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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.
| 1151. |
Why are `Mn^(2+)` compounds more stable than `Fe^(2+)` toward oxidation to their `+3` state?A. `Mn^(2+)` is more stable with high `3^(nd)` ionisation energyB. `Mn^(2+)` is bigger in sizeC. `Mn^(2+)` has completely filled d-orbitalsD. `Mn^(2+)` does not exist. |
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Answer» Correct Answer - A `Mn^(2+) (3d^(5))` is a statle since it has all half-filled orbitals. `Fe^(2+) (3d^(6))` is yet to acquire a stable configuration since it has one filled orbital. Therefore, `Fe^(2+)` ion has a tendency to get oxidised to `Fe^(3+)` ion whereas `Mn^(2+)` ion has no urge to participate in the oxidatioxn reaction. |
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| 1152. |
Explain briefly how `+2` state becomes more and more stable in the first half of the first row transition elements with increasing atomic number. |
| Answer» Except scandium ( which shows an oxidation state of `+3)` , all other first row transition elements show an oxidation state of `+2`. This is due to loss of two 4s electrons. In the first half, as we move from `Ti^(2+)` to `Mn^(2+)` , the electronic configuration changes from `3d^(2)` to` 3d^(5)` , i.e., more and more of d-orbitals are half- filled imparting greater and greater stability to `+2` state. In the second half, i.e., electrons in the 3d orbitals pair up and th enumber of half - filled orbital decreases. Hence, the stability of `+2` state decreases. | |
| 1153. |
Among the following, the compound that is both paramagnetic and coloured is :-A. `K_(20Cr_(2)O_(7)`B. `(NH_(4))_(2)[TiCl_(6)]`C. `VOSO_(4)`D. `K_(3)[Cu(CN)_(4)]` |
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Answer» Correct Answer - C `VOSO_(4)`is paramagnetic as well as coloured compound. The oxidation state of vanadium in `VOSO_(4) is +4`. `V[Z=23]=[Ar]3d^(3)4s^(2)` `V^(4+)[Z=23][Ar]3d^(1)4s^(0)` it has one unpaired electron. Hence, it is paramagnetic in nature. |
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| 1154. |
Which one of the following reactions will occure on heating `AgNO_(3)` above its melting point?A. `2AgNO_(3)rarr2Ag+2NO_(2)+O_(2)`B. `2AgNO_(3)rarr2Ag+N_(2)+3O_(2)`C. `2AgNO_(3)rarr2AgNO_(2)+O_(2)`D. `2AgNO_(3)rarr2Ag+2NO+2O_(2)` |
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Answer» Correct Answer - A Silver nitrate decomposes to silver nitrite on heating above its melting point `(212^(@))` `2AgNO_(3)overset(gt212^(@)C)rarr2AgNO_(2)+O_(2)` On heating above `450^(@)C`(red hot), silver nitrate decomposes to metallic silver, oxide of nitrogen and oxygen. `2AgNO_(3)overset(gt450^(@)C)rarr2Ag+2NO_(2)+O_(2)` |
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| 1155. |
In solution of `AgNO_(3)`.If Cu is added, the solution becomes blue due toA. oxidation of AgB. oxidation of CuC. reduction of AgD. reduction of Cu |
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Answer» Correct Answer - B Cu is oxidised which turns the solution blue. |
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| 1156. |
Assertion: `Ti^(3+)` salts are coloured whereas `Ti^(4+)` salts are white. Reason: `Ti^(3+)` is less stable than `Ti^(4+)`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 - b Correct R: `Ti^(3+)` salts have one unpaired electron in d-subshell `(3d^1)` whereas `Ti^(4+)` salts have no electron in the d-subshell `(3d^0)`. |
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| 1157. |
(a). The addition of NaOH solution to a solution of zinc chloride produces a white precipitate which dissolves on further addition of NaOH. (b). The addition of `NH_4OH` to `ZnSO_4` solution produces white precipitate but no precipitate is formed if it contains `NH_4Cl`. |
