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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.
| 1. |
Nitric oxide, though an odd electron molecule, is diamagenetic in liquid state. |
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Answer» Correct Answer - F Nitric oxide `(NO)` is paramegnetic, in liquid state due to presence of unpaired electron as show. `: dot N = ddot O:`. |
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| 2. |
In the reaction: `3Br_(2)+6OH^(Theta)rarr 5Br^(Theta)+BrO_(3)^(Theta)+3H_(2)O,Br_(2)` isA. OxidisedB. ReducedC. DisintegratedD. Disproportionted |
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Answer» Correct Answer - D `overset((0))3Br_(2)+overset(Theta)6OH rarr overset((-1))5Br^(Theta)+overset((+5))BrO_(3)^(Theta)+3H_(2)O` |
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| 3. |
Give balanced equations for the following : Phosphorous reacts with nitric acid to give equimolar ratio of nitric oxide and nitrogen dioxide. |
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Answer» (ii) Phosphoric acid is obtained. `4 P + 10 HNO_3 + H_2 O rarr 5 NO + 5 NO_2 + 4 H_3 PO_4` `[P rarr PO_4 ^(3-) , No_3 ^(Ө) rarr NO_2]`. |
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| 4. |
(a) Though nitrogen exhibits `+5` oxidation state, it does not form pentahalide. Give reason. (b) `PH_3` has lower boiling point than `NH_3`. Why ? |
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Answer» (a) Electronic configuration of nitrogen is `1 s^2 2 s^2 2 p^3`. Die to absence of d-orbitals in its valence shell, nitrogen cannot expand its coordination number beyond four. That is why nitrogen does not form pentahalides. (b) Nitrogen is more electronegative than phosphorous and hence capable of forming hydrogen bonds. Due to association of `NH_3` molecules by hydrogen bonding. ammonia has higher boiling point. On the other hand, due to absence of hydrogen bonding in `PH_3` molecules are not associated and `PH_3` has lower boiling point than `NH_3`. |
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| 5. |
In white phosphorous `/_ PPP = 90^@`. |
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Answer» Correct Answer - F In white phosphorous `PPP` and angle is `60^@`. |
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| 6. |
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?A. `HNO_3, NO, NH_4 Cl, N_2`B. `HNO_3, NO, N_2, NH_4 Cl`C. `HNO_3, NH_4 Cl, NO,N_2`D. `NO,HNO_3, NH_4 Cl, N_2` |
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Answer» Correct Answer - B `NHO_3 = +5, NO = + 2` `NH_4 Cl = -3, N_2 = 0`. |
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| 7. |
Why commercial `HNO_3` is generally yellow in colour ? |
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Answer» Commercial `HNO_3` due to impurity of dissolved `NO_2` in it, appears yellow on colour. `NO_2` is produced by photochemical decomposition of `HNO_3`. `4HNO_3 overset (hv)rarr 4NO_2 + 2H_2O + O_2`. |
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| 8. |
The product formed in the reaction of `SOCl_2` (thionyl chloride) with white phosphorous is.A. `PCl_3`B. `SO_2 Cl_2`C. `SCl_2`D. `POCl_3` |
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Answer» Correct Answer - A `P_4 + 8 SOCl_2 rarr 4 PCl_3 + 4 SO_2 + 2 S_2 Cl_2`. |
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| 9. |
Concentrated `HNO_3`, upon long standing, turns yellow-brown due to the formation ofA. `NO`B. `NO_2`C. `N_2 O`D. `N_2 O_4` |
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Answer» Correct Answer - B `4 HNO_3 rarr 2 H_2 O + 4 NO_2 + O_2` `NO_2` remains dissolved in `HNO_3` colouring it yellow or even red at higher temperature. |
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| 10. |
