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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.
| 151. |
Complete and //balance the following equation: `Ag_(2)S+CuCl_(2)+Hg…………….+…………………..+S+2Ag` |
| Answer» Correct Answer - `Ag_(2)S+2CuCl_(2)+2Hg to Hg_(2)Cl_(2)+2CuCl+S+2Ag`. | |
| 152. |
When `MnO_(2)` fused with `KOH`, a coloured compound is formed . The product and its colour is:A. `K_(2)MnO_(4)`, greenB. `Mn_(2)O_(3)`brownC. `Mn_(2)O_(4)`, blackD. `KMnO_(4)`, purple |
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Answer» Correct Answer - A Pyrolusite on fusion with `KOH` in air give green coloured maganate. `2MnO_(2)+4KOH+O_(2) to 2K_(2)MnO_(2)(green)+2H_(2)O` |
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| 153. |
Which of the following form an alloy?A. `Ag+Pb`B. `Fe+Hg`C. `Pt+Hg`D. `Fe+C` |
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Answer» Argentiferrous lead is an alloy contains `Ag & pb` `Fe` and `C` from an alloy called cementite, `Fe_(3)C` Therefore, (A) & (D) option are correct. |
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| 154. |
Which one of the ionic species will impart colour to an aqueous solution?A. `Ti^(4+)`B. `Cu^(+)`C. `Zn^(2+)`D. `Cr^(3+)` |
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Answer» Correct Answer - D In `Ti^(4+),Cu^(+)` and `Zn^(2+)` , all have electrons paired so all the diamagnetic. `Cr^(3+)` with electrons configuration `[Ar]^(18) 3d^(3)` has three unpaired electrons. So it undergoes d-d transition of electrons in presence of ligands according to CFT and thus it is coloured. |
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| 155. |
Statement-1: d-block metals have generally high M.P. Statement-2: Greater number of electrons from `(n-1)d` in addition to ns electrons are involved in the interatomin metallic bonding.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 156. |
Bordeaux used a fungicide is a mixture ofA. `CaSO_(4)+Cu(OH)_(2)`B. `CuSO_(4)+Ca(OH)_(2)`C. `CuSO_(4)+CaO`D. `CuO+CaO` |
| Answer» Correct Answer - B | |
| 157. |
Peacock ore is:A. `FeS_(2)`B. `CuFeS_(2)`C. `CuCO_(3)*Cu(OH)_(2)`D. `Cu_(5)FeS_(4)` |
| Answer» Correct Answer - D | |
| 158. |
The oxidation number of Mn in the product of alkaline oxidative fusion of `MnO_2` isA. `3` electrons in neutral mediumB. `5` electrons in neutral mediumC. `3` electrons in alkaline mediumD. `5` electrons in acidic medium |
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Answer» Correct Answer - ACD in acidic medium,`" "`MnO_(4)^(-)+8H^(+)+5e^(-) to Mn^(2+)+4H_(2)O` `(V.f=5)` `"in neutral//faintly alkaline medium"," " `MnO_(4)^(-)+2H_(2)O+3E^(-) to MnO_(2)+4OH^(-)` (v.f.=3) "In strong alkaline medium"," " MnO_(4)^(-)+e^(-) to MnO_(4)^(2-)` (v.f.=1) |
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| 159. |
Maximum oxidation state is shown by:A. `Os`B. `Mn`C. `Cr`D. `Co` |
| Answer» Correct Answer - AC | |
| 160. |
The less stable oxidation state of `Cr` are:A. `Cr^(2+)`B. `Cr^(3+)`C. `Cr^(4+)`D. `Cr^(6+)` |
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Answer» Correct Answer - ACD `Cr^(3+)` is most stable because in aqueous solution it has higher `CFSE` on account of half filled `t_(2g)^(3)` energy level of `3d` orbitals in octahedral spliting. |
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| 161. |
`CuSO_(4).5H_(2)Oundersetto(100)^(0)C[X](S) undersetto (250^(0))C[Y](S)`A. `X` and `Y` are `CuSO_(4).3H_(2)O` and `CUSO_(4)`B. `X` and `Y` are `CuSO_(4).3H_(2)O` and `CUSO_(4).H_(2)O`C. `X`and `Y` are `CuSO_(4).H_(2)O` and `CuSO_(4)`D. `X` and `Y` are `CuSO_(4)` and `CuO` |
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Answer» `CuSO_(4).5H_(2)Oundersetto(100)^(0)CuSO_(4).H_(2)O(s)undersetto(250)^(0)CuSO_(4)(s)` So, (C ) option is correct. |
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| 162. |