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Answer» First a white precipitate of `Zn(OH)_2` is formed which on further addition of NaOH forms soluble zincate `NaZnO_2`. (b). `NH_4OH` is a weak hydroxide. It is ionised slightly furnishing `OH^(ɵ)` ions which are sufficient to precipitate `Zn(OH)_2` because its low solubility product. However, in presence of `NH_4Cl`, the ionisation of `NH_4OH` is further suppressed and sufficient `OH^(ɵ)` are not available to cause precipitation. |
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| 1158. |
Which element`//` elements is`//` are alloyed with aluminium or tin in making air craft frames and jet engines and why? |
| Answer» Titaneium and niobium are used for alloying with aluminium or in making aircraft frames and jet engines. This is because the alloys obtained have high strength. | |
| 1159. |
Assertion (A) Cu cannot liberate hydrogen from acids. Reason (R ) Because it has positive electrode potential.A. Both assertion and reason are true, and reason is the correct explanation of the assertion.B. Both assertion and reason are true but reason is not the correct explanation of the assertion.C. Assertion is not true but reason is true.D. Both assertion and reason are false |
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Answer» Correct Answer - A (a) Reason is the correct explanation for assertion |
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| 1160. |
A certain metal will liberate hydrogen from dilute acids. It will react with to from hydrogen only when the metal is heated and water is in the from of steam . The metal is probablyA. ironB. Potassium manganateC. copperD. mercury |
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Answer» Correct Answer - A `Fe+underset(dil)(H_(2)SO_(4))rarrFeSO_(4)+H_(2)uarr` `underset("hot")(3Fe)+underset ("steam")(4H_(2)O)rarr4H_(2)uarr+Fe_(3)O_(4)` |
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| 1161. |
Which of the following compound is used in gas lighter?A. Pyrophoric Misch metalB. `CeO_(2)`C. NichromeD. Nitinol |
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Answer» Correct Answer - A Pyrophoric Misch metal, which is an alloy of Ce, La and some other rare earth elements, is extensively employed in making gas lighters and cigaratte lighters. It is also a powerful reductant. |
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| 1162. |
Potassium permaganate acts as oxidant in alkaline and acidic media. The final products formed from `KMnO_(4)` in the two conditions are respectivelyA. `MnO^(2-) and Mn^(3+)`B. `Mn^(3+) and Mn^(2+)`C. `Mn^(2+) and Mn^(3+)`D. `Mn^(2+), MnO_(2)` |
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Answer» Correct Answer - D In acidic medium, `overset (+7)(KMnO_(4))rarroverset(+2)(MnSO_(4)) ` In weak basic medium, `KMnO_(4)rarrMnO_(2)` |
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| 1163. |
The reaction of aqueus `KMnO_(4)` with `H_(2)O_(2)` in acidic conditions givesA. `Mn^(4+) and O_(2)`B. `Mn^(2+) and O_(2)`C. `Mn^(2+) and O_(3)`D. `Mn^(4+) and MnO_(2)` |
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Answer» Correct Answer - B The reaction of aqueous `KMnO_(4) with H_(2)O_(2)` in acidic medium is `3H_(2)SO_(4)+2KMnO_(4)+5H_(2)O_(2) rarr5O_(2)+2MnSO_(4)+8H_(2)O+K_(2)SO_(4)` In the above reaction, `KMnO_(4)" oxidises" H_(2)O_(2) to O_(2)` and itself `[MnO_(4)^(-)]` gets reduced to `Mn^(2+) ion as MnSO_(4)` . Hence, aqueous solution of `KMnO_(4)` with `H_(2)O_(2) yields Mn^(2+) and O_(2)` in acidic conditions. |
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| 1164. |
The reaction of aqueous `KMnO_(4)` with `H_(2)O_(2)` in acidic conditions givesA. `Mn^(4+)` and `O_(2)`B. `Mn^(2+)` and `O_(2)`C. `Mn^(2+)` and `O_(3)`D. `Mn^(4+)` and `MnO_(2)` |
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Answer» Correct Answer - B `H_(2)O_(2)` is oxidized to `H_(2)O` and `O_(2)` as under `:` `2KMnO_(4) +3H_(2)SO_(4) + 5H_(2)O_(2) rarr K_(2)SO_(4) + 2MnSO_(4) + 8 H_(2)O + 5O_(2)` or `MnO_(4)^(-) + 5H_(2)O + 6H^(+) rarr 2Mn^(2+) + 8 H_(2)O + 5O_(2)` |
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| 1165. |
