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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.
| 301. |
Oxidation state/s exhibited by Ce is / areA. `+2` onlyB. `+3,+4`C. `+4` onlyD. `+2,+3,+4` |
| Answer» Correct Answer - B | |
| 302. |
The titanium (atomic number 22) compound that does not exist isA. TiOB. `TiO_(2)`C. `K_(2)TiF_(6)`D. `TiCl_(3)` |
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Answer» Correct Answer - C `Ti(Z=22)` has configuration `[Ar]^(18)3d^(2)4s^(2)` ltbr It can show oxidation states of +1,+2,+3,+4 In `K_(2)TiO_(4)` : oxidation state of Ti=+6 Therefore, it does not exist. |
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| 303. |
In aqueous solutions `Eu^(2+)` acts asA. an oxidising agentB. an reducing agentC. either of theseD. redox agent |
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Answer» Correct Answer - B We know that, +2 O.S. in europium is stable in aqueous solution due to its `4f^(7)` configuration. But it is has tendency to lose electron and revert into `+3` O.S. Hence `Eu^(2+)` ion asct as reducing agents. |
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| 304. |
Consider the following statements: (I) La`(OH)_(3)` is the least basic among the hydroxides of Ianthanoids. (II) `Zr^(4+)` and `Hf^(4+)` possess almost same ionic radii. (III) `Cr^(4+)` can act as an oxidising agent . which of the above statement is/ are true?A. (I) and (II)B. (II) and (III)C. (II) OnlyD. (I) and (II) |
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Answer» Correct Answer - B Statement (II) and (III) are true. Statement (I) is wrong . |
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| 305. |
The atomic number of cerium `(Ce)` is 58. The correct electronic confguration of `Ce^(3+)` ions is :A. `[Xe]4f^(1)`B. `[Kr]4f^(1)`C. `[Xe]4f^(13)`D. `[Kr]4f^(1)` |
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Answer» Correct Answer - A Configuration of `Ce(Z = 58),` Configuration of `Ce^(3+)` ion `4f^(1)` `underset([Xe] 4f^(1))underset(darr-3e^(-))([Xe]4f^(1)5d^(1)6s^(2))` |
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| 306. |
Lanthanoids areA. 3rd group and 7th periodB. 3rd group and 6th periodC. 4th group and 7th periodD. 3rd group and 8th period |
| Answer» Correct Answer - B | |
| 307. |
Acidified `KMnO_(4)` solution is decolourized byA. tolueneB. `MnSO_(4)`C. `H_(2)O_(2)`D. `KI ` |
| Answer» Correct Answer - C | |
| 308. |
The most common and most stable oxidation state for all the elements of titanium family is |
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Answer» Correct Answer - 3 `+3` is most common oxidation state shown by the members of the lanthanoid family. |
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| 309. |
`Sm^(3+)` acts asA. an oxidising agentB. an reducing agentC. eitherD. neither |
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Answer» Correct Answer - B We know that, all `Ln^(3+)` ions act as reducing agents by loss of one electron. |
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| 310. |
Which of the following element show +2 oxidation state in `4f^(7)` configuration ?A. PmB. TbC. EuD. Ho |
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Answer» Correct Answer - C We know that `Eu^(2+)` hs electronic configuration `6s^(0), 4f^(0), 5d^(0).` |
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| 311. |
Consider the following statements, (I) La`(OH)_(3)` is the least basic among hydroxides of lanthanides. (II) `Zr^(4+) and Hf^(4+)` possess almost the same ionic radii (III) `Ce^(4+)` can act as an oxidising agent. Which of the above is/are true?A. I and IIIB. II and IIIC. Only IID. I and II |
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Answer» Correct Answer - B The basic character of hydroxide decreases from `La(OH)_(3) to Lu(OH)_(3)`.Due to smaller size of Lu, the Lu-OH bond attains more covalent character while `Zr^(4+) and Hf(4+)` has same ionic radii and `Ce^(4+)` is an oxidising agent. |
