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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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511 Medium
511 of 705
Analyzing +4 | -1
Given below are two statements:
Statement I : The boiling points of the following hydrides of group 16 elements increases in the order
H2O<H2S<H2Se<H2Te{H_2}O < {H_2}S < {H_2}Se < {H_2}Te
Statement II : The boiling points of these hydrides increase with increase in molar mass.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are incorrect because the boiling point of water is higher than the other hydrides due to hydrogen bonding, and boiling points do not simply increase with molar mass.
512 Medium
512 of 705
Understanding +4 | -1
Assertion (A): Deep sea divers use helium & oxygen mixture for respiration.

Reason (R): Helium is inert in nature.
View answer and explanation
Correct option: B
Both (A) and (R) are true and (R) is not the correct explanation of (A)
Option B (Both (A) and (R) are true and (R) is not the correct explanation of (A)) is correct because both the assertion and reason are true, but the reason doesn't explain why helium is used for deep-sea diving.
513 Easy
513 of 705
Remembering +4 | -1
When the mineral clevite is heated, it give off the inert gas
View answer and explanation
Correct option: A
Helium
Option A (Helium) is correct because clevite, a uranium mineral, releases helium gas upon heating due to alpha decay.
514 Easy
514 of 705
Understanding +4 | -1
Most acidic oxide among the following is
View answer and explanation
Correct option: D
Cl2O7C{l_2}{O_7}
Option D is correct because as the oxidation state of the central atom increases, the acidic character of the oxide also increases due to increased polarization of the O-H bond.
515 Hard
515 of 705
Understanding +4 | -1
Which of the following compounds gives chlorine dioxide when it reacts with SO2S{O_2} in the presence of acid?
View answer and explanation
Correct option: B
Sodium chlorate
Option B is correct because sodium chlorate reacts with SO2 in the presence of acid to produce chlorine dioxide.
516 Easy
516 of 705
Remembering +4 | -1
Sea-weeds are important sources of:
View answer and explanation
Correct option: C
Iodine
Option C (Iodine) is correct because seaweeds are known to be a significant source of iodine, which they accumulate from seawater.
517 Hard
517 of 705
Understanding +4 | -1
HI cannot be prepared by heating KI with conc.H2SO4{H_2}S{O_4} because:
View answer and explanation
Correct option: C
H2SO4{H_2}S{O_4} is an oxidizing agent
Option C is correct because concentrated sulfuric acid (H2SO4{H_2}S{O_4}) acts as an oxidizing agent and oxidizes HI to iodine.
518 Easy
518 of 705
Remembering +4 | -1
The compound used as refrigerant is:
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Correct option: D
CF2Cl2C{F_2}C{l_2}
Option D (CF2Cl2C{F_2}C{l_2}) is correct because dichlorodifluoromethane (Freon-12) is a common refrigerant.
519 Medium
519 of 705
Analyzing +4 | -1
MF+XeF4→A′(M+={MF}+{XeF}_4 \rightarrow {A}^{\prime}\left({M}^{+}=\right.Alkalimetalcation ))
The state of hybridisation of the central atom in 'AA' and shape of the species are:
View answer and explanation
Correct option: C
sp3 d3{sp}^3 {~d}^3, pentagonal planar
The correct answer is sp3 d3{sp}^3 {~d}^3, pentagonal planar because MF{MF} reacts with XeF4{XeF}_4 to form M+[XeF5−]{M}^{+}[{XeF}_5^{-}], which has a pentagonal planar structure and sp3 d3{sp}^3 {~d}^3 hybridization.
520 Medium
520 of 705
Understanding +4 | -1
Which of the following solutions does not change its colour on passing ozone through it?
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Correct option: C
Acidic solution of K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7
Option C is correct because acidic K2Cr2O7K_2Cr_2O_7 already contains Cr in its highest oxidation state (+6) and cannot be further oxidized by ozone, thus no color change occurs.
