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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.
| 1051. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr^(3+) and Cr_(2)O_(7)^(-) `formedB. `Cr_(2)O_(7)^(-) and H_(2)O` are formedC. `CerO_(4)^(2-)` is reduced tp `+3` state of CrD. None of the above |
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Answer» Correct Answer - B `2CrO_(4)^(2-)+2H^(+)rarrCr_(2)O_(7)^(2-)+H_(2)O` |
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| 1052. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr^(3+)`and `Cr_(2)O_(7)^(2-)` are formedB. `Cr_(2)O_(7)^(2-)` and `H_(2)O` are formedC. `CrO_(4)^(2-)` is reduced to `Cr^(3+)`D. `CrO_(4)^(2-)` is oxidised to `Cr_(2)O_(7)^(2-)` |
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Answer» Correct Answer - B The reaction of `K_(2)CrO_(4)` with dilute nitric acid is represented as `2CrO_(4)^(2-)+2H^(+)rarrCr_(2)O_(7)^(2-)+H_(2)O` |
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| 1053. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr^(3+)` and `Cr_(2)O_(7)^(2-)` are formedB. `Cr_(2)O_(7)^(2-)` and `H_(2)O` are formedC. `Cr_(2)O_(7)^(2-)`is reduced to `+3` state pf `Cr`D. `Cr_(2)O_(7)^(2-)` is oxidised to `+7` state of `Cr` |
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Answer» Correct Answer - B `CrO_(4)^(-2)overset(H^(+)(to)Cr_(2)O_(7)^(-2)` |
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| 1054. |
Which has same electronic configuration as expected?A. GdB. EuC. DyD. Er |
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Answer» Correct Answer - A `Gd (64) to [Xe] 4f^7 5d^1 6s^2 ` observed and expected configuration same |
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| 1055. |
f-block elements are calledA. Representative elementsB. Transition elementsC. Inner transition elementsD. Inert elements |
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Answer» Correct Answer - C Inner transition element as last `e^-` enters into (n-2) f orbital, which is perpenultimate shell. |
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| 1056. |
Transition elements having more tendency to form complex than representative elements (s and p-block elements) due to:A. availability of d-orbitals for bondingB. variable oxidation states are not shown by transition elementsC. all electrons are paired in d-orbitalsD. f-orbitals are available for bonding |
| Answer» Correct Answer - A | |
| 1057. |
What are inner-transition elements? Decide which of the following atomic number are the numbers of the inner transition elements: `29,59,74,95,102,104` |
| Answer» The inner transition elements also called f-block elements include the series of lanthanoids (Z=58 to 71) and actinoids (Z=90 to 103). This means that the elements with atomic numbers 59,95 and 105 belong to inner transition elements. | |
| 1058. |
(i). `ZnO` is usedj as a white paint inspite of the fact that it has less convering power than white lead. (ii). Hydrated `ZnCl_2` cannot be dehydrated on heating? (iii). Zn is used in galvanisation of Fe. |
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Answer» (i). `ZnO` is not blackened in atmosphere of `H_2S`. It can be used both as oil and water point. It is non-poisonous in nature. (ii). On heating it reacts with water of crystallisation and forms zinc oxycloride. `2ZnCl_2.2H_2Ooverset(Delta)toZn_2OCl_2+2HCl+3H_2O` (iii). Zinc coating protects the iron from corrosion as Zn is not affected by `O_2` and `H_2O`. |
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| 1059. |
Give reason for the following: (i). Silver ornament gets tarnished when exposed to atmosphere for a long time. (ii) Silver nitrate solution is kept in dark coloured bottles. (iii). Why does `AgNO_3` produce a black stain on the skin. |
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Answer» (i). Ag reacts slowly with trace of `H_2S` in air forming a black layer of `Ag_2S` on surface. `2Ag+H_2Stounderset(Black)(Ag_2S)+H_2` (ii). `AgNO_3` is photosensitive, It decomposes in presence of light. To prevent its decomposition, it is kept in coloured bottles as these do not permit light to pass through. (iii). In the presence of organic matter and light `AgNO_3`, Decomposes to give a black stain of metallic silver. `2AgNO_3to2Agdarr+2NO_2+O_2` |
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| 1060. |
`AgNO_3` gives red ppt. with.A. `NaI`B. `KCl`C. `NaNO_3`D. `Na_2CrO_4` |
