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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.
| 351. |
In which of the following compounds, Mn has highest oxidation state?A. `K_(2)MnO_(4)`B. `MnO_(2)`C. `KMnO_(4)`D. `Mn_(3)O_(4)` |
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Answer» Correct Answer - C It is +7. In others, (a) + 6 (b) + 4 (d) `+ 8//3` |
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| 352. |
Give reasons `:` (i) Mn shoes the highest oxidation state of `+7` with oxygen but with fluorine it shows the highest oxidation state of `+4` . (ii)Transition metals show variable oxidation state. (iii) Actinoids show iregularities in their electronic configuration. |
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Answer» (i) Manganese canform `p pi - d pi` bond with oxygen by utilizing 2p orbital of oxygen and 3d orbital of Mn . Hence, it can show highest oxidation state of `+7`. With fluorine, Mn cannote form such `p pi - d pi` bond. Hence, with fluorine it can show a maximum oxidaion state of `+4` . (ii Art. (iii) Art. |
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| 353. |
The elements which exhibit both vertical and horizontal similarities are:A. inert gas elementsB. representative elementsC. rare elementsD. transition elements |
| Answer» Correct Answer - D | |
| 354. |
The no. of unpaired electrons in `Mn^(7+)` ions (At. No of Mn =25) is |
| Answer» Correct Answer - A | |
| 355. |
In the `Cr^(+2), Mn^(+3), Fe^(+2)` and `CO^(+3)` ions, number of unpaired electrons and magnetic moment wiII beA. 3, 3.87B. 4, 4.90C. 3, 2.83D. 1, 1.73 |
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Answer» Correct Answer - B `Cr^(2+) , Mn^(3+),Fe^(2+) , CO^(3+)` has same no. of unpaired `e^-` in 3d orbitals, viz., 4. `therefore` Magnetic moment will be `sqrt(4(4+2))` =4.90 B.M. Formula is `sqrt(n(n+2))` B.M. where n is no. of unpaired `e^-` in 3d orbitals |
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| 356. |
The number of unpaired electrons in gaseous species of `Mn^(3+), Cr^(3+)` and `V^(3+)` respectively are………….and most stable species is…………..A. 4, 3 and 2 , `V^(3+)` is most stableB. 3, 3 and 2, `Cr^(3+)` is most stableC. 4, 3 and 2 , `Cr^(3+)` is most stableD. 3, 3 and 3, `Mn^(3+)` is most stable |
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Answer» Correct Answer - C `Mn^(3+) = 3d^4=4` unpaired electrons, `Cr^(3+) =3d^3=3` unpaired electrons, `V^(3+) =3d^2=2` unpaired electrons. `Cr^(3+)` is most stable out of these in aqueous solution because it has half-filled `t_(2g)` level (i.e., `t_(2g)^3`) |
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| 357. |
The number of unpaired electron in `Mn^(4+) (Z = 25)` is :-(a). Four(b). Two(c). Five(d). ThreeA. 5B. 4C. 3D. 2 |
| Answer» Correct Answer - A | |
| 358. |
The highest oxidation state of `Mn` is shown byA. `Kmno_(4)`B. `K_(2)MnO_(4)`C. `MnO_(2)`D. `Mn_(2)O_(4)` |
| Answer» Correct Answer - A | |
| 359. |
The number of unpaired electron in `Ni^(2+)` isA. ZeroB. 2C. 4D. 8 |
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Answer» Correct Answer - B Number of unpaired electrons in `Ni^(+2)` is 2 as electronic configuration of `Ni^(+2)` is `3d^8 4s^0` |
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| 360. |
Number of unpaired electrons present in `[Ni(H_(2)O)_(6)]^(2+)`A. zeroB. 2C. 4D. 8 |
| Answer» Correct Answer - B | |
| 361. |
The number of unpaired electrons in a nickel atom (ground state) are (At. No. of Ni=28)A. 2B. 5C. 3D. 7 |
| Answer» Correct Answer - A | |
| 362. |
