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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1851. |
The oxidation states of sulphur in the anions `SO_(3)^(2-), S_(2)O_(4)^(2-)`, and `S_(2)O_(6)^(2-)` follow the orderA. `S_(2)O_(4)^(2-) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)`B. `S_(2)O_(6)^(2-) lt S_(2)O_(4)^(2-) lt SO_(3)^(2-)`C. `S_(2)O_(4)^(2-) lt SO_(3)^(2-) lt S_(2)O_(6)^(2-)`D. `SO_(3)^(2-) lt S_(2)O_(4)^(2-) lt S_(2)O_(6)^(2-)` |
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Answer» Correct Answer - C Oxidation state of S in `SO_(3)^(2-)` `x + (-2 xx 3) = - 2`, `x = +6-2=+4` Oxidation state of `S` in `S_(2)O_(4)^(2-)`. `2x+(-2xx4) = -2` `2x=+8-2=+6` `x = (+6)/(2) = +3` Oxidation state of S in `S_(2)O_(6)^(2-)` `2x + (-2xx6) = -2` `2x = +12-2 =10` `x = 10/2 = +5` Hence, increasing order of oxidation states of S is `S_(2)O_(4)^(3-) lt SO_(3)^(2-) lt S_(2)O_(6)^(2-)`. |
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| 1852. |
Sulphur in `+3` oxidation state is present inA. dithionous acidB. sulphurous acidC. thionous acidD. pyrosulphuric acid |
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Answer» Correct Answer - A Dithionouos acid `(H_2S_2O_4)` has sulphur in +3 oxidation state. |
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| 1853. |
The oxidation states of sulphur in the anions `SO_(3)^(2-), S_(2)O_(4)^(2-)`, and `S_(2)O_(6)^(2-)` follow the orderA. `S_(2)O_(6)^(2-) lt SO_(4)^(2-) lt SO_(3)^(2-)`B. `S_(2)O_(4)^(2-) lt SO_(3)^(2-) lt S_(2)O_(6)^(2-)`C. `SO_(3)^(2-) lt S_(2)O_(4)^(2-) lt S_(2)O_(6)^(2-)`D. `S_(2)O_(4)^(2-) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)` |
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Answer» Correct Answer - B `O.S. "of S in" SO_(3)^(2-) =+4` `O.S. "of S in" SO_(4)^(2-) =+3` `O.S. "of S in" SO_(6)^(2-) =+5` (None of these conatains peroxide linkage). The correct order of incresasing `O.S` is , `S_(2)O_(4)^(2-) lt SO_(3)^(2-) lt S_(2)O_(6)^(2-)` |
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| 1854. |
`HCOOH underset(conc. H_(2)SO_(4)) overset(373K) rarr H_(2)O+(X)` `C(s)+H_(2)O overset(423-1273K) rarr (X)+H_(2)(g)` Mixture of (X) gas +`H_(2)` is called:A. water gas or synthesis gasB. producer gasC. methane gasD. None of the above |
| Answer» Correct Answer - A | |
| 1855. |
Sulphur in `+3` oxidation state is present inA. Sulphurous acidB. Pyrosulphuric acidC. Dithionous acidD. Thiosulphuric acid |
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Answer» Correct Answer - C Dithinous acid `(H_(2) S_(2)O_(4))` has sulphur in `+3` oxidation state. `HO - overset(O) overset(||)(S) - overset(O) overset(||)(S) - OH, 2(+1) + 2x + 4(-2) = 0` ltbrlt `2x = 8 - 2 = 6, x = +3` |
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| 1856. |
Bleaching action of `SO_(2)` is due toA. ReductionB. OxidationC. HydrolysisD. Its acidic nature |
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Answer» Correct Answer - A `underset("Colour flower "+2[H]rarr"Colour flower")(2H_(2)O+SO_(2)rarrH_(2)SO_(4)+2[H]("nascent hydrogen")` |
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| 1857. |
In the reaction `2Ag+2H_(2)SO_(4)rarrAg_(2)SO_(4)+2H_(2)O+SO_(2),H_(2)SO_(40`acts as `a//an`A. Reducing agentB. Oxidising agentC. Catalytic agentD. Dehydrating agent |
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Answer» Correct Answer - B `underset("agent")underset("Reducing")(2Ag) + underset("Agent") underset("Oxidising")(2H_(2) SO_(4)) rarr Ag_(2) SO_(4) + 2H_(2)O + SO_(2)` |
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| 1858. |
In the reaction `2Ag+2H_(2)SO_(4)rarrAg_(2)SO_(4)+2H_(2)O+SO_(2),H_(2)SO_(40`acts as `a//an`A. Reducing agentB. Oxidatising agentC. Catalytic agentD. Dehydrating agent |
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Answer» Correct Answer - B `underset("Reducing agent")(2Ag)+underset("Oxidising agent")(2H_(2)SO_(4))rarrAg_(2)SO_(4)+2H_(2)O+SO_(2)` |
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| 1859. |
Dilute solution of HF cannot be concentrated beyond 36% by distilling only because :A. HF is non volatilB. HF forms a constant boiling mixtureC. HF is least acidicD. It is bad conductor |