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Answer» Correct Answer - d `2AgNO_3+Na_2CrO_4to2NaNO_3+Ag_2CrO_4` (red ppt) |
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| 1061. |
Preparation of looking mirrors involves the use of ammonical `AgNO_3` red lead HCHO. Explain the funstion of each. |
| Answer» The process of deposition of a thin uniform silver layer on clean g | |
| 1062. |
Which of the following statement/s is / are true ?A. Actinides constitute second inner transition seriesB. Actinides are all radioactive in natureC. Elements from atomic number 93 to 103 are transuranic elementsD. All of them |
| Answer» Correct Answer - D | |
| 1063. |
Zr(Z=40) and Hf(Z=72) have similar atomic and ionic radii because of:A. Both belong to same groupB. Diagonal relationshipC. Lanthanoid contractionD. Having similar chemical properties |
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Answer» Correct Answer - C C It is the correct answer |
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| 1064. |
Which of the following statement/s is / are true?A. Lanthanides form oxo-ionsB. Actinides form oxo-ionsC. Both lanthanides and actinides form oxo-ionsD. Neither lanthanides nor actinides form oxo-ions |
| Answer» Correct Answer - B | |
| 1065. |
The properties of Zr and Hf are similar becauseA. both have same atomic radiiB. both belong to d blockC. both belong to same seriesD. both have same number of electrons |
| Answer» Correct Answer - A | |
| 1066. |
The roasting of `HgS` in air producesA. `HgO`B. `HgCl_2`C. `HgSO_4`D. `Hg` |
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Answer» Correct Answer - d `HgS+O_2overset(773-873K)toHg+SO_2` |
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| 1067. |
Which of the following is used as purgative in medicine?A. `ZnCl_2`B. `HgCl_2`C. `Hg_2Cl_2`D. `ZnSO_4.7H_2O` |
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Answer» Correct Answer - c `Hg_2Cl_2` is used as a purgative in medicine. |
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| 1068. |
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. `Hg_(2)O` |
| Answer» Correct Answer - A | |
| 1069. |
Nitriding is a process of hardening steel by treating it inan atmosphere ofA. `NH_(3)`B. `O_(3)`C. `N_(2)`D. `H_(2)S` |
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Answer» Correct Answer - A When steel is treated in the presence of `NH_(3)`, iron nitride on the surface of steel is formed which imparts a hard coating. This process is called nitriding. |
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| 1070. |
The colour of light absorbed by an aqueous solution of `CuSO_(4)` isA. orange-redB. blue-greenC. yellowD. voilet |
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Answer» Correct Answer - A The aqueous solution of `CuSO_(4)` consist of the complex `[Cu(H_(2)O_(4)]^(2+)` ion which absorb in orange-red region and impart deep blue colouration to solution. |
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| 1071. |
Which one of the following does not correctly represent the correct order of the property indicated against itA. `TiltVltCrltMn:`increasing number of oxidation statesB. `Ti^(3+)ltV^(3+)ltCr^(3+)ltMn^(3+):` increasing magnetic momentC. `TiltVltCrltMn:` increasing melting poitnsD. `TiltVltMnltCr:` increasing 2nd ionisation enthalpy |
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Answer» Correct Answer - C In 3d-series melting point increases as we move left to right .But Mn shows low melting point due to its complex formation nature.Hence the order is incorrect in this option. |
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| 1072. |
Which of the following statement (s) is /are correct? (I) In acidic solution dichromate ions are converted to chromate ions . (II) An acidified solution of `K_(2)Cr_(2)O_(7)` liberates iodine from iodides. (III) Potassium dichromate is used as a titrant for `Fe^(2+)` ions. (IV) Ammonium dichromate on heating undergo exothermic decomposition to give `Cr_(2)O_(3)`. Choose the correct option .A. I,II and IIIB. II,III, and IVC. I,II and IVD. II and III |
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Answer» Correct Answer - C Potassium dichromate is not used as a titrant for `Fe^(2+)` ion. Hence I,II and IV are correct. |
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| 1073. |
Explain why lanthanoids are paramagnetic in nature ? |
