Saved Bookmarks
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.
| 101. |
Higher oxidation state of manganese in fluoride is `+4 (MnF_(4))` but highest oxidation state in oxides is `+7(Mn_(2)O_(7))` becauseA. fluorine is more electronegative than oxygenB. fluorine does not posses d- orbitals.C. fluorine stabilises lower oxidation state.D. In covalent compounds fluorine can form single bond only while oxygen forms double bond. |
|
Answer» Correct Answer - D Fluorine can form single bond only white `O_(2)` forms double bonds. |
|
| 102. |
When `MnO_(2)` is fused with `KOH`, a coloured compound is formed. The product and its colour isA. `K_(2)MnO_(4)`,purple greenB. `KMnO_(4)` , purpleC. `Mn_(2)O_(3)`, brownD. `Mn_(3)O_(4)`, black |
|
Answer» Correct Answer - A (a) Stable oxidation state of Mn in alkaline medium is `+6` so, `MnO_(2)` is oxidised to `K_(2)MnO_(4)` ( purple green ) by atmospheric oxygen in KOH medium. `2MnO_(2)+4KOH+O_(2)rarrunderset(("purple"))(2K_(2)MnO_(4))+2H_(2)O` |
|
| 103. |
Match the compounds in column I with the products obtained on heating them in column II `:` A. `FeSO_(4)" "(PSO_(2)`B. `CuSO_(4)" "(Q)SO_(3)`C. `AgNO_(3)" "(R ) O_(2)`D. `{:(Zn(SO_(3))_(2),,(S)NO,,),(,,(T)NO_(2),,):}` |
|
Answer» Correct Answer - `A-P,Q;B-Q;C-R,T;D-R,T` Conceptual |
|
| 104. |
`K_(2)MnO_(4)` can be converted to `KMnO_(4)` byA. `Cl_(2)`B. `O_(3)`C. `CO_(2)`D. Anodic oxidation |
|
Answer» Correct Answer - A,B,C,D Al can oxidize `MnO_(4)^(2-)` to `MnO_(4)^(-)`. |
|
| 105. |
The diamagnetic compounds is/areA. `HgCl_(2)` is oxidised to `HgCl_(4)`B. `Hg_(2)Cl_(2)`C. `Cu_(2)Cl_(2)`D. `K_(2)Cr_(2)O_(7)` is oxidizing and hygroscopic |
|
Answer» Correct Answer - A,B,C,D Due to lakh of unpaired electrons they are diamagnetic compounds |
|
| 106. |
`KMnO_(4)` acts as on oxidising agent in alkaline medium. When alkaline `KMnO_(4)` is treated with `KI`, iodide ion is oxidised to `…………….`A. `I_(2)`B. `IO^(-)`C. `IO_(3)^(-)`D. `IO_(4)^(-)` |
|
Answer» Correct Answer - C `KMnO_(4)+KIrarrKIO_(3)` |
|
| 107. |
Copper becomes green when exposed to moist air for longer period because of the formation of a layer ofA. cupric oxide on the surface of copperB. basic carbonate of copper on the surface of copperC. cupric hydroxide on the surface of copperD. cupric acetate on the surface of copper |
|
Answer» Correct Answer - B `3Cu+H_(2)O+CO_(2)+O_(2)rarrCuCO_(3).Cu(OH)_(2)` ( green coloured basic carbonate of copper is formed ). |
|
| 108. |
Photography is based on the nature of silver halides. Except AgF, the silver halides are photosensitive. These undergo decomposition in light and turn black due to formation of free silver. `2AgBroverset(light)to2Ag+Br_2` The photography films are prepared by adding `20%` aqueous solution of `AgNO_3` to `NH_4Br` solution containing gelatic. When such a film is exposed, emulsion gets affected and a latent image is formed on the film. When this exposed film or plate is dipped ina developer which contains a reducing agent, the part affected most during exposure are reduced to the maximum. The image becomes visible. It is called a negative. The remaining sensitive emulsion on the negative is removed by dissolving it in hypo solution (fixer). Fihnally, a positive of the nagative already prepared is made on silver bromide paper. Q. The exposed part of the film or plate after developing containsA. silver metalB. silver oxideC. silver bromideD. silver nitrate |
|
Answer» Correct Answer - A Conceptual |
|
| 109. |
Which of the following reactions are disproportionation reactions? `(A)" " Cu^(+)rarrCu^(2+)+Cu` `(B)" "3 MnO_(4)^(2-)+4H^(+)rarr2MnO_(4)^(-)+MnO_(2)+2H_(2)O` `(C )" "2KMnO_(4)rarrK_(2)MnO_(4)+MnO_(2)+O_(2)` `(D)" "2MnO_(4)^(-)+3Mn^(2+)+2H_(2)O rarr5MnO_(2)+4H^(+)`A. a,bB. a,b,cC. b,c,dD. a,d |
|
Answer» Correct Answer - A both oxidation and reduction takes place by same element is called disproportionate reaction `Cu^(+1)rarrCu^(+2)+Cu` |
|
| 110. |
Incorrect StatementA. `SnCl_(2)` reduces Nitrobenzene to anilineB. `SnCl_(2)` reduces gold chloride `(AuCl_(3))` to metallic gold sol ( purple of cassius)C. When `SnCl_(2)` is treated with excess of `NaOH` an insoluble `Sn(OH)_(2)` which precipitate is formed.D. When `SnCl_(2)` reacts with conc. `HCL,` chlorostannic acid `H_(2)SnCl_(6)` is formed. |
