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NEET • CHEMISTRY • S-BLOCK ELEMENTS

S-block elements practice · Mixed difficulty

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31 Hard
31 of 113
Understanding +4 | -1
Sodium metabisulphite is not:
View answer and explanation
Correct option: C
An oxidizing agent
Option C is correct because sodium metabisulphite (Na2S2O5\mathrm{Na_2S_2O_5}) acts as a reducing agent and an antichlor, and it can also be used as a bleaching agent, but it does not act as an oxidizing agent.
32 Hard
32 of 113
Remembering +4 | -1
Which of the following is commercially known as oxone ?
View answer and explanation
Correct option: A
Na2O2+HClNa_2O_2+HCl
Option A (Na2O2+HClNa_2O_2+HCl) is correct because a mixture of sodium peroxide and hydrochloric acid is commercially known as oxone.
33 Medium
33 of 113
Analyzing +4 | -1
Select the correct statement among the following
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Correct option: D
Gypsum contains a lower percentage of Ca than in plaster of Paris
Option D (Gypsum contains a lower percentage of Ca than in plaster of Paris) is correct because plaster of Paris is obtained by heating gypsum, which removes some water and increases the relative percentage of calcium.
34 Medium
34 of 113
Analyzing +4 | -1
A certain compound(X) when treated with copper sulphate solution yields a brown precipitate. On adding hypo solution, the precipitate turns white. The compound is
View answer and explanation
Correct option: B
KIKI
Option B (KIKI) is correct because KIKI reacts with copper sulphate to form CuI2CuI_2 which decomposes to CuICuI (brown ppt) and I2I_2. I2I_2 reacts with hypo to form colorless solution.
35 Easy
35 of 113
Remembering +4 | -1
The element with strong electropositive nature is:
View answer and explanation
Correct option: B
Cs\mathrm{Cs}
Option D (H2O2 oxidisesPbSintoSO2{H_2}{O_2}\,oxidises\mathrm{ }PbS\mathrm{ }into\mathrm{ }S{O_2}) is correct because H2O2{H_2}{O_2} oxidizes PbSPbS into PbSO4PbSO_4, not SO2SO_2.
36 Hard
36 of 113
Understanding +4 | -1
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) Barium carbonate is insoluble in water and is highly stable.
Reason (R) The thermal stability of the carbonates increases with increasing cationic size.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A).
Option A is correct because the thermal stability of carbonates increases with increasing cationic size, and barium has a larger cationic size compared to other alkaline earth metals, making barium carbonate more stable and less soluble.
37 Hard
37 of 113
Understanding +4 | -1
Na2CO3\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3} can be manufactured by Solvay process but K2CO3{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3} cannot be prepared because:
View answer and explanation
Correct option: C
KHCO3\mathrm{KHC}{\mathrm{O}_3} is more soluble than NaHCO3\mathrm{NaHC}{\mathrm{O}_3}
Option C is correct because potassium bicarbonate (KHCO3\mathrm{KHCO_3}) is more soluble than sodium bicarbonate (NaHCO3\mathrm{NaHCO_3}), making it difficult to precipitate KHCO3\mathrm{KHCO_3} in the Solvay process.
38 Medium
38 of 113
Understanding +4 | -1
When CaC2Ca{C_2} reacts with H2O{H_2}O then, it gives
View answer and explanation
Correct option: B
C2H2{C_2}{H_2}
Option B is correct because the reaction of calcium carbide (CaC2CaC_2) with water produces acetylene (C2H2C_2H_2) and calcium hydroxide.
39 Hard
39 of 113
Analyzing +4 | -1
Which of the following statements are correct for alkali metal compounds?
Superoxides are paramagnetic in nature.
The basic strength of hydroxides increases down the group.
The conductivity of chlorides in their aqueous solutions decreases down the group.
The basic nature of carbonates in aqueous solutions is due to cationic hydrolysis.
View answer and explanation
Correct option: B
(i), and (ii), only
Option B ((i), and (ii), only) is correct because superoxides are paramagnetic, and the basic strength of hydroxides increases down the group for alkali metals.
40 Easy
40 of 113
Analyzing +4 | -1
A white solid reacts with dil. HCl\mathrm{HCl} to give colourless gas that decolourises aqueous bromine. The solid is most likely to be:
View answer and explanation
Correct option: D
Sodium thiosulphate
Option D (Sodium thiosulphate) is correct because it reacts with dilute HCl to produce sulfur dioxide, which decolorizes aqueous bromine.
41 Easy
41 of 113
Analyzing +4 | -1
Be Shows diagonal relationship with AlAl. Which of the following similarity incorrect?
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Correct option: B
BeBe like AlAl rendered passive by Con. HNO3HN{O_3}
Option C (Be(OH)2 like Al(OH)3 is basic) is incorrect because both Be(OH)2 and Al(OH)3 are amphoteric, not basic.