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Answer» Correct Answer - B HF forms a constant boiling mixture having 36% acid and boiling point of `120^@` |
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| 1860. |
It is because of inability of `ns^(2)` electrons of the valence shell to particular in bonding that:A. `Sn^(2+)` is oxidizing while `Pb^(4+)` is reducingB. `Sn^(2+)` and `Pb^(2+)` are both oxidizing and reducingC. `Sn^(4+)` is reducing while `Pb^(4+)` is oxidisingD. `Sn^(2+)` is reducing while `Pb^(4+)` is oxidising |
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Answer» Correct Answer - D `Sn^(+2)toSn^(+4)` `(R.A.) Sn^(2+) lt Sn^(+4)` Stability order `Pb^(+4)toPb^(+2)` `(O.A.) Pb^(+2)gtPb^(+2)` Stability order (inert pair effect) |
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| 1861. |
The increasing order of atomic radii of the following group 13 element isA. `Al lt Ga lt In lt Tl`B. `Ga lt Al lt In lt Tl`C. `Al lt In lt Ga lt Tl`D. `Al lt Ga ltTl lt In` |
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Answer» Correct Answer - B Due to ineffective shielding of 3d-orbitals , the effective nuclear charge of Ga is greater in magnitude than of Al. As a result, atomic radii of Fa is lower than that of Al. The atomic radii of remaining element increases regularly as we move down the group from ln to Tl. Thus, atomic radii of the four elements from Al to Tl increase in the order : `Ga lt Al lt In lt Tl`, i.e., option (b) is correct. |
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| 1862. |
In diborane, the two `H-B-H` angles are nearlyA. `60^(@),120^(@)`B. `95^(@),120^(@)`C. `95^(@) ,150^(@)`D. `120^(@),180^(@)` |
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Answer» Correct Answer - B is the correct answer. |
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| 1863. |
The increasing order of atomic radii of the following Group 13 element isA. `B lt Al lt In lt Ga lt TI`B. `B lt Al lt Ga lt In lt TI`C. `B lt Ga lt Al lt TI lt In`D. `B lt Ga lt Al lt In lt TI` |
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Answer» Correct Answer - D Due to poor screening effect of `4f` and `3d` electrons: (i) Size of `Ga` is less than `Al`. Hence the overall order is `: BltGaltAlltInltTI`. |
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| 1864. |
Assertion. Borax bead test is not stuitable for `Al (III)`. Reason. `Al_(2)O_(3)` is insoluble in `H_(2)O`.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason not is the true explanation of the assertion.C. If assertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Correct explanation. `Al_(2)O_(3)` being colourless does not form a coloured borax bead. |
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| 1865. |
Borax bead test is not suitable of for `Al(III)` `Al_(2)O_(3)` is insoluble in `H_(2)O`.A. If both assertion and reason are true and the reason is the correct explanation of teh assertion.B. If both assertion and reason are true but reason in not the correct explation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Borax bead test is used for coloured cations. `Al^(3+)` is white. |
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| 1866. |
Assertion: `Al(OH)_(3)` is insoluble in `NH_(4)OH` but soluble in `NaOH`. Reason: `NaOH` is a stronger base.A. If both assertion and reason are true and the reason is the correct explanation of teh assertion.B. If both assertion and reason are true but reason in not the correct explation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Both are correct statements but reason is not the correct explation of assertion. `Al(OH)_(3)` forms soluble complex with `NaOH` and not with `NH_(4)OH`. `Al(OH)_(3)+OH^(-)to[Al(OH)_(4)]^(-)` |
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| 1867. |
Assertion: Boron has unusually high melting point. Reason: Boron shows non-metallic character.A. If both assertion and reason are true and the reason is the correct explanation of teh assertion.B. If both assertion and reason are true but reason in not the correct explation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - B | |