The reaction of aqueus `KMnO_(4)` with `H_(2)O_(2)` in acidic conditions givesA. `Mn^(2+) and O_(2)`B. `Mn^(2) and O_(3)`C. `Mn^(4+) and MnO_(2)`D. `Mn^(4+) and O_(2)` |
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Answer» Correct Answer - A It is the correct answer `2KMnO_(4)+5H_(2)O_(2)+3H_(2)SO_(4)` `underset(Mn^(2+))(K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+SO_(2))` |
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| 1166. |
Lanthanoid contraction is caused due to: |
| Answer» Lanthanoid contraction is because of poor shielding effect of 4f electrons. | |
| 1167. |
Lanthanoid contraction is caused due to:A. the same effective nuclear charge from Ce to LuB. the imperfect shielding on outer electrons by 4f electrons fro the nuclear chargeC. the appreciable shielding on outer electrons by 4f electrons from the nuclear chargeD. the appreciable shielding on outer electrons by 5d electrons from the nuclear charge |
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Answer» Correct Answer - B Lanthanoid contraction is caused due to the imperfect shielding on outer electrons by `4f` electrons from the nuclear charge. |
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| 1168. |
Because of lanthanoid contraction :A. separation of the elements is possibleB. there is a very small difference in the atomic size of the transition metals of 5th and 6th period and present in the same group.C. there is a gradual decrease in the basic strengths of the hydroxides of lanthanoidsD. All are correct. |
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Answer» Correct Answer - D all the statements are correct. |
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| 1169. |
Lanthanoid contraction means :A. small atomic sizes of lanthanoid elementsB. small ionic size of lanthanoid ions.C. smaller atomic size of lanthanoids as compared to transition element in the same periodD. decrease in atomic and ionic radii of the element of lanthanoid series in moving from left to the right. |
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Answer» Correct Answer - D is the correct answer |
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| 1170. |
Pair of elements having almost equal size due to lanthanoid contraction areA. Al and GaB. Zr and HrC. Sb and BiD. Sc and Te |
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Answer» Correct Answer - B Zr (144 pm) and Hf (145 pm ) - same radius |
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| 1171. |
Which of the following lanthanoid ion is paramagnetic ?A. `Ce^(4+)`B. `Yb^(2+)`C. `Lu^(3+)`D. `Eu^(2+)` |
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Answer» `Ce= [Xe] 4f^(1) 5d^(1) 6s^(2), Ce^(4+)= [Xe]` `Yb= [Xe] 4f^(14) 6s^(2) , Yb^(2+) = [Xe] 4f^(14)` `Lu = [Xe] 4f^(14) 5d^(1) 6s^(2) ,Ly^(3+) = [Xe] 4f^(14)` `E u= [Xe]4f^(7)6s^(2) , Eu^(2+) = [Xe] 4f^(7)` Thus, only `Eu^(2+)` has unpaired electrons , vix.7. |
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| 1172. |
Among the following lanthanoid ions, the paramagnetic ion isA. `Yb^(2+)`B. `Eu^(2+)`C. `Lu^(3+)`D. `Ce^(4+)` |
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Answer» Correct Answer - B The valence shell electronic configuration of the given ions are `Yb=[Xe]4f^(14)6s^(2)impliesYb^(2+)=[Xe]4f^(14)implies` No unpaired `e^(-)` `Eu=[Xe]4f^(7)6s^(2)impliesEu^(2+)=[Xe]4f^(7)implies7` unpaired `e^(-)` `Lu=[Xe]4f^(14)5d^(1)6s^(2)impliesLu^(3+)=[Xe]4f^(14)implies` No unpaired `e^(-)` `Ce=[Xe]4f^(1)5d^(1)6s^(2)impliesCe^(4+)=[Xe]implies` No unpaired `e^(-)` Hence, `Eu^(2+)` is paramagnetic as it contains 7 unpaired electrons. |
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| 1173. |
Which of the following lanthanoid ion is diamagnetic?A. `Yb^(2+)`B. `Ce^(2+)`C. `Sm^(2+)`D. `TiCl_(4)` |
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Answer» Correct Answer - A `Yb^(2)` ion with configuration `[Xe]4F^(14)` is diamagntic |