| Answer» All lanthanoids except`La^(3+)` and `Lu^(3+)` contain unpaired electrons and hence are paramagnetic. | |
| 1074. |
Why is the separation of lanthanide elements difficult ? |
| Answer» Due to lanthanide contraction, the change in the atomic ionic radii of these element is very small. Hence, their chemical properties are similar . This makes their separation difficult. | |
| 1075. |
Ionisation enthalpies of Ce, Pr and Nd are higher than those of Th, Pa and U. why ? |
| Answer» The first three element belong to lanthanoids (4f-obitals are filled ) while the last three elements ar actinoids (5f orbital are filled ). We all know that 5f orbitals have less penetration in the innner core as compared to 4f orbitals. As a result, ionisation enthalpies of lanthanoids are expected to be more than those of actinoids. | |
| 1076. |
Give reason for the folloiwng : (a) `E_("value")^(@)` for `Mn^(3+)//Mn^(2+)` couple is much more positive than that of `Fe^(3+)//Fe^(2+)` couple. (b) Iron has higher enthalpy of atomization then that of copper. (c) `Sc^(3+)` is colourless in aqueous solution whereas `Ti^(3+)` is coloured. |
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Answer» (a) `underset("(Less stable)") underset(3d^(4))(Mn^(3+)) overset(+e^(-))to underset("(More stable)")underset(3d^(5))(Mn^(2+)),underset("(More stable)")underset(3d^(5))(Fe^(3+)) overset(+e^(-))to underset("(Less stable)")underset(3d^(6))(Fe^(2+))` Since `Mn^(3+)//Mn^(2+)` couple is more stable as compared to `Fe^(3+)//Fe^(2+)` couple, the `E_("value")^(@)` for the former is more. (b) In case of Fe (Z=26), number of unpaired electron is more than in case of Cu(Z=29). As a result, the metallic bond in iron is stronger and it has higher enthalpy of atomisation than that of Fe. (c) `Sc^(3+)(3d^(0))` has no unpaired electron while `Ti^(3+)(3d^(1))` has one unpaired electrons. Therefore, `Sc^(3+)` is colourless while `Ti^(3+)` is coloured in aqueous solution. |
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| 1077. |
Which of the folloiwng will not act as oxidising agents ?A. `CrO_(3)`B. `MoO_(3)`C. `WO_(3)`D. `CrO_(4)^(2-)` |
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Answer» Correct Answer - b,c A species can act as oxidizing agent only when metal ispresent in high oxidation state but lower oxidation state is more stable . As higher oxidation states of Mo and W are more stable, they will not act as oxidizing agent. |
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| 1078. |
Which of the following will not act as oxidising agents?A. `CrO_(3)`B. `MoO_(3)`C. `WO_(3)`D. `CrO_(4)^(2-)` |
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Answer» Correct Answer - B::C (b,c) The metal atoms in both `CrO_(3)` and `WO_(3)` cannot increase their oxidation state beyond +6. |
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| 1079. |
The lowest oxidation state of manganese is basic while the highest is acidic. Explain. |
| Answer» The lowest oxidation state of manganese is +2 `overset(II)((Mn))`. It has tendency to lose electrons in order to increase its oxidation state. Therefore, Mn is basic. In highest oxidation state of `+7 overset(VII)((Mn))`, there is no scope for the lose of any more electrons. It can rather accept the same. This shows that `Mn^(VII)` can act as acid by accepting electrons. | |
| 1080. |
Explain gives reason. (a) Transition metal and many of their compounds show paramagnetic behavior. (b) The enthalpies of atomisation of the transition metal are high . (c) The transition metals generally from coloured compounds. |
| Answer» (b) Enthalpy of atomisation may be defined as the amount of heat energy needed to break the metal lattice of a crystalline metal into free atoms. Greater the magnitude of lattice energy, more will be the value of enthalpy of atomisation. The transition metals have high enthalpies of atomisation because teh metallic bonds present are quite strong due to presence of large number of half-filled atomic orbitals. For more detail, consult text-part. | |
| 1081. |
Account for the following `:` (a) Europlum (II) is more stable than cerium (II). (b) Transition metal have high enthalpies of atomisation . (c ) Actinoid ions are generally coloured. |
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Answer» (a) Europlum has stable electronic configuration , i.e.,`[Xe] 4f^(7) 5d^(0)6s^(0)`. (b) Due to large number of unpaired electrons in their atom, there is larger and stronger metallic bonding. (c ) Unpaired electron are present in their ions whcih undergo f-f transition. |
