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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.
| 101. |
Lesser number of oxidation states at the extreme ends of the transition series, stems from:A. too few electrons to lose or shareB. too many d electrons for higher valenceC. greatest number of oxidation states for manganese +2 to +7D. presence of filled f-orbitals. |
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Answer» Correct Answer - A::B (a), `Sc,Ti` (b). Hence fewer orbitals available in which to share electrons with other (Cu,Zn). |
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| 102. |
Select the correct match for oxocation in acidic medium with respect to oxidation state.A. `VO_(2)^(+)toV^(V)`B. `VO^(2+)toV^(IV)`C. `TiO^(2+)toTi^(IV)`D. `FeO_(4)^(2-)toFe^(VI)`. |
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Answer» Correct Answer - A::B::C `FeO_(4)^(2-)` is an oxoanion formed in alkaline medium and readily decomposed to `Fe_(2)O_(3) and O_2`. |
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| 103. |
Conversion of manganous salt to `MnO_(2)` using `KMnO_(4)` in netural medium is catalysed by:A. `ZnSO_(4)`B. `ZnO`C. `Fe_(2)(SO_(4))_(3)`D. `FeO` |
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Answer» Correct Answer - A::B `2MnO_(4)^(2-)+3Mn^(2+)+2H_(2)Oto5MnO_(2)+4H^(+)`. |
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| 104. |
Cerium `(Z= 58)` is an important nember of the lanthanoids . Which of the following statements about cerium is incorrect ?A. The common oxidation state of cerium are `+3` and `+4`B. The `+3` oxidation state of cerium is more stable than `+4` oxidation state.C. The `+4` oxidation state of cerium is not known in solution.D. Cerium (Iv) acts as an oxidising agent. |
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Answer» Correct Answer - 3 (3) Cerium can also show the oxidation state of `+4` in solution as it leads to a noble gas configuration, from `[Xe]^(54) 4f^(1) 5d^(1) 6s^(2)` to `[Xe]^(54)`, after losing four electrons. It is only `Ce^(4+)` which exits in solution among the lanthanides. |
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| 105. |
The most common oxidation states of cerium areA. `+3,+4`B. `+2,+3`C. `+2,+4`D. `+3,+5` |
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Answer» Correct Answer - 1 Cerium`Ce_(58)[Xe]4f^(1)5d^(1)6s^(2)` its most stable oxidation state is `+3` but `+4` is also existing. |
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| 106. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr_(2)O_(7)^(2-)` and `H_(2)O` are formedB. `CrO_(4)^(2-)` is reduced to `+3` state of `Cr`C. `CrO_(4)^(2-) ` is oxidised to `0` state of `Cr`D. `Cr^(3+)` and `Cr_(2)O_(7)^(2-)` are formed |
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Answer» Correct Answer - 1 `2CrO_(4)^(2-)+2H^(+) to Cr_(2)O_(7)^(2-)+H_(2)O`. |
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| 107. |
Iron becomes passive by______due to formation of ______.A. dil. HCl, `Fe_(2)O_(3)`B. 80% conc. `HNO_(3),Fe_(3)O_(4)`C. conc. `H_(2)SO_(4),Fe_(3)O_(4)`D. conc. `HCl,Fe_(3)O_(4)` |
| Answer» Correct Answer - B | |
| 108. |
