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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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31 Hard
31 of 248
Remembering +4 | -1
Reactions of HNO3\mathrm{HNO}_3 (conc.) with I,S,P\mathrm{I}, \mathrm{S}, \mathrm{P} and C\mathrm{C} gives respectively:
View answer and explanation
Correct option: A
HIO3,H2SO4,H3PO4,CO2\mathrm{HIO}_3, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_3 \mathrm{PO}_4, \mathrm{CO}_2
Option A (HIO3,H2SO4,H3PO4,CO2\mathrm{HIO}_3, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_3 \mathrm{PO}_4, \mathrm{CO}_2) is correct because concentrated nitric acid oxidizes iodine to iodic acid, sulfur to sulfuric acid, phosphorus to phosphoric acid, and carbon to carbon dioxide.
32 Medium
32 of 248
Analyzing +4 | -1
Which of the following does not exist?
View answer and explanation
Correct option: B
[KrF3]−[SbF4]+{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }
Option B ([KrF3]−[SbF4]+{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }) is correct because the cation [SbF4]+{\left[ {Sb{F_4}} \right]^ + } is not stable and does not exist.
33 Hard
33 of 248
Analyzing +4 | -1
Pyro phosphorus acid (H4P2O5)\left( {{H_4}{P_2}{O_5}} \right)
View answer and explanation
Correct option: D
All of these
Option D is correct because Pyro phosphorus acid (H4P2O5)\left( {{H_4}{P_2}{O_5}} \right) is a dibasic acid, a reducing acid, and contains a P−O−PP - O - P bond.
34 Medium
34 of 248
Understanding +4 | -1
The element which evolves two gases on reacting with conc. H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} is:
View answer and explanation
Correct option: B
C\mathrm{C}
Option B (C\mathrm{C}) is correct because carbon reacts with concentrated H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} to produce CO2CO_2 and SO2SO_2 gases.
35 Medium
35 of 248
Remembering +4 | -1
Javelle water is:
View answer and explanation
Correct option: C
Both (1) and (2)
Option C is correct. Javelle water is an aqueous solution of sodium hypochlorite (NaOCl) and is used as a bleaching agent.
36 Easy
36 of 248
Analyzing +4 | -1
In which reaction does SO2S{O_2} act as oxidizing agent?
View answer and explanation
Correct option: D
H2S{H_2}S
Option D (H2S{H_2}S) is correct because in the reaction with H2S\mathrm{H_2S}, SO2\mathrm{SO_2} oxidizes H2S\mathrm{H_2S} to sulfur, thus acting as an oxidizing agent.
37 Medium
37 of 248
Application +4 | -1
XeF4\text{XeF}_{4} reacts with SbF5\mathrm{SbF}_{5} to form [XeFm]n+[SbFy]2−\left[\mathrm{XeF}_{\text{m}}\right]^{\text{n}+}\left[\mathrm{SbF}_{\text{y}}\right]^{2-}.