`CuSO_(4).5H_(2)O` is blue in colour becauseA. It contains water of crystallization.B. `SO_(4)^(2-)` ions absorbs red light.C. `Cu^(2+)` ion absorb red lightD. `Cu^(2+)` ions absorbs all colours except red from the white light. |
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Answer» Correct Answer - C `Cu^(2+)` ion `(3d^(9))` absorbs red light from the visible region, for the promotion of `3d` electrons, the ions reflect blue and appear blue. |
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| 163. |
When copper is placed in the atmosphere for sufficient time, a green crust is formed on its surface. The composition of the green crust is:A. `CU(OH)_(2)`B. `CuO`C. `CuCO_(3)`D. `CuCO_(3).Cu(OH)_(2)` |
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Answer» Correct Answer - D `Cu+H_(2)O+CO_(2) to CUCO_(3).Cu(OH)_(2)` Green crust of basic copper carbonate is formed. |
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| 164. |
The yellow colour of chromates changes to orange on acidification due to formation of:A. `Cr^(3+)`B. `Cr_(2)(CO_(3))_3`C. `Cr_(2)O_(7)^(2-) `D. `CrO_(4)^(-)` |
| Answer» Correct Answer - C | |
| 165. |
On heating `ZnXl_(2).2H_(2)O`, the compounds obtained is:A. `ZnCl_(2)`B. `Zn(OH)Cl`C. `Zn(OH)_(2)`D. `Zn ` |
| Answer» Correct Answer - B | |
| 166. |
`MnO_(4)^(-)` is of intense pink colour, though `Mn` is in`(+7)` oxidation state.It is due to:A. Oxygen gives colour to itB. charge transfer when `Mn` gives its electrons to oxygenC. charge transfer when oxygen its electron to `Mn` making it `Mn(+VI)` hence colouredD. None is correct |
| Answer» Correct Answer - C | |
| 167. |
Calomel `(H_(2)Cl_(2))` on reaction with ammonium hydroxide givesA. `HgNH_(2)Cl`B. `NH_(2)-Hg-Hg-Cl`C. `Hg_(2)O`D. `HgO` |
| Answer» Correct Answer - A | |
| 168. |
Colourless solutions of the following four salts are placed separately in four different test tubes and a strip of copper is dipped in each one of these. Which solution will turn blue?A. `KNO_(3)`B. `AgNO_(3)`C. `Zn(NO_(3))_(2)`D. `ZnSO_(4)` |
| Answer» Correct Answer - B | |
| 169. |
Which of the following increasing order of oxidising power is correct for the following species? `VO_(2)^(+),MnO_(4)^(-),Cr_(2)O_(7)^(2-)`A. `VO_(2)^(+)ltCr_(2)O_(7)^(2-)lt MnO_(4)^(-)`B. `VO_(2)^(+)ltMnO_(4)^(-)ltCr_(2)O_(7)^(2-)`C. `Cr_(2)O_(7)^(2-) lt VO_(2)^(+)ltMnO_(4)^(-)`D. `Cr_(2)O_(7)^(2-)ltMnO_(4)^(-)ltVO_(2)^(+)` |
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Answer» This is attributed to the increasing stability of the lower species to which they are reduced. `MnO_(4)^(-)` is reduced to `Mn^(2+)` which has stable half filled valency shell electron configuration `[3d^(5)]` `Cr_(2)O_(7)^(2-)`is reduced to `Cr_(3+)` which has half filled `t^(3)_(2g)` energy levels of `3d` orbitals in octahedral crystal field spliting `VO_(2)^(+)` is reduced to `V^(3+)` which has electonioc configuration `[Ar]^(18)3d^(2)4S^(0)`. So the order of increasing stability of the reduced species is `mn^(2+)gtCr^(3+)gtV^(3+)` and , therefore, the increasing order of oxidising power is `VO_(2)^(+)ltCr_(2)O_(7)^(2-)ltMnO_(4)^(-)` . Therefor, (a) option is correct. |
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| 170. |
Solder is an alloy of :A. Cu and PbB. Zn and CuC. Pb and SnD. Fe and Zn |
| Answer» Correct Answer - C | |
| 171. |
Metre scales are madeup of alloy:A. invarB. stainless steelC. electronD. magnalium |
| Answer» Correct Answer - A | |
| 172. |
Which physical characteristic distinguishes copper from brass (an alloy of copper and zinc)?A. Brass is a liquid at room temperature and copper is not.B. Brass is much less dense that copper.C. Brass is attracted to a magnetic but copper is not.D. Brass is a much poorer electrical conductor than copper. |
| Answer» Correct Answer - D | |
| 173. |
The correct statement(s) about transition element is/are.A. The most stable oxidation is +3 and its stability decreases across the period.B. transition elements of 3d- series have almost same atomic sizes from Cr to CuC. The stability of +2 oxidation state increases across the period.D. Some transition elements like Ni, Fe, Cr may show zero oxidation state in some of their compound. |