(a) Complete the following chemical equation: (i) `Cr_(2)O_(7)^(2-)(aq)+H_(2)S(g)+H^(+)(aq) to` , (ii) `Cu^(2+)(aq)+I^(-)(aq) to` (b) How would you account for the following: (i)The oxidising power of oxoanion are in the order`VO_(2)^(+)ltCr_(2)O_(7)^(2-)ltMnO_(4)^(-)`. (ii) The third ionization enthalpy of manganese `(Z=25)` is exceptionally high. (iii) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`. Or (a) Complete the following chemical equation: (i) `MnO_(4)^(-)(aq)+s_(2)O_(3)^(2-)(aq)+H_(2)O(I) to` (ii) `Cr_(2)O_(7)^(2-)(aq)+Fe^(2+)(aq)+H^(+)(aq) to` (b) Explain the following observation: (i) `La^(3+)(Z=57)` and `Lu^(3+)(Z=71)` do not show any colour in solutions. (ii) Among the divalent cation in the first series of transition elements, manganese exhibits the maximum paramagnetism. (iii) `Cu^(+)` ion is not known in aqueous solutions. |
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Answer» Correct Answer - (a) (i) `Cr_(2)O_(7)^(2-)(aq)+3H_(2)(S)(g)+8H^(+)(aq) to 2Cr^(3+)+3S+7H_(2)O` (ii) `2Cu^(2+)(aq)+2I^(-)(aq) to 2Cu^(+)(aq)+I_(2)(S)` (b) (i) It is because `V` in lower oxidation state is less stable than `Cr` which is less stable than `Mn`. That is why `MrO_(4)^(-)` is best oxidising agent and `VO_(2)^(+)` is least. (ii) ` Mn(25)` has electronic configuration `[Ar]4S^(2) 3d^(5)`,electronic configuration of `Mn^(2+) ` is `[Ar] 4s^(@) 3d^(5)`. After losing `2` electrons, it has half filled d-orbital, which is more stable that is why `Mn^(2+)` has exceptionally high third ionization energy .i.e., the energy required to remove third electrons is very high. (iii) it is because in `Cr^(3+), d^(3)` is more stable in aqueous solution than `Fe^(3+)(d^(5))` i.e., `Cr^(3+)` is more stable than `Fe`. `OR` (a) (i) `8MnO_(4)^(-)(aq)+3S_(2)O_(3)^(2-)(aq)+H_(2)O(I) to 8MnO_(2)(S)+2OH^(-)(aq)+6SO_(4)^(2-)(aq)` (ii `Cr_(2)O_(7)^(2-)(aq)+6Fe^(2+)(aq)14H^(+)(aq) to 2Cr^(3+)(aq)+6Fe^(3+)(aq)+7H_(2)O(I)` (b) (i) it is due to presence of unpaired electrons they cannot undergo f-f transition . (ii) it is because in `Cr^(3+)` gets oxidised to `Cu^(2+)` which is more stable and its hydration energy can overcome second ionsation energy. |
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| 109. |
Statement I : `Zn^(2+)` is diamagnetic Statement II : The electrons are lost from 4s orbital to from `Zn^(2+)`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 - B | |
| 110. |
The set of compounds which does not exist, isA. `PbI_(4),BiCl_(5),PH_(5)`B. `XeF_(2),XeF_(3)^(-),XeF_(4)`C. `FeI_(3),CuI_(2),CuCl_(2)`D. `Hg(OH)_(2),Ag_(2)O,V_(2)SO_(4)` |
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Answer» Correct Answer - A (a). `PbI_(4)toPbI_(2)+I_(2)` Due to inert pair effect, `Pb^(4+)` converts into `Pb^(2+)` ion, hence `PbI_(4)` does not exist, (b). `BiCl_(5)toBiCl_(3)+Cl_(2)` due to inert pair effect, `Br^(5+)` ion converts into `Bi^(3+)` hence `BiCl_(5)` does not exist. (c). `PH_(5)to` does not exist, because of absence of d-orbital contraction |
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| 111. |
`underset(("Acidified"))(KMnO_(4))+HCltoH_(2)O+X(g),X` is a:A. red liquidB. violet gasC. greenish-yellow gasD. yellow-brown gas |
| Answer» Correct Answer - C | |
| 112. |
Acidified chromic acid `+H_(2)O_(2)rarr underset(("blue colour"))(X+Y)`, X and Y areA. `CrO_(5) and H_(2)O`B. `Cr_(2)O_(3) and H_(2)O`C. `CrO_(2) and H_(2)O`D. `CrO` and `H_(2)O` |
| Answer» Correct Answer - A | |
| 113. |
Which of the following compounds is used as the startingn material for the preparation of potassium dichromate?A. `K_(2)SO_(4).Cr(SO_(4))_(3).24H_(2)O` (chrome alum)B. `PbCrO_(4)` (Chrome yellow)C. `FeCr_(2)O_(4)`(chromite)D. `PbCrO_(4).PbO`(chrome red) |
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Answer» Correct Answer - C `4FeO.Cr_(2)O_(3)("chromite ore")+8Na_(2)CO_(3)+7O_(2)undersetto("Roasting in air") 8Na_(2)CrO_(4)+2Fe_(2)O_(3)+8CO_(2)` `2Na_(2)CrO_(4)+H_(2)SO_(4) to Na_(2)Cr_(2)O_(7)+NaSO_(4)+H_(2)O,Na_(2)Cr_(2)O_(7)+2KCltoK_(2)Cr_(2)O_(7)toK_(2)Cr_(2)O_(7)+2NaCl` |
