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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.
| 1101. |
The alloy brass is made up ofA. Cu and ZnB. Cu and NiC. Cu and SnD. Cu, Cd |
| Answer» Correct Answer - A | |
| 1102. |
The chemistry of the actinoid elements is not so smooth as that of the lanthanoid. Justify this statement by giving some example from the oxidation state of these elementsA. 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. A ctinoids exhibit wide range of oxidation states. |
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| 1103. |
The chemistry of the actinoid elements is not so smooth as that of the lanthanoid. Justify this statement by giving some example from the oxidation state of these elements |
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Answer» The lanthanoids mostly exhibit +3 oxidation states in their compounds. Actinoid elements also normally show +3 oxidation states. But in their cases, 5f,6d and 7s energy levels are comparable and have very small energy difference in them. As a result, they can exhibit a number of oxidation states. For example., Neptunium (Np): +3,+4,+5,+6,+7 Plutonium (Pu) : +3,+4,+5,+6,+7 Americium (Am): +3,+4,+5,+6. |
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| 1104. |
Explain why `E^(@)` for `Mn^(3+)// Mn^(@+)` couple is more positive than that for `Fe^(3+) //Fe^(2+) ` ( At. Nos. `Mn= 25, Fe= 26)` ? Or Why is `+2` oxidation state of manganese quite stable while the same is not true for iron ? `[Mn= 25 , Fe = 26]` |
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Answer» `._(25)Mn^(2+) = [Ar] 3d^(5), . _(25) Mn^(3+) =[Ar] 3d^(4), ._(26) Fe^(2+) = [ Ar] 3d^(6) , . _(26) Fe^(3+)= [ Ar] 3d^(5)` Thus, `Mn^(2+)` has more stable configuration than `Mn^(3+)` while `Fe^(3+)` has more stable then `Fe^92+)`. Consequently, large third ionisation enthalpy is required to change `Mn^(2+)` to `Mn^(3+)`. As `E^(@)` is the sum of enthalpy of atomisation, ionization enthalpy and hydration enthalpy, therefore, `E^(@)` for `Mn^(3+)//Mn^(2+)` couple is more positive than `Fe^(3+) //Fe^(2+)` . Note `,` The large positive `E^(@)` for `Mn^(3+)//Mn^(2+)` means that `Mn^(3+)` can be easily reduced to `Mn^(2+)` , i.e. `Mn^(3+)` is less stable . `E^(@)` value for `Fe^(3+) //Fe^2+)` is positive but small, i.e., `Fe^(3+)` can also be reduced to `Fe^(2+)` but less easily. Thus,`Fe^(3+)` is more stable than `Mn^(3+)` . It also explains why `+3` state of Mn is of little importance. |
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| 1105. |
The `E^(@)` values in respect of the electrodes of chromium `(Z= =24),`manganese `( Z = 25)` and iron `( Z= 26 ) `are `: Cr^(3+) // Cr^(2+) = - 0.4 V, Mn^(3+)//Mn^(2+) = + 1.5 V, Fe^(3+) //Fe^(2+) = + 0.8 V`. On the basis of the above information compare the feasiblities of further oxidation of their `+2` oxidation states. |
| Answer» `-ve` value of `E^(@)` for `Cr^(3+) //Cr^(2+)` shows that `Cr^(2+)` is least stable. Greater `+ve` value for `Mn^(3+)// Mn^(2+)` than that for `Fe^(3+) //Fe^(2+)` shows that `Mn^(2+)` is more stable than `Fe^(2+)` . Hence, stability of `+2` oxidation state is in the order `: Cr^(2+) lt Fe^(2+) lt Mn^(2+)` or the oxidation of their `+2` state to `+3` state is in the order `: Cr^(2+) gt Fe^(2+) gt Mn^(2+)` | |
| 1106. |
Statement-1. Lanthanoids show a limited number of oxidation states whereas actinoids show a large number of oxidation states. Statement-2. Energy gap between4f, 5d and 6s subshells is small whereas that between 5f, 6d and 7s subshells is large.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 False because energy gap beween 4f, 5d and 6s is large whereas that between 5f, 6dand 7s is small. | |
| 1107. |
The colour of `FeF_3` isA. BrownB. Red BrownC. Light greenD. White |
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Answer» Correct Answer - d Because `Fe^(3+)` has `d^5` configuration and `F^(ɵ)` is a weak ligand so d-d transition does not occur. |
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| 1108. |
