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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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Practice format and chapter quiz

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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Showing 61-90 of 320 questions

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61 Medium
61 of 320
Analyzing +4 | -1
Assertion: Group 15 elements are less reactive than Group 16 elements.
Reason: Group 15 elements have a higher ionization energy and a lower electronegativity than Group 16 elements.
View answer and explanation
Correct option: B
Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion
Option B (Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion) is correct because while both statements are true, the reason doesn't fully explain the assertion.
62 Medium
62 of 320
Understanding +4 | -1
Which of the following(s) when heated give nitrogen gas?
View answer and explanation
Correct option: D
Both 1 and 2
Option D (Both 1 and 2) is correct because both ammonium dichromate and barium azide decompose upon heating to produce nitrogen gas.
63 Medium
63 of 320
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 C (CH3OCH2CH2CH3& CH3OCH(CH3)2\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \&\, \mathrm{CH}_3 \mathrm{OCH}(\mathrm{CH}_3)_2) is correct because these two compounds are chain isomers, not metamers, as they differ in the carbon chain branching.
64 Medium
64 of 320
Understanding +4 | -1
The substance which does not liberate oxygen on treatment with ozone is
View answer and explanation
Correct option: C
SO2S{O_2}
Option C (SO2S{O_2}) is correct because SO2S{O_2} reacts with ozone to form SO3S{O_3} without liberating oxygen.
65 Medium
65 of 320
Analyzing +4 | -1
Assertion (A): Noble gases have very high boiling point.
Reason (R): Noble gases have strong dispersion forces hence they liquify at low temperature, hence they have high boiling point.
View answer and explanation
Correct option: C
Assertion and Reason are false
Option C (Assertion and Reason are false) is correct because noble gases have very low boiling points due to weak dispersion forces, and the reason provided is also incorrect.
66 Medium
66 of 320
Analyzing +4 | -1
Consider NH3,  PH3,  AsH3  &  SbH3N{H_3},\,\,P{H_3},\,\,As{H_3}\,\,\& \,\,Sb{H_3} find which statements are correct for NH3N{H_3}
A) Most basic
B) Highest bond angle
C) Strongest Reducing agent
D) Highest B.P.
E) Most −ve ΔHf0- ve\,\Delta H_f^0
View answer and explanation
Correct option: D
A, B, E
Option D (A, B, E) is correct because NH3 is the most basic, has the highest bond angle, and the most negative enthalpy of formation among the given hydrides.
67 Medium
67 of 320
Analyzing +4 | -1
Assertion (A): (NH4)2Cr2O7,\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7, on heating doesn't form any residue.