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| 114. |
The final product obtained for the following reaction is: `KMnO_(4)("excess") +H_(2)SO_(4)("concentrated and cold") to`A. `Mn_(2)O_(7)`B. `MnO`C. `Mn_(3)O_(4)`D. `MnO_(3)^(+)` |
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Answer» Correct Answer - A `2KMnO_(4)+3H_(2)SO_(4)to2KHSO_(4)+(MnO_(3))_(2)SO_(4)+2H_(2)O` `(MnO_(3))_(2)SO_(4)+H_(2)O toMn_(2)O_(7)+H_(2)SO_(4)` |
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| 115. |
A white substance (A) reacts with dilute `H_(2)SO_(4)` to produce a colourless suffocating gas (B) and a colourless solution (C )/ The reaction of gas (B) with potassium iodate and starch solution produce a blue colour solution. Aqueous Solution of (A) gives a white precipitate with `BaCl_(2)` solution which is soluble in dilute `HCl`. Addition of aqueous `NH_(3)` or `NaOH` to (C ) produced first a precipitate which dissolves in excess of the respective reagent to poduced a clear solution. Similarly addition of excess of potassium ferrocyanide to (C ) produced a precipitate (D) which also dissolves in aqeous `NaOH` giving a clear solution. Identify (A),(B),(C ) and (D ). Write the equation of the reactions involved. |
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Answer» Correct Answer - `ZnSO_(3)+H_(2)SO_(4) to ZnSO_(4)+SO_(2)downarrow+H_(2)O` (A) ,(C) ,(B) `5 SO_(2)+2IO_(3)^(-)+4H_(2)O to I_(2)+5SO_(4)^(2-)+8H^(+)` `I_(2)`+Starch solution to blue colour solution. `SO_(3)^(-)+2H^(+) to Ba^(2+)+SO_(2)+H_(2)O` `Zn^(2+)+2OH^(-) to Zn(OH)_(2)downarrow("white")` `Zn(OH)_(2)+2OH^(-) to [Zn(OH)_(4)]^(2-)` `Zn^(2+)+2NH_(3)+2H_(2) to Zn(OH)_(2) downarrow+2NH_(4)^(+)` `Zn(OH)_(2)downarrow+4NH_(3) to [Zn(NH_(3))_(4)]^(2+)+2OH^(-)` `3Zn^(+)+2K^(+)+2[Fe(CN)_(6)^(4-) to K_(2)Zn_(3)[Fe(CN)_(6)]_(2)downarrow` `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)+12OH^(-) to 2[Fe(CN)_(6)]^(4-)+3[Zn(OH)_(4)]^(2-)+2K^(+).` |
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| 116. |
What happen when(i) a small amount of `Kmno_(4)` is added to concentrated `H_(2)SO_(4)` (ii) an excess amount of `KMnO_(4)` is added to concentrated `H_(2)SO_(4)` solution. |
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Answer» Correct Answer - (i) `KMnO_(4)+3H_(2)SO_(4) to K^(+) + MnO_(3)^(+) (green)+3HSO_(4)^(-)+H_(3)O^(+)` `2MnO_(3)^(+)+Na_(2)CO_(3) to 2MnO_(3)^(+)+Na_(2)CO_(3) to 2MnO_(3)+CO_(2)+1/2O_(2)+2Na^(+)` (ii) `2KMnO_(4)+3H_(2)SO_(4) to Mn_(2)O_(7)+H_(2)SO_(4)` |
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| 117. |
Why it is not advisable to dissolve `KMnO_(4)` in cold and concentrated `H_(2)SO_(4)` |
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Answer» Correct Answer - Because explosive `Mn_(2)O_(7)` is formed. `2KMnO_(4)+2H_(2)SO_(4)toMn_(2)O_(7)+2KHSO_(4)+H_(2)O` `2Mn_(2)O_(7)to4MnO_(7) to 4MnO_(2)+3O_(2)` |
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| 118. |
Why is `AgBr` used in photography ? |
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Answer» Correct Answer - Out of all silver halidas, `AgBr` is most sensitive to light and under goes photo reduction to metallic silver instantaneously on exposure to light. `2AgBrto2Ag+Br_(2)` Unexposed `AgBr` can be dissolved out in hypo `(Na_(2)S_(2)O_(3))` solution. `AgBr+2Na_(2)S_(2)O_(3)toNa_(3)[Ag(S_(2)O_(3))]+NaBr`. |
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| 119. |
Which type of reaction `MnO_(4)^(2-)`show with acid , dilute -alkali or water. |