Statement-1. `Ce^(3+)` has the tendency to change to `Ce^(4+)` Statement-2. `Ce^(3+)` has the tendency to change to `Ce^(4+)`.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» Correct Statement-2. `Ce^(4+)` has the tendency to change to `Ce^(3+)` ( as `+3` oxidation state is more stable). | |
| 1109. |
A mixed oxide of iron and chromium `FeO.Cr_(2)O_(3)` is fused with sodium carbonate in the presence of air to form a yellow coloured compound (A). On acidification , the compound (A) forms an orange coloured compound (B) which is a strong oxidising agent. Identify (i) the compound A and B (ii) Write balanced chemical equation for each step. |
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Answer» The mixed oxide `FeO.Cr_(2)O_(3)` is the chromatic ore of iron also called ferro-chrome. It is written as `FeCr_(2)O_(4)`. Upon fusing with sodium carbonate in the presence of air, it give as yellow compound (A) which is sodium chromate. . `2FeCr_(2)O_(4)+8Na_(2)CO_(3)+underset((Air))(7O_(2)) overset("Fuse") to underset("(Yellow)")underset("Sod. chromate (A)")(8Na_(2)CrO_(4))+2Fe_(2)O_(3)+8CO_(2)` Sodium chromate (A) upon acidification with concentrated sulphuric acid changes to sodium dichromate (B) which has an orange colour. `underset("Sod.chromate")(2Na_(2)CrO_(4))+H_(2)SO_(4) to underset("(orange)")underset("Sod. dichromate (B)")(Na_(2)Cr_(2)O_(7))+Na_(2)SO_(4)+H_(2)O` |
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| 1110. |
Which among `FeO` and `Fe_2O_3` is more basic?A. `FeO`B. `Fe_2O_3`C. both have same basic lengthD. None of them is basic |
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Answer» Correct Answer - a Iron is extracted from haematite `(Fe_2O_3)`. Al is extracted from bauxite `(Al_2O_32H_2O)`. |
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| 1111. |
Statement-1. MnO is basic whereas `Mn_(2)O_(7)`is acidic. Statement -2. Higher the oxidation state of a transition metal in itsoxide, greater is the acidic character.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. | |
| 1112. |
`Fe_2O_4` is a mixed oxide of `FeO` and `F_2O_3`. |
| Answer» Correct Answer - True. | |
| 1113. |
`Mn_2O_7` is a basic oxide. |
| Answer» Correct Answer - False. Acidic | |
| 1114. |
How many of the transition elements are called Platinum metals. |
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Answer» Second and third triad of 4d and 5d series of group VIII or group, 8,9 and 10 are collectively called platinum metals, these are: 4d series: Ru Rh Pd `5d` series: Os Ir Pt |
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| 1115. |
How many of the transition elements are called coinage metals. |
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Answer» First triad of 3d series of group VIII or group 8,9 and 10 are called coinage metals, these are: `3d` series: `Fe,Co,Ni` |
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| 1116. |
Which of the following group of transition metals is called coinage metals?A. Ag, Cu, NiB. Zn, Ag, AuC. Ag, Fe, CuD. Cu, Ag, Au |
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Answer» Correct Answer - A Coinage alloy contains Ag,Cu,Ni |
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| 1117. |
Solder is an alloy of :A. Cu, Ag, ZnB. Cd, Ag, ZnC. Ni, Ag, ZnD. Cd, Zn, Ni |
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Answer» Correct Answer - A Silver solder contain Cu,Ag,Zn |
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| 1118. |
Which of the following group of transition metals is called coinage metals?A. Fe, Co, NiB. Pt, Au, AgC. Cu, Ag, AuD. Pt, Ir, Pd |
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Answer» Correct Answer - C Cu,Ag, Au are coinage metals as all of them are being used for making coins. |
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| 1119. |
Coinage metals show the properties of:A. Typical elementsB. Normal elementsC. Transition elementsD. Coin elements |
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Answer» Correct Answer - C Coinage metals have property of transition elements as they are made from transition metals |
