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

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

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

The bond angle in ${\rm{P}}{{\rm{H}}_3}$ is:

Much lesser than ${\rm{N}}{{\rm{H}}_3}$

Equal to that in ${\rm{N}}{{\rm{H}}_3}$

Much greater than in${\rm{N}}{{\rm{H}}_3}$

Slightly more than in ${\rm{N}}{{\rm{H}}_3}$

Back to P-BLOCK ELEMENTS

Your Progress

0 / 93 Answered

Correct

Incorrect

Accuracy

NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →

Showing 31-60 of 93 questions

31 Medium

31 of 93

Application +4 | -1

{SbF}_5

reacts with

{XeF}_4

and

{XeF}_6

to form ionic compounds

\left[{XeF}_3^{+}\right]\left[{SbF}_6\right]^{-}

and

\left[{XeF}_5^{+}\right]\left[{SbF}_6^{-}\right].

The geometry of

{XeF}_3{ }^{+}

ion and

{XeF}_5{ }^{+}

ion, respectively, is:

View answer and explanation

Correct option: B

Bent T-shaped, square pyramidal

Option B (Bent T-shaped, square pyramidal) is correct because the geometries of

\mathrm{XeF}_3^+

and

\mathrm{XeF}_5^+

are bent T-shaped and square pyramidal, respectively, based on VSEPR theory.

32 Medium

32 of 93

Understanding +4 | -1

At ordinary temperature and pressure, among halogens, the chlorine is a gas, bromine is a liquid and iodine is a solid. This is because:

View answer and explanation

Correct option: A

The specific heat is in the order

\mathrm{C}{\mathrm{l}_2} > B{\mathrm{r}_2} > {\mathrm{I}_2}

Option B (Intermolecular forces among molecules of chlorine are the weakest and those in iodine are the strongest) is correct because the physical state of halogens at room temperature depends on the strength of intermolecular forces, which increase with molecular size.

33 Medium

33 of 93

Remembering +4 | -1

The substance used in Holme’s signal of the ship is a mixture of

View answer and explanation

Correct option: A

\mathrm{Ca}{\mathrm{C}_2} + \mathrm{C}{\mathrm{a}_3}{\mathrm{P}_2}

Option A (

\mathrm{Ca}{\mathrm{C}_2} + \mathrm{C}{\mathrm{a}_3}{\mathrm{P}_2}

) is correct because the mixture of calcium carbide and calcium phosphide produces acetylene and phosphine, which ignite spontaneously and produce a bright light, used as a signal on ships.

34 Medium

34 of 93

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+}

35 Medium

35 of 93

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.

36 Medium

36 of 93

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.

37 Medium

37 of 93

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.

38 Medium

38 of 93

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.

39 Medium

39 of 93

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.

40 Medium

40 of 93

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.

41 Medium

41 of 93

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.

42 Medium

42 of 93

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

43 Medium

43 of 93

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.

44 Medium

44 of 93

Analyzing +4 | -1

Which of the following statements is not valid for oxoacids of phosphorous

View answer and explanation

Correct option: B

Hypophosphorous acid is a diprotic acid

Option B is correct because Hypophosphorous acid is a monoprotic acid, not a diprotic acid, due to the presence of two P-H bonds and one P-OH bond.

45 Medium

45 of 93

Understanding +4 | -1

The bond angle in

\mathrm{P}{\mathrm{H}_3}

is:

View answer and explanation

Correct option: A

Much lesser than

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

Option A (Much lesser than

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

) is correct because the bond angle in

P{H_3}

is closer to 90 degrees due to less s-character in the bonding orbitals compared to

N{H_3}

Explore Other Chapters in CHEMISTRY

View All →

CHEMICAL BONDING & MOLECULAR STRUCTURE 595 MCQs THERMODYNAMICS 583 MCQs STRUCTURE OF ATOM 499 MCQs CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 429 MCQs Equilibrium 412 MCQs STATES OF MATTER 410 MCQs

46 Medium

46 of 93

Remembering +4 | -1

The function of

Fe{\left( {OH} \right)_3}

in the contact process is

View answer and explanation

Correct option: A

To remove arsenic impurity

Option A (To remove arsenic impurity) is correct because

Fe(OH)_3

is used to adsorb arsenic impurities in the Contact process for sulfuric acid production.

47 Medium

47 of 93

Application +4 | -1

If two litre of air is passed repeatedly over heated copper and heated Mg till no further reduction in volume takes place, the volume finally obtained will be approximately:

View answer and explanation

Correct option: B

20 mL

Option A (200 mL) is correct because oxygen and nitrogen are removed, leaving mostly noble gases and some residual gases.

48 Medium

48 of 93

Remembering +4 | -1

The gases absorbed by alkaline pyrogallol and oil of turpentine respectively are:

View answer and explanation

Correct option: B

{\mathrm{O}_2},\mathrm{\;}{\mathrm{O}_3}

Option B (

{\mathrm{O}_2},\;{\mathrm{O}_3}

) is correct because alkaline pyrogallol absorbs oxygen (O2) and oil of turpentine absorbs ozone (O3).