| 1868. |
Select the correct statement (s).A. `XCl_(3).6H_(2)O` exists as `[X (H_(2)O))_(6)]Cl_(3)`B. With moist air, anhydrous `XCl_(3)` produce HCl gasC. When X is treated with aqueous caustic soda, hydrogen gas is liberated.D. all of the above |
| Answer» Correct Answer - D | |
| 1869. |
`XeF_(2)+H_(2)O rarr` Product (Set-I) `XeF_(4)+H_(2)O rarr` Product (Set-II) `XeF_(6)+H_(2)O rarr` Product (Set-III) Set-III experiment was performed by three different students and they got three set of products, (based on quantity of water added) set A, B and C. Which product was present and which was absent respectively, in all three cases?A. `HF,O_(2)`B. `O_(2),HF`C. `Xe,O_(2)`D. `O_(2),Xe` |
| Answer» Correct Answer - A | |
| 1870. |
Consider the following compounds. (P) `XeF_(2)` (Q) `XeF_(4)` (R ) `XeF_(6)` The correct order of tendancy of accept `F^(-)` ion is :A. `P gt Q gt R`B. `Q gt P gt R`C. `P gt R gt Q`D. `R gt Q gt P` |
| Answer» Correct Answer - D | |
| 1871. |
`XeF_(2)+H_(2)O rarr` Product (Set-I) `XeF_(4)+H_(2)O rarr` Product (Set-II) `XeF_(6)+H_(2)O rarr` Product (Set-III) Which product is not obtained in Set I?A. `Xe`B. HFC. `O_(2)`D. `XeO_(3)` |
| Answer» Correct Answer - D | |
| 1872. |
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Given reason for your choice. State the nature of bonding also. (i) `TlCl_(3), TlCl`, (ii) `AlCl_(3), AlCl` , (iii) `InCl_(3), InCl` |
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Answer» (i) Due to strong inert pair effect , +1 oxidation state of Tl is more stable than `+3`. Since compounds in lower oxidation state are ionic but covalent in higher oxidation state, therefore `TlCl_(3)` is less stable and covalent in nature but `TlCl` is more stable and is ionic in nature. (ii) Due to absence ofd-orbitals, Al does not show inert pair effect, Therefore its most stable oxidation state is +3. Thus, `AlCl_(3)` is much more stable than `AlCl`. Further in the solid or the vapour state, `AlCl_(3)` is covalent in nature but in aqueous solution , it ionizes to form `Al^(3+) (aq)` and `Cl^(-) (aq)` ions. (iii) Due to inert pair effect, indium exists in both +1 and +3 oxidation states out of which +3 oxidation state is more stable than +1 oxidation state.In other words,`InCl_(3)` is more stable than `InCl`. Being unstable, ICl undergoes disproportionation reaction. `3ICl (aq) rarr 2 In (s) + In^(3+) (aq) +3Cl^(-) (aq)` , |
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| 1873. |
Which of the following statements(s) is/are true for `XeF_(6)`?A. Its partial hydrolysis gives `XeOF_(4)`B. Its reaction with silica gives `XeOF_(4)`C. It is prepared by the reaction of `XeF_(4)` and `O_(2)F_(2)`D. Its reaction with `XeO_(3)` gives `XeOF_(4)`. |
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Answer» Correct Answer - ABCD |
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| 1874. |
Which of the following product is are obtained when `Cl_(2)O` reacts with `NH_(3)`?A. `NO_(2)`B. `N_(2)`C. `NCl_(3)`D. `NH_(4)Cl` |
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Answer» Correct Answer - BD |
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| 1875. |
Thermal decomposition product(s) of `XeF_(6)` is/are :A. XeB. `XeF_(2)`C. `XeF_(4)`D. `F_(2)` |
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Answer» Correct Answer - BCD |
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| 1876. |
Select the incorrect order .A. `He gt Ar gt Kr gt Ne gt Xe`-(abundance in air).B. `He lt Ne lt Ar lt Kr lt Xe`-(boiling point).C. `XeF_(6) gt XeF_(4) gt XeF_(2)`- (melting point)D. `XeF_(6) lt XeF_(4) lt XeF_(2)-` (Xe-F bond length) |
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Answer» Correct Answer - AC |
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| 1877. |
Which among the following statements is/are correct ?A. `XeF_(4)` and `SbF_(5)` combine to form salt.B. He and Ne do not form clathrateC. He diffuses through rubber and polyvinyl chlorideD. He has lowest boiling point in its group. |