m+n+y+z=? \mathrm{m}+\mathrm{n}+\mathrm{y}+\mathrm{z}=\text {? }

View answer and explanation
Correct option: A
11
Option A (11) is correct because the reaction forms [XeF3+][SbF6−][XeF_3^+][SbF_6^-], so m=3, n=1, y=6, and z=1, thus m+n+y+z = 3+1+6+1 = 11.
38 Easy
38 of 248
Application +4 | -1
Mg on heating to redness in an atmosphere of N2{N_2} and then on treating with H2O{H_2}O gives:
View answer and explanation
Correct option: A
NH3N{H_3}
Option A (NH3N{H_3}) is correct because magnesium reacts with nitrogen at high temperatures to form magnesium nitride, which then hydrolyzes to ammonia.
39 Easy
39 of 248
Application +4 | -1
In which of the following compound central atom has +4+4 oxidation state?
View answer and explanation
Correct option: B
H2SO3\mathrm{H}_2 \mathrm{SO}_3
Option B (H2SO3\mathrm{H}_2 \mathrm{SO}_3) is correct because, in this compound, the oxidation state of sulfur is +4, calculated by considering the oxidation states of hydrogen (+1) and oxygen (-2).
40 Medium
40 of 248
Understanding +4 | -1
On heating silver nitrate strongly ……. is obtained finally:
View answer and explanation
Correct option: D
All
Option D (All) is correct because on heating silver nitrate strongly, it decomposes to form silver metal, nitrogen dioxide, and oxygen gas.
41 Easy
41 of 248
Application +4 | -1
Which Cl2\mathrm{Cl}_2 gas reacts with cold and dilute sodium hydroxide solution, the oxidation number of chlorine changes from:
View answer and explanation
Correct option: D
Zero to -1 and zero to +1
Option D (Zero to -1 and zero to +1) is correct because chlorine undergoes disproportionation in cold and dilute NaOH, changing from 0 to -1 and +1.
42 Easy
42 of 248
Understanding +4 | -1
Which of the following halogen acids is least basic?
View answer and explanation
Correct option: D
HI
Option D (HI) is correct because HI is the strongest acid among the halogen acids due to the large size of the iodide ion and weak H-I bond, making it the least basic.
43 Hard
43 of 248
Understanding +4 | -1
FeCl3FeC{l_3} solution on reaction with SO2S{O_2} changes to:
View answer and explanation
Correct option: A
FeCl2FeC{l_2}
Option A (FeCl2FeCl_2) is correct because FeCl3FeCl_3 is reduced to FeCl2FeCl_2 by SO2SO_2, which acts as a reducing agent.
44 Easy
44 of 248
Remembering +4 | -1
Major credit for the discovery of noble gases is given to:
View answer and explanation
Correct option: B
Ramsay
Option B (Ramsay) is correct because Sir William Ramsay is credited with the discovery of noble gases.
45 Medium
45 of 248
Remembering +4 | -1
Atomicity of sulphur in rhombic sulphur is
View answer and explanation
Correct option: A
8
Option A (8) is correct because rhombic sulfur exists as S8 molecules, meaning each molecule contains 8 sulfur atoms.
46 Hard
46 of 248
Analyzing +4 | -1
Assertion: P4O10\mathrm{P}_4 \mathrm{O}_{10} dehydrates HNO3\mathrm{HNO}_3.

Reason: In P4O10\mathrm{P}_4 \mathrm{O}_{10} oxidation state of P\mathrm{P} is +5
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both statements are true, the high affinity of P4O10P_4O_{10} for water is what drives the dehydration, not the oxidation state of P.
47 Hard
47 of 248
Understanding +4 | -1
In the compounds of type ECl3E\mathrm{C}{\mathrm{l}_3}, Where E=B,PE = \mathrm{ }B,\mathrm{ }P, As,orBi,As,\mathrm{ or }Bi, the angle Cl−E−Cl\mathrm{Cl} - E - \mathrm{Cl} for different EE are in the order :
View answer and explanation
Correct option: B
B>P>As>Bi\mathrm{B > P > As > Bi}
Option B (B>P>As>Bi\mathrm{B > P > As > Bi}) is correct because as the electronegativity of the central atom decreases down the group, the bond pair-bond pair repulsion decreases, leading to smaller bond angles.
48 Easy
48 of 248
Understanding +4 | -1
Which of the following is a nitric acid anhydride
View answer and explanation
Correct option: C
N2O5{N_2}{O_5}
Option C (N2O5{N_2}{O_5}) is correct because dinitrogen pentoxide reacts with water to form nitric acid, making it the anhydride of nitric acid.
49 Easy
49 of 248
Understanding +4 | -1
PCl5\mathrm{PC}{\mathrm{l}_5} exists but NCl5\mathrm{NC}{\mathrm{l}_5} does not because:
View answer and explanation
Correct option: A
Nitrogen has no vacant 2d-orbitals
Option A (Nitrogen has no vacant 2d-orbitals) is correct because nitrogen, being a second-period element, does not have available 2d orbitals to expand its valence shell and form five bonds, unlike phosphorus.
50 Medium
50 of 248
Analyzing +4 | -1
The perhalate ion with maximum oxidizing power is:
View answer and explanation
Correct option: B
BrO4−BrO_4^ -
Option B (BrO4−BrO_4^ -) is correct because among the perhalate ions, BrO4−BrO_4^- has the highest reduction potential and thus the maximum oxidizing power.
51 Easy
51 of 248
Analyzing +4 | -1
Assertion (A) : The atomic radii of the elements of the oxygen family are smaller than the atomic radii of the corresponding elements of the nitrogen family.

