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NEET • CHEMISTRY • HYDROGEN

Hydrogen practice · Medium

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61 Medium
61 of 94
Application +4 | -1
Calculate the normality of 10 volume H2O2{H}_2 {O}_2 ?
View answer and explanation
Correct option: A
1.7 N1.7 {~N}
Option A (1.7 N1.7 {~N}) is correct. The normality of 10 volume H2O2 is calculated using the formula: Normality = Volume strength / 5.6.
62 Medium
62 of 94
Remembering +4 | -1
In solid hydrogen the intermolecular bonding is
View answer and explanation
Correct option: B
Vander waals
Option B is correct because in solid hydrogen, the molecules are held together by weak van der Waals forces, which are intermolecular forces arising from temporary fluctuations in electron distribution.
63 Medium
63 of 94
Understanding +4 | -1
H2O2{\mathrm{H}_2}{\mathrm{O}_2} restores the colour of old lead paintings,blackened by the action of H2S{\mathrm{H}_2}\mathrm{S} gas,by:
View answer and explanation
Correct option: B
Oxidising PbS\mathrm{PbS} to PbSO4\mathrm{PbS}{\mathrm{O}_4}
Option B (Oxidising PbS\mathrm{PbS} to PbSO4\mathrm{PbS}{\mathrm{O}_4}) is correct because H2O2{\mathrm{H}_2}{\mathrm{O}_2} oxidizes black lead sulfide (PbS) to white lead sulfate (PbSO4\mathrm{PbS}{\mathrm{O}_4}), restoring the painting's color.
64 Medium
64 of 94
Remembering +4 | -1
The catalyst used in Bosch process of manufacture of H2{H_2} is
View answer and explanation
Correct option: D
Fe2O3+Cr2O3F{e_2}{O_3} + C{r_2}{O_3}
Option D (Fe2O3+Cr2O3F{e_2}{O_3} + C{r_2}{O_3}) is correct because it is the catalyst used in the Bosch process for the manufacture of H2{H_2}.
65 Medium
65 of 94
Application +4 | -1
When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of Hydrogen evolved is
View answer and explanation
Correct option: A
1:11:1
Option A (1:11:1) is correct because zinc reacts with both sulfuric acid and sodium hydroxide to produce hydrogen gas in a 1:1 molar ratio.
66 Medium
66 of 94
Analyzing +4 | -1
Assertion : We feel cold on touching the ice.

Reason : Ice is a solid form of water.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain why we feel cold.
67 Medium
67 of 94
Understanding +4 | -1
On shaking H2O2{\mathrm{H}_2}{\mathrm{O}_2} with acidified potassium dichromate and ether, ethereal layer becomes
View answer and explanation
Correct option: C
Blue
Option C (Blue) is correct because H2O2{\mathrm{H}_2}{\mathrm{O}_2} oxidizes Cr3+Cr^{3+} to CrO5CrO_5 which has a blue color when dissolved in ether.
68 Medium
68 of 94
Understanding +4 | -1
Both temporary and permanent hardness are removed on boiling water with
View answer and explanation
Correct option: B
Na2CO3N{a_2}C{O_3}
Option B is correct because sodium carbonate reacts with both calcium and magnesium ions to form insoluble carbonates, removing both temporary and permanent hardness.
69 Medium
69 of 94
Understanding +4 | -1
Metal which does not react with cold water but evolves H2{H_2} with steam is:
View answer and explanation
Correct option: D
FeFe
Option D (FeFe) is correct because iron reacts with steam to produce hydrogen gas, but does not react with cold water.
70 Medium
70 of 94
Understanding +4 | -1
A molten ionic hydride on electrolysis gives
View answer and explanation
Correct option: C
H2{H_2} liberated at anode
Option C (H2{H_2} liberated at anode) is correct because in molten ionic hydrides, the hydride ion (H−{H^-}) migrates to the anode and gets oxidized to form hydrogen gas.
71 Medium
71 of 94
Understanding +4 | -1
What is formed when calcium carbide reacts with Heavy water
View answer and explanation
Correct option: A
C2D2{C_2}{D_2}
Option A (C2D2{C_2}{D_2}) is correct because calcium carbide reacts with heavy water to produce deuterated acetylene (C2D2{C_2}{D_2}).