When sulphur is boiled is boiled with `Na_2 SO_3`, a compound `(X)` is produced, `(X)` with excess of `AgNO_3` solution gives a compound `(Y)` which is soluble in water and produces a black coloured sulphide `(Z)`. Identify compounds `(X),(Y)` and `(Z)`. |
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Answer» Sulphur on boiling with `Na_(2)SO_(3)` gives `Na_(2) S_(2) O_(3) (X) S + Na_(2) S_(2) O_(3)` reacts with excess of `AgNO_(3)` solution to give `Na_(2) S_(2) O_(3) + 2 AgNO_(3) rarr 2NaNO_(3) + underset((Y)) (Ag_(2) S_(2) O_(3))` `Ag_(2) S_(2) O_(3)` is water soluble and a black ppt of `Ag_(2) S` is formed. `Ag_(2) S_(2) O_(3) + H_(2) O rarr Ag_(2) S + H_(2) SO_(4)` Hence `(X) = Na_(2) S_(2) O_(3) , (Y) = Ag_(2) S_(2) O_(3) , (Z) = Ag_(2) S`. |
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| 11. |
Among the oxides gives below, how many are acidic ? `CrO_3, Mn_2 O_7, CO, SO_2`. |
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Answer» Correct Answer - `(3)` `CrO_(3), MnO_(2) O_(7)` and `SO_(2)` are acidic oxides. |
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| 12. |
When an inorganic compound reacts with `SO_2` in aqueous medium produces `(A). (A)` on reaction with `Na_2 CO_3` gives the compound `(B)` which with sulphur gives a substance `(c)` used in photography. The compound `(c)` is.A. `Na_2 S_2 O_3`B. `Na_2 SO_4`C. `Na_2 S`D. `Na_2 S_2 O_7` |
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Answer» Correct Answer - A Inorganic compound + `SO_2` in aq. Medium rarr `(A)`. `(A) + Na_2 CO_3 rarr (B)` `(B) + S rarr (c)` (Used in photography) `Na_2 CO_3 + 2SO + H_2 O rarr 2 NaHSO_4 + CO_2` `2 NaHSO_4 + Na_2 CO_3 rarr underset ((B)) (2 Na_2 SO_3) + H_2 O + CO_2` `Na_2 SO_3 + S rarr underset((C)) (Na_2 S_2 O_3)`. |
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| 13. |
Why dry `SO_2` cannot bleach dry flowers ? |
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Answer» `SO_(2)` acts as a bleaching agen only in the presence of moisture, as the nascent hydrogen which is responsible for bleaching is produced. `SO_(2) + H_(2) O rarr H_(2) SO_(4) + 2[H]` Coloured flower `+ [H] rarr` colourless flower. |
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| 14. |
Sulphur reacts with chlorine in `1:2` ratio and forms X hydrolysis of X gives a sulphure compound Y. What is the hybridisation state od central atom in the compound?A. `sp`B. `sp^3`C. `sp^2`D. `sp^2 d` |
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Answer» Correct Answer - B `S + 2Cl_2rarr underset ((X)) (SCl_4) overset(4 H_2 O)rarr underset(underset(underset((Y))(H_2 SO_3 + H_2 O))(darr))(S(OH)_(4) + 4 HCl)` |
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| 15. |
An inorganic halide `(A)` reacts with water to form two acids `(B)` and `(c). (A)` also reacts with `NaOH` to form two salts `(D)` and `( E)` which remain in solution. The solution gives white precipitate with both `AgNO_3` and `BaCl_2` solutions respectively. `(A)` is a useful organic reagent. Identify `(A)` to `( E)`. |
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Answer» `underset("Inorganic halide") (A + H_(2) O) rarr (B) + (C)` `(A) + NaOH rarr underset(("Salts which are water soluble")) ubrace((D)+(E))` `(D) + AgNO_(3) rarr` White ppt `( E) + BaCl_(2) rarr` White ppt `AgNO_(3)` gives white ppt. on reaching with `Cl^(Ө)` ions, hence (D) is `NaCl`, `AgNO_(3) + NaCl rarr AgCl darr + NaNO_(3)` `Ag^(oplus) + Cl^(Ө) rarr AgCl darr` `BaCl_(2)` gives white ppt. with `SO_(4)^(2-)` ions, hence (D) is `Na_(2) SO_(4)`. `BaCl_(2) + Na_(2) SO_(4) rarr BaSO_(4) darr + 2 NaCl` `Ba^(2+) + SO_(4)^(2-) rarr BaSO_(4) darr` Hence (A) can be `SO_(2) Cl_(2)`, since this on reacting with water gives acids namely `H_(2) SO_(4)` and `HCl (A)` reacts with `NaOH` to form `Na_(2) SO_(4)` and `NaCl` `underset((A)) (SO_(2) CL_(2)) + 2H_(2) O rarr underset((B)) (H_(2)SO_(4)) + underset((C))(2 HCl)` `underset((A)) (SO_(2) Cl_(2)) + NaOH rarr underset((D))(Na_(2) SO_(4)) + underset((E))(NaCl) + 2 H_(2) O` Both (D) and (E) are water soluble. Hence `(A) SP_(2) Cl_(2), (B)` is `H_(2) SO_(4), (C) is HCl, (D)` is `Na_(2) SO_(4)` and (E) is `NaCl`. |
