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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.
| 1001. |
The common oxidation state of actinides isA. `+4`B. `+5`C. `+3`D. `+7` |
| Answer» Correct Answer - C | |
| 1002. |
The free gaseous `Cr` atom has six unpaired electrons. Half-filled s-orbital has greater stability.A. 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. All the six orbitals (one 4s and five 3d) are unpaired. Therefore, configuration is symmetrical as well as stable. |
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| 1003. |
Predict which of the following will be coloured in aqueous solution ? `Ti^(3+), V^(3+),Cu^(+),Sc^(3+), Mn^(2+), Fe^(3+), Co^(2+)`. |
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Answer» The electronic configuration of cations are listed : `{:(Sc^(3+)(Z=21)[Ar]^(18),V^(3+)(Z=23)[Ar]^(18)3d^(2)),(Ti^(4+)(Z=22)[Ar]^(18),Mn^(2+)(Z=23)[Ar]^(18)3d^(6)):}` |
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| 1004. |
Inner transition elements belongs to which block of the periodic table ?A. s - blockB. p - blockC. d - blockD. f - block |
| Answer» Correct Answer - D | |
| 1005. |
Which of the following ion posses six unpaired electrons ?A. `Eu^(3+), Ce ^(3+), Gd^(3+)`B. `Eu^(3+), Tb ^(3+), Sm ^(2+)`C. `Tb ^(3+), Nd ^(3+), Sm ^(2+)`D. `Sm ^(2+) , Er ^(3+) , Eu ^(2+)` |
| Answer» Correct Answer - B | |
| 1006. |
Which of the following cations are coloured in aqueous solution and why ? `Sc^(3+), V^(3+), Te^(4+), Mn^(2+)` ( At.Nos. `Sc= 21, V = 23, Ti= 22, Mn =25)` |
| Answer» `Sc^(3+)= 3d^(0)4s^(0)`( No unpaired electron), `V^(3+)= 3d^(2) 4s^(0) `( 2 unpaired electrons).Thus, only `V^(3+)` and `Mn^(2+)` are coloured because they have unpaired electrons. | |
| 1007. |
`f-`block elements (sometimes called inner transition elements) are the elements in which the last electron enters the_____________orbitals.A. electrons are edded in `(n-3)` f shellB. electrons are added in `(n-1)` f shellC. electrons are added in `(n-2)` f shellD. electrons are added in `(n-2)` d shell |
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Answer» Correct Answer - C We know that, in f-block elements electrons are in antipenultimate shall, i.e. third from outer most shell i.e.(n-2) f orbitals. |
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| 1008. |
The number of unpaired electrons in ferrous ion isA. 4B. 2C. 3D. 0 |
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Answer» Correct Answer - A Number of unpaired electrons in Ferrous `Fe^(+2)` ions is 4 as the electronic configuration of `Fe^(+2)` is `3d^6 4s^0` |
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| 1009. |
Which one of the following forms a colourless solution in aqueous medium ? (Atomic number : `Sc =21, Ti = 22, V = 23`, and `Cr = 24`)A. `Ti^(3+)`B. `Sc^(3+)`C. `V^(3+)`D. `Cr^(3+)` |
| Answer» Correct Answer - B | |
| 1010. |
The first and last element of the second transition series respectively areA. Y, CdB. La, HgC. Cd, YD. Y, Hg |
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Answer» Correct Answer - A Second transition series Y (39) to Cd (48) |
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| 1011. |
Transition metals:A. show diamagnetismB. show catenationC. do not form alloyD. show variable oxidation states |
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Answer» Correct Answer - D Various oxidation states in transition metals is due to participation of `e^-` from ns and (n-1) d orbitals |
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| 1012. |
Among the following outermost configurations of transitionn metals, which shows the highest oxidation stateA. `3d^2 4s^2`B. `3d^4 4s^1`C. `3d^5 4s^2`D. `3d^8 4s^2` |
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Answer» Correct Answer - C `3d^5 4s^2` shows oxidation state of +7 , which is highest oxidation states of other configuration is `3d^2 4s^2 (+4), 3d^4 4s^1 (+5), 3d^8 4s^2 (+4)` |
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| 1013. |
The pair of compounds having metals in their highest oxidation state is .A. `MnO_(2),FeCl_(3)`B. `[NiCl_(4)]^(2-),[CoCl_(4)]^(-)`C. `[MnO_(4)]^(-),CrO_(2)Cl_(2)`D. `[Fe(CN)_(6)]^(3-),[Co(CN)_(3)]` |