The last three orbitals are partially filled in which of the following ?A. transition elementsB. s-block elementsC. p-block elementsD. f-block elements |
| Answer» Correct Answer - D | |
| 363. |
In lanthanides, the differential electron is filled inA. sf-orbitalB. 4d-orbitalC. 6f-orbitalD. 4f-orbital |
| Answer» Correct Answer - D | |
| 364. |
4f and 5f-series has total number of elementsA. 13B. 12C. 14D. 28 |
| Answer» Correct Answer - D | |
| 365. |
Out of the following how many of them are coloured compounds. `MnO_4^(ɵ),Cr_2O_7^(2-),CrO_4^(2-),Sc^(3+),Ti^(4+),Zn^(2+),Mn^(3+),Cu^(2+),Fe^(2+),Fe^(3+)` |
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Answer» `MnO_4^(ɵ),Cr_2O_7^(2-),CrO_4^(2-)` (they are `3d^0` configuration yet they are coloured due to charge transfer theory) `Mn^(3+)(3d^4),Cu^(2+)(3d^9),Fe^(2+)(3d^6),Fe^(3+)(3d^5)`, all are coloured due to the presence of unpaired electrons. colourless ions are: `Sc^(3+),Ti^(4+)(3d^0),Zn^(2+)(2d^(10))` |
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| 366. |
Which of the following ion has the maximum magnetic moment?A. `Sc^(3+)`B. `Ti^(3+)`C. `Cu^(+)`D. `Zn^(2+)` |
| Answer» Correct Answer - B | |
| 367. |
Which of the following oxidation states is the most common among the lanthanoids ?A. `+2`B. `+3`C. `+4`D. `+5` |
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Answer» Correct Answer - B Lanthanoids show common oxidation state of `+3` Some of which also show `+2` and `+4` stable oxidation state alongwith `+3` oxidation state. These are shown by those elements which by losing 2 or 4 electrons acquire a stable configuration of `f^(0), f^(7)` or `f^(14)`, e.g., `Eu^(2+)" is "[Xe]4f^(7), Yb^(2+) " is "[Xe]4f^(14), Ce^(4+)" is "[Xe]4f^(0) " and "Tb^(4f) " is "[Xe]4f^(7)`. |
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| 368. |
What happens when `FeSO_(4)` solution reacts with acidified `KMnO_(4)` solution?A. Iron (II) is oxidisedB. `KMnO_(4)` is oxidisedC. Iron (II) is reducedD. Iron (III) is reduced |
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Answer» Correct Answer - A `5Fe^(2+)+MnO_(4)^(-)+8H^(+)rarrMn^(2+)+4H_(2)O+5Fe^(3+)` |
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| 369. |
When `KMnO_(4)` reacts with acidified `FeSO_(4)`A. Only `FeSO_(4)` is oxidisedB. Only `KMnO_(4)` is oxidisedC. `FeSO_(4)` is oxidised and `KMnO_(4)` is reducedD. none of these |
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Answer» Correct Answer - C `FeSO_(4)` is oxidised and `KMnO_(4)` is reduced. `({:(2KMnO_(4) +3H_(2)SO_(4)toK_(2)SO_(4)+2MnSO_(4)),([2FeSO_(4)+ H_(2)SO_(4)toFe_(2)(SO_(4))_(3)+2H]xx5):})/(2KMnO_(4)+ 8H_(2)SO_(4)+10Fe SO_(4)to K_(2)SO_(4)+ 2MnSO_(4)+ 5Fe_(2) (SO_(4))_(3)+8H_(2)O)` In this reaction oxidation state of Mn is changing from +7 to +2 while oxidation state of Fe is changing from +2 to +3. |
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| 370. |
Copper is the most noble of the first row transition elements it occurs in small deposits in several countries. Ores of Copper include chalcanthite `(CuSO_4.5H_2O)`, atacamite `[Cu_2Cl(OH)_3]`, cuprite `(Cu_2O)`, copper glance `(Cu_2S)`, and malachite `[Cu_2(OH)_2CO_3]`. However, `80%` of the world copper production comes from the ore chalcopyrite `(CuFeS_2)`. Extraction of copper from chalcopyrite includes roasting, iron removal, and self-reduction. Q. Partial roasting of chalcopyrite producesA. `Cu_2S` and `FeO`B. `Cu_2O` and `FeO`C. `CuS` and `Fe_2O_3`D. `Cu_2O` and `Fe_2O_3` |