Reason (R): In Cr2O72−\mathrm{Cr}_2 \mathrm{O}_7^{2-} there are six equivalent Cr−O\mathrm{Cr}-\mathrm{O} bonds.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because both the assertion and the reason are false. Ammonium dichromate on heating leaves a residue of Cr2O3, and the Cr-O bonds in dichromate are not all equivalent.
68 Medium
68 of 320
Understanding +4 | -1
KF{KF} combines with HF{HF} to form KHF2{KHF}_2. The compound contains the species:
View answer and explanation
Correct option: C
K+{K}^{+}and [HF2]−\left[{HF}_2\right]^{-}
Option C (K+{K}^{+}and [HF2]−\left[{HF}_2\right]^{-}) is correct because KHF2{KHF}_2 dissociates into K+{K}^{+} and [HF2]−\left[{HF}_2\right]^{-} ions in solution.
69 Medium
69 of 320
Understanding +4 | -1
The hydrides of the first elements in groups 15-17 namely NH3,H2O\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O} and HF\mathrm{HF} respectively shows abnormally high values for melting and boiling points. This is due to
View answer and explanation
Correct option: B
the ability to form extensive intermolecular H\mathrm{H}-bonding
Option B (the ability to form extensive intermolecular H\mathrm{H}-bonding) is correct because the high melting and boiling points are due to the strong intermolecular hydrogen bonds in NH3,H2O\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O} and HF\mathrm{HF}.
70 Medium
70 of 320
Understanding +4 | -1
Which of the following acid posses oxidising, reducing and complex forming properties?
View answer and explanation
Correct option: C
HNO2HN{O_2}
Option C (HNO2HN{O_2}) is correct because nitrous acid can act as both an oxidizing and reducing agent, and it can also form complexes.
71 Medium
71 of 320
Understanding +4 | -1
The products obtained when I2\mathrm{I}_2 reacts with hot and concentrated aqueous NaOH\mathrm{NaOH} are
View answer and explanation
Correct option: D
I−\mathrm{I}^{-} and IO3−\mathrm{IO}_3^{-}
Option D is correct because the reaction of iodine with hot and concentrated sodium hydroxide produces iodide (I−I^−) and iodate (IO3−IO_3^−) ions.
72 Medium
72 of 320
Understanding +4 | -1
Hydrolysis of ‘XX’ is an example of disproportionation. XX is
View answer and explanation
Correct option: C
XeF4Xe{F_4}
Option C (XeF4XeF_4) is correct because its hydrolysis results in disproportionation, where Xenon is both oxidized and reduced.
73 Medium
73 of 320
Understanding +4 | -1
Which one of the following correctly represents the order of stability of oxides, X2O{X_2}O (X = halogen)?
View answer and explanation
Correct option: D
I>Cl>BrI > Cl > Br
Option D (I>Cl>BrI > Cl > Br) is correct because iodine forms more stable oxides due to greater polarity, and chlorine can form multiple bonds with oxygen.
74 Medium
74 of 320
Remembering +4 | -1
Which of the following is a double base propellant?
View answer and explanation
Correct option: B
Nitroglycerine and nitrocellulose
The correct answer is B (Nitroglycerine and nitrocellulose) because double-base propellants contain both the fuel and oxidizer in the same molecule, and this combination is a common example.
75 Medium
75 of 320
Understanding +4 | -1
Which statement is not correct?
View answer and explanation
Correct option: A
White and red phosphorus react with chlorine at room temperature
Option A (White and red phosphorus react with chlorine at room temperature) is not correct because red phosphorus is less reactive and does not react with chlorine at room temperature.
76 Medium
76 of 320
Analyzing +4 | -1
The spectrum of helium is similar to:
View answer and explanation
Correct option: B
Na
Option C is correct because Li+ and He have the same number of electrons and similar electronic structure, resulting in similar spectra.
77 Medium
77 of 320
Analyzing +4 | -1
Statement I: PF5\mathrm{PF}_5 and IF have similar shapes

Statement II: PF5\mathrm{PF}_5 has to two types of P−F\mathrm{P}-\mathrm{F} bond lengths
View answer and explanation
Correct option: D
Statement I if false but statement II is correct
Option D is correct because PF₅ has trigonal bipyramidal shape, IF is unknown, and PF₅ has two different P-F bond lengths (axial and equatorial).
78 Medium
78 of 320
Remembering +4 | -1
Weldon mud is:
View answer and explanation
Correct option: C
2CaO.MnO22CaO.Mn{O_2}
Option C (2CaO.MnO22CaO.Mn{O_2}) is the correct answer because 2CaO.MnO22CaO.Mn{O_2} is the chemical formula for Weldon mud.
79 Medium
79 of 320
Application +4 | -1
Oxidation states of P{P} in H4P2O5,H4P2O6,H4P2O7{H}_4 {P}_2 {O}_5, {H}_4 {P}_2 {O}_6, {H}_4 {P}_2 {O}_7 are respectively
View answer and explanation
Correct option: A
+3,+4,+5+3,+4,+5
Option D (17.325 mm Hg) is correct because it is calculated using Raoult's law for vapor pressure lowering.
80 Medium
80 of 320
Analyzing +4 | -1
[Fe(H2O)5NO]2+\left[{Fe}\left({H}_2 {O}\right)_5 {NO}\right]^{2+} is a complex formed during the brown ring test for NO3−{NO}_3^{-}ion. In this complex.
View answer and explanation
Correct option: D
All of the above statements are correct
Option D (All of the above statements are correct) is correct because the complex has three unpaired electrons, the color is due to charge transfer, and the electronic configuration of Fe and NO are as described.
81 Medium
81 of 320
Understanding +4 | -1
Assertion (A) Elementary phosphorus exists in three principal allotropic forms, i.e. white (or yellow), red (or violet) and black.