521 Medium
521 of 705
Analyzing +4 | -1
Assertion (A) : The reaction XeO3+6H++6e−⇌Xe+3H2O,E∘=2.1 V\mathrm{XeO}_3+6 \mathrm{H}^{+}+6 e^{-} \rightleftharpoons \mathrm{Xe}+3 \mathrm{H}_2 \mathrm{O}, E^{\circ}=2.1 \mathrm{~V} is not feasible thermodynamically to oxidise H2O\mathrm{H}_2 \mathrm{O}

Reason (R) : There is a high activation energy barrier between oxidation state +6 and 0 of Xe\mathrm{Xe}.
View answer and explanation
Correct option: C
A\mathbf{A} is true but R\mathbf{R} is false.
Option C is correct. Assertion (A) is true because XeO3XeO_3 is a strong oxidizing agent with a high reduction potential. Reason (R) is false because the kinetics of the reaction, not a high activation energy barrier, prevents it from oxidizing water readily.
522 Easy
522 of 705
Understanding +4 | -1
Phosphine on reaction with hydrobromic acid gives:
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Correct option: B
PH4BrP{H_4}Br
Option B is correct. Phosphine (PH3) reacts with hydrobromic acid (HBr) to form phosphonium bromide (PH4Br), a salt.
523 Medium
523 of 705
Analyzing +4 | -1
Which of the following cannot as an oxidising agent?
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Correct option: C
N3−\mathrm{N}^{3-}
The correct answer is C (N3−\mathrm{N}^{3-}) because it is already in its lowest oxidation state and cannot be further reduced to oxidize others.
524 Easy
524 of 705
Remembering +4 | -1
Which element is not considered as ‘chalcogens’?
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Correct option: D
Polonium
Option D (Polonium) is correct because Polonium is a radioactive element and is not generally considered a chalcogen, which are ore-forming elements.
525 Hard
525 of 705
Application +4 | -1
When one mole of bleaching powder is completely decomposed in presence of CO2\mathrm{C}{\mathrm{O}_2} then the mass of chlorine gas that is liberated will be:
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Correct option: B
70.90 g
Option B (70.90 g) is correct because it is the mass of chlorine gas liberated when one mole of bleaching powder decomposes.
526 Medium
526 of 705
Analyzing +4 | -1
Oxide of nitrogen which has smallest ' N−O\mathrm{N}-\mathrm{O} ' bond length among the following is
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Correct option: B
NO\mathrm{NO}
Option B (NO) is correct because NO has a triple bond character due to resonance, resulting in the shortest N-O bond length.
527 Medium
527 of 705
Understanding +4 | -1
XeO2F2Xe{O_2}{F_2} is obtained by partial bydrolysis of
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because XeO2F2XeO_2F_2 can be obtained by partial hydrolysis of both XeOF4XeOF_4 and XeF6XeF_6.
528 Easy
528 of 705
Remembering +4 | -1
98%   H2SO4  98\% \;\,{H_2}S{O_4}\; is:
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Correct option: C
Azeotropic mixture
Option C (Azeotropic mixture) is correct because 98% H2SO4H_2SO_4 forms an azeotropic mixture with water.
529 Medium
529 of 705
Remembering +4 | -1
Iodide of Millon's base is :
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Correct option: C
NH2HgO⋅HgI{NH}_2 {HgO} \cdot {HgI}
Option C (NH2HgO⋅HgI{NH}_2 {HgO} \cdot {HgI}) is correct because Iodide of Millon's base is NH2HgO⋅HgI{NH}_2 {HgO} \cdot {HgI}, which is formed when Nessler's reagent reacts with NH3{NH}_3.
530 Medium
530 of 705
Understanding +4 | -1
When copper is heated with conc. HNO3\mathrm{HN}{\mathrm{O}_3} it produces:
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Correct option: A
Cu(NO3)2  and NO2\mathrm{Cu(N}{\mathrm{O}_3}{\mathrm{)}_2}\,\,\mathrm{and}\,\mathrm{N}{\mathrm{O}_2}
Option A is correct because copper reacts with concentrated nitric acid to produce copper nitrate and nitrogen dioxide.
