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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.
| 51. |
Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured ?A. `KMnO_(4)`B. `Ce(SO_(4))_(2)`C. `TiCl_(4)`D. `Cu_(2)Cl_(2)` |
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Answer» Correct Answer - a,b `KMnO_(4)` is coloured not because of unpaired electrons but due to charge transfer . Similarly , in `Ce(SO_(4))_(2)` , Ce is in `+4` oxidation state with `4f^(0)` configuration . It is again coloured ( yellow) not due to f-f transition but due to charge transfer. |
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| 52. |
The electronic configuration of few transition metal ions are as under (A)`Ti^(+3) (d^1)` , (B) `Co^(2+) (d^7)` , (C ) `Ni^(2+) (d^8)` Increasing order of paramagnetic character isA. A,C,BB. B,C,AC. A,B,CD. C,B,A |
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Answer» Correct Answer - A `{:(,Ti^(3+)to,3d^1,Co^(2+) to , 3d^7 , Ni^(3+) to , 3d^8),("unpaired",e^(-) ,1,,2,,3) :}` |
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| 53. |
Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?A. `KMnO_(4)`B. `Ce(SO_(4))_(2)`C. `TiCl_(4)`D. `Cu_(2)Cl_(2)` |
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Answer» Correct Answer - A::B (a,b) are coloured. |
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| 54. |
Assertion : Change in colour of the acidic solution of breath is used as a test for drunken driver. Reason : Change in colour is due to complexation of alcohol with potassium dichromate.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - C (c) Correct reason. Acidified `K_(2)Cr_(2)O_(7)` is an oxidising agent and oxideses primary alcohol (ethyl alcohol) to acid (acetic acid). At the same time, it is reduced to chromium sulphate (green). `underset("(orange)")underset()(K_(2)Cr_(2)O_(7))+4H_(2)SO_(4)rarrK_(2)SO_(4)+underset("(green)")underset()(Cr_(2)(SO_(4))_(3))+4H_(2)O+3(O)` If the colour of orange solution changes to green, this means that breath has alcohol or the driver has consumed alcohol. If the colour re1nains unchanged, this means that breath has no alcohol. |
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| 55. |
The penultimate shell of f-block elements contains how many electrons ?A. 19 to 32B. 8 to 9C. 8 to 14D. 19 to 36 |
| Answer» Correct Answer - B | |
| 56. |
Transition metal ions containing `(n-1)d^(0)` or `(n-1)d^(10)` configuration are only coloured. |
| Answer» False. Transition metal ion having `(n-1)d^0` or `(n-1)d^(10)` configuration are colourless. | |
| 57. |
Steel contains more carbon than wrought iron |
| Answer» Correct Answer - True. | |
| 58. |
Assertion: Actinoids show greater number of oxidation states than lanthanoids. Reason: Actinoids are radioactive.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - B (b) Correct explanation. In actinoids, the 5f, 6d and 7s sub-shells are close in energy whereas in lanthanoids, these sub-shells have comparatively large energy gaps. |
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| 59. |
In lanthanide and actinide, the penultimate shell isA. sB. pC. dD. f |
| Answer» Correct Answer - C | |
| 60. |
Assertion: Manganese shows a number of oxidation states. Reason: The difference of energy between 3d and 4s subshells is large.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - C (c) Correct reason. The difference of energy between 3d and 4s sub-shells is small. |
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| 61. |
The transition elements are those elements which have partially filled penultimate d-subshells in their elementary form or in their commonly occurring oxidation states. |
| Answer» Correct Answer - true. | |
| 62. |