|
Answer» Correct Answer - C `SnCl_(2)+NaOHrarrSn(OH)_(2)darr` `SnCl_(2)+underset(("excess"))(NaOH)rarrNa_(2)SnO_(2)("soluble")` |
|
| 111. |
Which of the following give `O_(2)` on strong heating ?A. `KMnO_(4)`B. `K_(2)Cr_(2)O_(7)`C. `AgNO_(3)`D. `ZnO` |
|
Answer» Correct Answer - A,B,C,D All decomposes to give `O_(2)` |
|
| 112. |
Which of the following statement is wrong regarding copper sulphate.A. It reacts with NaOH and Glucose to give `Cu_(2)O`B. It reacts with KCl to give `Cu_(2)O`C. It gives CuO on strong heating in airD. It reacts with KI to give brown colouration |
|
Answer» Correct Answer - A,C,D `underset((M))(Z)overset(dlHNO_(3))(rarr)underset("Colourless solution")(Zn(NO_(3))_(2))underset(NaOH)overset(Ag)(rarr)underset("White ppt")(Zn(OH)_(2))underset(NaOH)overset("Excess")(rarr)underset("Soluble")(Na_(2)[Zn(OH)_(4)])overset(H_(2)S)(rarr)underset("Whiteppt")(Zns)` |
|
| 113. |
Identify the statement which is correct for copper sulphateA. It reacts with NaOH and Glucose to give `Cu_(2)O` ( brick red ppt)B. It reacts with Kci to give `Cu_(2)Cl`C. It gives CuO ono strong heatingD. It reacts with KI to gives `I_(2)` |
|
Answer» Correct Answer - A,C,D `CuSO_(4)+NaOH//"glucose"rarr underset(("brick red ppt"))(Cu_(2)O)` `CuSO_(4)overset(Delta)(rarr)CuO+SO_(3),CuSO_(4)+KIrarrI_(2)` |
|
| 114. |
To the `CuSO_(4)` solution, `NaHSO_(3)` and HCl were added followed by ammonium thiocanate solution. The compound formed isA. `Cu(SCN)_(2)`B. `CuSCN`C. `Cu(SO_(3))_(2)`D. `CuSO_(4)` |
|
Answer» Correct Answer - B `Cu^(2+)` is reduced to `Cu^(+1)` by `SO_(2)`. CuSCN get precipitated. |
|
| 115. |
When caustic soda is added in excess to a mercurous chloride in HCl then what all are obtained ?A. HgB. `Hg_(2)O`C. HgOD. `HgO_(2)` |
|
Answer» Correct Answer - A,C `NaOH+HgCl_(2)+HClrarrHg+HgO` |
|
| 116. |
`Pb_(3)O_(4)+HNO_(3)rarr A+B+H_(2)O.`A is soluble in water. The oxidation state of lead in B is |
|
Answer» Correct Answer - 4 `Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)darr+2H_(2)O` |
|
| 117. |
Zinc and cold dil. `HNO_(3)` reacts to produceA. `NH_(4)NO_(3)`B. `N_(2)O`C. `NO`D. `NO_(2)` |
|
Answer» Correct Answer - A `3Zn+8HNO_(3)("dil")rarr3Zn(NO_(3))_(2)+2NO+4H_(2)O` |
|
| 118. |
In the oxidation reactions of `KMnO_(4)` , in acidic medium, only `H_(2)SO_(4)` is used to produce acidic medium but not HCl(or ) `HNO_(3)` becauseA. HCl reacts with `KMnO_(4)` and produces `Cl_(2)`, `HNO_(3)` itself is oxidizing agent.B. `HNO_(3)` reacts with `KMnO_(4)` and produces `Cl_(2)`, HCl itself acts as oxidizing agent.C. Both `HNO_(3)` and HCl liberates `Cl_(2)` from `KMnO_(4)`D. Both `HNO_(3)` and HCl liberates `O_(2)` from `KMnO_(4)` |
|
Answer» Correct Answer - A HCl reacts with `KMnO_(4)` and produce `Cl_(2) . HNO_(3)` acts as oxidising agent. |
|
| 119. |
For the given aqueous reactions, which of the statements(s) is ( are) true ? Excess `KI+K_(3)[Fe(CN)_(6)]overset("dilute"H_(3)SO_(4))(rarr)` brownish yellow solution `overset(ZnSO_(4))(rarr) white ppt + brownish yellow fil trate overset(Na_(2)S_(2)O_(3))(rarr)` colourless solution`A. The first reaction is a redox reactionB. White precipitate is `Zn_(3)[Fe(CN)_(6)]_(2)`C. Addition of filtration to starch solution gives blue colourD. White precipitate is soluble in NaOH solution |
|
Answer» Correct Answer - A,C,D Brownish yellow solution is `K_(4)[Fe(CN)_(6)]` White ppt is `K_(2)Zn[Fe(CN)_(6)]` or `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)&` Brownish yellow filterate is `I_(2)`. |
|
| 120. |
Copper sulphate is prepared by blowing a current of air through copper scrap and dilute `H_(2)SO_(4)`. Dilute `HNO_(3)` is also added`:`A. to oxidise copper to `Cu^(2+)` which then form `CuSO_(4)` with dilute `H_(2)SO_(4)`B. to oxidise `Fe^(2+)` to iron `(III)` sulphate, which remains in solution after crystallisation of `CuSO_(4)`C. to speed up the ionisation of `H_(2)SO_(4)` to give `SO_(4)^(2-)` ionsD. which combines with `H_(2)SO_(4)` to give a very strong oxidising mixture and oxidise Cu to `Cu^(2+)` |
|
Answer» Correct Answer - A `Fe(OH)_(3) & Al(OH)_(3)` can be separated by NaOH |
|