42 Easy
42 of 113
Understanding +4 | -1
The correct order of density of following elements is :- (Be, Mg, Ca, Sr)
View answer and explanation
Correct option: C
Ca < Mg < Be < Sr
The correct answer is C (Ca < Mg < Be < Sr) because density generally increases down the group, but Calcium is an exception due to its larger atomic size and weaker metallic bonding compared to Magnesium.
43 Medium
43 of 113
Remembering +4 | -1
For which one of the following minerals, the composition given is incorrect?
View answer and explanation
Correct option: B
Borax – Na2B4O7.7H2O\mathrm{N}{\mathrm{a}_2}{\mathrm{B}_4}{\mathrm{O}_7}.7{\mathrm{H}_2}\mathrm{O}
Option B (Borax – Na2B4O7.7H2O\mathrm{N}{\mathrm{a}_2}{\mathrm{B}_4}{\mathrm{O}_7}.7{\mathrm{H}_2}\mathrm{O}) is incorrect. The correct formula for Borax is Na2B4O7.10H2O\mathrm{N}{\mathrm{a}_2}{\mathrm{B}_4}{\mathrm{O}_7}.10{\mathrm{H}_2}\mathrm{O}.
44 Easy
44 of 113
Understanding +4 | -1
Which is most basic in character?
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Correct option: A
CsOH\mathrm{CsOH}
Option A (CsOH\mathrm{CsOH}) is correct because Cesium hydroxide is the most basic among the given options due to Cesium's lowest ionization energy and largest size, leading to the weakest bond with hydroxide.
45 Hard
45 of 113
Understanding +4 | -1
KI and CuSO4\mathrm{CuS}{\mathrm{O}_4} solution when mixed, give
View answer and explanation
Correct option: C
K2SO4+Cu2I2+I2{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{C}{\mathrm{u}_2}{\mathrm{I}_2} + {\mathrm{I}_2}
Option C is correct because potassium iodide (KI) reacts with copper sulfate (CuSO4\mathrm{CuSO_4}) to form cuprous iodide (Cu2I2\mathrm{Cu_2I_2}), potassium sulfate (K2SO4\mathrm{K_2SO_4}), and iodine (I2\mathrm{I_2}).
46 Easy
46 of 113
Understanding +4 | -1
Which of the following statements is correct for CsBr3\mathrm{CsB}{\mathrm{r}_3} ?
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Correct option: D
It contains Cs+\mathrm{C}{\mathrm{s}^ + } and Br3−\mathrm{Br}_3^ - ions
Option D is correct because CsBr3\mathrm{CsBr}_3 is an ionic compound containing Cs+\mathrm{Cs}^+ cations and tribromide Br3−\mathrm{Br}_3^- anions.
47 Hard
47 of 113
Understanding +4 | -1
Acidified solution of sodium thiosulphate is unstable because in thiosulphate:
View answer and explanation
Correct option: B
The two sulphur atoms are in different oxidation states of +5 and -1
Option B is correct because in acidified solution, thiosulphate disproportionates due to the two sulphur atoms being in different oxidation states (+5 and -1), leading to its instability.
48 Medium
48 of 113
Analyzing +4 | -1
Which pair cannot exist together in solution?
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Correct option: C
NaHCO3andNa2CO3\mathrm{NaHC}{\mathrm{O}_3}\mathrm{and N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}
Option A (NaHCO3andNaOH\mathrm{NaHC}{\mathrm{O}_3}\mathrm{and NaOH}) is correct because NaHCO3\mathrm{NaHC}{\mathrm{O}_3} reacts with NaOH\mathrm{NaOH} to form Na2CO3\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3} and H2O{\mathrm{H}_2}\mathrm{O}, hence they cannot coexist.
49 Hard
49 of 113
Understanding +4 | -1
A metal carbonate is sparingly soluble in water and evolves CO2\mathrm{C}{\mathrm{O}_2} on heating. The metal is:
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Correct option: C
An alkaline earth metal
Option C is correct because alkaline earth metal carbonates are sparingly soluble and decompose on heating to evolve CO2CO_2.
50 Medium
50 of 113
Remembering +4 | -1
The characteristic colour exhibited by Rb atom to an oxidizing flame is
View answer and explanation
Correct option: C
Red violet
The correct answer is C (Red violet) because Rubidium (Rb) exhibits a red-violet color in an oxidizing flame.
51 Medium
51 of 113
Understanding +4 | -1
Which of the following compounds on reaction with NaOH\mathrm{NaOH} and H2O2{\mathrm{H}_2}{\mathrm{O}_2} gives yellow colour?
View answer and explanation
Correct option: B
Cr(OH)3\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3}
Option B is correct because chromium(III) hydroxide reacts with sodium hydroxide and hydrogen peroxide to form sodium chromate, which has a yellow color.