Reason (R): The members of the oxygen family are more electronegative and thus have lower values of nuclear charge than those of the nitrogen family.
View answer and explanation
Correct option: C
A\mathbf{A} is true but R\mathbf{R} is false.
Option C (A\mathbf{A} is true but R\mathbf{R} is false) is correct because the atomic radii decrease across a period due to increasing effective nuclear charge, not due to lower nuclear charge.
52 Hard
52 of 248
Application +4 | -1
SO2+2x→2H2O+3S,S{O_2} + 2x \to 2{H_2}O + 3S, the compound “xx” is
View answer and explanation
Correct option: A
H2S{H_2}S
Option A (H2S{H_2}S) is correct because it reduces SO2S{O_2} to elemental sulfur, forming water as a byproduct.
53 Hard
53 of 248
Understanding +4 | -1
In the reaction K+SO2→K + S{O_2} \to...., the products are:
View answer and explanation
Correct option: B
K2SO3+K2S2O3{K_2}S{O_3} + {K_2}{S_2}{O_3}
Option B (K2SO3+K2S2O3{K_2}S{O_3} + {K_2}{S_2}{O_3}) is correct because potassium reacts with sulfur dioxide to form potassium sulfite and potassium thiosulfate.
54 Easy
54 of 248
Understanding +4 | -1
General decreasing order of stability of oxides is
View answer and explanation
Correct option: B
I >Cl> Br
Option B (I >Cl> Br) is correct because the stability of oxides generally decreases as you move up the halogen group due to decreasing bond strength.
55 Medium
55 of 248
Understanding +4 | -1
In ramsay and rayleigh’s isolation of noble gases from air, the nitrogen of the air is finally converted into
View answer and explanation
Correct option: D
NaNO2NaN{O_2} and NaNO3NaN{O_3}
Option D (NaNO2NaNO_2 and NaNO3NaNO_3) is correct because in Ramsay and Rayleigh's method, nitrogen is converted into sodium nitrite and sodium nitrate.
56 Medium
56 of 248
Application +4 | -1
In the reaction,

HNO3+P4O10→4HPO3+XHN{O_3} + {P_4}{O_{10}} \to 4HP{O_3} + X

the product X is
View answer and explanation
Correct option: B
N2O5{N_2}{O_5}
Option B (N2O5{N_2}{O_5}) is correct because the reaction of nitric acid with phosphorus pentoxide produces metaphosphoric acid and dinitrogen pentoxide.
57 Medium
57 of 248
Remembering +4 | -1
The gas which is absorbed by ferrous sulphate solution giving blackish brown colour is:
View answer and explanation
Correct option: D
NO
Option D (NO) is correct because nitric oxide (NO) reacts with ferrous sulfate to form a complex, resulting in a blackish-brown color.
58 Medium
58 of 248
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 A (+3, +4, +5) is correct because it accurately represents the oxidation states of P in each compound, calculated based on electronegativity differences and charge neutrality.
59 Hard
59 of 248
Analyzing +4 | -1
Assertion (A): Nitrogen is restricted to a maximum covalency of 4 .

Reason (R):Nitrogen exists as a diatomic molecule (N2)with a triple bond.
View answer and explanation
Correct option: B
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is not the correct explanation of A.
Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. Nitrogen's maximum covalency is 4, and it exists as a diatomic molecule with a triple bond, but these are independent facts.
60 Easy
60 of 248
Analyzing +4 | -1
Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by other
View answer and explanation
Correct option: B
Used as a food preservative
Option B (Used as a food preservative) is correct because only SO2SO_2 is used as a food preservative, while NO2NO_2 is not.
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