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| 1082. |
Explain giving reasons : The transition metals generally form coloured compounds. |
| Answer» Formation of coloured compounds by transition metals is due to partial adsorption of visible light. The electron absorbs the radiation of a particular frequency (of visible region) and Jumps into next orbital | |
| 1083. |
Explain giving reasons : Transition metals and many of their compounds show paramagnetic behaviour. |
| Answer» Transition elements have unpaired electrons. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. This is the reason of paramagnetism in transition metals. | |
| 1084. |
Explain giving reasons : The enthalpies of atomisation of the transition metals are high. |
| Answer» The reason for the high enthalpy of atomisation is the presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them. | |
| 1085. |
Why do transition metals exhibit higher enthalpies of atomisation ? |
| Answer» Enthalpy of atomisation is the amount of heat required to break the metal lattice of get free atoms. As transition metals contain a large number of unpaired electrons, they have strong interatomic attractions ( metallic bonds). Hence, they have high enthalpies of atomisation. | |
| 1086. |
Assertion: The metals of 4d and 5d greater entalpies of atomisation than the corresponding elements of the 3d series. Reason: The metal-metal bond in 4d and 5d series are stronger than those in the 3d series.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 | |
| 1087. |
First I.E. 5d elements are higher than those of 3d and 4d elements. This is due to 1) greater effective nuclear charge action on outer valence electrons. 2) greater effective nuclear charge is experience because of weak shielding effect of 4-f orbitalsA. only 1B. only 2C. 1 and 2D. none is correct |
| Answer» Correct Answer - C | |
| 1088. |
Statements-1. The metals of 4dand 5d transition series have greater enthalpies of atomisation than the corresponding elementsof the 3d series. Statements-2. The metal-metal bond in 4d and 5d series are stronger than those in the 3d series.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement -1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation of Statement-1C. Statement-1 is True, Statement-2 is falseD. Statement-1 is False, Statement-2is True |
| Answer» Statement-2 is the correct explanation of Statement-1 | |
| 1089. |
The first ionisation energies of the elements of the first transition series `(TitoCu)`A. Increases as the atomic number increasesB. Decreases as the atomic number increasesC. Do not show any change as the addition of electrons takes place in the inner `(n-i)` d-orbitals.D. Increases from Ti to Mn and then decreases from Mn to Cu. |
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Answer» Correct Answer - a `IE_1` of 3d-series increases with increase in atomic number due (i) increase in cuclear charge and (ii) decrease in atomic size. |
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| 1090. |
How would you account for the increasing oxidising power in the series `VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)` ? |
| Answer» This is due to the increasing stability of the lower species to which they are reduced, | |
| 1091. |
Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^(4)` configuration ? |
| Answer» `Cr^(2+)` is reducing as its configuration changes from `d^(4)" to "d^(3)` , the latter having a half-filled `t_(2g)` level. On the other hand, the change from `Mn^(2+)" to "Mn^(3+)` results in the half-filled `(d^(5))` configuration which has extra stability. | |
| 1092. |
Which of the following statement are correct about `Cr^(2+)` (Z = 24) and `Mn^(3+)` (Z = 25) ? (i) `Cr^(2+)` is a reducing agent (ii) `Mn^(3+)` is an oxidizing agent (iii) Both `Cr^(2+)` and `Mn^(3+)` exhibit `d^(4)` configuration (iv) When `Cr^(2+)` is used as a reducing agent, the chromium ion attains `d^(5)` electronic configurationA. `Cr^(2+)` is a reducing agentB. `Mn^(3+)` is an oxidizing agentC. Both `Cr^(2+)` and `Mn^(3+)` exhibit `d^(4)` electronic configurationD. When `Cr^(2+)` is used as a reducing agent, the chromium ion attains `d^(5)` electronic configuration |
| Answer» Correct Answer - A::B::C | |
| 1093. |