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Answer» Correct Answer - C The compounds along with their metals, oxidaton statesare `[MnO_(2)]rarrMn^(4+)," " FeCl_(3)rarrFe^(3+)` `[NiCl_(4)]^(2-)rarrNi^(2+), " " [CoCl_(4)]^(-)rarrCo^(3+)` `[MnO_(4)]^(-)rarr Mn^(7+)," " CrO_(2)Cl_(2)rarrCr^(6+),` `[Fe(CN)_(6)]^(3-)rarr Fe^(3+)," " [Co(CN)_(3)]rarrCo^(3+)` |
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| 1014. |
The colour of zinc sulphide isA. WhiteB. blackC. brownD. red |
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Answer» Correct Answer - A The colour of ZnS is white as zinc as completely filled 3d-orbitals |
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| 1015. |
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» The electronic configuration of both the ions are : `Mn^(2+): [Ar]3d^(5),Fe^(2+):[Ar]3d^(6)` The `Mn^(2+)` ion has more symmetrical configuration than `Fe^(2+)` ion and is therefore, more stable. Thus, +2 oxidation of manganese is quite stable while that of iron is not. |
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| 1016. |
For the four successive transition elements (Cr, Mn, Fe, and Co), the stability of `+2` oxidation state will be there in which of the following order ? `(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)`A. `CrgtMngtCogtFegt`B. `MngtFegtCrgtCo`C. `FegtMngtCogtCr`D. `CogtMngtFegtCr` |
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Answer» Correct Answer - B The correct order is `MngtFegtCrgtCo` |
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| 1017. |
For the four successive transition elements (Cr, Mn, Fe and Co) ,the stability of `+2` oxidation state will be there in which of the following order ?A. `Cr gt Mn gt Co gt Fe`B. `Mn gtFe gt Cr gt Co`C. `Fe gt Mn gt Co gt Cr`D. `Co gt Mn gt Fegt Cr` |
| Answer» `Mn^(2+)(d^(5))` gets highest stabilization due to half filled configuration .In case of`Fe^(2+)(d^(6))` , addition of one extra electron in the subshell destabilizes it. Addition of two electrons incase of `Co^(2+)(d^(7))` destabilizes itmore . `Cr^(2+)(d^(4))` has one vacant subshell. `Fe^(2+)` gets more stabilizationas compared to `Cr^(2+)` through exchange energy . So, the correct order is`Mn gt Fe gt Cr gtCo`. | |
| 1018. |
Assign reasons for each of the following `:` (i) Manganese exhibits the highest oxidation state of `+7` . (ii) Unlike `Cr^(3+), Mn^(2+), Fe^93+)` and the subseuent `M^(3+) ` ions of 3d series of elements , the 4d an 5d series metals generally do not form stable cationic species. |
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Answer» (i) Electronic configuration of `._(25)Mn` is `3d^(5) 4s^(2)` . As all the 7 electrons of 3d and 4s can participate in bonding , it exhibits highest oxidation state of `+7` (ii) In the 4d and 5d series higher oxidation states are more stable ( because difference of energy between 4d and 5s or 5d and 6s is smaller than between 3d and 4s). In the higher oxidation states, bonds formed are mostly covalent .Hence they do not form cationic species. |
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| 1019. |
The correct order of `E_(M^(2+)//M)^(@)` Values with negative sign for the four successive elements `Cr, Mn, Fe` and `Co` is:A. `Cr gt Mn gt Fr gt Co`B. `Mn gt Cr gt Fe gt Co`C. `Cr gt Fe gt Mn gt Co`D. `Fe gt Mn gt Cr gt Co` |
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Answer» Correct Answer - B (b) is the correct answer `E^(@)Mn^(2+)//Mn= -1.18 V` `E^(@) Cr^(2+)//Cr= -0.91 V` `E^(@) Fe^(2+)//Fe = -0.44 V` `E^(@) CO^(2+)//Co= -0.28 V` |
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| 1020. |
Which of the 3d series of the transition metals exhibits the largest number of oxidation , states and why ? |
| Answer» Manganese (atomic no.= 25) has electronic configuration `[Ar] 3d^(5) 4s^(2)` It shows maximum number of oxidation states, i.e., from +2 to +7 (+2, +3, +4, +5, +6, +7) in its compounds. | |
| 1021. |
What may be the stable oxidation state of the transition elements with the following d electron configurations in the ground state of their atoms:`3d^(3),3d^(5),3d^(4)?` |
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Answer» The maximum oxidation state of reasonable stability in the transition metal of 3d series correspond to the sum of s and d-electrons upto Mn. However, after Mn there is an abrupt decrease in oxidation states. For details, consult Section. In the light of this, most stable oxidation states of the elements are : `3d^(3):3d^(3)4s^(2)(+5),3d^(5):3d^(5)4s^(1)(+6)` and `3d^(5)4s^(2)(+7)` `3d^(8):3d^(8)4s^(2)(+2),3d^(4):3d^(4)4s^(2)` or `3d^(5)4s^(1)(+6)` |