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Answer» Correct Answer - It is unstable in acidic medium and disproportionates `3MnO_(4)^(2-)+4H^(+) to MnO_(2)+2MnO_(4)^(-)+2H_(2)O` `3MnO_(4)^(2-)+2H_(2)O to 2MnO_(4)^(-)+MnO_(2)+4OH^(-)` |
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| 120. |
Which of the following group of transition metals is called coinage metals?A. `Cu,Ag,Au`B. `Ru,Rh,Pb`C. `Fe,Co,Ni`D. `Os,Ir,Pt` |
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Answer» Correct Answer - 1 `Cu,Ag,Au` group of element are called coinage metals as these are used in minting coins. |
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| 121. |
Choose the correct statement(s):A. Ions of the same charge in a given transition series show progressive decrease in radius with increasing atomic number.B. Same trend as (a) is observed in the atomic radii.C. there is occurrence of much more frequent metal-metal bonding in compounds of heavy transition metals.D. Metals of 2nd and 3rd transition series have greater enthalpies of atomization than the corresponding element of first series. |
| Answer» Correct Answer - A::B::C::D | |
| 122. |
Which metal(s) has/have a typical metallic structure?A. `Zn`B. `Cd`C. `Hg`D. `Mn` |
| Answer» Correct Answer - A::B::C::D | |
| 123. |
Which one of the following species is stable in ferromagnetic?A. `MnO_(4)^(3-)`B. `MnO_(4)^(2-)`C. `Cu^(+)`D. `Cr^(2+)` |
| Answer» Correct Answer - B | |
| 124. |
Which of the following compounds is metallic and ferromagnetic ?A. `MnO_(2)` dissolves in conc. HCl, but does not form `Mn^(4+)` ionsB. `TiO_(2)`C. `CrO_(2)`D. `VO_(2)` |
| Answer» Correct Answer - C | |
| 125. |
Galvanization of iron denotes coating with `…………………………`A. `Zn`B. `Pg`C. `Cr`D. `Cu` |
| Answer» Correct Answer - A | |
| 126. |
Explain the following cases giving appropriate reason: (i) Nickel does not from low spin octahedral complexes: (ii) The `pie`-Complexes are known for the transition metals only. (iii) `CO^(2+)` is easily oxidised to `Co^(3+)` in the presence of a strong ligand. |
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Answer» Correct Answer - (i) `N(28):[Ar]4s^(2) 3d^(8)` ; `Ni^(2+)(28):[Ar]4s^(@)3d^(8)` In octahedral comples, Pairing of electrons will not take place even if we have strong field lingand, therefore `Ni` does not from low spin octahedral complex. (ii) in `pi`-complex ,`sigma` bond is formed by donation of `pi` electrons or lone pair to vacant d-orbital of transition metal and p-bond is formed by back donation of pair of electrons from transition metal to vacant antibonding orbital of alkene or carbon monoxide. These condition are found only in transition are found only in transition metals. (iii) `Co^(2+)` is oxidised to `Co^(3+)` in presence of strong field ligand because energy needed for oxidation is provided by strong field ligand and `Co^(3+)` is more stable than `Co^(2+)`. |
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| 127. |
Ion(s) having non zero magnetic moment (spin only) is/areA. `Sc^(3+)`B. `Ti^(3+)`C. `Cu^(2+)`D. `Zn^(2+)` |
| Answer» Correct Answer - B::C | |
| 128. |
Which of the following ion has the maximum magnetic moment?A. `Mn^(2+)`B. `Fe^(2+)`C. `Ti^(2+)`D. `Cr^(2+)` |
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Answer» Correct Answer - 1 (1) Valence shell electron configuration of `Mn^(2+)`is `3d^(5)`, therefore , has the maximum number of unpaired electrons equal to `5` and , therefore ,has maximum magnetic moment. (2) Valence shell electron configuration of `Fe^(2+)` is `3d^(6)` , therefore, has the maximum number of unpaired electrons equal to `4` Valence shell electron configuration of `Ti^(2+)` is `3d^(2)`, therefore , has the maximum number of unpaired electron equal to `2` (4) Valence shell electron configuration of `Cr^(2+)` is `3d^(4)`, therefore, has the maximum number of unpaired electron equal to `4` |