Assertion: `K_2Cr_2O_7` is used as primary standard in volumetric analysis. Reason: It has a good solubilityin water.A. If both (A) and (R) are correct and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
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Answer» Correct Answer - c Correct R: `[email protected][email protected]_7` is not much soluble in cold water, howerver, it is obtained in pure state and is not hygroscopic in nature. |
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| 63. |
Assertion: Manganese shows a number of oxidation states. Reason: The difference of energy between 3d and 4s subshells is large.A. If both (A) and (R) are correct and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
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Answer» Correct Answer - c Correct R: The difference of energy between 3d and 4s subshells is small. |
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| 64. |
Assertion. Mn atom loses ns electrons first during ionization as compared to (n-1) d electrons. Reason. The effective nuclear charge experienced by ( n-1) d electrons is greater than that of ns electrons. |
| Answer» R is the correct explanation of A. | |
| 65. |
Assertion: Tungsten has very high melting point. Reason: Tungsten is a covalent compound.A. If both (A) and (R) are correct and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
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Answer» Correct Answer - c Correct R: W has large number of unpaired electrons. Hence, metallic bonding in it is quite strong. |
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| 66. |
Assertion. Tungsten has very high melting point. Reason. Tunsten is a rare earth element. |
| Answer» Correct R `//`explanation. W has large number of unpaired electron . Hence, metallic bonding in it is quite strong. | |
| 67. |
Statement-I: Tungsten has a very high melting point. Statement-II: `Ag_(2)S_(2)O_(3)` is soluble in excess of Hypo solution.A. If both Statement-I `&` Statement-II are True `&` the Statement-II is a correct explanation of the Statement-IB. If both Statement-I `&` Statement-II are True but Statement-II is not a correct explanation of the Statement-IC. If statement-I is True but the Statement-II is FalseD. If Statement-I is False but the but the Statement-II is True |
| Answer» Correct Answer - C | |
| 68. |
Statement_I: `Na_(2)O_(3)` is used in Photography. Statement_II: `Ag_(2)S_(2)O_(3)` is soluble in excess of Hypo solution.A. If both Statement-I `&` Statement-II are True `&` the Statement-II is a correct explanation of the Statement-IB. If both Statement-I `&` Statement-II are True but Statement-II is not a correct explanation of the Statement-IC. If statement-I is True but the Statement-II is FalseD. If Statement-I is False but the but the Statement-II is True |
| Answer» Correct Answer - B | |
| 69. |
When an oxide of manganese(A) is fused with KOH in the presenceof an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound(B) disproportionates in neutral or acidic solution to give purple compound (C ). An alkaline solution of compound (C ) oxides potassium iodide solutionto a compound (D) and compound (A) is also formed.Identity compounds A to D andalso explain the reactions involved. |
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Answer» `A=MnO_(2), B=K _(2)MnO_(4) , C=KMnO_(4) , D = KIO_(3)` Reactions `: underset((A))(2MnO_(2))+4KOH + O_(2) rarr underset((B))(2K_(2)MnO_(4)) + 2H_(2)O` `3MnO_(4)^(2-) + 4H^(+)rarr underset((C ))(2MnO_(4)^(-)) + MnO_(2)+2H_(2)O` `2MnO_(4)^(-) + H_(2)O + KI rarr underset((A))(2MnO_(2))+ 2OH^(-) + underset((D))(KIO_(3))` |
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| 70. |
Oil paintings turn blackish after some time. The salt formed isA. SnSB. CuSC. PbSD. CdS |
