When copper is heated with conc. ${\rm{HN}}{{\rm{O}}_3}$ it produces:

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,\,{\rm{and}}\,{\rm{N}}{{\rm{O}}_2}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,\,{\rm{and}}\,{\rm{NO}}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2},\,{\rm{NO}}\,\,{\rm{and}}\,{\rm{N}}{{\rm{O}}_2}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,{\rm{and}}\,{{\rm{N}}_2}{\rm{O}}$

${{\rm{I}}_2}$ is stirred in between two liquids, ${{\rm{C}}_6}{{\rm{H}}_6}$ and water. It:

Dissolves more in ${{\rm{C}}_6}{{\rm{H}}_6}$

Dissolves more in ${{\rm{H}}_2}{\rm{O}}$

Dissolves in neither ${{\rm{C}}_6}{{\rm{H}}_6}$ nor water

Back to P-BLOCK ELEMENTS

Your Progress

0 / 248 Answered

Correct

Incorrect

Accuracy

NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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 211-240 of 248 questions

Preparing practice and saved progress…

211 Medium

211 of 248

Understanding +4 | -1

The correct order of reducing power of halide ions is :

View answer and explanation

Correct option: D

{\mathrm{I}^ - } > B{\mathrm{r}^ - } > \mathrm{C}{\mathrm{l}^ - } > {\mathrm{F}^ - }

Option D (

{\mathrm{I}^ - } > B{\mathrm{r}^ - } > \mathrm{C}{\mathrm{l}^ - } > {\mathrm{F}^ - }

) is correct because reducing power increases down the group due to increasing size and decreasing electron gain enthalpy.

212 Easy

212 of 248

Remembering +4 | -1

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

is manufactured by:

View answer and explanation

Correct option: A

Birkeland and Eyde’s process

Option A (Birkeland and Eyde’s process) is correct because it is the industrial process specifically used for the manufacture of nitric acid (HNO3).

213 Medium

213 of 248

Analyzing +4 | -1

Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and

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

(in 4: 1 ratio) is

View answer and explanation

Correct option: A

acidic

Option A (acidic) is correct because the reaction between concentrated nitric acid and

P_4O_{10}

produces

N_2O_5

, which is the anhydride of nitric acid and thus acidic in nature.

214 Medium

214 of 248

Understanding +4 | -1

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

View answer and explanation

Correct option: B

C{l_2},\;KCl{O_3}

Option B (

Cl_2, KClO_3

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

215 Medium

215 of 248

Remembering +4 | -1

Nitric acid is generally light yellow due to the presence of:

View answer and explanation

Correct option: C

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

Option C (NO2) is correct because nitric acid decomposes slightly, especially upon exposure to light, forming nitrogen dioxide, which imparts a yellow color.

216 Easy

216 of 248

Application +4 | -1

In the reaction,

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

the product

x

View answer and explanation

Correct option: B

{N_2}{O_5}

Option B (

{N_2}{O_5}

) is correct because when nitric acid reacts with phosphorus pentoxide, it forms metaphosphoric acid and dinitrogen pentoxide.

217 Hard

217 of 248

Understanding +4 | -1

Presence of sulphide ion cannot be confirmed by:

View answer and explanation

Correct option: A

\mathrm{BaC}{\mathrm{l}_2}

Option A (

BaCl_2

) is correct because barium chloride does not react with sulfide ions to produce a characteristic precipitate or gas that confirms the presence of sulfide ions.

218 Medium

218 of 248

Understanding +4 | -1

When copper is heated with conc.

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

it produces:

View answer and explanation

Correct option: A

\mathrm{Cu(N}{\mathrm{O}_3}{\mathrm{)}_2}\,\,\mathrm{and}\,\mathrm{N}{\mathrm{O}_2}

Option A (

\mathrm{Cu(N}{\mathrm{O}_3}{\mathrm{)}_2}\,\mathrm{and}\,\mathrm{N}{\mathrm{O}_2}

) is correct because the reaction of copper with concentrated nitric acid produces copper nitrate and nitrogen dioxide.

219 Medium

219 of 248

Analyzing +4 | -1

Which of the following statements is wrong?

View answer and explanation

Correct option: A

Single N – N bond is stronger than the single P – P bond

Option A is correct because the single N-N bond is weaker than the single P-P bond due to smaller size and higher interelectronic repulsion in nitrogen.

220 Hard

220 of 248

Analyzing +4 | -1

P-P bond is present in

View answer and explanation

Correct option: A

{H_4}{P_2}{O_6}

Option A is correct because

{H_4}{P_2}{O_6}

(Hypophosphoric acid) contains a P-P bond in its structure.

221 Medium

221 of 248

Remembering +4 | -1

Fluorine gas is stored in:

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because fluorine gas can be stored in both copper and steel vessels due to the formation of a protective fluoride layer on the metal surface.

222 Hard

222 of 248

Understanding +4 | -1

The oxide insoluble in water is:

View answer and explanation

Correct option: C

\mathrm{Po}{\mathrm{O}_2}

Option C (

\mathrm{Po}{\mathrm{O}_2}

) is correct because as metallic character increases down the group, the oxides become less soluble in water. Polonium is the most metallic in the group.

223 Medium

223 of 248

Understanding +4 | -1

The correct shape and hybridization for

Xe{F_2}

are

View answer and explanation

Correct option: C

Linear,s{p^3}d

Option C (

Linear, sp^3d

) is correct because

XeF_2

has a linear shape with three lone pairs on Xe, resulting in

sp^3d

hybridization.

224 Medium

224 of 248

Application +4 | -1

Which of the following is linear?

View answer and explanation

Correct option: A

\mathrm{XeF}_2

Option A (

\mathrm{XeF}_2

) is correct because it has a linear structure due to the arrangement of bonding pairs and lone pairs around the central Xe atom.