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| 16. |
Sulphur on reaction with concentrated `HNO_3`. Gives `(A)` which reacts with reacts with `NaOH` gives `(B).(A)` and `(B)` areA. `H_2 SO_3, Na_2 S_2 O_3`B. `NO_2, Na_2 S`C. `H_2 SO_4, Na_2 SO_4`D. `H_2 S_2 O_3, Na_2 S_2 O_3` |
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Answer» Correct Answer - C `S + HNO_(3) rarr (A)` `(A) + overset(conc.) (NaOH) rarr (B)` `S + 6 HNO_(3) rarr underset((A))(H_(2) SO_(4) + 6 NO_(2) + 2 H_(2) O` `underset((A)) (H_(2) SO_(4)) + 2 NaOH rarr Na_(2) SO_(4) + 2 H_(2) O`. |
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| 17. |
What happens when (i) Concentrated `H_2 SO_4` is added to calcium fluoride. (ii) `SO_3` is passed through water ? |
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Answer» (i) `underset ("Calcium fluoride") (CaF_2) + underset ("Sulphuric acid") (H_2 SO_4) rarr underset ("Calcium sulphate") (CaSO_4) + underset ("Hydorgen fluoride")(2HF)` (ii) `underset ("Sulphur trioxide")(SO_3) + underset ("Water") (H_2O) rarr underset ("Sulphuric acid")(H_2 SO_4)` `SO_3` dissolves on water to form `H_2SO_4`. |
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| 18. |
Which concentrated `H_2 SO_4` can be used to dry the gas ?A. `H_2 S`B. `CO_2`C. `NH_3`D. All |
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Answer» Correct Answer - B `Conc. H_(2) SO_(4)` reacts with `H_(2) S` and `NH_(3)`, while `CO_(2)` remains unaffected, hence `conc. H_(2) SO_(4)` can be used to dry `CO_(2)` gas. `H_(2) S + H_(2) SO_(4) rarr S darr + 2H_(2) O + SO_(2)` `2 NH_(3) + H_(2) SO_(4) rarr (N_(4))_(2) SO_(4)`. |
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| 19. |
Which one of the following is wrong ?A. Oxygen and sulphur belong to the same group of periodic tableB. Oxygen is a gas while sulphur is solidC. Both oxygen and sunphur show `+2, +4` and `+6` oxidation states.D. `H_2 S` has no hydrogen bonding |
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Answer» Correct Answer - C Oxygen does not show oxidation state `+4` and `+6`. |
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| 20. |
A green coloured solution of a salt changes its colour to light pink on passing ozone through it. Which of the following species represent pink and green colour respectively.A. `Mn^(2+)` and `MnO_2`B. `MnO_4^(Ө)` and `MnO_4^(2-)`C. `Co^(2+)` and `Co^(3+)`D. `MnO_4^(2-)` and `MnO_4^(Ө)` |
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Answer» Correct Answer - D `[MnO_(4)^(2-) rarr MnO_(4)^(Ө) + e^(Ө)] xx 2` `O_(3) 2 H^(oplus) + 2 e^(Ө) rarr O_(2) + H_(2) O` `ulbar(underset(("Green")) (2MnO_(4)^(Ө)) + O_(3) + 2 H_(oplus) rarr underset(("Pink"))(2MnO_(4)^(Ө)+ O_(2) + H_(2) O))`. |
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| 21. |
On heating ozone, its volume.A. Increase to `1.5` timesB. Decreases to halfC. Remain unchargedD. Becomes double |
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Answer» Correct Answer - A `O_3 overset (Delta) rarr O_2 + [O]` `1` mole of `O_3` on heating produces `1` mole of `O_2` and `1` mole of `[O]`, hence its volume increases to `1.5` times. |
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| 22. |
Which of the following solutions does not change its colour on passing ozone through it ?A. Starch iodide solutionB. Alcoholic solidium of benzidineC. Acidic solution of `FeSO_4`D. Acidified solution of `K_2Cr_2O_7` |
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Answer» Correct Answer - C Ozone does not react with acidified solution of `K_2 Cr_2 O_7`. |