52 Medium
52 of 113
Analyzing +4 | -1
In the following reaction,

NaOH+S→A+Na2S+H2O;Ais\mathrm{NaOH} + \mathrm{S} \to A + \mathrm{N}{\mathrm{a}_2}\mathrm{S} + {\mathrm{H}_2}\mathrm{O};A\mathrm{is}
View answer and explanation
Correct option: D
Na2S2O3\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}
Option D (Na2S2O3\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}) is correct because the reaction of NaOH with S produces sodium thiosulfate (Na2S2O3\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}) along with Na2SNa_2S and water.
53 Medium
53 of 113
Application +4 | -1
One mole of magnesium nitride on the reaction with an excess of water gives
View answer and explanation
Correct option: B
Two moles of NH3\mathrm{N}{\mathrm{H}_3}
Option B (Two moles of NH3) is correct because magnesium nitride reacts with water to produce magnesium hydroxide and ammonia.
54 Hard
54 of 113
Analyzing +4 | -1
Mg→HeatAirX+Y→H2OZ→H2Osolution→CuSO4AMg\xrightarrow[{{\text{Heat}}}]{{{\text{Air}}}}X + Y\xrightarrow{{{H_2}O}}Z\xrightarrow{{{H_2}O}}{\text{solution}}\xrightarrow{{CuS{O_4}}}A
Blue coloured solution, substances X,  Y,   ZX,\,\,Y,\,\,\,Z and AA are respectively.
View answer and explanation
Correct option: C
MgO,   Mg3N2,   NH3,   (Cu(NH3)4)SO4MgO,\,\,\,M{g_3}{N_2},\,\,\,N{H_3},\,\,\,\left( {Cu{{\left( {N{H_3}} \right)}_4}} \right)S{O_4}
Option C (MgO,  Mg3N2,  NH3,  (Cu(NH3)4)SO4MgO,\,\,M{g_3}{N_2},\,\,N{H_3},\,\,\left( {Cu{{\left( {N{H_3}} \right)}_4}} \right)S{O_4}) is correct because it accurately represents the products formed during the reaction of Magnesium with air, followed by subsequent reactions with water and copper sulfate.
55 Easy
55 of 113
Analyzing +4 | -1
A metal MM forms water soluble MSO4M\mathrm{S}{\mathrm{O}_4} and inert MO.MOMO.MO in aqueous solution forms insoluble M(OH)2M{\left( {\mathrm{OH}} \right)_2} soluble in NaOH. Metal MM is
View answer and explanation
Correct option: A
Be
Option A (Be) is correct because Beryllium forms a water-soluble sulfate, an inert oxide, and an amphoteric hydroxide.
56 Easy
56 of 113
Remembering +4 | -1
Carnallite is
View answer and explanation
Correct option: D
KCl.MgCl2.6H2O\mathrm{KCl }\mathrm{. MgC}{\mathrm{l}_2}.6{\mathrm{H}_2}\mathrm{O}
Carnallite (Option D) is a hydrated double chloride of potassium and magnesium, with the formula KCl⋅MgCl2⋅6H2OKCl \cdot MgCl_2 \cdot 6H_2O.
57 Easy
57 of 113
Analyzing +4 | -1
Which substance gives a different flame colouration from the others?
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Correct option: D
Table salt
Option D (Table salt) is correct because table salt (NaCl) contains sodium, which imparts a yellow color to the flame, while the others contain potassium which imparts a violet/lilac color.
58 Easy
58 of 113
Analyzing +4 | -1
Na2SO3\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_3} and NaHCO3\mathrm{NaHC}{\mathrm{O}_3} may be distinguished by treating their aqueous solution with:
View answer and explanation
Correct option: D
MgSO4\mathrm{MgS}{\mathrm{O}_4}
Option D (MgSO4\mathrm{MgS}{\mathrm{O}_4}) is correct because Na2SO3Na_2SO_3 reacts with MgSO4MgSO_4 to form a precipitate of MgSO3MgS{O_3}, while NaHCO3NaHCO_3 does not form a precipitate.
59 Medium
59 of 113
Remembering +4 | -1
Electric cookers have a coating of ….that protects them against fire.
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Correct option: B
Magnesium oxide
The correct answer is B (Magnesium oxide) because Magnesium oxide (MgO) is a refractory material with high thermal resistance, making it suitable for coating electric cookers to protect against fire.
60 Medium
60 of 113
Analyzing +4 | -1
The compound XX on heating gives a colourless gas. The residue is dissolved in water to obtain YY . Excess CO2\mathrm{C}{\mathrm{O}_2} is bubbled through aqueous solution of Y,ZY,Z is formed. ZZ on gentle heating gives back XX . The compound XX is:
View answer and explanation
Correct option: A
CaCO3\mathrm{CaC}{\mathrm{O}_3}
Option A (CaCO3\mathrm{CaC}{\mathrm{O}_3}) is correct because it decomposes to give CO2CO_2, and the residue dissolves in water to form Ca(OH)2Ca(OH)_2, which reacts with excess CO2CO_2 to form Ca(HCO3)2Ca(HCO_3)_2, which on heating gives back CaCO3CaCO_3.
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