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| 1174. |
The expression for effective magnetic moment `(mu_(eff))` isA. `mu _("eff")=sqrt(n(n+2)BM)`B. `mu _("eff")=sqrt(2(n+2)BM)`C. `mu _("eff")=sqrt(n(2n+2)BM)`D. `mu _("eff")=sqrt(n(n+1)BM)` |
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Answer» Correct Answer - A A it is the correct answer |
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| 1175. |
Catalytic activity of the transition metals and their compounds can be ascribed to:A. their magnetic behaviorB. their unfilled d-orbitalsC. their ability to adopt variable oxidation statesD. the chemical ractivity |
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Answer» Correct Answer - C C it is the correct answer |
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| 1176. |
Most stable oxidation state of iron isA. `+1`B. `+3`C. `-2`D. `-3` |
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Answer» Correct Answer - B `Fe(26) to [Ar] 4s^2 3d^6` `Fe^(3+) to [Ar] 4s^0 3d^5` |
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| 1177. |
How do transition metals exhibit catalytic activity? |
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Answer» Catalytic properties : `rarr` Transition metals and their compounds form important catalysts in industry and in biological systems. `rarr` The catalytic activity of transition metals depends on their ability to exist in different oxidation states of oxidation (or) to form coordination compounds. Eg. : 1) `V_(2)O_(5)` is used as catalyst in manufacturing of `SO_(3)" from SO_(2)` `2SO_(2_(g))+O_(2_(g))overset(V_(2)I_(5))hArr2SO_(3_(g))` 2) Fe is used as catalyst in manufacturing of `NH_(3)`. `N_(2_(g))+3H_(2_(g))overset(Fe_(5))hArr2NH_(3_(g))` |
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| 1178. |
Which one of the following characteristics of transition metals is associated with their catalytic activity?A. high heat of atomisationB. paramagnetic behaviorC. colour of hydrated ionsD. variable oxidation states |
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Answer» Correct Answer - D Variable oxidation state and ability to form complexes with the reactant molecules help in catalytic activity of transition metals |
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| 1179. |
Which is/are not the true statement(s) about `KMnO_(4)`?A. Its solution is unstable in acidic mediumB. It gets reduced to `MnO_(2)` in neutral mediumC. `MnO_(4)^(-)` changes to `Mn^(2+)` in basic mediumD. It is a self indicator in `FeSO_(4)` and oxalic acid titrations. |
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Answer» Correct Answer - C `MnO_(4)^(-)` changes to `Mn^(2+)` in acidic medium `MnO_(4)^(-)+8H^(+)+5e^(-) to Mn^(2+)+4H_(2)O`. |
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| 1180. |
The reaction of `KMnO_(4) and HCI` results in:A. Oxidation of Mn in `KMnO_(4)` and production of `Cl_(2)`B. reduction of Mn in `KMnO_(4)` and production of `H_(2)`C. oxidation of Mn in `KMnO_(4)` and production of `H_(2)`D. reduction of Mn in `KMnO_(4)` and production of `Cl_(2)` |
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Answer» Correct Answer - D `{:(2KMnO_(4),+,16HCl,,,,),(" "darr,,,,,,),(2KCl,+,2MnCl_(2),+,8H_(2)O,+,5Cl_(2)):}` In this reaction `Mn^(+7)` in `KMnO_(4)` is reduced to `Mn^(2+)` in `MnCl_(2)` while `Cl^(-)` is oxidised to `Cl_(2)`. |
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| 1181. |
The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is becauseA. the 5f orbital extend farther from the nucleus than the 4f orbitalsB. the 5f orbitalare more buried than the 4f orbitalC. there is a similarity between 4f and 5f orbitals in their angular part of the wave functionsD. The actinoids are more reactive thean the lanthanoids. |