| Answer» Correct Answer - C | |
| 71. |
`E^(Theta)` of Cu is +0.34V while that of Zn is -0.76 V. Explain. |
| Answer» `E^(Theta)` value of Cu is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert Cu(s) to `Cu^(2+)` (aq) is so high that it is not compensate by its hydration enthalpy. `E^(Theta)` value for Zn is negative because of the fact that after removal of electrons from 4s orbital, stable `3d^(10)` configuration is obtained. | |
| 72. |
While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why? |
| Answer» During filling up of electrons follow (n+l) rule. Here 4s has lower energy than 3d orbital. After the orbitals are filled 4s goes beyond 3d, i.e., 4s is farther from nucleus than 3d. So, electron from 4s is removed earlier than from 3d. | |
| 73. |
While filling up electron in the atomic orbitals, 4s orbital is filled before 3d orbital but reverse happens during the ionisation of the atom. Explain why ? |
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Answer» According to (n+1) rule governing the filling of electrons in a multielectron atom, for 4s orbital, n+l=4+0=4 for 3d orbital,m+l=3+2=5. Therefore, 4s orbital fille dbefore electron filling takes place in 3d orbital since it has lower energy. In case of ionisation of the atom, the electrons present in 4s orbital are loosely held by the nucleus as compared to 3d orbital. Therefore 4s electron is released prior to 3d electron. |
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| 74. |
Assertion: An aqueous solution of `FeCl_3` is acidic. Reason: It is due to cationic hydrolysis of `Fe^(3+)` ion to give `Fe(OH)_3` and `H^(o+)` ions.A. If both (A) and (R) are correct and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
| Answer» Correct Answer - a | |
| 75. |
`(NH_(4))_(2)Cr_(2)O_(7)(` Ammonium dichromate ) is used in fire works. The green coloured powder blown in air isA. `Cr_(2)O_(3)`B. `CrO_(2)`C. `Cr_(2)O_(4)`D. `CrO_(3)` |
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Answer» Correct Answer - A `(NH_(4))Cr_(2)O_(7) overset(Delta)(to) N_(2)+Cr_(2)O_(3)+H_(2)O` `Cr_(2)O_(3)` is used as a green pigment in fire-works. |
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| 76. |
Explain the following : (i). `FeCl_3` and `FeBr_3` are well known, but `FeI_3` has doubtful existence. (ii). Anhydrous `FeCl_3` cannot be obtained by heating hydrated ferric chloride. (iii). Cast iron is hard but pure iron is soft in nature. (iv) . A ferrous salt decolourises `KMnO_4` solution. (v). A ferrous salt turns brouwn air. |
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Answer» (i). `I^(ɵ)` ion being reducing agent reduces `Fe^(3+)` iions into `Fe^(2+)` as: `Fe^(3+)+I^(ɵ)toFe^(2+)+(1)/(2)I_2` (ii). On heating hydrated ferric chloride, anhydrous ferric chloride is not formed as water of crystallision reacts to form `Fe2O_3` and HCl. (iii). Cast iron contains carbon `2.5%` to `5%` which is responsible for its hardness. (iv). Ferrous salt acts as a reducing agent. It reduces acidified `KMnO_4` into `K_2SO_4` and `MnSO_4` which form colourless solution i.e., declourisation of `KMnO_4` solution takes place. `2KMnO_4+10FeSO_4+8H_2SO_4toK_2SO_4+2MnSO_4+5Fe_2(SO_2)_3+8H_2O` (v). A ferrous salt turns brows in air due to oxidation to ferric salt. |
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| 77. |
Iron forms iron (II) chloride, `FeCl_2` and iron (III) chloride `FeCl_3`. One of these chlorides is a dark down solid melting `306^@C`. The other is a white crystalline solid with greenish tint, melts at `674^@C` Which description best fits iron (II) chloride Explain. |
| Answer» Anhydrous `FeCl_3` is a dark brown solid with `306^@C` as melting point. Hence second description best fits for `FeCl_2`. | |