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| 129. |
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. |
| Answer» Correct Answer - With atomic number `25` the divalent ion in aqueous solution will have `d^(5)` configuration (Five unpaired electrons)The magnetic moments,muis | |
| 130. |
Select correct statement(s).A. `PH_(3)` reduce `AgNO_(3)` to metallic AgB. Organic tissues turn `AgNO_(3)` black by reducing it to `Ag`C. `AgCN` is soluble in `KCN.`D. `Zr` and `Ta` have almost similar size due to lanthanide contraction |
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Answer» Correct Answer - ABC (A) `PH_(3)+6AgNO_(3)+3H_(2)Oto6Ag+6HNO_(3)+H_(3)PO_(3)` (B ) It possessses powerful corrosive action action on organic tissue, which it turn black especially in presence of light. The blackening is due to finely divided metallic silver,reduced by organic tissue. (C ) `AgCN(s)+CN(-) (aq) to [Ag(CN)_(2)]^(-) ("soluble complex")`. (D) `Zr` is the element of the `4d` series `4^(th)` group while Ta is a element of `5d` series `5^(th) ` group so have differ atomic sizes. |
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| 131. |
The `f`-block elements of the periodic table contains those element in whichA. Only `4f` orbitals are progressively filled in `6th` period.B. Only `5f` orbitals are progressively filled in `7th` period.C. `4f` and `5f` orbitals are progressivley filled in `6th` and `7th` period respectively.D. none |
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Answer» Correct Answer - C Electronic configuration of f-block element`(n-2)f^(1-14),(n-1),ns^(2)` f-block start with `6^(th)` period `(n=6)` |
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| 132. |
What is the effect of increasing `pH` on `K_2Cr_2O_7` solution? |
| Answer» Correct Answer - In aqueous solution, we have `Cr_(2)O_(7)^(2-)+H_(2)O to 2CrO_(4)^(2-)+2H^(+)`. When `pHlt` (acidic medium),it exists as `Cr_(2)O_(7)^(2-)` and colour is orange. When `pHgt7` (basic medium ),it exits as `CrO_(4)^(2-)` and colour is yellow. | |
| 133. |
Sodium thiosulphate is used in photography because of its:A. oxidising behaviourB. reducing behaviourC. complexing behaviourD. photochemical behaviour |
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Answer» Correct Answer - C `Ag^(+)+2S_(2)SO_(3)^(2-) to [Ag(S_(2)O_(3))_(2)]^(3-)`, soluble complex is formed. |
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| 134. |
Which of the following statement(s) is/are not correct.A. The blue colour of aqueous `CuCl_(2)` is due to `[Cu(H_(2)O)_(4)]^(2+)`B. The yellow colour of aqueous `CuCl_(2)` is due to `[CuCl_(4))]^(2-)`C. The green colour of aqueous `CuCl_(2)` is due to `[CuCl_(2)` is due to presence of both `[Cu(H_(2)O_(4))]^(2+)` and `[CuCl_(4)]^(2-)`D. The blue colour of aqueous`CuCl_(2)` is due to `[CuCl_(4)]^(2-)` |
| Answer» Correct Answer - D | |
| 135. |
Select correct statement(s).A. `MnO_(4)^(-)` is intensive pink colour due to d-d electron.B. `Cu(I)` is diamagnetic while `Cu(II)` is paramagnetic .C. `CrO_(3)` is amphoteric oxide.D. `[Ti(H_(2)O)_(6)]^(3+)` and `[Sc(H_(2)O)_(6)]^(3+)` both are coloured in aqueous soloution. |