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| 23. |
Which of the following is not correct ?A. `3 O_2 overset ("Sileny electric") underset ("Discharge") hArr 2 O_3 , Delta H = -284.5 kJ`.B. Ozone undergoes addition reaction with unsaturated carbon compoundsC. Sodium thiosulphate reacts with `I_2` to form sodium tetrathionate and sodium iodide.D. Ozone oxidises lead sulphide to lead sulphate |
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Answer» Correct Answer - A Reaction is endothermic. |
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| 24. |
Xenon florides are colourless and at room temperature areA. SolidB. LiquidC. GasesD. Superifuil |
| Answer» Correct Answer - a | |
| 25. |
Which of the possible following florides of xenon is impossible ?A. `XeF_(2)`B. `XeF_(4)`C. `XeF_(6)`D. `XeF_(3)` |
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Answer» Correct Answer - d Electronic configuration of `Xe` is `5s^(2)5p^(6)`. After unpaired element in `5p` orbital number of unpaired element can be `2,4,6`. Hence `XeO_(6)` can form `XeF_(2), XeF_(4)` or `Xef_(6)` but not `XeF_(3)` |
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| 26. |
The rare gases areA. monoatomicB. diatomicC. triatomicD. polyatomic |
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Answer» Correct Answer - a `C_(p)//C_(v) = 1.66` for rare gases and thus they are monatomic |
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| 27. |
Bleaching of flowers by chlorine is permanet, while that by sulphur dioxide is tempory. Explain. |
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Answer» Chlorine bleaches coloured material by oxidation: `Cl_(2)+H_(2)O to 2HCl+[O]` Coloured `+[O] to "colourless material"` Hence, the bleaching is permanent. Whereas, sulphur dioxide, `SO_(2)` bleaches coloured material by reduction and hence bleaching is temporary. When the bleached colourless material is exposed to air, it gets oxidised and the colour is restored. `SO_(2)+2H_(2)Oto 2H_(2)SO_(4)+2[H]` `underset("material")("Coloured")+[H]underset("material")("Colourless")overset("Aerial")underset("Oxidation")(rarr)underset("material")("Coloured")`. |
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| 28. |
Why fluorine does not exhibit any positive oxidation state? |
| Answer» Electronic configuration of `F` is `1s^(2)2s^(2)2p^(5)`. `F` has only one unpair electorn. As `F` is the highest electronegative element, the possibility of sharing its electron with more electronegative element than itself is not there, hence it cannot show an oxidation state of `+1` . further, due to absence of d-orbitals in its valence shell, the paired electrons present in `2s` or `2p` electrons cannot be unpaired, hence `F` does not shown `+3,+5` or `+7` oxidation state too. Thus, fluorine does not exhibt any positive oxidation state. | |
| 29. |
A colourless gas `X` forms a brown coloured gas when mixed with air. The gas `X` isA. `N_2 O`B. `NO`C. `NH_3`D. `NO_2` |
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Answer» Correct Answer - B `underset ("Colourless gas")(NO) + NO_2 rarr underset ("Brown coloured gas") (NO_2)`. |
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| 30. |
Give reason : Urea is better nitrogenous fertiliser than ammonium sulphate. |
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Answer» Urea on hydrolysis gives `NH_3` and `CO_2`. `H_2 NCONH_2 + H_2 O rarr CO_2 + 2 NH_3`. `NH_3` is assimilated by plants whereas `CO_2` goes into the atmosphere, i.e., `NO` residue is left in the soil. On the other hand, ammonium sulphate on hydrolysis gives `NH_3` and `H_2SO_4` `(NH_4)_2 SO_4 + H_2 O rarr 2 NH_3 + H_2 SO_4` `NH_3` is assimilated by plants whereas `H_2 SO_4` is retained in the soil making it acidic. |
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| 31. |