| 78. |
The deep red colour of `Fe(SCN)_3` and `Fe(SCN)_4^(ɵ)` is destroyed by addition of :A. `F^(ɵ)`B. `CN^(ɵ)`C. `SCN^(ɵ)`D. `Fe^(ɵ)` |
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Answer» Correct Answer - a Because the complex `[FeF_6]^(3-)` is formed. |
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| 79. |
The types of bonds present in `CuSO_4.5H_2O` are onlyA. Electrovalent and covalentB. Electrovalent and coordinate covalentC. Electrovalent, covalent,and coordinate covalentD. Covalent and coordinate covalent. |
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Answer» Correct Answer - C `[Cu(H_2O)_4]SO_4.H_2O` There are electrovalent, covalent and coordinate bonds. |
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| 80. |
Iron is rendered passive by treatment with concentratedA. `H_2SO_4`B. `H_3PO_4`C. `HCl`D. `HNO_3` |
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Answer» Correct Answer - D The inertness exhibited by metals under conditions in which chemical activity is expected is known as passivity. The phenomenon of passivity is explained by assuming the formaion of a thin film of oxide on the surface of the metal which prevents the action of the reagent. The concentrated nitric acid makes metals such as `Fe,Co,Ni,Cr,Al`. etc., passive. |
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| 81. |
Amongst the following, the lowest degree of paramgnetism per mole of the compound at `298 K` will be shown byA. `MnSO_4 . 4H_2O`B. `CuSO_4 . 5H_2O`C. `FeSO_4 . 6H_2O`D. `NiSO_4 . 6H_2O` |
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Answer» Correct Answer - B `Mn^(2+)` in `MnSO_4 . 4H_2O - d^5` configuration 5 unpaired `e^-` `Cu^(2+)` in `CuSO_4. 5H_2O - d^9` configuration 1 unpaired `e^-` `Fe^(2+)` in `FeSO_4 . 6H_2O - d^6` configuration 4 unpaired `e^-` `Ni^(2+)` in `NiSO_4 . 6H_2O - d^8` configuration 2 unpaired `e^-` `therefore CuSO_4 . 5H_2O` - lowest degree of paramagnetism |
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| 82. |
From a solution of `CuSO_4` the metal used to recover copper isA. `Na`B. `Ag`C. `Hg`D. `Fe` |
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Answer» Correct Answer - d Fe can be used to recover Cu from `CuSO_4` solution. |
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| 83. |
Explain why is `Ce^(4+)` ion a strong oxidising agent ? |
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Answer» `underset("+4 oxidation state")(Ce^(4+)+e^(-)) to underset("+3 oxidation state")(Ce^(3+))` `Ce^(4+)` ion is a strong oxidising agent because +3 oxidation state is more stable than +4 oxidation state . |
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| 84. |
A man made white silvery metal,radioactive innature has a strong tendency to form oxocations and complexes. It is used as nuclear fuel in atomic reactors. The metal is aA. LanthanideB. ActinideC. Transition metal representative elementD. |
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Answer» Correct Answer - b Actinides are man made and radioactive in nature. They are used in nuclear fuels and have a strong tendency to forms complexes. |
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| 85. |
In netural of faintly alkaline medium, thiosulphate is quantitatively oxidized by `KMnO_(4)` toA. `So_(3)^(2-)`B. `So_(4)^(2-)`C. `So_(5)^(2-)`D. `S_(2)O_(8)^(2-)` |
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Answer» Correct Answer - B In netural or weakly alkaline medium thiosulphtae is quantitatively oxidized by `KMnO_(4)` to sulphate. According to the equation, `8MnO_(4)^(-) + 3S_(2)O_(3)^(2-) + H_(2)O to 8MnO_(2) + 6SO_(4)^(2-) + 2OH^(-)` |
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| 86. |