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| 1022. |
Discuss the relative stability in aqueous solution of +2 oxidation states among the elements: Cr, Mn, Fe and Co |
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Answer» The relative stability of the +2 oxidation states ( decreasing order ) can be compared in terms of `E^(@)` values. `(M^(2+)//M)` `Mn^(2+)(-1.18V) gt Cr^(2+)(-0.90 V) gt Fe^(2+)(-0.44 V) gt CO^(2+)(-0.28 V)` |
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| 1023. |
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : `3d^(3) 3d^(5) 3d^(8) and 3d^(4)` ? |
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Answer» i) `3d^(3)` -stable oxidation states are +2, +3, +4 and +5 (V) ii) `3d^(5)` -stable oxidation states are +3, +4 and +6 (Cr) iii) `3d^(5)` -stable oxidation states are +2, +4, +6 and +7 (Mn) iv) `3d^(8)` -stable oxidation states are +2, +3 v) `3d^(4)` -This configuration does not exist. |
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| 1024. |
To what extent do the electronic configurations, decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples. |
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Answer» The presence of half filled or completely filled orbitals impart stability to a particular elemtn/ion. Greater the number of such orbitals, more will be the relative stability. For example, let us write the different oxidation states of Mn(Z=25) along with the electronic configuration. `Mn:[Ar]3d^(5)4s^(2),Mn^(2+),[Ar]3d^(5),Mn^(3+),[Ar]3d^(4),Mn^(4+),[Ar]3d^(3)` +2 oxidation state of the element is likely to be the most stable because the corresponding electronic configuration of `Mn^(2+)` is highly symmetrical ( all the five 3d- orbitals are half filled). |
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| 1025. |
Mention the type of compounds formed when small atoms like H,Cand N get trapped inside the crystal lattice of transition metals, Also give physical and chemical characteristic of these compounds. |
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Answer» Interstitial compounds Characteristics . |
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| 1026. |
Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following are the characteristic properties of interstitial compounds? I. They have high melting points in comparison to pure metals. II. They are very hard. III. They retain metallic conductivity. IV. They are chemically very reactive.A. They have high melting points in comparison to pure metal.B. They are vary hard.C. They retain metallic conductivityD. They are chemically very reactive. |
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Answer» Correct Answer - D (d) It is the correct answer. |
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| 1027. |
Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following are the characteristic properties of interstitial compounds? I. They have high melting points in comparison to pure metals. II. They are very hard. III. They retain metallic conductivity. IV. They are chemically very reactive.A. They have high melting points in comparison to pure metalsB. They are very hardC. They retain metallic conductivityD. The are chemically very reactive |
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Answer» Correct Answer - D Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Some of their important characteristics are as follows (i) They are very hard and rigid. (ii) They have high melting point which are higher than those of the pure metals. (iii) They show conductivity like that of the pure metal. (iv) They acquire chemical inertness. |
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| 1028. |
Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following are the characteristic properties of interstitial compounds? I. They have high melting points in comparison to pure metals. II. They are very hard. III. They retain metallic conductivity. IV. They are chemically very reactive.A. I, II and IIIB. II, III and IVC. I, III and IVD. I, II and IV |