72 Medium
72 of 94
Remembering +4 | -1
Tailing of mercury is a laboratory test for:
View answer and explanation
Correct option: A
O3{\mathrm{O}_3}
Option A is correct because the tailing of mercury is a laboratory test specifically for O3{\mathrm{O}_3}.
73 Medium
73 of 94
Remembering +4 | -1
Which of the following is the laboratory method of preparation of dihydrogen?
View answer and explanation
Correct option: C
Reaction of zinc with aqueous alkali
Option C is correct because the reaction of zinc with aqueous alkali produces dihydrogen in the laboratory.
74 Medium
74 of 94
Understanding +4 | -1
Hydrogen gas will not reduce:
View answer and explanation
Correct option: D
Heated aluminium oxide
Option D (Heated aluminium oxide) is correct because aluminium oxide is very stable and cannot be reduced by hydrogen gas.
75 Medium
75 of 94
Remembering +4 | -1
The O−O−HO - O - H bond angle in H2O{H_2}O is
View answer and explanation
Correct option: D
970{97^0}
Option D is correct because the O-O-H bond angle in H2O is approximately 97 degrees due to the two lone pairs on the oxygen atom causing greater repulsion and reducing the bond angle from the ideal tetrahedral angle.
76 Medium
76 of 94
Remembering +4 | -1
Which one of the following is used for laboratory preparation of D2O2{D_2}{O_2}
View answer and explanation
Correct option: A
K2S2O8{K_2}{S_2}{O_8}
Option A is correct because K2S2O8{K_2}{S_2}{O_8} is used for the laboratory preparation of D2O2{D_2}{O_2}.
77 Medium
77 of 94
Analyzing +4 | -1
Hydrogen peroxide oxidises [Fe(CN)6]4−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} to [Fe(CN)6]3−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} in acidic medium but reduces [Fe(CN)6]3−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} to [Fe(CN)6]4−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} in alkaline medium. The other products formed are, respectively.
View answer and explanation
Correct option: C
H2O   and  (H2O+O2){H_2}O\mathrm{ }\,\,\,\mathrm{and}\,\,\mathrm{ }\left( {{H_2}O + {O_2}} \right)
Option D (H2O(v){\mathrm{H}_2}\mathrm{O}\left( v \right)) is correct because washing soda (sodium carbonate decahydrate) loses water of crystallization upon heating.
78 Medium
78 of 94
Remembering +4 | -1
Ordinary hydrogen at room temperature is a mixture of
View answer and explanation
Correct option: A
75% Ortho Hydrogen + 25% para Hydrogen
Option A (75% Ortho Hydrogen + 25% para Hydrogen) is correct because ordinary hydrogen at room temperature exists as a mixture of 75% Ortho and 25% para hydrogen.
79 Medium
79 of 94
Understanding +4 | -1
The best explanations for not placing hydrogen with the group of alkali metals or halogens is:
View answer and explanation
Correct option: C
The ionization energy of hydrogen is too high for group of alkali metals but too low for halogen group
Option C (The ionization energy of hydrogen is too high for group of alkali metals but too low for halogen group) is correct because hydrogen's ionization energy falls between that of alkali metals and halogens.
80 Medium
80 of 94
Analyzing +4 | -1
To an aqueous solution containing ions such as Al3+,Zn2+, Ca2+, Fe3+,Ni2+,Ba2+ and  Cu2+A{l^{3 + ,}}Z{n^{2 + ,\,}}C{a^{2 + ,\,}}F{e^{3 + ,}}N{i^{2 + ,}}B{a^{2 + }}\,and\,\,C{u^{2 + }} conc. HClHCl was added followed by H2S{H_2}S . The total number of cations precipitated during this reaction is/are
View answer and explanation
Correct option: A
1
Option A (1) is correct. Only Cu2+Cu^{2+} precipitates as CuSCuS when H2SH_2S is passed through an acidic solution.
81 Medium
81 of 94
Understanding +4 | -1
The in correct statement(s) about heavy water is (are)
A) used as a moderator in nuclear reactor.
B) obtained as a by-product in fertiliser industry.
C) used for the study of reaction mechanism.
D) has a higher dielectric constant than water.
Choose the correct answer from the options given below:
View answer and explanation
Correct option: C
(4) only
Option C ((4) only) is correct because heavy water has a lower dielectric constant than water, not a higher one.