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Answer» Correct Answer - D In the electron orbitals involved in actinoid family (5d,6f and 7s) the energy difference is less as compared to orbitals (5f, 4d and 6s) involved in the lanthanoid family. As a result more electrons becomes available for bond formation in the actinoid family as compared to lanthanoid family, Therefore, the elements belonging to actinoid family exhibited more variable oxidation states as compared to those belonging to lanthanoid family. |
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| 1182. |
The basic character of the transition metal monoxide follows the orderA. `VO lt CrO gt TiO gt FeO`B. `CrO gt VO gt FeO gt TiO`C. `TiO gt FeO gt VO gt CrO`D. `TiO gt VO gt CrO gt FeO` |
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Answer» Correct Answer - D Metal oxide with more ionic character is expected to be more basic. Ionic radii of a metal ion decrease from `Ti^(2+)` to `Fe^(2+)`. As a result the basic strengths of their oxides decreases from TiO to FeO. Thus, (d) is the correct option |
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| 1183. |
Variable valency is generally shown by-A. `s-`block elementsB. `p-`block elementsC. Transition elementsD. All elements in periodic table |
| Answer» One of the most striking features of the transition elements, is that they exhibit variable valency. The variable valency occurs to some extent in the `p-`block elements also. In this case the valency changes usually in units of two. Some example are `SnCl_(2),SnCl_(4),PCl_(3),PCl_(5) etc`. But in transition elements the valency changes in one unit `e.g Cu^(+),Cu^(+2),Fe^(+2),Fe^(+3) etc`. | |
| 1184. |
A mixture consist of a yellow solid [A] and colourless solid [B] which given lilac colour in flame. The miture given a black precipiate [C] on passing `H_(2)S` gas. The precipitate [C] is soluble in aqura region and on evaporating and on adding `SnCl_(2)`, gives greyish black precipitate [D]. The misture solution on reacing with `NH_(4)OH` gives a brown precipiates. The sodium extract of the mixture with `C Cl_(4)//FeCl_(3)` given violet layer. The extract also given yellow precicitate with `AgNO_(3)` solution which is insoluble in `NH_(3)`. The black precipitate [c] is of :A. Mercurous sulphideB. Mercuric sulphide.C. Bismuth sulphideD. Lead sulphide. |
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Answer» Correct Answer - B (b) `Hgl_(2)` reacts with `H_(2)S` to give a black precipitate [C] of merucric sulphide (HgS) `HgI_(2)+H_(2)Srarrunderset([C])underset("Black ppt.")(HgS)+2HI` |
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| 1185. |
A mixture consist of a yellow solid [A] and colourless solid [B] which given lilac colour in flame. The miture given a black precipiate [C] on passing `H_(2)S` gas. The precipitate [C] is soluble in aqura region and on evaporating and on adding `SnCl_(2)`, gives greyish black precipitate [D]. The misture solution on reacing with `NH_(4)OH` gives a brown precipiates. The sodium extract of the mixture with `C Cl_(4)//FeCl_(3)` given violet layer. The extract also given yellow precicitate with `AgNO_(3)` solution which is insoluble in `NH_(3)`. The greyish black precipitate [D] is ofA. mercuryB. silverC. leadD. None of these |
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Answer» Correct Answer - A (a) In aqua regia, `HgI_(2)` is converted to `HgCl_(2)` which is then reduced by `SnCl_(2)` to mercury [D] which is greyish black in colour `HgCl_(2)+SnCl_(2)rarrunderset("black[D]")underset("Greyish")(Hg)+SnCl_(4)` |
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| 1186. |
An inorganic compound [A] yellowish green in colour loses its water of crystallisation upon heating . When heated furthr, it changes to blackish brown powder [B] and also gives two oxides of sulphure. The power [B] on boiling with hydrochloric acid given a yellowish solution which gives a bloodred colouration on reacting with potassium thiocyanate. The yellowish green compound [A] isA. `CuSO_(4). 5H_(2)O`B. `FeSO_(4).7H_(2)O`C. `FeSO_(4).(NH_(4)) SO_(4). 6H_(2)O`D. `Fe_(2)(SO_(4))_(3).7H_(2)O` |