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Answer» Correct Answer - A Interstitial compounds have high melting points in comparison to pure metals. They are very hard and retain metallic conductivity. They are chemically inert. |
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| 1029. |
`underset("(B) white fumes")underset(darr)(MCl_(4)"(colourless liquid) M = transition metal moist air") overset(Zn+H_(2)O)(rarr) underset((A))("Purple coloured compound")` Identify (A), (B) and `MCl_(4)` . Also explain colour differnece between `MCl_(4)` and (A) . |
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Answer» `MCl_(4) = TiCl_(4)` ( colourless, diamagnetic, covalent fuming liquid ) `TiCl_(4)(aq) underset(Delta"(Reduction)") overset(Zn) (rarr)TiCl_(3)overset(H_(2)O) (rarr) underset("Purple coloured(A))")([Ti(H_(2)O)_(6)]Cl_(3))` `TiCl_(4)+ H_(2)O rarrTiOCl_(2) + underset("White fumes (B)") (2HCl)` Hence, `A=[Ti(H_(2)O)_(6)] Cl_(3) ,B = HCl,MCl_(4) = TiCl_(4)` Coloure difference. `TiCl_(4)` is colourless because Ti(IV) has empty d-subhsell. Hence, no d-d transition is possible, Ti(III) has `d^(1)` configuration . Hence, Ti(III) is coloured due to d-d transition . `Ti^(3+)` absorbs greenish yellow component of white light . The complementary coloure is purple. Hence, aqueous solution containing `T^(3+)` ions is purple. |
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| 1030. |
What may the possible oxidation state of transition metals with the following electronic configurations in the ground state of their atoms ? (i) `3d^(3)4s^(2)` (ii) `3d^(5)4s^(2)` (iii) `3d^(6)4s^(2)`. |
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Answer» (i) The element with `3d^(3)4s^(2)` configuration (18+3+2+=23) is vanadium. The possible oxidation states are : +2,+3,+4,+5. (ii) The element with `3d^(4)4s^(2)` configuration (18+5+2+=25) is manganese. The possible oxidation states are : +2,+3,+4,+5,+6,+7. (iii) The element with `3d^(6)4s^(2)` configuration (18+6+2+=23) is iron . The possible oxidation states are : +2,+3,+4,+5,+6 |
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| 1031. |
Assertion . The purple colour of `KMnO_(4)` is due to the charge transfer transition.Reason. The intense colour of most of the transition metal complexesis due to d-d transition,. |
| Answer» Correct explanation.The intense colourof `KMnO_(4)` is not due to d-d transition but due to charge transfer from O to M ( Ligand to Metal) reducing the oxidation state of Mn from `+7` to `+6 ` temporarily. | |
| 1032. |
Match the following: `{:("Column A"," Column B"),((1)V^(4+),(a)" Colourless"),((2)Ti^(3+),(b)" Pink"),((3)Ti^(4+),(c) " Purple"),((4)Mn^(2+),(d)" Violet"):}`A. 1-(d), 2-(c ) , 3-(a) , 4-(e )B. 1-(d) , 2-( c ) , 3-(a) , 4-(b)C. 1-(d) , 2- (e ) , 3-(a ) , 4-(b)D. 1-(e ) , 2-(c ) , 3-(b) , 4-(a) |
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Answer» Correct Answer - B Colours are due to presence of unpaired `e^-` in (n-1) d-orbitals and colourless in `Ti^(4+)` is due to no `e^-` in 3d orbitals |
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| 1033. |
Transition metals in which vacent spaces are occupied by small atoms such as hydrogen, carbon etc. are called `"….................."` |
| Answer» Correct Answer - interstitial compounds | |
| 1034. |
`[Ti(H_(2)O)_(6)]^(3+)` complex ion has purple colour due to absorption of `"…............"` coloured light and causing transition from `"…......."` to `"…............."`orbitals. |
| Answer» Correct Answer - yellow,`t_(2g) ` to `e_(g)` | |
| 1035. |
The lanthanum exhibits which of the following oxidation state ?A. only `+2`B. `+2 and +3`C. only `+3`D. only `+3 and +4` |
| Answer» Correct Answer - D | |
| 1036. |
Lanthanum is the first element of the lanthanide series hasA. only 6s-orbital is filledB. 4d and 4f-orbitals are unfilledC. 3d and 5d-orbitals are unfilledD. 4f-and 5d-orbitals are filled |
| Answer» Correct Answer - A | |
| 1037. |
Write the formula of different oxides of manganese. What is the oxidation state of manganese in each of them ? Arrange them in order of their decreasing acidic character. |
| Answer» `underset("Basic")overset(+2)(MnO)" "ubrace(overset(+8//3)(Mn_(3))O_(4)" "overset(+3)(Mn_(2))O_(3)" "overset(+4)(Mn)O_(2))_(" Amphoteric")" "underset("Acidic")(overset(+7)(Mn_(2))O_(7))` | |
| 1038. |