| Answer» Correct Answer - A::B::C::D | |
| 174. |
The number of transition series is:A. twoB. threeC. fourD. five |
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Answer» There are four transition series , `3d,4d,5d and 6d` series Therefore, (c ) option is correct. |
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| 175. |
The element with the electronic configuration `[Ze]^(54) 4f^(14)5d^(1)6S^(2)` is aA. representative elementB. d-block elementC. lanthanoidD. actinoid |
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Answer» After achieving `4f^(14),5d^(0),6s^(2)` configuration , the next electron goes to `5d` and this is the case of `Lu(Z=71)` which is the last element of lanthanoid series Therefore , (C ) option is correct |
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| 176. |
Assertion : Silver chloride dissolves in execss ammonia. Reason : `AgCl` forms a soluble complex, `[Ag (NH_(3))_(2)] Cl` with ammonia.A. Statement-1 is true , Statement-2 is true and Statement-2 is correct explanation for Statement-1B. Statement-1 is true, Statement-2 is True and Statement-2 is Not correct explanation for statement-1C. Statement-1 is true, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 177. |
Copper sulphate dissolves in `Nh_(4)OH` solution but `FesO_(4)` does not. |
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Answer» Correct Answer - Copper sulphate dissolve in the ammonium hyroxide due to formation of a complex. Ferrous sulphate reacts with `NH_(4)OH` to form insoluble `Fe(OH)_(2)` . It does not any complex with `NH_(4)OH`. `CuSO_(4)+4NH_(4)OH to [Cu(NH_(3))_(4)]So_(4)+4H_(2)O` Deep blue solution `FeSO_(4)+2NH_(4)OH to Fe(OH)_(2)+(NH_(4))_(2)SO_(4)` Insoluble |
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| 178. |
Complete and balance the following reactions: (a) `FeSO_(4)+H_(2)O+O_(2) to` , (b) `FeCl_(3).6H_(2)O+CH_(3)-underset(OCH_(3))underset(|)overset(OCH_(3))overset(|)C-CH_(3) to` (C ) `CU(OH)_(2)+(NH_(4))_(2)SO_(4)to`,(d)`CuCl_(2).2H_(2)Ounderset(Strong)undersettoDelta` (e ) `AgNO_(3)(excess)+I_(2)+H_(2)O to ` , (f) `KMnO_(4)+Na_(2)S_(2)O+H_(2)O to` (g) `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+SO_(2)+H_(2)O undersetto("Temptlt70^(@)C)` |
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Answer» Correct Answer - (a)`FeSO_(4)+2H_(2)O+Oto4Fe(OH). SO_(4)` (b) `FeCl_(3).6H_(2)O+6CH_(3)-underset(OCH_(3))underset(|)overset(OCH_(3))overset(|)(C)-CH_(3)toFeCl_(3)+12Ch_(3)OH+6CH_(3)COOCH_(3)` `(C )Cu(OH)_(2)+2nH_(4)OH(NH_(4))_(2)SO_(4)to[Cu(NH_(3))_(4)]SO_(4)("switzer reagent")+4H_(2)O` (d) `3CuCl_(2).2H_(2)Ounderset("Strong")undersettoDelta CuO+Cu_(2)Cl+2HCI+Cl_(2)+H_(2)O` (e ) `6AgNO_(3)+3I_(2)+3H_(2)OtoAgIO_(3)+5AgI+6HNO_(3)` (f) `8MnO_(4)^(-)+3S_(2)O_(3)^(2-)+H_(2)Oto8MnO_(2)+6SO_(4)^(2-)+2OH^(-)` (g) `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2)+23H_(2)Oundersetto(Temptlt70^(@)C) K_(2)SO_(4).Cr_(2)(SO_(4))_(3).24H_(2)O` |
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| 179. |
Coinage metals show the properties of:A. typical elementsB. normal elementsC. inner-transition elementsD. transition element. |
| Answer» Correct Answer - D | |
| 180. |
`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. `2//3` mole of `MnO_(4)^(-)` and `1//3` mole of `MnO_(2)`B. `1//3` mole of `MnO_(4)^(-)` and `2//3` moles of `MnO_(2)`C. `1//3` mole of `Mn_(2)O_(7)` and `1//3` mole of `MnO_(2)`D. `2//3` mole of `Mn_(2)O_(7)` and `1//3` moles of `MnO_(2)` |
| Answer» Correct Answer - A | |
| 181. |
Transition metals:A. exhibit only diamagnetismB. undergo inert pair effectC. do not form alloysD. show variable oxidation states. |
