosphoric acid.

23 Easy

23 of 248

Understanding +4 | -1

Oxygen is paramagnetic. The unpaired electrons are present in :

View answer and explanation

Correct option: A

Antibonding orbitals

Option A (Antibonding orbitals) is correct. Oxygen's paramagnetism arises from the presence of unpaired electrons in its antibonding molecular orbitals.

24 Medium

24 of 248

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.

25 Easy

25 of 248

Remembering +4 | -1

A clathrate is defined as a:

View answer and explanation

Correct option: A

Cage compound

Option A (Cage compound) is correct because a clathrate is a chemical substance consisting of a lattice that traps or contains molecules.

26 Hard

26 of 248

Remembering +4 | -1

Pearl white is:

View answer and explanation

Correct option: A

BiOCl

Option A (

BiOCl

) is correct because Bismuth oxychloride is known as pearl white due to its pearlescent appearance and is used as a pigment.

27 Easy

27 of 248

Analyzing +4 | -1

A gas does not turn lime water milky, supports the combustion of burning magnesium. It has no smell and is colourless. It extinguishes a glowing splint but under some circumstances reacts with oxygen and hydrogen. It is not poisonous. The gas is likely to be :

View answer and explanation

Correct option: B

Nitrogen

Option B (Nitrogen) is correct because nitrogen does not turn lime water milky, supports combustion of burning magnesium, is odorless and colorless, extinguishes a glowing splint, and is not poisonous.

28 Easy

28 of 248

Understanding +4 | -1

Which factor is most responsible for the increase in boiling points of noble gases from

He\,\,\mathrm{to}\,\,Xe

View answer and explanation

Correct option: D

Increase in Polarisibility

Option D (Increase in Polarisibility) is correct because as the atomic size and number of electrons increase down the group, the polarizability of the noble gas atoms increases, leading to stronger London dispersion forces and higher boiling points.

29 Hard

29 of 248

Analyzing +4 | -1

Consider the following statements
(a) In acidic medium nitrous acid disproportionate to nitric acid and nitrogen gas
(b) Ammonium nitrate on heating gives nitrous oxide
(c) N – N bond is present both in

{N_2}{O_3}\mathrm{ }and\mathrm{ }{N_2}{O_4}

gases
The correct statements are

View answer and explanation

Correct option: B

(b) and (c) only

Option B is correct because ammonium nitrate decomposes to nitrous oxide and N-N bond is present in both

{N_2}{O_3}

and

{N_2}{O_4}

30 Hard

30 of 248

Remembering +4 | -1

The element present in combined state in Laminaria stenophylla is:

View answer and explanation

Correct option: B

Iodine

Option B (Iodine) is correct because Laminaria stenophylla, a type of seaweed, is known to accumulate iodine in combined form.

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Key Takeaways

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