At `pH =4. Cr_(2)O_(7)^(2-)` existasA. `CrO_(2)^(2+)`B. `CrO_(4)^(2-)`C. `CrO_(4)^(2+)`D. `Cr_(2)O_(7)^(2-)` |
| Answer» Correct Answer - D | |
| 1039. |
At `pH=12,Cr_(2)O_(7)^(2-)` changes to:A. `CrO_(3)`B. `Cr O_(2)^(2+)`C. `CrO_(4)^(2-)`D. `CrO_(4)^(2+)` |
| Answer» Correct Answer - C | |
| 1040. |
`Cr_(2)O_(7)^(2-)overset(pH=x)rarrCrO_(4)^(2-)overset(pH=y)rarrCr_(2)O_(7)^(2-)` x and y can be :A. 4 and 5B. 4 and 8C. 8 and 4D. 8 and 9 |
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Answer» Correct Answer - C (c) `Cr_(2)O_(7)^(2-)` ion changes to `CrO_(4)^(2-)` ion in alkaline medium. `CrO_(4)^(2-)` changes to `Cr_(2)O_(7)^(2-)` ion in acidic medium. |
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| 1041. |
What is the effect of pH on the solution of `K_(2)Cr_(2)O_(7)` solution? |
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Answer» In aqueous solution `:` ` underset("Dichromate ion ( orange red)")(Cr_(2)O_(7)^(2-))+H_(2)O hArr underset("Chromate ion ( yellow)") (2CrO_(4)^(2-))+ 2H^(+)` In acidic medium (i.e., decreasing pH ) , equilibrium shifts backward and the colour is organe red. In basic medium (i.e., increasin pH ), equilibrium will shift forward and the solution is yellow . |
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| 1042. |
What is the effect of increasing pH on the colour of `K_(2)Cr_(2)O_(7)` solution ? |
| Answer» When pH is less than 7 (acidic medium), `K_(2)Cr_(2)O_(7)(Cr_(2)O_(7)^(2-))` has orange colour. In the basic medium with pH more than 7, the colour of the solution changes to yellow due to the formation `K_(2)CrO_(4)(CrO_(4)^(2-) " ion")`. | |
| 1043. |
The addition of acidic `K_(2)Cr_(2) O_(7)` to NaCl produces a colourA. greenB. pinkC. redD. violet |
| Answer» Correct Answer - C | |
| 1044. |
When acidified `K_(2)Cr_(2)O_(7)` solution is added to `Sn^(2)` salts , then `Sn^(2+)` changes toA. SnB. `Sn^(3+)`C. `Sn^(4+)`D. `Sn^(+)` |
| Answer» Correct Answer - c | |
| 1045. |
Transition metals show variable oxidation states because of :A. participation of nd electrons along with np electronsB. participation of (n-1)d and ns electronsC. participation of (n-1)d and np electronsD. None of these is correct |
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Answer» Correct Answer - B is the correct answer |
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| 1046. |
On adding NaOH solution to the aqueous solution of `K_(2)Cr_(2)O_(7)`, the colour of the solution changes from :A. orange to yellowB. yellow to orangeC. orange to redD. yellow to pink. |
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Answer» Correct Answer - A Colour change from orange to yellow. |
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| 1047. |
The radius of `La^(+)` (at no 57) is `1.06Å`. What may be the radius of `Lu^(3+)`(at no.71)?A. 1.40Ã…B. 1.06Ã…C. 0.85Ã…D. 1.60Ã… |
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Answer» Correct Answer - C `(r_(1))/(r_(2)) = (z_(2))/(z_(1))` `r_(2) = (z_(1))/(z_(2)) xxr_(1) =(57)/(11) xx 1.06 = 0.85Ã…` |
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| 1048. |
Arrange `Ce^(3+),La^(3+), Pm^(3)` and `Yb^(3+)` in increasing order of their size -A. `Yb^(3+)ltPm^(3+)ltCe^(3+)ltLa^(3+)`B. `Ce^(3+)gtYb^(3+)ltPm^(3+)La^(3+)`C. `Yb^(3+)Pm^(3+)ltLa^(3+)ltCe^(3+)`D. `Pm^(3+)ltLa^(3+)ltCe^(3+)ltYb^(3+)` |
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Answer» Correct Answer - A Due to lanthnide contraction left to right ionic size decreases. |
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| 1049. |
Arrange `Ce^(3+),La^(3+), Pm^(3)` and `Yb^(3+)` in increasing order of their size -A. `Yb^(3+) lt Pm^(3+) lt Ce^(3+) lt La^(3+)`B. `Ce^(3+) lt Yb^(3+) lt Pm^(3+) lt La^(3+)`C. `Yb^(3+) lt Pm^(3+) lt La^(3+) lt Ce^(3+)`D. `Pm^(3+) lt La^(3+) lt Ce^(3+) lt Yb^(3+)` |
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Answer» Correct Answer - A Ionic radii decreases from left to right |
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| 1050. |
`Cu^(+2)` forms halides like `CuF_(2), CuCl_(2) and CuBr_(2)` but not `CuI_(2)`. Why ? |
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Answer» `Cu^(+2)` forms halides like `CuF_(2), CuCl_(2) and CuBr_(2)` but not `CuI_(2)` because `Cu^(+2)` oxidises `I^(-)" to "I_(2))` `2Cu^(+2)+4I^(-)rarrCu_(2)I_(2)+I_(2).` |
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