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Answer» Correct Answer - B (A) coloured due to charge transfer from ligand to metal ion. (B) Valence shell electron configuration of `Cu(I)` is `3d^(10)` and , therefore , all electron are paired, Valence shell electron configuration of `Cu(II)` is `3d^(9)` and , therefore , one electron is unpaired. (C ) `CrO_(3)` is an acid anhydried of chromic acid. (D ) Valence shell electrons configuration of `Ti^(3+)` is `3d^(1)` and , therefore , has one unpaired electron, So it is coloured . But valence shell electron configuration of `Sc^(3+) is `3d^(0)` and , therefore , it has no unpaired electron, so it is colourless. |
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| 136. |
Correct statements about transition metals are that they:A. form complexB. show variable oxidation statesC. show magnetic propertiesD. do not from coloured compounds |
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Answer» Correct Answer - ABC the great tendency of transition material ion to form complex is due ,(i) small size to the atoms and ions(ii) high nuclear charge and (iii) availability of vacant d-suitable energy to accept lone pair of electrons donates by ligands. (B) The existence of the transition element in different oxidation state is due to the participation of inner `(n-1)` d-electrons in addition to outer ns-electrons because , the energies of the ns and (n-1) d-sub-shells are nearly same. (C ) The transition metals show magnetic properties due to presence of number of unpaired electrons in d-orbital. (D) Most the transition metal ion are coloured, because they have unpaired electrons which can undergo d-d transition in presence of ligands. |
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| 137. |
An ornament of gold having 75% of gold, is of ______carat.A. 18B. 16C. 24D. 20 |
| Answer» Correct Answer - A | |
| 138. |
Transition elements have greater tendency to form complex because they have:A. vacant d-orbitalsB. small sizeC. higher nuclear chargeD. variable oxidation states. |
| Answer» Correct Answer - ABC | |
| 139. |
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. orbitals are available for bonding. |
| Answer» Correct Answer - A | |
| 140. |
which of the following group of ion is paramagnetic in nature:A. `Cu^(+),Zn^(2+)Sc^(2+)`B. `Mn^(2+),Fe^(3+),Ni^(2+)`C. `Cr^(2+),Mn^(3+),Sc^(3+)`D. `Cu^(2+),Ni^(2+),Ti^(4+)` |
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Answer» Correct Answer - B (A)` Cu^(+)[Ar]^(18)3d^(10),Son=0,Zn^(2+)[Ar]^(18)3d^(10),So n=0,Sc^(3+)[Ar]^(18)3d^(0),So n=0` (N ) (B)` Mn^(2+)[Ar]^(18)3d^(10),Son=5,Fe^(3+)[Ar]^(18)3d^(5),So n=5,Ni^(2+)[Ar]^(18)3d^(8),So n=2` (C )` Cr^(2+)[Ar]^(18)3d^(4),Son=4,Mn^(3+)[Ar]^(18)3d^(4),So n=4,Sc^(3+)[Ar]^(18)3d^(0),So n=0` (D ) ` Cu^(2+)[Ar]^(18)3d^(9),Son=1,Ni^(2+)[Ar]^(18)3d^(8),So n=2,Ti^(4+)[Ar]^(18)3d^(0),So n=0` |
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| 141. |
Compound that is both paramagnetic and coloured is:A. `K_(2)Cr_(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 (A) in `Cr_(2)O_(7)^(2-)` the valenc shell electron configuration of `Cr(VI)` is `3d^(0)`. Thus `Cr(VI)` is diamagnetic but coloured due to the charge transfer spectrum. (B) In `(NH_(4))_(2)[TiCl_(6)]`, the valence shell electrons configuration of `Ti(IV)` is `3d^(0)` . Thus `Ti(IV)` is diamagnetic and colourless. (C ) In `VOSO_(4)` , then valence shell electron configuration of `V(IV)` is `3d^(1)`. Thus `V(IV)` is paramagnetic and blue coloured due to d-d transition. (D) In `K_(3)[Cu(CN)_(4)]`, the valence shell electron configuration of `CU(I)` is `3d^(10)`. Thus `Cu(I)` is diamagnetic and colourless. |