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| 1120. |
Why `KMnO_(4)` is used in cleaning surgical instruments in hospitals? |
| Answer» This is because `KMnO_(4)` has agermicidal action. | |
| 1121. |
The first elements of first, second and third transition series respectively areA. Zn, Cd, HgB. Sc, Y, LaC. Cu, Ag, AuD. Sc, Y, La |
| Answer» Correct Answer - B | |
| 1122. |
Which of the following represents the incoorect order of the properties indicated?A. `Ni^(2+)gtCr^(2+)gtFe^(2+)gtMn^(2+)` (size)B. `ScgttIgtCrgtMn` (size)C. `Ni^(2+)ltCo^(2+)ltFe^(2+)ltMn^(2+)` (unpaired electron)D. `H_3AsO_4gtH_3PO_4` (acidic strength order) |
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Answer» Correct Answer - a,d (a). Same order for neutral atom and ion having same charge. (a) and (d) are incorrect. `overset(ScgtTigtVgtCrgtMn)to` Therefore `Cr^(2+)gtMn^(2+)gtFe^(2+)gtCo^(2+)gtNi^(2+)` (d). `H_2AsO_4ltH_2PO_4` as P is more electronegative than As. |
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| 1123. |
The first elements of first, second and third transition series respectively areA. Zn, Cd, HgB. Sc, Y, LaC. Cu, Ag, AuD. Cr, Mo, W |
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Answer» Correct Answer - B Sc-3d, Y-4d , La-5d series |
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| 1124. |
Which among the following is not an element of the first transition series?A. ZnB. VC. TiD. Ag |
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Answer» Correct Answer - D Ag belongs to second transition series (4d) |
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| 1125. |
What will be the correct representation of quatum numbers f the last electron entered into Ce?A. `n=1,l=3,m=-3,s=(-1)/(2)`B. `n=4,l=3,m=0,s=(-1)/(2)`C. `n=4,l=2,m=-3,s(+1)/(2)`D. `n=4,l=3,m=+2,s=(-1)/(2)` |
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Answer» Correct Answer - a,b,d `Ce(Z=58)` f-block element: Last electron enters in f-orbital and highest value of `n=6` but the last electron will enter in 4f-orbitals filled would be `[Xe]6s^24f^15d^1` |
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| 1126. |
The first transition element is__________.A. AcB. TiC. ScD. Pt |
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Answer» Correct Answer - C Sc (21) `to 4s^2 3d^1 to` belonging to 3d series |
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| 1127. |
Which of the following electronic structures refers to transition elements?A. `2,8,18,5`B. `2,8,14,2`C. `2,8,18,32,18,8,1`D. `2,8,5` |
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Answer» Correct Answer - b Transition elements have unpaired electrons in `(n-1)` d-orbitals. |
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| 1128. |
The first transition element is__________.A. CrB. ScC. ZnD. Cu |
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Answer» Correct Answer - B Sc (21) -`4s^2 3d^1 ` - first transition element |
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| 1129. |
Which one of the following refers to configuration of transition elements ?A. 2,8,18,3B. 2,8,18,8C. 2,8,14,2D. 2,8,6 |
| Answer» Correct Answer - C | |
| 1130. |
Which of the following electronic configuration is that of a transitional element?A. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6`B. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^1`C. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2`D. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9` |
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Answer» Correct Answer - D `4s^2 3d^9` - last `e^-` enters d orbital `therefore` d-block |
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| 1131. |
The outer electronic configuration of chromium isA. `4s^2 3d^5`B. `4s^1 3d^5`C. `4s^2 3d^4`D. `4s^2 3d^6` |
| Answer» Correct Answer - B | |
| 1132. |
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` |
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Answer» Correct Answer - D Correct ground state configuration of Cr is `[Ar] 3d^5 4s^1` Anamolous configuration |
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| 1133. |