531 Easy
531 of 705
Remembering +4 | -1
Non-combustible hydride is:
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Correct option: D
NH3N{H_3}
Option D (NH3N{H_3}) is correct because ammonia (NH3N{H_3}) is a non-combustible hydride, while the other hydrides are combustible.
532 Medium
532 of 705
Remembering +4 | -1
The substance used in Holme’s signal of the ship is a mixture of
View answer and explanation
Correct option: A
CaC2+Ca3P2\mathrm{Ca}{\mathrm{C}_2} + \mathrm{C}{\mathrm{a}_3}{\mathrm{P}_2}
Option A (CaC2+Ca3P2\mathrm{Ca}{\mathrm{C}_2} + \mathrm{C}{\mathrm{a}_3}{\mathrm{P}_2}) is correct because Holmes' signal is a mixture of calcium carbide and calcium phosphide.
533 Easy
533 of 705
Remembering +4 | -1
Which of the light effective in the formation of chlorophyll?
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Correct option: B
Neon lamp
Option B (Neon lamp) is correct because neon lamps emit orange-red light, which is effective in chlorophyll formation and used in greenhouses to promote plant growth.
534 Medium
534 of 705
Understanding +4 | -1
Which among the following is least covalent in nature?
View answer and explanation
Correct option: B
BiF3{BiF}_3
Option B (BiF3{BiF}_3) is correct because, according to Fajan's rule, larger cations with lower charge tend to form more ionic compounds, making BiF3 the least covalent among the given options.
535 Easy
535 of 705
Remembering +4 | -1
Beilstein test is used for the detection of :
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Correct option: D
Cl{Cl}
Option D (Cl{Cl}) is correct because the Beilstein test is used to detect the presence of halogens, particularly chlorine.
536 Medium
536 of 705
Remembering +4 | -1
Fertilizer having the highest nitrogen percentage is:
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Correct option: B
Urea
Option B (Urea) is correct because urea has the highest nitrogen percentage (around 46%) among the given fertilizers.
537 Medium
537 of 705
Application +4 | -1
In qualitative analysis, when H2 S\mathrm{H}_2 \mathrm{~S} is passed through an aqueous solution of salt acidified with dil. HCl\mathrm{HCl}, a black precipitate is obtained. On boiling the precipitate with dil. HNO3\mathrm{HNO}_3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives ......... .
View answer and explanation
Correct option: B
deep blue solution of [Cu(NH3)4]2+\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}
Option B (deep blue solution of [Cu(NH3)4]2+\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}) is correct because copper ions react with excess ammonia to form a tetraammine complex, which has a deep blue color.
538 Easy
538 of 705
Remembering +4 | -1
Formula of calcium chlorite is:
View answer and explanation
Correct option: B
Ca(ClO2)2Ca{\left( {Cl{O_2}} \right)_2}
Option B (Ca(ClO2)2Ca{\left( {Cl{O_2}} \right)_2}) is correct because calcium has a +2 charge and chlorite has a -1 charge, requiring two chlorite ions to balance the charge.
539 Easy
539 of 705
Analyzing +4 | -1
Statement-I: F2{F}_2 is better oxidising agent than Cl2{Cl}_2
Statement-II: HIHI is strongest acid and strongest reducing agent among hydrides of VIIA
View answer and explanation
Correct option: A
Statement-I and II are correct
Option A is correct because both statements are correct. Fluorine is a stronger oxidizing agent than chlorine, and HI is the strongest acid and reducing agent among the hydrides of Group 17.
540 Medium
540 of 705
Remembering +4 | -1
Red P\mathrm{P} can be obtained by white P\mathrm{P} by
View answer and explanation
Correct option: A
Heating it with a catalyst in an inert atmosphere
Option A (Heating it with a catalyst in an inert atmosphere) is correct because red phosphorus is obtained from white phosphorus by heating it with a catalyst in an inert atmosphere.
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