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Answer» Correct Answer - B (b) The compound [A] is sulphate of some metal as it give two oxides of sulphur `(SO_(2)" and "SO_(3))` on heating. Its yellowish green colour suggests that (A) if `FeSO_(4).7H_(2)O` (green vitriol). |
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| 1187. |
An aqueous solution of `FeSO_4.Al_2(SO_4)_3` and chrome alum is heated with excess of `Na_2O_2` and filtered. The materials obtained areA. A colourless filtrate and a green residueB. A yellow filtrate and a green residueC. A yellow filtrate and a brown residueD. A green filtrate and a brown residue |
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Answer» Correct Answer - C In the presence of peroxide,chromium ions are oxidised to chromate ions which give a yellow filtrate. Ferric ions form brown precipitate of `Fe(OH)_3`. |
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| 1188. |
The blue colour produced on adding `H_2O_2` to acidified `K_2Cr_2O_7` is due to the formation ofA. `CrO_5`B. `Cr_2O_3`C. `CrO_4`D. `CrO_3` |
| Answer» Correct Answer - a | |
| 1189. |
When `SO_2` is passed through acidified `K_2Cr_2O_7` solutionA. the solution turns blueB. the solution is discolourisedC. `SO_2` is reducedD. green `Cr_2(SO_4)_3` is formed |
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Answer» Correct Answer - D `K_2Cr_2O_7 + H_2SO_4 + 3SO_4 to K_2SO_4 + underset"Green"(Cr_2(SO_4)_3) + H_2O` |
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| 1190. |
The addition of high proportions of maganese makes steel useful in making rails or railroads because manganese useful in making rails or railroads because maganeseA. Gives hardness to steelB. Helph in the formation of oxides of ironC. Can remove oxygen and sulphurD. Can show the highest oxidation state of `+7` |
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Answer» Correct Answer - A::C Manganese gives hardness to steel. It also removes oxygen and sulphur from steel by forming slag. `Fe_2O_3+3Mnto3MnO+2Fe` `MnO+SiO_2toMnSiO_3` |
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| 1191. |
Acidified `K_2Cr_2O_7` solution turns green when sodium sulphite is added to it. Explain. |
| Answer» `Cr_2O_7^(2-)+8H^(ɵ)+3SO_3^(2-)tounderset(green)(2Cr^(3+)+3SO_4^(2-)+4H_2O` | |
| 1192. |
Which of the following statements is/are correct, when a mixture of `NaCl` and `K_2Cr_2O_7` is gently warmed with concentrated `H_2SO_4`?A. Deep red vapours are evolved.B. The vapours when passed into NaOH solution give a yellow solution of `Na_2CrO_4`.C. Chlorine gas is evloved.D. Chromyl chloride is formed. |
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Answer» Correct Answer - A::B::D `K_2Cr_2O_7+6H_2SO_4+4NaClto2KHSO_4+4NaHSO_4+2CrO_2Cl_2+3H_2O` Orange-red vapours of `CrO_2Cl_2` are produced. When these vapours are passed through `NaOH`, a yellow solution of `Na_2CrO_4` is produced. `CrO_2Cl_2+4NaOHtoNa_2CrO_4+2H_2O+2NaCl` |
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| 1193. |
Identify A, B, C, D, and X. When a white crystalline compound X is heated with `K_2Cr_2O_7` and concentrated `H_2SO_4`, a reddish brown gas A is evolved. On passing A into caustic soda solution a yellow coloured solution of B is obtained Neutralising the solution of B with acetic acid and on subsequent addition of lead acetate, a yellow precipitate C is obtained. When X is heated with NaOH solution a colourless gas is evolved and on passing this gas into `K_2Hgl_4` solution a reddish brown precipitate D is formed. |
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Answer» `X` is `NH_4Cl` `4Cl^(ɵ)+K_2Cr_2O_76H_2SO_4tounderset((gas A))(2CrO_2Cl_2+2KHSO_4+4HSO_4^(ɵ)+3H_2O` `CrO_2Cl_2+4NaOHtounderset((B))(Na_2CrO_4+2NaCl+2H_2O` `Na_2CrO_4+(CH_3COOH)_2PbtoPbCrO_4+2NaCl+2H_2O` `NH_4^(o+)+4NaOHoverset(Delta)tounderset((Gas))(NH_3)+Na^(o+)+H_2O` `NH_3+2[HgI_4]^(2-)+H_2OtoH_2N-Hg-O-Hg-Idarr` |