Assertion. Mercury vapour have shining silvery appearance. Reason. Mercury isa metal with shining silvery appearance. |
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Answer» Correct A. Mercury vapours are invisible and no metallic bonding is possible in vapour state. CorrectR. Mercury is a liquid metal with shining silvery appearance. |
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| 87. |
Describe the oxidising property of `KMnO_(4)` in neutral or faintly alkaline medium for its reactions with iodide and thiosulphate ions. |
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Answer» (i) `2MnO_(4)^(-)+I^(-)+H_(2)O to 2MnO_(2)+2OH^(-)+IO_(3)^(-)` (ii) `2MnO_(4)^(-)+3S_(2)O_(3)^(2-)+H_(2)O to 2OH^(-)+2MnO_(2)+3SO_(4)^(2-) +3S` |
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| 88. |
Assertion. Aufbau rule is violated in writing electronic configuration of Pd. Reason. Pd shows diamagnetic nature. |
| Answer» Pd has `4d^(10)5s^(0)` configuration rather than `4d^(8) 5s^(2)` configurationand hence is diamagnetic in nature. | |
| 89. |
Assertion . In `Cr_(2)O_(7)^(2-)` ion, all the Cr-O bond lengths are equal. Reason. In `Cr_(2)O_(7)^(2-)` ion, all the O-C-O bond angles are equal. |
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Answer» Correct A. Two Cr-O bonds have different bond lengths than the remaining six Cr-O bonds. Correct R. Cr-O-Cr bond in the middle is different from other O-Cr-O bond angles. |
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| 90. |
1 mole of `FeSO_(4)` (Atomic mass of `Fe = 55.84g mol^(-1)`) is oxidised to `Fe_(2)(SO_(4))_(3)`. The equivlent mass of `Fe^(2+)` ions is:A. 55.84B. 27.92C. 83.76D. 111.68 |
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Answer» Correct Answer - A `overset(+2)(2FeSO_(4)) to overset(+3)(2Fe_(2))(SO_(4))_(3)` Change in oxidation state per Fe atom =1 `:.` Equivalent mass = Atomic mass =55.84 g `"equiv"^(-1)` |
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| 91. |
Assertion. Equivalent mass of `KMnO_(4)` is same in the neutral as well alkaline medium. Reason. The product formed in both cases in `MnO_(2)`. |
| Answer» R is the correct explanation of A. | |
| 92. |
For redox reaction `{:(xMnO_(4)^(-)+yH_(2)C_(2)O_(4)+ZH^(+)),(" "darr),(mMn^(2+)+nCO_(2)+pH_(2)O):}` The value of x,y,m and n are:A. 10,2,5,2B. 2,5,2,10C. 6,4,2,4D. 3,5,2,10 |
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Answer» Correct Answer - B The balanced chemical equation is : `{:(2KMnO_(4)+5H_(2)C_(2)O_(4)+6H^(+)),(" "darr),(2Mn^(2+)+10CO_(2)+8H_(2)O):}` x=2, y=5,m=2 and n=10 |
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| 93. |
Two colourless gas A and B are enclosed in two separate jars. When passed through acidified `K_(2)Cr_(2)O_(7)` solution, turn the solution green. But gas A turns lead acetate paper black whereas if both the gases are passed into water, the solution develops a yellowish white turbidity. Identify the gases A and B and write balanced equations for the chemical reactions taking place. |
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Answer» As both gases are colourless and each of these turns acidified `K_(2)Cr_(2)O_(7)` solution green, the gases expected are `H_(2)S` and `SO_(2)` . Further, as gas A turns lead acetate paper black, it must be `H_(2)S` `underset("(Gas A)") (H_(2)S) + (CH_(3)COO)_(2)Pb rarr underset("(Black ppt.)")(PbS) + 2CH_(3)COOH` `H_(2)S` and `SO_(2)` together in aqueous solution react to produce S which gives a yellowish white turbidity `underset("(Gas A)")(2H_(2)S)+ underset("(Gas B )")(SO_(2)) rarr 2H_(2)O + underset("(Yellowish turbidity)")(3S)` |
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| 94. |