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Answer» Correct Answer - ABCD |
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| 1878. |
Which of the following inert gas(es) form(s) clatharate compound(s) with quinol ?A. HeliumB. XenonC. KryptonD. Neon |
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Answer» Correct Answer - BC |
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| 1879. |
What are the oxidation states of phosphorus in the following ? `"Ca"_(3)P_(2)` |
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Answer» `"Ca"_(3)P_(2)` `3(+2)+2x=0` `x=-3` |
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| 1880. |
Select the incorrect order of stability:A. `Tl^(+) gt In^(+) gt Ga^(+)`B. `TI^(3+) gt In ^(3+) gt Ga ^(3+)`C. `Pb^(2+) gt Sn ^(2+) gt Ge ^(2+)`D. `Bi^(3+) gt Sb ^(3+) gt As^(3+)` |
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Answer» Correct Answer - B `TI^(3+)` is unstable due to inert pair effect |
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| 1881. |
Which reaction does not produce ammonia?A. `NH_(4)I+KOH rarr`B. `N_(2)+3H_(2) underset(Fe//Mo)overset(Delta) rarr`C. `(NH_(4))_(2)Cr_(2)O_(7) overset(Delta) rarr`D. `Mg_(3)N_(2)+HCl(aq) rarr` |
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Answer» Correct Answer - C `(NH_(4))_(2)Cr_(2)O_(7) overset(Delta)rarr N_(2)+4H_(2)O+Cr_(2)O_(3)` |
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| 1882. |
The reaction that doses not produce nitrogen is :A. heating `(NH_(4))_(2) Cr_(2)O_(7)`B. `NH_(3)`+ excess of `Cl_(2)`C. heating of `NaNH_(3)`D. heating of `NH_(4)NO_(3)` |
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Answer» Correct Answer - B `NH_(3)+underset("(excess)")(3Cl_(2))NCl_(3)+HCl` |
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| 1883. |
Ammonica can be dried by:A. Conc. `H_(2)SO_(4)`B. `P_(4)O_(10)`C. CaOD. anhydrous `CaCl_(2)` |
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Answer» Correct Answer - C `NH_(3)` being basic,should be dried with reagents that are basic. With `CaCl_(2)` (anhyd.)`iff CaCl_(2).8NH_(3)` is formed. |
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| 1884. |
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe. |
| Answer» Correct Answer - Platinum (Pt). | |
| 1885. |
Which of the following is a mixed anhydride (A)`P_4O_10` , (B)`SO_3` , ( C)`Cl_2O_6` , (D)`SO_2` |
| Answer» ( C)`Cl_2O_6` when dissolves in water produces `HCIO_3 and HCIO_4` , therefore, it is mixed anhydride. | |
| 1886. |
`H_2SO_4` cannot be used for obttaining HBr from KBr because :A. HBr oxidises `H_2SO_4`B. HBr reduces `H_2SO_4`C. HBr undergoes disproportionation.D. KBr reacts very slowly. |
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Answer» Correct Answer - B |
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| 1887. |
The number of `sigma` bonds in `P_(4)O_(10)` isA. 6B. 16C. 20D. 7 |
| Answer» Correct Answer - B | |
| 1888. |
Name the fluoro carbon used in refrigerators. |
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Answer» Correct Answer - Freon `(CF_(2)Cl_(2))` |
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| 1889. |
Euchlorine is: |
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Answer» Correct Answer - `2KCIO_(3)+4HCito2CIO_(2)+Cl_(2)uarr+ 2K^(+)+2Cl^(-)+2H_(2)O`.The mixture of various gases thus evolved is known as euchlorine. |
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| 1890. |
Name the compound which is used to obtain fluorine gas on electrolysis. At which electrode does `F_2` appears? |
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Answer» Correct Answer - Fused anhydrous potassium hydrogen fluoride `(KHF_(2))` `CaF_(2)+H_(2)SO_(4)toCaSO_(4)+2HF ; KF + HFtoK[HF_(2)] overset("electrolysis")toH_(2)+F_(2)` On Electrolysis : Cathode : `2H^(+) + 2e^(-)toH_(2)(g)` Anode : `2F^(-)toF_(2) + 2e^(-)` |
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| 1891. |
`PH_(3)`, the hydrine of phosphorus isA. metallicB. ionicC. non-metallicD. covalent |
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Answer» Correct Answer - D `PH_(3)` is a covalent hydride . |