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Answer» Correct Answer - A Partial roasting: `2CuFeS_2+O_2overset(Delta)toCu_2S+2FeS+SO_2` `2FeS+3O_2to2FeO+2SO_2` |
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| 371. |
`La^(3+)(Z=57)` and `Lu^(3+)(Z=71)` do not show any colour in solution. Assign reason. |
| Answer» `La^(3+)` ion has the configuration of Xe while `Lu^(3+)` ion `[Xe]4f^(14)` has completely filled orbitals and there is no transition. Therefore, both these ions donot show any colour. | |
| 372. |
The metal that cannot obtained by electrolysis of an aqueous solution of its salts is `:`A. AgB. caC. CuD. Cr |
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Answer» Correct Answer - B Higher the position of lelement in the electro-chemical series more difficult is the reduction of its cations. `If Ca^(2+) (aq)` is electrolysed , water is reduced in perference to it . Hence, it cannot be reduced electrolytically from their aqueous solution. `Ca^(2+)(aq)+H_(2)OrarrCa^(2+)+OH^(-)+H_(2)uarr` |
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| 373. |
What is the general electronic configuration for second row transition series?A. `[Ne]3d^(1-10),4s^(2)`B. `[Ar]3d^(1-10),4s^(1-2)`C. `[Kr]4d^(1-10),5s^(1-2)`D. `[Xe]5d^(1-10),5s^(1-2)` |
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Answer» Correct Answer - C Transition elements have three series 3d, 4d and 5d, 3d series has elements 21 (Sc) to 30 (Zn) atomic number. 4d series has elements 39 (Y) to 48 (Cd) atomic number.` [Kr]4d^(1-10),5s^(1-2)` is the general configuration of Iind row series. |
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| 374. |
What is the general molecular formula of the products obtained on heating lanthanoid (Ln) with sulphur?A. LnSB. `LnS_(3)`C. `Ln_(3)S_(2)`D. `Ln_(2)S_(3)` |
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Answer» Correct Answer - D When lanthanoids (Lu) are heated with sulphur , then `Ln_(2)S_(3)` are formed their sulphides `(Ln_(2)S_(3))` are highly stable because Ln is `+3` oxidation state. |
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| 375. |
What is the general molecular formula of the products obtained on heating lanthanoid (Ln) with sulphur?A. `LnS`B. `LnS_3`C. `Ln_3S_2`D. `Ln_2S_3` |
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Answer» Correct Answer - B Lanthanoids possess common oxidation state +3 |
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| 376. |
Number of moles of `SnCl_(2)` required for the reduction of `1`mole of `K_(2)Cr_(2)O_(7)` into `Cr_(2)O_(3)` is `(`in acidic medium`)`A. `3`B. `2`C. `1`D. `1//3` |
| Answer» Correct Answer - A | |
| 377. |
Zinc dust is obtained byA. Grinding Zinc metalB. Melting Zinc and then atomising it with a blast of airC. Buming Zinc metal in airD. Reducing its ore |
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Answer» Correct Answer - B Preparation of Zn dust |
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| 378. |
Zinc is used to protect corrosion of iron becauseA. ZnSB. ZnOC. `Zn(OH)_2`D. `ZnCO_3 · Zn(OH)_2` |
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Answer» Correct Answer - D Zn surface is covered with layer of `ZnCO_3`. `Zn(OH)_2` due to reaction of Zn with moisture and `CO_2` |
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| 379. |
When Kl (excess) is added to (I)`CuSO_(4)` , (II) `HgCl_(2)` `(III) `Pb(NO_(3)_(2))`A. a white ppt. of cul in I, an orange ppt. of `Hgl_(2)` in II and a yellow ppt. of `Pbl_(2)`in IIIB. a white ppt. of `Cul_(2),Hgl_(2)` and a yellow ppt. of `Pbl_(2)` in IIIC. a white ppt. of `Cul_(2), Hgl_(2)` and `Pbl_(2)`D. None of the above |