Reason (R) Of the three forms, white phosphorus is the most important and most reactive.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Option A (If both Assertion and Reason are true and Reason is correct explanation of Assertion) is correct because white phosphorus is indeed the most reactive allotrope due to its unstable tetrahedral structure.
82 Medium
82 of 320
Analyzing +4 | -1
Statement I: The dipole moment of NF3\mathrm{NF}_3 is less than NH3\mathrm{NH}_3

Statement II: The presence of lone pair of electrons on nitrogen atom shows an additive contribution in dipole moment of NH3\mathrm{NH}_3 whereas it shows negative contribution towards dipole moment of NF3\mathrm{NF}_3
View answer and explanation
Correct option: A
Statement I and statement II are correct and statement II is the correct explanation of statement I
Option A (Statement I and statement II are correct and statement II is the correct explanation of statement I) is correct because the dipole moment of NF3 is less than NH3 due to the opposing directions of bond dipoles and lone pair dipole.
83 Medium
83 of 320
Understanding +4 | -1
What is the oxidising agent chlorine water?
View answer and explanation
Correct option: C
HOCI
Option C (HOCI) is correct because hypochlorous acid (HOCl) decomposes to release nascent oxygen, which acts as the oxidizing agent.
84 Medium
84 of 320
Remembering +4 | -1
Iodine exists in which form in iodised salt
View answer and explanation
Correct option: B
NaIO3{NaIO}_3
Option B (NaIO3{NaIO}_3) is correct because iodized salt typically contains sodium iodate (NaIO3{NaIO}_3) to provide iodine.
85 Medium
85 of 320
Understanding +4 | -1
Hydrolysis of PCl5{PCl}_5 gives:
View answer and explanation
Correct option: C
H3PO4{H}_3 {PO}_4
Option C (H3PO4{H}_3 {PO}_4) is correct because the complete hydrolysis of PCl5{PCl}_5 results in the formation of orthophosphoric acid (H3PO4{H}_3 {PO}_4) and hydrochloric acid.
86 Medium
86 of 320
Analyzing +4 | -1
In a reaction, 4 mole of electrons are transferred to 1 mole of HNO3{HNO}_3, the possible product obtained due to reduction is :-
View answer and explanation
Correct option: B
0.5 mole of N2O{N}_2 {O}
The correct answer is B (0.5 mole of N2O{N}_2 {O}) because the reduction of HNO3{HNO}_3 by 4 moles of electrons results in nitrogen having a +1 oxidation state, corresponding to N2O{N}_2 {O}.
87 Medium
87 of 320
Understanding +4 | -1
The stability of +2 oxidation state among Si, Ge, Sn and Pb increase in the sequence
View answer and explanation
Correct option: D
Si < Ge < Sn < Pb
Option D (Si < Ge < Sn < Pb) is correct because the stability of the +2 oxidation state increases down the group due to the inert pair effect.
88 Medium
88 of 320
Analyzing +4 | -1
Which is bad conductor of electricity?
View answer and explanation
Correct option: A
H2F2{H_2}{F_2}
Option A (C22−C_2^{2 - } is expected to be diamagnetic) is correct because C22−C_2^{2-} has 14 electrons, resulting in a completely filled molecular orbital diagram and thus diamagnetism.
89 Medium
89 of 320
Analyzing +4 | -1
Statement I: All the halogens exhibit -1 oxidation state.

Statement II: The oxidation states of +4 and +6 occur in the oxides and oxoacids of chlorine and bromine.
View answer and explanation
Correct option: A
Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
Option D (Statement II is correct but Statement I is incorrect) is correct because not all halogens exhibit only -1 oxidation state, and the oxides and oxoacids of chlorine and bromine can have +4 and +6 oxidation states.
90 Medium
90 of 320
Application +4 | -1
Hot conc. H2SO4{H_2}S{O_4} act as moderately strong oxidizing agent. It oxidises both metals and non-metals. Which of the following element is oxidized by conc. H2SO4{H_2}S{O_4} into two gaseous products.
View answer and explanation
Correct option: C
C
Option C is correct because carbon is oxidized by concentrated H2SO4 into two gaseous products: CO2 and SO2.
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