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Answer» Correct Answer - D (A) According to the definition of transition metals, they have partially filled `(n-1)` d orbitals excepts copper and Zinc thus mostly show paramagnetism. (B) It is property of heavier p-block elements. (C ) Transition metals from a large number of alloys. The transition metals are quite similar in size and , therefore , the atoms of one metal can substitute the atom of other metal can substitute the atoms of other metal in its crystal lattice. Thus , on cooling a mixture , solution of two or more transition metals, solid alloys are formed. (D ) Show variable oxidation states as `(n-1)` d and ns orbitals have nearly the same energy and , thus ns as well as (n-1) d orbital can be lost giving variable oxidation state. |
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| 182. |
The correct ground state electronic configuration of chromium atom(Z=24) is :A. `[Ar]4d^(5)4s^(1)`B. `[Ar]3d^(4)4s^(2)`C. `[Ar]3d^(6)4s^(0)`D. `[Ar]3d^(5) 4s^(1)` |
| Answer» Correct Answer - D | |
| 183. |
How would you account for the following? (i) The atomic radii of the metals of the third`(5d)` series of transition elements are virtually the same as those of the corresponding member of the second `(4d)` serie (ii) The `E^(@)` value for the `Mn^(3+)//Mn^(2+)` couple is much more positive than that for `Cr^(3+)//Cr^(2+)` couple of `Fe^(3+)//Fe^(2+)` couple. (iii) The highest oxidation state of a metal is exhibited in its oxides or fluoride. |
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Answer» Correct Answer - (i) In the second transition series , the number of shells increases and so does the atomic size. However , in the third series `(5d series)` , the atomic size is more or loss the same . This is known as lanthanide contraction. This Is due to the poor shielding effect by f-electrons. (ii) The large positive `E^(@)` value for `Mn^(3+)//Mn^(2+)` show that `Mn^(2+)` is much more stable than `Mn^(3+)` . This is because of the fact that `Mn^(2+)` has `3d^(5)` configuration. This means that the d-orbital is half -filled . Half -filled and fully-filled configuration are very stable. Thus , the third ionization energy of `Mn` will be very high. (iii) Oxygen and fluorine act as strong oxidising agents from various metals . In other words, a metal exhibits highest oxidation state in its oxides and fluorides. For example , in `OsF_(6)` and `V_(2)O_(5)`, the oxidation states of `Os` and `V` are `+6` and `+5` respectively. |
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| 184. |
Which one of the transittion metal ions is coloured in aqueous solution?A. `Cu^(+()`B. `Zn^(2+)`C. `Sc^(3+)`D. `V^(4+)` |
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Answer» Correct Answer - D Valence shell electron configuration of `Cu^(+) is [Ar]^(18) 3d^(10),so n=0` (B) Valence shell electron configuration of `Zn^(2+) is [Ar]^(18) 3d^(10),so n=0` (C ) Valence shell electron configuration of `SC^(3+) is [Ar]^(18) 3d^(0),so n=0` (D) Valence shell electron configuration of `V^(4+) is [Ar]^(18) 3d^(1),so n=1` As `V^(4+)` has one unpaired electrons and so in presence of water as ligand it will undergo d-d transition of electrons. Hence `V^(4+)` ions will produce colour in the solution |
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| 185. |
Arrange the following ions in their magnetic moment: (P). `V^(4+)` (Q). `Mn^(4+)` (R). `Fe^(3+)` (S) `Ni^(2+)`A. QgtRgtPgtSB. RgtSgtRgtPC. RgtQgtSgtPD. PgtSgtRgtQ |
| Answer» Correct Answer - C | |
| 186. |
Beyond Mn no metal has a trihalide except:A. CoB. FeC. NiD. Cu |
| Answer» Correct Answer - A::B | |
| 187. |
`MnO_(2)+KOHoverset(O_(2))toX`. X must be:A. `KMnO_(4)`B. `K_(2)MnO_(4)`C. `MnO`D. `Mn_(3)O_(4)` |
| Answer» Correct Answer - B | |
| 188. |
Beyond Mn which trifluoride exists?A. `CoF_(3)`B. `FeF_(3)`C. `FeCl_(3)`D. `FeBr_(3)` |
| Answer» Correct Answer - A::B::C::D | |
| 189. |
`CuSO_(4)(aq)+4NH_(3)toX`, then X isA. `[Cu(NH_(3))_(4)]^(2+)`B. paramagneticC. colouredD. of a magnetic moment of 1.73 BM |