225 Hard

225 of 248

Understanding +4 | -1

When KBr is treated with concentrated

{H_2}S{O_4}

reddish brown gas evolved, gas is

View answer and explanation

Correct option: C

Bromine

Option C (Bromine) is correct because concentrated sulfuric acid oxidizes bromide ions to bromine, which is a reddish-brown gas.

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

226 Easy

226 of 248

Understanding +4 | -1

{\mathrm{I}_2}

is stirred in between two liquids,

{\mathrm{C}_6}{\mathrm{H}_6}

and water. It:

View answer and explanation

Correct option: A

Dissolves more in

{\mathrm{C}_6}{\mathrm{H}_6}

Option A (Dissolves more in

{\mathrm{C}_6}{\mathrm{H}_6}

) is correct because iodine is non-polar and dissolves better in non-polar solvents like benzene.

227 Medium

227 of 248

Remembering +4 | -1

Dinitrogen pentoxide a colourless solid is prepared by

View answer and explanation

Correct option: C

Dehydrating

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

with

{\mathrm{P}_\mathrm{4}}{\mathrm{O}_{\mathrm{10}}}

Option C is correct because dinitrogen pentoxide is prepared by dehydrating nitric acid with phosphorus pentoxide.

228 Easy

228 of 248

Analyzing +4 | -1

The

M-Cl

bond energies are different in:

View answer and explanation

Correct option: A

PC{l_5}

Option A (

PC{l_5}

) is correct because in

PCl_5

, the axial and equatorial

P-Cl

bonds have different bond lengths and thus different bond energies.

229 Medium

229 of 248

Remembering +4 | -1

Orthophosphoric acid is ionized in…….steps.

View answer and explanation

Correct option: C

Option C (3) is correct because orthophosphoric acid (H3PO4) is a triprotic acid, meaning it can donate three protons (H+) in three ionization steps.

230 Hard

230 of 248

Understanding +4 | -1

Acidified iodates are reduced to … by

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

View answer and explanation

Correct option: C

Iodine

Option C (Iodine) is correct because sulfur dioxide reduces iodates to iodine in acidic conditions.

231 Hard

231 of 248

Remembering +4 | -1

The word neon signifies:

View answer and explanation

Correct option: A

New

Option A (New) is correct because the word \"neon\" is derived from the Greek word \"neos,\" meaning \"new.\"

232 Hard

232 of 248

Analyzing +4 | -1

A dark brown solid (X) reacts with

N{H_3}

to form a mild explosive which decomposes to give a violet coloured gas. (X) also reacts with

{H_2}

to give an acid (Y).(Y) can also be prepared by heating its salt with

{H_3}P{O_4}

X and Y are

View answer and explanation

Correct option: D

{I_2},HI

Option D is correct because iodine reacts with ammonia to form

NI_3.NH_3

which is explosive and decomposes to give violet iodine vapor, and iodine reacts with hydrogen to form HI.

233 Easy

233 of 248

Remembering +4 | -1

Mixture used in Holme’s signals is

View answer and explanation

Correct option: C

Ca{C_2} + C{a_3}{P_2}

Option C (

Ca{C_2} + C{a_3}{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.

234 Hard

234 of 248

Analyzing +4 | -1

Which reaction represents the oxidizing behaviour of

{H_2}S{O_4}

View answer and explanation

Correct option: D

2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O

Option D (

2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O

) is correct because

H_2SO_4

oxidizes

HI

I_2

, demonstrating its oxidizing behavior.

235 Hard

235 of 248

Remembering +4 | -1

A cold, green flame can be made by passing

C{O_2}

over warm:

View answer and explanation

Correct option: B

White P

Option B (White P) is correct because white phosphorus reacts with carbon dioxide to produce a cold, green flame due to chemiluminescence.

236 Hard

236 of 248

Understanding +4 | -1

{\mathrm{F}_2}

is formed by the reaction of

{\mathrm{K}_2}\mathrm{Mn}{\mathrm{F}_6}

with:

View answer and explanation

Correct option: A

\mathrm{Sb}{\mathrm{F}_5}

Option A (

\mathrm{Sb}{\mathrm{F}_5}

) is correct because it acts as a fluoride ion acceptor, leading to the formation of

F_2

237 Medium

237 of 248

Application +4 | -1

Find the number of lone pairs of electrons on

\mathrm{Xe}

\mathrm{XeOF}_4

View answer and explanation

Correct option: A

Option A (1) is correct because in

\mathrm{XeOF}_4

, Xenon has one lone pair of electrons.

238 Medium

238 of 248

Analyzing +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: C

Both show +2, +4 and +6 oxidation states

Option C (Both show +2, +4 and +6 oxidation states) is wrong because Oxygen does not show +2, +4 and +6 oxidation states.

239 Easy

239 of 248

Analyzing +4 | -1

S{O_2}

reduces:

View answer and explanation

Correct option: C

KMn{O_4}

Option C (

KMn{O_4}

) is correct because

SO_2

reduces

KMn{O_4}

by decolorizing it, acting as a reducing agent itself.

240 Easy

240 of 248

Understanding +4 | -1

Helium –oxygen mixture is used by deep by sea divers in preference to nitrogen oxygen mixture because

View answer and explanation

Correct option: A

Helium is much less soluble in blood than nitrogen

Option A (Helium is much less soluble in blood than nitrogen) is correct because the lower solubility of helium in blood at high pressures prevents nitrogen narcosis, a dangerous condition for deep-sea divers.

Revision notes and key takeaways

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

Showing 211-240 of 248 questions

Your Progress

P-block elements practice · Mixed difficulty

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice