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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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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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Showing 151-180 of 320 questions

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151 Medium
151 of 320
Analyzing +4 | -1
When a mixture of NaCl\mathrm{NaCl} and K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 is gently warmed with conc. H2SO4\mathrm{H}_2 \mathrm{SO}_4
(i) an orange-red vapour is evolved
(ii) the vapour formed when passed into NaOH\mathrm{NaOH} solution gives yellow solution of Na2CrO4\mathrm{Na}_2 \mathrm{CrO}_4
(iii) Chlorine gas is not evolved at all
(iv) chromyl chloride gas is formed
(v) SnCl4\mathrm{SnCl}_4 used instead of NaCl\mathrm{NaCl} also gives same observations
View answer and explanation
Correct option: C
3 statements are correct
Option C (3 statements are correct) is correct because statements (i), (ii), and (iv) are correct regarding the reaction of NaCl and K2Cr2O7 with conc. H2SO4.
152 Medium
152 of 320
Analyzing +4 | -1
Assertion (A): Cl2\mathrm{Cl}_2 on reaction with NaOH (Cold and dilute) gives NaClO3.\mathrm{NaClO}_3.

Reason (R): Cl2\mathrm{Cl}_2 get oxidized only in this reaction.
View answer and explanation
Correct option: D
Both (A) and (R) are false
The correct answer is D because chlorine reacts with cold and dilute NaOH to form NaCl and NaClO, not NaClO3, and chlorine can be oxidized in other reactions as well.
153 Medium
153 of 320
Understanding +4 | -1
Which blue liquid is obtained on reacting equimolar amounts of two gases at −30∘C- {30^ \circ }C ?
View answer and explanation
Correct option: B
N2O3{N_2}{O_3}
The correct answer is B (N2O3{N_2}{O_3}) because equimolar amounts of NO(g) and NO2N{O_2}(g) react at -30°C to form liquid N2O3N_2O_3, which is blue.
154 Medium
154 of 320
Remembering +4 | -1
The compound that attacks pyrex glass is:
View answer and explanation
Correct option: C
XeF6Xe{F_6}
Option C (XeF6Xe{F_6}) is correct because xenon hexafluoride reacts with silica in glass to form volatile silicon tetrafluoride and other xenon compounds, thus attacking the glass.
155 Medium
155 of 320
Analyzing +4 | -1
Assertion (A) : Equivalent mass of N2H4\mathrm{N}_2 \mathrm{H}_4 is 8 in acidic medium or basic medium.

Reason (R) : It is oxidised to N2\mathrm{N}_2 with loss of four electron per N2H4\mathrm{N}_2 \mathrm{H}_4 unit.
View answer and explanation
Correct option: A
Both A\mathbf{A} and R\mathbf{R} are true and R\mathbf{R} is the correct explanation of A\mathbf{A}.
Option A is correct because the equivalent mass of N2H4 is indeed 8, and this is because it is oxidized to N2 with a loss of four electrons per N2H4 unit, making the reason a correct explanation of the assertion.
156 Medium
156 of 320
Remembering +4 | -1
In the ring test for NO3−\mathrm{NO}_3^{-}ion, a brown ring is formed due to the formation of
View answer and explanation
Correct option: C
[Fe(H2O)5(NO)]2+\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}
Option C ([Fe(H2O)5(NO)]2+\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}) is correct because the brown ring in the nitrate test is due to the formation of the pentaaquanitrosyliron(II) complex.
157 Medium
157 of 320
Remembering +4 | -1
A colourless gas with the smell of rotten fish is
View answer and explanation
Correct option: B
PH3{PH}_3
Option B (PH3{PH}_3) is correct because phosphine gas has a characteristic smell of rotten fish.
158 Medium
158 of 320
Remembering +4 | -1
Ammonia is used in detection of Cu2+\mathrm{Cu}^{2+} ion because
View answer and explanation
Correct option: A
Aqueous solution of NH3\mathrm{NH}_3 reacts with Cu2+\mathrm{Cu}^{2+} ion to form deep blue coloured complex
Option A (Aqueous solution of NH3\mathrm{NH}_3 reacts with Cu2+\mathrm{Cu}^{2+} ion to form deep blue coloured complex) is correct because this is the basis of the test for copper(II) ions.
159 Medium
159 of 320
Understanding +4 | -1
Statement I : CIF undergoes hydrolysis as below,
CIF+H2O→HF+HOClCIF + {H_2}O \to HF + HOCl
Statement II : Fluorine containing interhalogen compounds are very useful as fluorinating agents
View answer and explanation
Correct option: A
Both Statement I and II are correct
Option A (Both Statement I and II are correct) is correct because CIF undergoes hydrolysis to form HF and HOCl, and fluorine-containing interhalogen compounds are indeed useful as fluorinating agents.
160 Medium
160 of 320
Remembering +4 | -1
Cl2  C{l_2}\; is used in the manufacture of:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because chlorine (Cl2Cl_2) is used in the manufacture of chloroform (CHCl3CHCl_3), carbon tetrachloride (CCl4CCl_4), and Westron (C2H2Cl4C_2H_2Cl_4).
161 Medium
161 of 320
Understanding +4 | -1
Assertion (A): Ammonia cannot be dried over concentrated H2SO4{H}_2 {SO}_4.

Reason (R): Conc. H2SO4{H}_2 {SO}_4 reacts with NH3{NH}_3.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because concentrated sulfuric acid reacts with ammonia, making it unsuitable as a drying agent for ammonia.
162 Medium
162 of 320
Remembering +4 | -1
Oxygen can be obtained from bleaching powder by:
View answer and explanation
Correct option: C
Heating with a cobalt salt
Option C (Heating with a cobalt salt) is correct because cobalt salts act as catalysts in the decomposition of bleaching powder to produce oxygen.
163 Medium
163 of 320
Understanding +4 | -1
The reaction of NONO with N2O4{N_2}{O_4} at 250 K gives
View answer and explanation
Correct option: C
N2O3{N_2}{O_3}
Option C is correct because the reaction of NONO with N2O4N_2O_4 at 250 K results in the formation of dinitrogen trioxide (N2O3N_2O_3).
164 Medium
164 of 320
Analyzing +4 | -1
S-I: H2O{H}_2{O} is most volatile and H2S{H}_2 {S} is least volatile hydrides of group 16 elements.

S-II: H2O{H}_2{O} has the highest and H2S{H}_2 {S} has the lowest boiling point.
View answer and explanation
Correct option: D
S-I is false, S-II is true
Option D (S-I is false, S-II is true) is correct because water is less volatile and has the highest boiling point due to hydrogen bonding.
165 Medium
165 of 320
Understanding +4 | -1
S-I: XeF2{XeF}_2 react with PF5{PF}_5 gives [XeF+][PF6]−\left[{XeF}^{+}\right]\left[{PF}_6\right]^{-}.

S-II: He{He} is used to produce NMR spectrometers and MRI systems for clinical diagnosis.
View answer and explanation
Correct option: A
Both S-I and S-II are true
Option A (Both S-I and S-II are true) is correct because XeF2 does react with PF5 to form [XeF+][PF6]-, and Helium is indeed used in NMR spectrometers and MRI systems.
166 Medium
166 of 320
Understanding +4 | -1
Boiling of dil. HClHCl acid does not increase its concentration beyond 20.24 per cent because hydrochloric acid:
View answer and explanation
Correct option: C
Forms a constant boiling mixture
Option C is correct because hydrochloric acid forms a constant boiling azeotropic mixture with water at a concentration of 20.24%.
167 Medium
167 of 320
Analyzing +4 | -1
Assertion (A): PCl5(s){PCl}_5(\mathrm{s}) is an example of molecular solid.

Reason (R): PCl5(s){PCl}_5(\mathrm{s}) exist as (PCl4)+(PCl6)−\left({PCl}_4\right)^{+}\left({PCl}_6\right)^{-}
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (If both the assertion and reason are false) is correct because the correct IUPAC name is pent-1-en-4-yne, and the triple bond is not always given a lower locant.
168 Medium
168 of 320
Analyzing +4 | -1
Which is not the disproportionation reaction?
View answer and explanation
Correct option: C
NH4NO3→N2O+2H2ON{H_4}N{O_3} \to {N_2}O + 2{H_2}O
Option C (NH4NO3→N2O+2H2ON{H_4}N{O_3} \to {N_2}O + 2{H_2}O) is correct because it is not a disproportionation reaction as it involves two different nitrogen atoms with different oxidation states reacting to form a product.
169 Medium
169 of 320
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 the incorrect statement because oxygen typically exhibits -2, -1, and +2 (in OF2) oxidation states, but not +4 or +6.
170 Medium
170 of 320
Analyzing +4 | -1
Which of the following reactions will not produce chlorine gas?
View answer and explanation
Correct option: D
K2Cr2O3+H2SO4\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{SO}_4 (conc.) +KCl→Δ+\mathrm{KCl} \stackrel{\Delta}{\rightarrow}
Option D (K2Cr2O3+H2SO4\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{SO}_4 (conc.) +KCl→Δ+\mathrm{KCl} \stackrel{\Delta}{\rightarrow}) will not produce chlorine gas, instead, it produces chromyl chloride (CrO2Cl2CrO_2Cl_2).
171 Medium
171 of 320
Analyzing +4 | -1
What is X{X} and Y{Y} in the given reactions ?

2X2(g)+2H2O(l)→4H+(aq)+4X−(aq)+O2(g)Y2(g)+H2O(l)→HY(aq)+HOY(aq)\begin{aligned} & 2 {X}_2(g)+2 {H}_2 {O}(l) \rightarrow 4 {H}^{+}(a q)+4 {X}^{-}(a q)+{O}_2(g) \\ & {Y}_2(g)+{H}_2 {O}(l) \rightarrow {HY}(a q)+{HOY}(a q) \end{aligned}
View answer and explanation
Correct option: C
X=F,Y=Cl{X}={F}, {Y}={Cl}
Option C is correct because X is F and Y is Cl, which satisfy the given reactions.
172 Medium
172 of 320
Analyzing +4 | -1
A certain compound when burnt gave three oxides. The first turned lime water milky, the second turned cobalt chloride paper pink and the third formed an aqueous solution of pH 3 nearly. The elements present in the compound are:
View answer and explanation
Correct option: C
C, H, S
Option C is correct because carbon dioxide turns lime water milky, water turns cobalt chloride paper pink, and sulfur dioxide forms an acidic solution.
173 Medium
173 of 320
Analyzing +4 | -1
Find A,  BA,\,\,B and CC in the following reactions:
NH3 + A + CO2→(NH4)2CO3(NH4)2CO3 + H2O + B→NH4HCO3NH4HCO3 + NaCl→NH4CI + C\begin{array}{l} N{H_3}\, + \,A\, + \,C{O_2} \to {(N{H_4})_2}C{O_3}\\ {(N{H_4})_2}C{O_3}\, + \,{H_2}O\, + \,B \to N{H_4}HC{O_3}\\ N{H_4}HC{O_3}\, + \,NaCl \to N{H_4}CI\, + \,C \end{array}
View answer and explanation
Correct option: D
A−H2O,  B−CO2,  C−NAHCO3A - {H_2}O,\,\,B - C{O_2},\,\,C - NAHC{O_3}
Option D (A−H2O, B−CO2, C−NAHCO3A - {H_2}O,\,B - C{O_2},\,C - NAHC{O_3}) is correct because it accurately represents the reactants involved in the Solvay process for the production of sodium bicarbonate.
174 Medium
174 of 320
Understanding +4 | -1
Phosphorous forms two chlorides which have different compositions. The oxidation number of which element remains constant?
View answer and explanation
Correct option: D
Chlorine
Option D (Chlorine) is correct because chlorine always has an oxidation number of -1 in both PCl3PCl_3 and PCl5PCl_5.
175 Medium
175 of 320
Understanding +4 | -1
Consider the following statements for lead storage battery.

a. It consists of a grid of lead packed with PbO2\mathrm{PbO}_2 as anode and a lead cathode

b. 38%38 \% solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 is used as an electrolyte

c. It is commonly used in automobiles and invertors

Which of the above statements are correct?
View answer and explanation
Correct option: B
b and c
Option B (b and c) is correct because lead storage batteries use a 38% H2SO4 solution as an electrolyte and are commonly used in automobiles and inverters, but the anode is lead and the cathode is lead packed with PbO2.
176 Medium
176 of 320
Analyzing +4 | -1
Statement I: Triple phosphate is more soluble in water than super phosphate of lime

Statement II: The CaSO4\mathrm{CaSO}_4 in super phosphate of lime is an insoluble waste product and no value to the plants
View answer and explanation
Correct option: B
Statement I and statement II are correct but statement II is not the correct explanation of statement I
The correct answer is B because both statements are correct, but statement II does not explain statement I. Triple phosphate is more soluble, and the calcium sulfate in superphosphate is an insoluble byproduct.
177 Medium
177 of 320
Analyzing +4 | -1
S2−{S}^{2-} and SO32−{SO}_3{ }^{2-} can be distinguished by using
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because both lead(II) acetate and sodium nitroprusside can distinguish between S²⁻ and SO₃²⁻ based on the formation of distinct precipitates or colored complexes.
178 Medium
178 of 320
Understanding +4 | -1
Starch paper moistened with KI\mathrm{KI} solution turns blue in ozone because of:
View answer and explanation
Correct option: A
Iodine liberation
Option A (Iodine liberation) is correct because ozone oxidizes KI to liberate iodine, which turns starch paper blue.
179 Medium
179 of 320
Understanding +4 | -1
Dipole moment of the hydrides of group 15 elements decreases in the order -
View answer and explanation
Correct option: C
NH3>PH3>AsH3>SbH3>BiH3\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3>\mathrm{BiH}_3
Option C (NH3>PH3>AsH3>SbH3>BiH3\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3>\mathrm{BiH}_3) is correct because dipole moment decreases down the group due to decreasing electronegativity difference.
180 Medium
180 of 320
Remembering +4 | -1
Bromine is obtained on a commercial scale from:
View answer and explanation
Correct option: B
Carnallite
Option B (Carnallite) is correct because bromine is commercially extracted from carnallite, a hydrated potassium magnesium chloride mineral.
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