49 Medium

49 of 93

Remembering +4 | -1

Orthophosphoric acid on heating gives:

View answer and explanation

Correct option: D

Metaphosphoric acid

Option D (Metaphosphoric acid) is correct because orthophosphoric acid (

H_3PO_4

) loses water upon heating to form metaphosphoric acid (

HPO_3

50 Medium

50 of 93

Analyzing +4 | -1

Which of the following halogen oxides is ionic?

View answer and explanation

Correct option: A

{\mathrm{I}_4}{\mathrm{O}_9}

Option A (

{\mathrm{I}_4}{\mathrm{O}_9}

) is the correct answer because iodine can exhibit a +9 oxidation state only in ionic compounds.

51 Medium

51 of 93

Understanding +4 | -1

Which is the correct thermal stability order for

{H_2}E\,(E\, = \,O,\,S,\,Se,\,Te\,\mathrm{and}\,Po)?

View answer and explanation

Correct option: C

{H_2}PO\, < \,{H_2}Te\, < \,{H_2}Se\, < \,{H_2}S\, < \,{H_2}O

Option C (

{H_2}PO < {H_2}Te < {H_2}Se < {H_2}S < {H_2}O

) is correct because thermal stability decreases down the group due to decreasing bond dissociation energy.

52 Medium

52 of 93

Analyzing +4 | -1

The correct order of acidic nature is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > \mathrm{S}{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}

Option A is correct because acidic nature of oxides increases with the electronegativity of the central atom and oxidation state.

53 Medium

53 of 93

Understanding +4 | -1

Product which is not obtained on the reaction of white phosphorous with thionyl chloride is

View answer and explanation

Correct option: D

POC{l_3}

Option D (

POC{l_3}

) is correct because the reaction of white phosphorus with thionyl chloride (

SOCl_2

) produces

PC{l_3}

S{O_2}

, and

{S_2}C{l_2}

, but not

POC{l_3}

54 Medium

54 of 93

Analyzing +4 | -1

The electronic configurations of four elements are given below. Which element does not belong to the same family as others?

View answer and explanation

Correct option: C

\left[ {\mathrm{Ne}} \right]3{s^2},\mathrm{\;}3{p^5}

Option C ([Ne]3s², 3p⁵) is correct because it represents a halogen, while the others belong to the same family (transition metals or group 2).

55 Medium

55 of 93

Application +4 | -1

4N{H_3} + 5{O_2} \to 6{H_2}O + 4NO.

On taking one mole of

N{H_3}

and one mole of oxygen.

View answer and explanation

Correct option: A

Oxygen is completely reacted

Option A (Oxygen is completely reacted) is correct because, based on the stoichiometry of the reaction, oxygen is the limiting reagent when one mole of each reactant is taken.

56 Medium

56 of 93

Remembering +4 | -1

Chlorine acts as a bleaching agent only in presence of

View answer and explanation

Correct option: B

Moisture

Option B is correct. Chlorine acts as a bleaching agent only in the presence of moisture because it forms hypochlorous acid (HOCl), which is the active bleaching agent.

57 Medium

57 of 93

Application +4 | -1

Iodine reacts with concentrated

HN{O_3}

to yield Y along with other products. The oxidation state of iodine in

Y,

View answer and explanation

Correct option: D

Option C (7) is correct because iodine reacts with concentrated nitric acid to form iodic acid (

HIO_3

), where the oxidation state of iodine is +5. The question asks for the oxidation state of I in Y, where Y is

HIO_3

. The reaction is

I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O

. Thus, Y is

HIO_3

and the oxidation state of I is +5.

58 Medium

58 of 93

Understanding +4 | -1

The final product of hydrolysis of

Xe{F_6}

View answer and explanation

Correct option: C

Xe{O_3}

Option C (

XeO_3

) is correct because

XeF_6

undergoes complete hydrolysis to form

XeO_3

and HF.

59 Medium

59 of 93

Analyzing +4 | -1

The spectrum of helium is similar to:

View answer and explanation

Correct option: B

Option C (

L{i^ + }

) is correct because both Helium and

Li^+

have 2 electrons and thus have similar electronic structures and spectra.

60 Medium

60 of 93

Remembering +4 | -1

Which of the following pairs is obtained on heating ammonium dichromate?

View answer and explanation

Correct option: A

{\mathrm{N}_\mathrm{2}}\,\mathrm{and}\,\mathrm{ }{\mathrm{H}_\mathrm{2}}\mathrm{O}

Option A is correct. Heating ammonium dichromate (NH4)2Cr2O7 produces nitrogen gas (N2) and water (H2O) along with chromium(III) oxide (Cr2O3).

Revision notes and key takeaways

Review P-BLOCK ELEMENTS in your textbook or class notes, then use this set to check your understanding.

Showing 31-60 of 93 questions

Continue Your Practice

📚 Switch Set (MEDIUM)

Chapter Sets

ALL 248 MCQs EASY 88 MCQs HARD 67 MCQs MEDIUM 93 MCQs

0 /93

0✓ 0✗

0% Accuracy