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| 312. |
Consider the following statements I `La(OH)_(3)` isleastbasic among hydroxides of lanthanides II. `Zx^(4+)` and `Hf^(4+` possess almost the same ionic radii III. `Ce^(4+)` can actas an oxidizing agent.A. I and IIB. II and IIIC. II onlyD. I and II |
| Answer» `La(OH)_(3)` is most basic. Hence, (I) is wrong. (II) is correct due to lanthanoid contraction. (III) is correct because `Ce^(4+)` tends to change to stable `Cr^(3+)` | |
| 313. |
The most stable oxidation state of cerium (Z=58 ) isA. `4+`B. `2+`C. `1+`D. `5+` |
| Answer» Correct Answer - A | |
| 314. |
Which of the following statement is correct?A. Paramagnetic nature of lanthanides can be explained easilyB. Paramagnetic nature of actinides can be explained easilyC. Paramagnetic nature of actinides cannot be explained easilyD. Both (a) and (c) |
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Answer» Correct Answer - D Comparison between lanthanide and actinides |
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| 315. |
Equivalent weight of `KMnO_(4)` when it is convert into `MnSO_(4)` isA. M/5B. M/6C. M/3D. M/2 |
| Answer» Correct Answer - A | |
| 316. |
The number of moles of acidified `KMn0_(4)` required to convert one mole of sulphite ion into sulphate ion is, x/y. The value of X + y = 0 is |
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Answer» Correct Answer - 7 `{:(2MnO_(4)^(-)+6H^(+)+5SO_(3)^(2-)),(" 2 mole 5 moles"),(" "darr),(2Mn^(2+)+5SO_(4)^(2-)+3H_(2)O):}` 5 moles of `SO_(3)^(2-)-= `2 moles of `MnO_(4)^(-)` 1 mole of `SO_(3)^(2-)-=2//5` moles of `MnO_(4)^(-)` THerefore, `(x)/(y)=(2)/(5)` or x+y=2+5=7. |
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| 317. |
How many electrons are accepted when `KMn0_(4)` acts as an oxidising agent in the presence of dilute sulphuric acid? |
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Answer» Correct Answer - 5 `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` |
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| 318. |
Atomic number `""_(64)Gd, ""_(66)Dy, ""_(69)Tm, ""_(71)Lu.` which of the following does not have unpaired electron ?A. `Gd^(3+)`B. `Dy^(3+)`C. `Tm^(3+)`D. `Lu^(3+)` |
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Answer» Correct Answer - D We know that, `Gd ^(3+)` has `4f^(7)` configuration containing 7 unpaired electrons. `Dy ^(3+)` has `4f ^(9)` congiguration, containing five unpaired electrons. `Tm^(3+)` has `4f ^(12)` configuration, containing two unpaired electrons. `Lu^(3+)` has `4f^(14)` configuration does not contains unpaired electrons. |
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| 319. |
For hydroxides of lanthanidesA. basicity decreases from La to LuB. basicity increases from La to LuC. acidity decreases from La to LuD. basicity remains same from La to Lu |
| Answer» Correct Answer - A | |
| 320. |
When `KMnO_(4)` solution is added to hot oxalic acid solution the decoloursitation is slow in the beginning but becomes instantaneous after some time. This is becauseA. `MnSO_(4)`B. `MnO_(2)`C. `MnO_(4)^(2-)`D. `MnO_(3)^(-)` |
| Answer» Correct Answer - A | |
| 321. |
Assertion : `KMn0_(4)` is an oxidising agent in neutral, acidic and alkaline medium. Reason : Equivalent mass of `KMn0_(4)` in acidic medium is 31 ·6.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 - B (b) Correct explanation. `KMn0_(4)` undergoes decrease in oxidation state in neutral acidic and alkaline medium. |
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| 322. |
(i) Write an oxidising reaction of `K_(2)Cr_(2)O_(7)` in the acidic medium. (ii) Write one oxidising reaction of `KMnO_(4)` in the basic medium. |
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Answer» (i) `K_(2)Cr_(2)O_(7)` oxidises `H_(2)S` to a sulphur (yellow ppt.) in the acidic medium. `K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)+3H_(2)S to K_(2)SO_(4) +Cr_(2)(SO_(4)_(3)+7H_(2)O+3S` (ii) `KMnO_(4)` oxidises KI to `KIO_(3)` in the basic medium. `2KMnO_(4)+KI+H_(2)O to 2MnO_(2)+2KOH+KIO_(3)`. |