The second and third rows of transition elements resemble each other much more than they resemble the first row element. Explain why ? |
| Answer» This is because of lanthanoid contraction experienced by the elements presen in third row. But the same is not noticed in the element present in second row. | |
| 1134. |
Resemblance between Nb and Ta is because theyA. belong to same group of perioidc tableB. have the same mineral sourceC. have almost same ionic and covalent radiiD. are transition metals |
| Answer» Correct Answer - C | |
| 1135. |
Ziegler-Natta catalyst is:A. `Pt//PtO`B. `TiCl_(4)//Al(C_(2)H_(5))_(3)`C. `Pt//Rh`D. Pt |
| Answer» Correct Answer - B | |
| 1136. |
Zeigler-Natta catalyst is usedA. in the polymerization of ethene to produce polyetheneB. for oxidizing alcohols to aldehydesC. in the polymerization of alkyne to give benzeneD. in the Ostwald process for converting `NH_(3)` into NO |
| Answer» Correct Answer - A | |
| 1137. |
The atomic radii of the metals of the third (5d) series of transitions elements are virtually the same as those of the corresponding members of the second (4d) series. Explain. |
| Answer» In between-second transition series (4d series ) of transition series (5 series ), fourteen elements (La with Z=57 to Hf with Z=72) are present. Since these elements are involved in the lanthanoids contraction,this effect is obviously experienced by the element of 5d series also. As a result, the difference in the atoic sizes of the element belonging to the two series is negligible. These sizes are virtually the same. | |
| 1138. |
Metallic radii of some transitions element are given below. Which of these elements will have highest density ? `{:("Element",Fe,Co,Ni,Cu),("Metallic",126,125,125,128),("radii/pm",,,,):}`A. FeB. NiC. CoD. Cu |
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Answer» Correct Answer - D On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal. Hence, among the given four choices Cu belongs to right side of Periodic Table in transition metal, and it has highest density `(89 g//cm^(3))`. |
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| 1139. |
Gold dissolves in aqua regia formingA. `Au(NO_3)_2`B. `H[AuCl_4]`C. `AuCl`D. `AuNO_3` |
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Answer» Correct Answer - b `3HCl+4HNO_3toNOCl+2H_2O+2[Cl]` `Au+3[Cl]toAuCl_3` `AuCl_3+3"HCl" toH[AuCl_4]` (complex) |
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| 1140. |
Why first ionisation enthalpy of Cr is lower than that of Zn? |
| Answer» Ionisation enthalpy of Cr is less than that of Zn because Cr has stable `d^(5)` configuration. In case of zinc, electron comes out from completely filled 4s-orbital. So, removal of electron from zinc requires more energy as compared to the chromium. | |
| 1141. |
Why does copper not replace hydrogen from acids? |
| Answer» Copper not replace hydrogen from acids because Cu has positive `E^(@)` value, i.e., less reactive than hydrogen which has electrode potential 0.00V. | |
| 1142. |
Why does copper not replace hydrogen from acids ? |
| Answer» Copper has positive electrode potential `(E^(@))` and lies below hydrogen in electrochemical series. | |
| 1143. |
Why `E^(-)` values for Mn, Ni and Zn are more negative than expected? |
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Answer» Negative values of `Mn^(2+)` and `Zn^(2+)` are related to stabilities of half-filled and completely filled configuration respectively. But for `Ni^(2+)`, `E^(@)` value is related to the highest negative enthalpy of hydration Hence, `E^(Theta)` values for Mn, Ni and Zn are more negative than expected. |
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| 1144. |
Why `E^(@)` values for Mn, Ni and Zn are more negative than expected ? |
| Answer» The values of `E^(@)` for Mn and Zn are more negative than expected because `Mn^(2+)` and `Zn^(2+)` have stable exactly half- filled and fully filled configuration respectively. The high negative value of `E^(@)` for Ni is due to its highest negative enthalpy of hydration | |
| 1145. |