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| 1892. |
`PH_(3)`, the hydride of phosphorous isA. metallicB. ionicC. non-metallicD. covalent |
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Answer» Correct Answer - D `PH_(3)` is a covalent hydride . |
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| 1893. |
White phosphorus on reaction with `NaOH` gives `PH_(3)` as one of the products. This is aA. dimerisation reactionB. disproportionation reactionC. condensation reactionD. precipitation reaction |
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Answer» Correct Answer - B White phosphorus on reaction with NaOH gives `PH_3` as one of the product in disproportionation reactions. ` P_(4)+3NaOH+3H_2O to 3NaH_(2)PO_(2)+PH_(3)` |
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| 1894. |
it is possible to obtain oxygen from air by fractional distillation becauseA. oxygen is in a different group of the periodic table from nitrogenB. oxygen in more reactive than nitrogenC. oxygen has higher boiling point than nitrogenD. oxygen has a lower density than nitrogen |
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Answer» Correct Answer - C Oxygen gas is prepared by fractional distillation of air. During this process, dinitrogen with less boiling point `(78K)` distills as vapours while dioxygen with higher boiling point 90 K) remains in the liquid state and can be separated. |
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| 1895. |
The gases respectively absorbed by alkaline pyrogallon and oil of cinnamon is.A. `O_(3).CH_(4)`B. `O_(2).O_(3)`C. `SO_(2).CH_(4)`D. `N_(2)O.O_(3)` |
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Answer» Correct Answer - B Pyrogallol absorb the oxygen gas and oil of cinnamon absorb the ozone `(O_(3))`. |
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| 1896. |
Which of the following elements does not show allotropy ?A. nitrogenB. phosphorusC. arsenicD. bismuth |
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Answer» Correct Answer - A The only element which does not show allotropy is nitrogen. |
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| 1897. |
How does `SO_(2)` react with the following ? `Na_(2)SO_(3)("aq")` |
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Answer» Sodium sulphite (aq) reacts with `So_(2)` to form sodium hydrogen sulphite. `Na_(2)SO_(3)+H_(2)O+SO_(2)to2NaHSO_(3)` |
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| 1898. |
Which halogen produces `O_(2)` and `O_(3)` on passing through water ? |
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Answer» Fluorine produces `O_(2)` and `O_(3)` on passing through water. `3" F"_(2)+3" H"_(2)O to6" HF"+O_(3)` `2" F"_(2)+2" H"_(2)O to4" HF"+O_(2)` |
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| 1899. |
Which of the following reactions does not produce oxygen ?A. `O_(3)+Kl+H_(2)Oto`B. `H_(2)O_(2)+Cl_(2) to`C. `KO_(2)(s)+CO_(2)(g) to`D. None |
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Answer» Correct Answer - D `(A) O_(3)+2KL+H_(2)O to l_(2)+2KOH+O_(2)` `(B) H_(2)O_(2)+Cl_(2) to 2Cl^(-)+2H^(+)+O_(2)` `(C) 4KO_(2)(s)+2CO_(2) to 2K_(2)CO_(3)(s)+3O_(2)(s)` |
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| 1900. |
Write a short note on the allotropy of sulphur. |
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Answer» The important allotropes of sulphur are a) yellow rhombic (`alpha`. Sulphur) b) Monoclinic (`beta` - sulphur) `to` The stable from is `alpha`-sulphur (at room temperature) Rhombic sulphur (`alpha`-Sulphur) : `to` Colour : Yellow. `to` Melting point : 385.8K. `to` Specific gravity : 2.06. `to` It is insouble in water and partially soluble in alcohol, benzene etc. and readily soluble in `CS_(2)`. Monoclinic sulphur (`beta` - Sulphur) : `to` Melting point : 392K `to` Specific gravity : 1.98. `to` It is soluble in `CS_(2)`. `to` Rhombic sulphur transforms to monoclinic sulphur by heating above 369K. This temperature is called transition temperature. |
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