Thionyl chloride can be synthesised by chlorinating `SO_2` using `PCl_5`. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from its hexahydrated salt by treating with 2,2-dimethoxypropane. Discuss all this using balanced chemical equations. |
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Answer» Thionyl chloride is `SOCl_2` `SO_2+PCl_5toSOCl_2+POCl_3` `FeCl_3.6H_2O+6SOCl_2toFeCl_3+12KCl+6SO_2` `FeCl_3.6H_2O+6CH_3C(OCH_3)_2CH_3toFeCl_3+13CH_3OH+6CH_3COCH_3` |
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| 1094. |
The element of 3d transition series are given as : Sc Ti V Cr Mn Fe Co Ni Cu Zn Answer the following : (i) Which element has the highest melting point and why ? (ii) Which element is a strong oxidizing agent in +5 oxidation state and why ? (ii) Which element is soft and why ? |
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Answer» (i) The element Cr has the highest melting point because the metallic bonding is strongest due to the presence of five unpaired electron (`d^(5)` configuration ) (ii) The element Mn acts as a strong oxidising agent because it has `3d^(5)` configuration ( very stable) in +2 oxidatin state. (iii) The element Zn is soft since the metallic bond is very weak due to teh absence of any unpaired electrons. |
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| 1095. |
The magnetic moments of few transition metal ions are given below : `{:("Metal ion":,Sc^(3+),Cr^(2+),Ni^(2+),Ti^(3+)),("Megnetic moment (BM):",0.00,4.90,2.84,1.73):}` (at no. Sc=21, Ti =22, Cr=24,Ni =28) Which of the given metals ions: (i) has the maximum number of unpaired electrons ? (ii) forms colourless aqueous solution ?(iii) exhibits most stable +3 oxidation state ? |
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Answer» (i) `Cr^(2+)` ion (`3d^(4)` configuration) has maximum number of unpaired electrons. (ii) `Sc^(3+)` ion ( Ar-configuration) forms colourless solution. (iii) `Sc^(3+)` ion (Ar-configuration) exhibits most stable +3 oxidation state. . |
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| 1096. |
A comapound in which a metal ion `M^(x+)(Z=25)` has a spin only magnetic moment of `sqrt24 `BM The number of unpaired electrons in the compound and the oxidation state of the metal ion respectively areA. 4 and 2B. 5 and 3C. 3and 2D. 4 and 3 |
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Answer» Correct Answer - A Spin only magnetic moment , `mu=sqrt(n(n+2))=sqrt24BM`or `n^(2)+2n-24=0` `implies(n+6)(n-4)=0` `therefore " "n=4" " [becausen=-6" not possible"]` Here, n is the number of unpaired electrons. Since four unpaired electrons are present, the oxidation state must be `+2`. `therefore Z+(25)=1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),3d^(4)4s^(2)` |
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| 1097. |
If n is the number of unpaired electrons, the magnetic moment (in BM) of transition metal/ion is given byA. `sqrt(n(n+2))`B. `sqrt(2n(n+1))`C. `sqrt(n(n-2))`D. `sqrt(n2(n-1))` |
| Answer» Correct Answer - A | |
| 1098. |
There are three unpaired electrons in `[Co(H_(2)O)_(6) ]^(2+)` and calculated value of magnetic moment is 3 · 87 BM which is quite different from the experimental value of 4·40 BM. This is because of :A. increase in number of unpaired electronsB. some contribution of the orbital motion of the electron to the magnetic momentC. Change in orbital spin of the electronD. d-d transition. |
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Answer» Correct Answer - B (b) It is the correct answer. |
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| 1099. |
Although `Cr^(3+)` and `Co^(2+)` ions have same number of unpaired electrons but the magnetic moment of `Cr^(3+)` is `3.87 B.M`. and that of `Co^(2+)` `4.87` B.M. because…. |
| Answer» Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in `Cr^(3+)` ion. However, appreciable orbital contribution takes place in `Co^(2+)` ion. | |
| 1100. |
Although zirconium belongs to 4d transition series and hafnium to 5d transition series even then they show similar physical and chemical properties because …….. .A. belong to d-blockB. have same number of electronsC. have similiar atomic radiusD. belongs to the same group of the periodic table |
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Answer» Correct Answer - C Due to lanthanoid contraction Zr and Hf have nearly equal size |
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