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| 142. |
Which of the following contains the maximum number of unpaired electrons?A. `TiCl_(3)`B. `MnCl_(2)`C. `FeSO_(4)`D. `CuSO_(4)` |
| Answer» Correct Answer - B | |
| 143. |
Four elements A (with one valence electron), B (with three valence electrons), C (with five valance electrons) and D (with seven valance electrons) are lying in the second period of periodic table. Which of the following does not exist at room temperature?A. `C_(2)`B. `A_(2)`C. `D_(2)`D. None of these |
| Answer» Correct Answer - B | |
| 144. |
Super conductors are derived from compound of:A. p-block elementsB. lanthanoidesC. actinoidesD. transition elements |
| Answer» Correct Answer - A | |
| 145. |
The aqueous solution of `CuCrO_(4)` is green because it contains:A. green `Cu^(2+)` ionsB. green `CrO_(4)^(2-)` ionsC. blue `Cu^(2+)` ions and green `CrO_(4)^(2-)` ionsD. blue `Cu^(2+)` ions and yellow `CrO_(4)^(2-)` ions. |
| Answer» Correct Answer - D | |
| 146. |
Which of the following pair of compounds is expected to exhibit same colour in aqueous solution?A. `FeCl_(3),CuCl_(2)`B. `VOCl_(2),CuCl_(2)`C. `CoCl_(2),FeCl_(2)`D. `FeCl_(2),MnCl_(2)` |
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Answer» Correct Answer - B The colour of the transition metal ion is mainly attributed to d- transition take place because of the presence of unpaired electrons. The metal ion having same number of unpaired electrons may give same colour in their aqueous solution. In `VOCl_(2)`,V^(IV)` ion having `3d^(1)` ion having `3d^(1)` configuration has one unpaired . SImilarly in `CuCl_(2)`, the `Cu^(2+)` ion having `3d^(9)` configuration also has one unpaired electron. SO `VOCl_(2)` and CuCl_(2)` are expected to exhibit same colour `i.e ` blue colour. `Mn^(2+)` has five unpaired and `Fe^(2+)` has four unpaired electrons. |
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| 147. |
The pair of compounds having metals in their highest oxidation state is:A. `MnO_(2),FeCl_(3)`B. `[MnO_(4)]^(-),CrO_(2)Cl_(2)`C. `[Fe(CN)_(6)}^(4-),[Co(CN)_(6)]^(3-)`D. `[NiCl_(4)]^(2-),[Ni(CO)_(4)]` |
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Answer» Correct Answer - B `MnO_(4)^(-),X+4(-2)=-1` or X=+7, `CrO_(2)Cl_(2),X+2(-2)+2(-1)=+6` or `X=+6`. |
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| 148. |
Write the balanced chemical equation for the following reaction:- Nitrrogen is obtained in the reaction of aqueous ammonia with potassium permanganate |
| Answer» Correct Answer - `2KMnO_(4)+2NH_(3) to 2MnO_(2)+2KOH+2H_(2)O+N_(2)` | |
| 149. |
In dilute alkaline solution,`MnO_(4)^(-)` changes to:A. `KMnO_(4)^(2-)`B. `MnO_(2)`C. `Mn_(2)O_(3)`D. `MnO` |
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Answer» Correct Answer - B `MnO_(4)^(-)+2H_(2)O+3e^(-) to MnO_(2)+4OH^(-)`(weak alkaline medium)` |
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| 150. |
`Cl_2` gas is obtained by various reactions select the reactions from the following (s):A. `KMnO_(4)(s)+Conc Hclundersetto(Delta)`B. `KCl(s)+K_(2)CR_(2)O_(7)(s)+Conc H_(2)SO_(4)undersetto(Delta)`C. `MnO_(2)(s)+Conc Hclundersetto(Delta)`D. `Kcl(s)+f_(2)(l) to` |
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Answer» Correct Answer - B (A) `2MnO_(4)^(-)+16H^(+)+10Cl^(-) to 2Mn^(2+)+8H_(2)O+5Cl_(2)` (B ) Chromyl chloride test,`Cr_(2)O_(7)^(2-)+4Cl^(-)+6H^(+) to 2CrO_(2)uparrow("deep red")+3H_(2)O`. (C ) `MnO_(2)+4HCl to MnCl_(2)+Cl_(2)+2H_(2)O` (D ) `2Cl^(-) +Fe to Cl_(2)+2F^(-)` |
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