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Answer» Correct Answer - B (I)`2CuSO_(4)+4Klrarrunderset("White ppt.")(Cu_(2)l_(2))+l_(2)+2K_(2)SO_(4)` (II) `HgCl_(2)+2Klrarr2KCl+underset("Orange ppt.")(Hgl_(2)darr``Hgl_(2)+2Klrarrunderset("soluble")(K_(2)Hgl_(4))` (III) `Pb(NO_(3))_(2)+2Klrarrunderset("yellow ppt")(Pbl_(2)darr)+2KNO_(3)` |
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| 380. |
When `CuSO_(4)` reacts with aqueous Kl, the products areA. `Cu_(2)l_(2)+K_(2)SO_(4)`B. `Cu+K_(2)SO_(4)+l_(2)`C. `Cu l_(2)+K_(2)SO_(4)`D. `Cu_(2)l_(2)+K_(2)SO_(4)+l_(2)` |
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Answer» Correct Answer - D `2[CuSO_(4)+2KlrarrCu l_(2)+K_(2)SO_(4)]` `2Cu l_(2)rarrCu_(2)l_(2)+l_(2)` `2Cu SO_(4)+4Kl rarrCu_(2)l_(2)+2K_(2)SO_(4)+l_(2)` |
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| 381. |
Which of the following is formed, when copper (II) sulphate is treated with excess ammonia?A. A black precipitateB. A red precipitateC. A deep blue solutionD. A white precipitate turning black |
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Answer» Correct Answer - C Tetraamino cupric ion is formed. `Cu^(+)+4NH_(3)rarr underset("Deep blue")([Cu(NH_(3))_(4)]^(2+))` |
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| 382. |
`A_1` and `A_2` are two ores of metal `M.A_1` on calcination gives black precipitate, `CO_2` and water. Identify `A_1` and `A_2`. |
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Answer» `A_1` is malachite `CuCO_3.Cu(OH)_3` `CuCO_3Cu(OH)_2overset(Delta)toCuO+CO_2uarr+H_2O` `CuO` is the black solid. `CuCO_3.Cu(OH)_2overset(dil.HCl)toCuCl_2+CO_2+H_2O` `CuCl_2overset(KI)toCu_2I_2+KCl+I_2` `A_2` is copper glance `Cu_2S`, a sulphide ore. `Cu_2Sunderset(Delta)overset(O_2)toCu_2O+SO_2` `Cu_2S+Cu_2OtoCu+SO_2` The gas `SO_2` gives green colour with acidified `K_2Cr_2O_7.K_2Cr_2O_7+H_2SO_4+3SO_4toK_2SO_4+Cr_2(SO_4)_3+4H_2O` |
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| 383. |
Which of the followig ions will finally give a black precipitate with `Ag^(o+)` ion?A. `SO_(3)^(2-)`B. `Br^(-)`C. `CrO_(4)^(2-)`D. `S_(2)O_(3)^(2-)` |
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Answer» Correct Answer - D `2Ag+S_(2)O_(3)^(2-)rarrAg_(2)S_(2)O_(3)`, white ppt which readily changes to yellow, orange, brown and finally black due to the formation of silver sulphide. `Ag_(2)S_(2)O_(3)+H_(2)OrarrH_(2)SO_(4)+Ag_(2)S ("black")` |
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| 384. |
Out of `Fe^(2+), Fe^(3 +), Mn^(2+)` and `Ti^(2+)` the least size is ofA. `Mn^(2+)`B. `Ti^(2+)`C. `Fe^(2+)`D. `Fe^(3+)` |
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Answer» Correct Answer - D Size increases in the order `Fe^(3+) lt Fe^(2+) lt Mn^(2+) lt Ti^(2+)` as nuclear changes decreases. |
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| 385. |
The equilibrium `Cr_(2)O_(7)""^(2-)hArr2CrO_(4)""^(2-)` is shifted to right in-A. an acidic mediumB. a basic mediumC. a neutral mediumD. it does not exist |
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Answer» Correct Answer - B Shifted to right in basic medium |
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| 386. |
The shape of `MnO_(4)^(-)` ion and the hybridisation of Mn in `MnO_(4)^(-)` isA. `sp^(3)`B. `sp^(2)`C. `sp`D. `sp^(3)d^(2)` |
| Answer» Correct Answer - A | |
| 387. |
The reaction,` MnO_(4)^(2-) +e^(-)hArr MnO_(4)^(-)` is shifted to right inA. an acidic mediumB. a basic mediumC. a netural mediumD. both acidic and bisic media |