82 Medium
82 of 94
Application +4 | -1
The molarity of pure water at 4∘C4^\circ C is:
View answer and explanation
Correct option: D
55.5  M55.5\;M
The molarity of pure water at 4°C is 55.5 M because it is calculated by dividing the density of water by its molar mass.
83 Medium
83 of 94
Application +4 | -1
XH2{XH}_2 (ion exchange resin) can replace Ca+2{Ca}^{+2} ions in hard water as XH2+Ca+2⟶XCa+2H+{XH}_2+{Ca}^{+2} \longrightarrow {XCa}+2 {H}^{+}. If 1 L1 \mathrm{~L} hard water after passing through XH2{XH}_2 has a pH=4\mathrm{pH}=4 then hardness in parts per million of Ca+2{Ca}^{+2} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the pH of 4 corresponds to a specific H+ concentration, which relates to the Ca2+ concentration.
84 Medium
84 of 94
Analyzing +4 | -1
When a substance 'AA' reacts with water, it produces a combustible gas ' B{B} ' and a solution of substance ' C{C} ' in water. while another substance ' DD ' reacts with solution of ′C′'C' to produce the same gas ′B′'B' on warming while ' D{D} ' can produce gas ' B{B} ' on reaction with dilute H2SO4{H}_2 {SO}_4 at room temperature. 'AA' imparts a deep golden yellow colour to a smokeless flame on Bunsen burner. Identify ′A′,′B′,′C′'A', 'B', 'C' and ' D{D} ' respectively are
View answer and explanation
Correct option: B
Na,H2,NaOH,Zn{Na}, {H}_2, {NaOH}, {Zn}
Option B (Na,H2,NaOH,Zn{Na}, {H}_2, {NaOH}, {Zn}) is correct because sodium reacts with water to produce hydrogen gas and sodium hydroxide, and zinc reacts with sodium hydroxide to produce hydrogen gas.
85 Medium
85 of 94
Remembering +4 | -1
The starting material for convenient preparation of deuterated hydrogen peroxide (D2O2)\left(\mathrm{D}_{2} \mathrm{O}_{2}\right) in laboratory is
View answer and explanation
Correct option: D
K2S2O8\mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{8}
Option D (K2S2O8\mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{8}) is correct because deuterated hydrogen peroxide (D2O2D_2O_2) is conveniently prepared in the laboratory by the reaction of potassium peroxodisulfate (K2S2O8K_2S_2O_8) with heavy water (D2OD_2O).
86 Medium
86 of 94
Understanding +4 | -1
Given below are two statements.
One is labelled as assertion (A) and the other is labelled as Reason (R).
Assertion (A) Heavy water is used for the study of reaction mechanism.
Reason (R) The rate of reaction for the cleavage of O−HO - H bond is slower than the of O−DO - D bond.
Choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: D
(A) is true but (R) is false.
Option A (Both (A) and (R) are true but (R) is not the correct explanation of (A)) is correct because while both the assertion and reason are true statements, the reason does not explain why heavy water is used to study reaction mechanisms.
87 Medium
87 of 94
Understanding +4 | -1
Which of the following statements is correct? Dielectric constant of H2O2{\mathrm{H}_2}{\mathrm{O}_2}
View answer and explanation
Correct option: A
Increases with dilution
Option A (Increases with dilution) is correct because the dielectric constant of H2O2{\mathrm{H}_2}{\mathrm{O}_2} increases as its concentration decreases due to the increased relative proportion of water, which has a higher dielectric constant.
88 Medium
88 of 94
Understanding +4 | -1
The ionization energy of hydrogen is:
View answer and explanation
Correct option: D
Closer to halogens
Option D (Closer to halogens) is correct because hydrogen's ionization energy is more similar to that of halogens than alkali metals.
89 Medium
89 of 94
Understanding +4 | -1
Ortho and parahydrogen differ in
View answer and explanation
Correct option: D
Direction of spin of nuclei
Option D (Direction of spin of nuclei) is correct because ortho and parahydrogen differ in the relative spin of their nuclei.
90 Medium
90 of 94
Application +4 | -1
If percentage strength of H2O2{H_2}{O_2} solution is 17% then the volume strength of H2O2{H_2}{O_2} solution will be
View answer and explanation
Correct option: C
56 V
Option C (56 V) is correct because the volume strength of H2O2 is calculated by multiplying the percentage strength by 5.6.
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