The product `A` in the following reaction : `2KMnO_4 rarr A + KMnO_2 + O_2` isA. `K_2 Mn_2 O_7`B. `K_2 MnO_4`C. `K_2 O`D. `K_2 O_2` |
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Answer» Correct Answer - B `2KMnO_(4) rarr underset((A))(K_(2)MnO_(4))+MnO_(2)+O_(2)` |
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| 32. |
`HCIO_(4)+P_(2)O_(5)rarr(A)+(B)` (A) and (B) areA. `HCIO_(3),H_(3)PO_(4)`B. `CI_(2)O_(6),HPO_(3)`C. `CIO_(2)H_(3)PO_(4)`D. `CI_(2)O_(7),HPO_(3)` |
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Answer» Correct Answer - D `2HCIO_(4)+P_(2)O_(5)rarr CI_(2)O_(7)+2HPO_(3)` |
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| 33. |
Sea weads are important source of bromine. |
| Answer» Sea-weads are important source of iodine. | |
| 34. |
Bromine can be liberated form potassium bromide solution by the action ofA. Iodine solutoinB. Chlorine waterC. Sodium chlorideD. Potassium iodide |
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Answer» Correct Answer - B Potassium bromide is oxidised by chlorine water and bromine is evolveld. `CI_(2)+2Br^(Theta)rarr 2CI^(Theta)+Br_(2)` |
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| 35. |
Assertion (A): F-F bond in `F_(2)` melocule is weak. F atom is small in size. |
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Answer» Correct Answer - C Correct Reason: F-F bond is weaker than CI-CI bond. |
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| 36. |
Why fluorine never acts as the centreal atom in polyatomic interhalogen compounds? |
| Answer» Fluorine never acts as the central atom is polyatomic interhalogen compounds, electronic configuration of fluorine is `2s^(2)2p^(5)` and due to absence of vacant d-orbital, fluorine cannot expand its coordination number more than 1(one). | |
| 37. |
`AgCOP_(3) + (A) to (B) + (C ) + (D)` The substances (A), (B), (C ) and (D) areA. `CI_(2),AgCI,CIO_(2)O_(2)`B. `CI_(2)Ag,CI_(2)O_(6),O_(2)`C. `H_(2),AgCI,H_(2)O,O_(2)`D. `HCIO,AgCI,CIO_(2),O_(2)` |
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Answer» Correct Answer - A `2AgCIO_(3)+underset((A))CI_(2)rarrunderset((B))2AgCI+underset((C ))2CIO_(2)+underset((D))O_(2)` |
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| 38. |
Select the correct stattement(s):A. `CI_(2)O` and `CIO_(2)` are used as bleaching agents and as germicides.B. `CIO_(2)` is the anhydride of `HCIO_(2)` and `HCIO_(3)`.C. `I_(2)O_(5)` is used in the quantitative estimation of CO.D. All of the above are correct. |
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Answer» Correct Answer - D `2CIO_(2)+H_(2)O rarr HCIO_(2)+HCIO_(3)` `I_(2)O_(5)CO rarr I_(2)+5CO_(2)` |
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| 39. |
One atom of …….. combines with one atom of bromine.A. ArB. RbC. MgD. HCI |
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Answer» Correct Answer - B `Rb^(o+)+Br^(Theta)rarr RbBr` |
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| 40. |
Which one of the following pairs of reactantas does not form oxygen when they react with each other ?A. `F_(2),NaOH` solution (hot, conc.)B. `F_(2),H_(2)O`C. `CI_(2),NaOH` solution (cold, dilute)D. `CaOCI_(2),H_(2)SO_(4)` (dilute, small amount) |
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Answer» Correct Answer - C `2F_(2)+underset((Hot, conc))4NaOH rarr 4NaF+2H_(2)O+O_(2)` `2F_(2)+2H_(2)Orarr 4HF+O_(2)` `CI_(2)+2NaOH rarr NaCI+NaCIO+H_(2)O` `2CaOCI_(2)+H_(2)SO_(4)rarr CaCI_(2)+CaSO_(4)+2HCI+O_(2)` |
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| 41. |
Which one of the following oxidises water to oxygen with large evolution of heat ?A. ChlorineB. BromineC. IodineD. Fluorine |
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Answer» Correct Answer - D `2F_(2)+2H_(2)O rarr O_(2)+4HF` |
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| 42. |