The correct basicity order of the following lanthanoid ions isA. `La^(3+) gt Lu^(3+) gt Ce^(3+) gt Eu^(3+)`B. `Ce^(3+) gt Lu^(3+) gt La^(3+)gt Eu^(3+)`C. `Lu^(3+) gt Ce^(3+) gt Eu^(3+) gt La^(3+)`D. `Lu^(3+) gt Ce^(3+) gt Eu^(3+) gt Lu^(3+)` |
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Answer» Correct Answer - D It is the correct order and is due to Lanthanoid contraction . For details, consult section 11. |
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| 95. |
Identify the cases(s) where there in change in oxidation number.A. Acidified soltuion of `Cr_(2)O_(4)^(2-)`B. `SO_(2)` gas bubbled throught an acidic solution of `Cr_(2)O_(7)^(2-)`C. Alkaline solution of `Cr_(2)O_(7)^(2-)`D. Ammoniacal solution of `CrO_(4)^(2-)` |
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Answer» Correct Answer - B (b) (a) `overset(+6)(2Cr_(2))O_(4)^(2-)+2H^(+)+ to overset(+6)(Cr_(2))O_(7)^(2-)+H_(2)O` (b) `overset(+6)(Cr_(2))O_(7)^(2-)+2H^(+)+3SO_(2) to overset(+3)(Cr_(2))(SO_(4))_(3)+H_(2)O` (c) `overset(+6)(Cr_(2))O_(7)^(2-)+2OH^(-)+ to overset(+6)(2Cr_(2))O_(4)^(2-)+H_(2)O` (d) `2NH_(4)OH+ overset(+6)(Cr)O_(4)^(2-) to (NH_(4))_(2)overset(+6)(CrO_(4))+2OH^(-)` (e) `overset(+6)(Cr)O_(2)Cl_(2)+4NaOH to Na_(2)overset(+6)(Cr)O_(4)+2H_(2)O+2NaCl` |
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| 96. |
Identify the product and its colour when `MnO_(2)` is fused with solid KOH in the presence of `O_(2)` `:`A. `KMnO_(4)`, purpleB. `K_(2)MnO_(4)` , dark greenC. MnO, colourlessD. `Mn_(2)O_(3)` ,brown |
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Answer» Correct Answer - B `2MnO_(2) + 4KOH + O_(2) rarr underset("Pot. manganate (Green)")(2K_(2)MnO_(4))+ 2H_(2)O` |
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| 97. |
When `MnO_(2)` is fused with `KOH,` a coloured compound is formed, the product and its colour is:A. `K_2MnO_4`, purple greenB. `K_2 MnO_4`, purpleC. `Mn_2O_3`, brownD. `Mn_3O_4`, black |
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Answer» Correct Answer - A In alkaline medium the stable oxidation state of Mn is +6 . Hence `MnO_2` is oxidised to `K_2MnO_4` (purple green) by atmoshere oxygen in KOH medium |
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| 98. |
Acidified `K_(2)Cr_(2)O_(7)`, solution turns green when `Na_(2)SO_(3)` is added to it. Thus is due to the formation ofA. `Cr(SO)_(4))_(3)`B. `CrO_(4)^(2-)`C. `Cr_(2)(SO_(3))_(3)`D. `CrSO_(4)` |
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Answer» Correct Answer - A `K_(2)Cr_(2)o_(7)+3Na_(2)SO_(4)to` `3Na_(2)SO_(4)+K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O` |
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| 99. |
When `MnO_(2)` fused with `KOH`, a coloured compound is formed . The product and its colour is:A. `K_(2)MnO_(4)` greenB. `KMnO_(4)` purpleC. `Mn_(2)O_(3)`, brownD. `MN_(3)O_(4)` black |
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Answer» Correct Answer - A When `MNO_(2)` is fused with KOH in the presence of oxidising agent such as potassium nitrate `(KNO_(3))`, a green coloured compound, potassium manganate, `(K_(2) MNO_(4))` is formed. `MnO_(2) + 2KOH + KNO_(3) to underset(("Green"))(K_(2) MnO_(4)) + KNO_(2) +H_(2)O` |
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| 100. |
When `MnO_(2)` fused with `KOH`, a coloured compound is formed . The product and its colour is:A. `K_(2)MnO_(4)`, purple greenB. `KMnO_(4)`, purpleC. `Mn_(2)O_(3)`, borwnD. `Mn_(3)O_(4)`, black. |
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Answer» Correct Answer - A (a) `MnO_(2)` is oxidised to `K_(2)MnO_(4)` in alkaline medium to acquire a stable oxidation state of + 6. `2MnO_(2)+4KOH+O_(2) ""^(3//4®)" "2K_(2)MnO_(4)+H_(2)O` The colour of the solution is light pruple green. |
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