Why `E^(@)` value of Mn, Ni and Zn are more negative than expected ? |
| Answer» The statement alos implies that why these transition metal are stronger reducing agents than expected. This is related to the stabilities of the divalent cations which are formed by the loss of electrons. The extr- stabilities of `Mn^(2+)` and `Zn^(2+)` ions is due to the presence of five half filled `(3d^(5))` and fully filled orbitals `(3d^(10))`. However, the extra stability of `Ni^(2+)` ion is because of its high negative enthalpy of hydration because of the small ionic size. | |
| 1146. |
Which of the following statement regarding lanthandes is false?A. All lanthanoide ar solid at room temperature.B. Their usual oxidation state is `+3`.C. They can be seprated from one anther by ion-exchange method.D. Inoic radii of trivalent lanthanoids steadily increases with increases in atomic number. |
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Answer» Correct Answer - D It is the correct answer. |
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| 1147. |
The bivalent metal ion having maximum paramagnetic behaviour among the first transition series elements isA. `Mn^(2+)`B. `Cu^(2+)`C. `Sc^(2+)`D. `Cu^(+)` |
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Answer» Correct Answer - A (a) `Mn^(2+)(3d^(9)4s^(@)) to` 5 unpaired electrons. (b) `Cu^(2+)(3d^(9)4s^(@)) to ` 1 unpaired electrons. (c) `Sc^(2+)(3d^(1)4s^(@)) to ` 1 unpaired electron. (d) `Cu^(+)(3d^(10)4s^(@)) to ` 0 unpaired electron. `Mn^(2+)` ion is maximum paramagnetic in nature. |
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| 1148. |
Write down the electronic configuration of: (i). `Cr^(3+)` (ii). `Prn^(3+)` (iii). `Cu^(o+)` (iv). `Ce^(4+)` (v). `Co^(2+)` (iv). `Lu^(2+)` (vii). `Mn^(2+)` (viii). `Th^(4+)` |
| Answer» `{:((a) Cr^(3+),:,[Ar]3d^(3),(b)Cu^(+),:,[Ar]3d^(10)),((c) Co^(2+),:, [Ar]3d^(7),(d) Mn^(2+),:,[Ar]3d^(5)),((e)Pm^(3+),:,[Xe]"4f"^(4),(f)Ce^(4+),:,[Xe]),((g)Lu^(2+),:,[Xe]4f^(14)5d^(1),(h)Th^(4+),:,[Rn]):}` | |
| 1149. |
(a). Why `Mn^(2+)` compounds are more stable than `Fe^(2+)` towards oxidation to their `+3` state? (b). Calculate the magnetic moment of `V^(3+)` ion. (c). `[Ti(H_2O)_6_^(3+)` gives violet coloured aqueous solution but `[Mg(H_2O)_6]^(2+)` solution is colourless. (d). `[Ti(H_2O)_6]^(3+)` is coloured whereas `[Sc(H_2O)_6]^(3+)` is colourless. Why? |
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Answer» (a). Electronic configuration of `Mn^(2+)` is `3d^5` which is half filled and hence stable. Therefore `IF_3` is very high i.e., 3rd electron cannot be lost easily. In case of `Fe^(2+)`, electronic configuration is `3D^6`. (b). `V(Z=23)implies3d^34s^2,V^(3+)=3d^2(n=2)` `mu=sqrt(n(n+2))BM=sqrt(2(2+2))=sqrt(8)=2.73BM`. (c). In `[Ti(H_2O)_6]^(3+)`, the `d^1` electron occupies `t_(2g)` orbital in octahedral field. On irradiation with light, the `t_(2g)` electron is promoted to `e_g` orbital and the resulting absorption band goves violet colour. In case of `[Mg(H_2O)_6]^(2+)` the electronic configuration of `Mg^(2+)` is `1s^2`, `2p^6` which does not permit any electronic transition `(2pto3s)` as the energy gap is very large or there does not occur d-d transition of electron. Hence, it gives colourless solution. (d). A in `[Ti(H_2O)_6]^(3+)`, the titanium ion is present in `Ti^(3+` form with `3d^1`, `4s^0` configuration. The single upaired electron of `3d^1` orbital makes the compound to show colour. In `[Sc(H_2O)_6]^(3+)` scandium is present as `Sc^(3+)` state with `3d^0`,`4s^0` configuration. Since no electron is present in the 3d and 4 s orbitals, it remains colourless. |
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| 1150. |
Why are `Mn^(2+)` compounds more stable than `Fe^(2+)` toward oxidation to their `+3` state? |
| Answer» electronic configuration of `Mn^(2+)` is `3d^(5)` while that of `Fe^(2+)` is `3d^(6)`. This shows that `Fe^(2+)` ion has an urge to change to `Fe^(3+)` ion by losing an electron whereas `Mn^(2+)` ion has no such tendency. Thus, +2 oxidation state of Mn is more stable as compared to +2 oxidation state of Fe. | |