| Answer» Correct Answer - A | |
| 388. |
`MnO_(4)^(2-)` has `"_____"` geometryA. TrigonalB. TetrahedralC. OctahedralD. Pyramidal |
| Answer» Correct Answer - B | |
| 389. |
Following ions are equilibrium at `Cr_(2) O_(7)^(2-) hArr CrO_(4)^(2-)`A. `pH-4`B. `pH-9`C. `pH-7`D. `pH-10` |
| Answer» Correct Answer - C | |
| 390. |
`MnO_(4)^(2-)` isA. DiamagneticB. FerromagneticC. paramagneticD. nonmagnetic |
| Answer» Correct Answer - C | |
| 391. |
The mineral from which potassium permagnate is manufacturedA. pyrolusite, `MnO_(2)`B. braunite, `Mn_(2)O_(3)`C. hausmannite, `Mn_(3)O_(4)`D. manganite, `Mn_(2). H_(2)O.` |
| Answer» Correct Answer - A | |
| 392. |
Statement I : `Zn^(2+)` is diamagnetic Statement II : The electrons are lost from 4s orbital to from `Zn^(2+)`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 `Zn^(2+)` ion (`d^(10)` configuration), all electron orbitals are paired. |
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| 393. |
Which of the following types of metal form the most efficient catalysts?A. Transition metalsB. Alkali metalsC. Alkaline earth metalsD. Inert elements |
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Answer» Correct Answer - A Transition metals shows catalytic properties |
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| 394. |
Diamagnetic ion isA. `Cu^(2+)`B. `Cr^(+3)`C. `Ti^(+3)`D. None of these |
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Answer» Correct Answer - D `Cu^(2+) , Cr^(3+) , Ti^(3+) to` have unpaired `e^(-) " " therefore` Paramagnetic |
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| 395. |
Amongst the ion of 3d transition series paramagnetic character increases from `Ti^(2+)` toA. `Cr^(+2)`B. `Mn^(+2)`C. `Ni^(+2)`D. `Fe^(+2)` |
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Answer» Correct Answer - B `Cr^(2+) to [Ar] 4s^0 3d^3 , Ni^(2+) to [Ar] 4s^0 3d^8`, `Fe^(2+) to [Ar] 4s^0 3d^6 , Mn^(2+) to [Ar] 4s^0 3d^5` No. of unpaired `e^-` increases from `Fe^(2+)` (1) to `Cr^(3+)` (3), `Ni^(2+)` (3) to `Mn^(2+)` (5) |
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| 396. |
The salts of Cu in +1 oxidation state are unstable becauseA. `Cu^(+)` has `3d^10` configurationB. `Cu^+` disproportionates easily to Cu(0) and `Cu^(2+)`C. `Cu^+` disproportionates easily to `Cu(2+)` and `Cu^(2+)`D. `Cu^+` is easily reduced to `Cu^(2+)` |
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Answer» Correct Answer - B `Cu^+` ions undergo disproportionation . `2Cu^(+) to Cu^(2+) + Cu` |
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| 397. |
Depositive zinc oxhibits paramagnetism due to loss of two electrons from 3d-orbitals of neutral atom. |
| Answer» Correct Answer - False. | |
| 398. |
Ferric chloride gives dark blue colour with ammonium thiocyanate solution. |
| Answer» False. Gives deep red colour. | |
| 399. |
In an aqueous solution, `Cu(+1)` salts are unstable becauseA. `Cu(+1)` has a `3d^(10)`configurationB. they disproportionate easily to the `Cu and (+2)` statesC. they disproprtionate easily to the `Cu(+2) and Cu(+3)`statesD. the change in free energy of the overall is zero |
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Answer» Correct Answer - B `Cu^(+)` salts are unstable because `Cu^(+)` salts disproportionate into `Cu and Cu^(2+)` salts. |
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| 400. |
What happens when `SO_2` gas is bubbled through an aqueous solution of copper sulphate in the presence of potassium thiocyanate? |
| Answer» `2CuSO_4+SO_2+2KCNS+2H_2Oto2CuCNSdarr+K_2SO_4+2H_2SO_4` | |