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| 1194. |
Which of the following statements is/are correct, when a mixture of `NaCl` and `K_2Cr_2O_7` is gently warmed with concentrated `H_2SO_4`?A. `CrO_2Cl_2`B. `CrCl_3`C. `Cr_2(SO_4)_2`D. `Na_2CrO_4` |
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Answer» Correct Answer - A `K_2Cr_2O_7 + 6H_2SO_4+4 NaCl to 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O` |
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| 1195. |
Mixture of `K_2Cr_2O_7` and conc. `H_2SO_4` is calledA. Perchromic acidB. Chromic acidC. Chromium sulphateD. None of these |
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Answer» Correct Answer - B Solution of `K_2Cr_2O_7` in conc. `H_2SO_4` is called chromic acid |
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| 1196. |
Assertion (A) Separation of Zr and Hf is difficult. Reason (R ) Because Zr and Hf lie in the same group of the Periodic Table.A. Both assertion and reason are true, and reason is the correct explanation of the assertion.B. Both assertion and reason are true but reason is not the correct explanation of the assertion.C. Assertion is not true but reason is true.D. Both assertion and reason are false |
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Answer» Correct Answer - B (b) Correct explanation. It is because of lanthanoids contraction. |
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| 1197. |
Assertion (A) Separation of Zr and Hf is difficult. Reason (R ) Because Zr and Hf lie in the same group of the Periodic Table.A. Both assertion and reason are true, and reason is the correct explanation of the assertion.B. Both assertion and reason are true but reason is not the correct explanation of assertion.C. Assertion is not true but reason is true.D. Both assertion and reason are false. |
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Answer» Correct Answer - B Assertion and reason are true but reason is not correct explanation of assertion. Separation of Zr and Hf is difficult, it is not because of they lie in the same group of Periodic Table. This is due to lanthanoid contraction which causes almost similar radii of both of them. |
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| 1198. |
Assertion `:` Separation of Zr and Hf is difficult. Reason `:` Because Zr and Hf lie in the same group of the periodic table.A. Both assertion and reason are true and reason is the correct explanation of the assertion .B. Both assertion and reason are true but reason is not the correc explanation of assertion.C. Assertion is not true for reason is trueD. Both assertion and reason are false. |
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Answer» Correct Answer - b Correct explanation . Zr and Hf have nearly the same size. |
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| 1199. |
A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C ) of manganese are formed. Compound (C ) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C ) with conc. `H_(2)SO_(4)` and NaCl, chlorine gas is liberate and a compound (D) of manganese alongwith other products is formed. Identify compounds A to D and also explain the reaction involved. |
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Answer» Since, compound (C ) on treating with conc. `H_(2)SO_(4)` and NaCl gives `Cl_(2)` gas, so it is manganese dioxide `(MnO_(2))`. It is obtained alongwith `MnO_(4)^(2-)` when `KMnO_(4)` (violet) is heated. Thus, (A) `=KMnO_(4)" (B)"=K_(2)MnO_(4)` (C ) `=MnO_(2)" (D)"=MnCl_(2)` `underset([A])(KMnO_(4))overset(Delta)rarrunderset([B])(K_(2)MnO_(4))+underset([C])(MnO_(2))+O_(2)` `2MnO_(2)+4KOH + O_(2) rarr2K_(2)MnO_(4)+2H_(2)O` `MnO_(2) + 4NaCl+4H_(2)SO_(4)rarrunderset([D])(MnCl_(2))+4NaHSO_(4) + 2H_(2)O + Cl_(2)` |
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| 1200. |
In the 5f series highest oxidation state is exhibited byA. U and NpB. Np and PuC. Pu and AmD. Cm and Fm |
| Answer» Correct Answer - B | |