The correct statement for the molecule `CsI_(3)` isA. It contains `Cs^(o+)` and `I^(Theta)` ionsB. It contains `Cs^(o+)` and `I^(Theta)` and lattice `I_(2)` moleculeC. It is a covalent moleculeD. It contains `Cs^(o+)` and `I_(3)^(Theta)` ions |
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Answer» Correct Answer - D `CsI_(3)rarr Cs^(o+)+i_(3)^(Theta)` i. Cs cannot show `+3` oxidation state. ii. `I_(2)` molecules are too large to be accommodated in lattice. |
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| 43. |
Which of the following does not liberate `Br_(2)` form KBr ?A. `I_(2)`B. `CI_(2)`C. conc. `H_(2)SO_(4)`D. `F_(2)` |
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Answer» Correct Answer - A `I_(2)` being weaker oxidising than `Br_(2)`, will not liberate `Br_(2)` form KBr. |
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| 44. |
which one of the following salts will evolve halogen on treatment with conc. `H_(2)SO_(4)` ?A. `NaCI`B. `CaCI_(2)`C. `NaBr`D. `KI` |
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Answer» Correct Answer - C::D `2NaBr+underset(conc.)H_(2)SO_(4)rarr 2NaHSO_(4)+Br_(2)` `2KI+H_(2)SO_(4)rarr 2KHSO_(4)+I_(2)uarr` |
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| 45. |
Which of the following exists as an associated molecule even in the vapour state ?A. HCIB. HBrC. HFD. HI |
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Answer» Correct Answer - C HF due to hydrogen boinding is associated and exists as `H_(2)F_(2)` even in the vapour state. |
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| 46. |
Halogen that is most easily reduced is flurine. |
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Answer» Correct Answer - 1 due to highest reduction potential fluorine acts as strongest oxidising agent. |
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| 47. |
Which of the following halogen acid is a liquid ?A. HFB. HCIC. HBrD. HI |
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Answer» Correct Answer - A HF due to hydrogen boinding exists as a liquid. |
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| 48. |
Which one of the followign reactions does not occur ?A. `F_(2)+2CI^(Theta)rarr 2F^(Theta)+CI_(2)`B. `CI_(2)+2F^(Theta)rarr 2CI^(Theta)+F_(2)`C. `Br_(2)+2I^(Theta)rarr 2Br^(Theta)+I_(2)`D. `I_(2)+2Br^(Theta)rarr Br_(2)^(Theta)+2I^(Theta)` |
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Answer» Correct Answer - B::D `CI_(2)` is weaker oxidise agents than `F_(2)` and hence would not oxidice `F^(Theta)` to `F_(2)`. Similarly `I_(2)` is weaker oxidising agent than `Br_(2)` and hence will not oxidse `Br^(Theta)` to `Br_(2)`. |
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| 49. |
Aqueous solution of `Na_(2)S_(2)O_(3)` on reaction with `CI_(2)`, givesA. `Na_(2)S_(4)O_(6)`B. `NaHSO_(4)`C. `HCI`D. `NaOH` |
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Answer» Correct Answer - B::C `Na_(2)S_(2)O_(3)+4CI_(2)rarr 2NaHSO_(4)+8HCI+5H_(2)O` |
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| 50. |
Fill in teh blanks. i. `H_(2)SO_(4)+HI rarr …..+SO_(2)+2H_(2)O` ii. `CaOCI_(2)+NaI+HCI rarr ….. +CaCI_(2)+H_(2)O+NaCI` iii. `NH_(3)+CI_(2)(excesse) rarr …. + HCI` iv. `KMnO_(4)+KCI+H_(2)SO_(4) rarr K_(2)SO_(4)+MnSO_(4)+…..+…..` v. `K_(2)Cr_(2)O_(7)+HCI rarr KCI+Cr4CI_(3)+.....+....` vi. `CuSO_(4)+KI rarr Cu_(2)I_(2)+.....+....` vii. `NH_(3)+NaOCI rarr N_(2)+NaCI+.....` viii. `CI_(2)+H_(2)O+HgO rarr HgCI_(2).HgO+.....` ix. `P_(4)+I_(2)+H_(2)O rarr H_(3)PO_(3)+.......` x. `NaBr+MnO_(2)+H_(2)S_(4)rarr NaHSO_(4)+......+H_(2)O+.....` |
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Answer» (i) `I_(2)`, (ii) `I_(2)` (iii) `NCl_(3)`, (iv) `Cl_(2),H_(2)O` (v)`H_(2)O,Cl_(2)`, (vi) `K_(2)SO_(4),I_(2)` (vii) `H_(2)O` , (viii) `HOCl` (ix)`HI`, (x)`MnSO_(4),Br_(2)`. |
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