| Answer» Correct Answer - A::B::C::D | |
| 190. |
Amphoteric oxide(s) of Mn is/areA. `MnO_(2)`B. `Mn_(3)O_(4)`C. `Mn_(2)O_(7)`D. `MnO` |
| Answer» Correct Answer - A::B | |
| 191. |
Statement-1: the highest oxidation state of transition metal is exhibited in its oxide or fluoride only. Statement-2: Small size and high electronegativity of O and F can oxidise the metal to its highest oxidation state.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 192. |
Amphoteric oxide(s) is/areA. `Al_(2)O_(3)`B. `SnO`C. `ZnO`D. `Fe_(2)O_(3)` |
| Answer» Correct Answer - A::B::C | |
| 193. |
Catalytic activity of the transition metals and their compounds can be ascribed to:A. Their ability to adopt multiple oxidation states their ability to form complexesB. Their ability to form complexes.C. catalytic adsorption by utilization of 3d and 4s electrons for bonding.D. all of the above |
| Answer» Correct Answer - D | |
| 194. |
The catalytic activity of transition elements is related to their:A. Variable oxidation statesB. Surface areaC. complex formation abilityD. magnetic moment. |
| Answer» Correct Answer - A::B::C | |
| 195. |
Which one of the following characteristics of the transition metals is associated with their catalytic activity?A. Colour of hydrated ions.B. Variable oxidation states.C. High enthalpy of atomization.D. paramagnetic behaviour. |
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Answer» Correct Answer - B (A) Associated with d--d transition of electrons. (B) The transition metals from the reaction intermediates due to the presence of vacant orbitals or their tendency to form variable oxidation states. (C ) Associated with the number of unpaired electrons participating in metallic bondig (D) As`mu=sqrt(n(n+2))`,So, it associated with number of unpaired electron. |
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| 196. |
Cementite is:A. interstitial compound of iron and carbonB. an alloy of `Fe` and `Cr`C. a compound resembling cementD. an ore of iron |
| Answer» Correct Answer - A | |
| 197. |
Ammonium dichromate is used in some fireworks. The green-coloured powder blown in the air isA. `CrO_(3)`B. `Cr_(2)O_(3)`C. `Cr`D. `CrO(O_(2))` |
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Answer» Correct Answer - B `(NH_(4))_(2)Cr_(2)O_(7)toN_(2)+Cr_(2)O_(3)+4H_(2)O`. Green coloured powder blown in air is `Cr_(2)O_(3)`. |
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| 198. |
Statement-1: Promethium is a man mae element. Statement-2: It is radioactive and has been prepared by artifical means.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 199. |
Which one of the following show highest magnetic moment?A. `V^(3+)`B. `Cr^(3+)`C. `Fe^(3+)`D. `CO^(3+)` |
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Answer» Correct Answer - C Valence shell electron configuration of `V^(3+) is [Ar]^(18)3d^(2),so n=2` (B) Valence shell electron configuration of `Cr^(3+) is [Ar]^(18) 3d^(3),so n=3` (C ) Valence shell electron configuration of `Fe^(3+) is [Ar]^(18) 3d^(5),so n=5` (D) Valence shell electron configuration of `Co^(3+) is [Ar]^(18) 3d^(6),so n=4` |
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| 200. |
The colour of the transition metal ions is//are due to:A. d-d transition of electrons in presence of ligandsB. charge transfer from ligand to metal ion.C. change in the geometryD. Polarisation of anion by cation |
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Answer» Correct Answer - ABD (A) The colour of the transition metal ion/compound is attributed to d-d transition of electrons e.g. in `[Cu(CN_(3))_(4]^(2+)`. Charge transger spectrum e.g. in `MnO_(4)^(-)`(no. d electron present). The colour change is not because of change in the geometry of the molecules. Yellow colour of the `Agl` is due to the polarisation of `l^(-)` by `ag^(+)`. |
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