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

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

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

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

Acidic solution of ${{\rm{S}}_2}{\rm{O}}_3^{2 - }$ is converted to ……… in presence of ${{\rm{I}}_2}$

${{\rm{S}}_4}{\rm{O}}_6^{2 - } + {{\rm{I}}^ - }$

${\rm{SO}}_4^{2 - } + {{\rm{I}}^ - }$

${\rm{S}}{{\rm{O}}_3} + {{\rm{I}}^ - }$

${{\rm{S}}_4}{\rm{O}}_6^{2 - } + {\rm{I}}_3^ - $

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 · Hard

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Showing 1-30 of 67 questions

1 Hard

1 of 67

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 Hard

2 of 67

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.

3 Hard

3 of 67

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.

4 Hard

4 of 67

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

5 Hard

5 of 67

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.

6 Hard

6 of 67

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.

7 Hard

7 of 67

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.

8 Hard

8 of 67

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}

9 Hard

9 of 67

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.

10 Hard

10 of 67

Remembering +4 | -1

Reactions of

\mathrm{HNO}_3

(conc.) with

\mathrm{I}, \mathrm{S}, \mathrm{P}

and

\mathrm{C}

gives respectively:

View answer and explanation

Correct option: A

\mathrm{HIO}_3, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_3 \mathrm{PO}_4, \mathrm{CO}_2

Option A (

\mathrm{HIO}_3, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_3 \mathrm{PO}_4, \mathrm{CO}_2

) is correct because concentrated nitric acid oxidizes iodine to iodic acid, sulfur to sulfuric acid, phosphorus to phosphoric acid, and carbon to carbon dioxide.

11 Hard

11 of 67

Analyzing +4 | -1

Pyro phosphorus acid

\left( {{H_4}{P_2}{O_5}} \right)

View answer and explanation

Correct option: D

All of these

Option D is correct because Pyro phosphorus acid

\left( {{H_4}{P_2}{O_5}} \right)

is a dibasic acid, a reducing acid, and contains a

P - O - P

bond.

12 Hard

12 of 67

Understanding +4 | -1

FeC{l_3}

solution on reaction with

S{O_2}

changes to:

View answer and explanation

Correct option: A

FeC{l_2}

Option A (

FeCl_2

) is correct because

FeCl_3

is reduced to

FeCl_2

SO_2

, which acts as a reducing agent.

13 Hard

13 of 67

Analyzing +4 | -1

Assertion:

\mathrm{P}_4 \mathrm{O}_{10}

dehydrates

\mathrm{HNO}_3

.

Reason: In

\mathrm{P}_4 \mathrm{O}_{10}

oxidation state of

\mathrm{P}

is +5

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both statements are true, the high affinity of

P_4O_{10}

for water is what drives the dehydration, not the oxidation state of P.

14 Hard

14 of 67

Understanding +4 | -1

In the compounds of type

E\mathrm{C}{\mathrm{l}_3}

, Where

E = \mathrm{ }B,\mathrm{ }P

As,\mathrm{ or }Bi,

the angle

\mathrm{Cl} - E - \mathrm{Cl}

for different

E

are in the order :

View answer and explanation

Correct option: B

\mathrm{B > P > As > Bi}

Option B (

\mathrm{B > P > As > Bi}

) is correct because as the electronegativity of the central atom decreases down the group, the bond pair-bond pair repulsion decreases, leading to smaller bond angles.

15 Hard

15 of 67

Application +4 | -1

S{O_2} + 2x \to 2{H_2}O + 3S,

the compound “

x

” is

View answer and explanation

Correct option: A

{H_2}S

Option A (

{H_2}S

) is correct because it reduces

S{O_2}

to elemental sulfur, forming water as a byproduct.

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16 Hard

16 of 67

Understanding +4 | -1

In the reaction

K + S{O_2} \to

...., the products are:

View answer and explanation

Correct option: B

{K_2}S{O_3} + {K_2}{S_2}{O_3}

Option B (

{K_2}S{O_3} + {K_2}{S_2}{O_3}

) is correct because potassium reacts with sulfur dioxide to form potassium sulfite and potassium thiosulfate.

17 Hard

17 of 67

Analyzing +4 | -1

Assertion (A): Nitrogen is restricted to a maximum covalency of 4 .

Reason (R):Nitrogen exists as a diatomic molecule (N2)with a triple bond.

View answer and explanation

Correct option: B

Both

\mathrm{A}

and

\mathrm{R}

are correct;

\mathrm{R}

is not the correct explanation of A.

Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. Nitrogen's maximum covalency is 4, and it exists as a diatomic molecule with a triple bond, but these are independent facts.

18 Hard

18 of 67

Understanding +4 | -1

Reaction of solid

\mathrm{KMn}{\mathrm{O}_4}

with conc.

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

produces manganese heptoxide

\left( {\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}} \right)

in:

View answer and explanation

Correct option: A

Solution state

Option A (Solution state) is correct because manganese heptoxide is formed as a dark green oily liquid in solution when solid potassium permanganate reacts with concentrated sulfuric acid.

19 Hard

19 of 67

Remembering +4 | -1

Which is the true covalent oxide of iodine?

View answer and explanation

Correct option: B

{I_2}{O_5}

Option B (

{I_2}{O_5}

) is correct because iodine pentoxide is a stable covalent oxide of iodine.

20 Hard

20 of 67

Application +4 | -1

What volume of the

0.1 \mathrm{~M} \mathrm{~KMnO}_4

\mathrm{ml}

will be required to react with

0.158 \mathrm{~gm}

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

\mathrm{S}_2 \mathrm{O}_3^{--}+\mathrm{MnO}_4^{-}+ \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MnO}_2+\mathrm{SO}_4^{--}+\mathrm{OH}^{-}

View answer and explanation

Correct option: B

26.7 ml

Option B (26.7 ml) is correct because the volume of

KMnO_4

required is calculated using stoichiometry and the balanced redox reaction.

21 Hard

21 of 67

Understanding +4 | -1

Sometimes a yellow turbidity appears while passing

{H_2}S

gas even in the absence of II group radicals. This is because:

View answer and explanation

Correct option: C

Of the oxidation of

{H_2}S

gas by some acid radicals

Option C (Of the oxidation of

{H_2}S

gas by some acid radicals) is correct because the yellow turbidity is due to the oxidation of

{H_2}S

to elemental sulfur by oxidizing agents present in the solution.

22 Hard

22 of 67

Understanding +4 | -1

Acidic solution of

{\mathrm{S}_2}\mathrm{O}_3^{2 - }

is converted to ……… in presence of

{\mathrm{I}_2}

View answer and explanation

Correct option: A

{\mathrm{S}_4}\mathrm{O}_6^{2 - } + {\mathrm{I}^ - }

Option A (

{S_4}{O_6}^{2-} + I^−

) is correct because thiosulfate ion is oxidized to tetrathionate ion by iodine in acidic solution.

23 Hard

23 of 67

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})]

24 Hard

24 of 67

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

25 Hard

25 of 67

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.

26 Hard

26 of 67

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.

27 Hard

27 of 67

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.

28 Hard

28 of 67

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.

29 Hard

29 of 67

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.

30 Hard

30 of 67

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.

Revision notes and key takeaways

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Key Takeaways

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