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| 323. |
Some statements are given below about lanthanides : (A)Their tripositive ions are all coloured (B) With increase in atomic number, their atomic radii increases (C) Gadolinium exhibit oxidation state of +3 only (D) Last electrons in them enters into 5f orbitals Among the aboveA. Only A is falseB. A, B and C are falseC. A, B and D are falseD. B, C and D are false |
| Answer» Correct Answer - C | |
| 324. |
As atomic number increases from 57 to 71 in tripositive lanthanon, the number of unpaired electronsA. increases regularly from 0 to 14B. first increases from 0 to 7 and then fall to zeroC. increases from 0 to 5 and then fall to zeroD. increase from 0 to 6 and then fall all zero |
| Answer» Correct Answer - B | |
| 325. |
Which of the following statement is not correct ?A. Nitrates and sulphates of lanthanide are water solubleB. Nitrates and sulphates of actinides are water solubleC. Nitrates and sulphates of lanthanide and actinides are water solubleD. Both (a) and (b) |
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Answer» Correct Answer - C Water insoluble |
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| 326. |
Why is HCl not used to make the medium acidic in oxidation reactions of `KMnO_(4)` in acidic medium ?A. Both HCl and `KMnO_(4)` act as oxidising agents.B. `KMnO_(4)` oxidises HCl into `Cl_(2)` which is also an oxidising agent.C. `KMnO_(4)` is a weaker oxidising agent than HClD. `KMnO_(4)` acts as a reducing agents in the presence of HCl. |
| Answer» Correct Answer - b | |
| 327. |
The equivalent weights of `KMnO_(4)` as an oxidising agent in acidic and neutral media will be respectively (M = molecular weight of `KMnO_(4)` )A. M/7 and M/2B. M/5 and M/3C. M/4 and M/5D. M/2 and M/4 |
| Answer» Correct Answer - B | |
| 328. |
Which of the following tripositive ion has half filled 4 f subshell ?A. GdB. AmC. PrD. Yb |
| Answer» Correct Answer - A | |
| 329. |
Calculate the oxidation number of Mn in `KMnO_(4)` molecule.A. `2+`B. `4+`C. `6+`D. `7+` |
| Answer» Correct Answer - D | |
| 330. |
In acidic medium oxidation state of Mn changes fromA. `+7` to `+2`B. `+6` to `+2`C. `+3` to `+2`D. `-5` to `+2` |
| Answer» Correct Answer - A | |
| 331. |
The metal ion which does not form coloured compound isA. CrB. Fe C. ZnD. Mn |
| Answer» Correct Answer - C | |
| 332. |
Which metal common in brass and bronze ?A. ZnB. MgC. CuD. Al |
| Answer» Correct Answer - C | |
| 333. |
In inner transition elements, which f subshell gets filled up with electrons ?A. (n - 1) f subshellB. (n + 1) f subshellC. (n - 2) f subshellD. (n + 2) f subshell |
| Answer» Correct Answer - C | |
| 334. |
Which of the following properties are similar for Ianthan ides and actinides?A. Formation of oxo-ionsB. Tendency to form complexesC. Filling of (n - 2) f orbitalsD. acidic character |
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Answer» Correct Answer - C Lanthanide 4f orbital, Actinide - 5f orbital both are (n-2) f orbital |
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| 335. |
Which of the following properties are different for Ianthan ides and actinides?A. Their radioactive natureB. Their tendency to form oxo-ionsC. Their tendency to form complexesD. All of them |
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Answer» Correct Answer - D Comparison between lanthanide and actinides |
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| 336. |
The complex ion which has no `""d""` electrons in the central atom is `(at. Cr=24, Mn=25, Fe=26,Co=27)`A. `[MnO_(4)]^(-)`B. `[Co(NH_(3))_(6)]^(3+)`C. `[Fe(CN)_(6)]^(3-)`D. `[Cr(H_(2)O)_(6)]^(3+)` |
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Answer» Correct Answer - A `{:(overset(+7)(Mn)O_(4)^(" "-)),(Mn=(Ar)4s^(@)3d^(@)):}` |
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| 337. |
Amongst `Cu^(+1), Fe ^(+2)and Cr^(+3)` (At. No. `Cu =29, Fe =26, Cr= 24`)A. `Cu^(+1)` is colourless, `Fe ^(+2) and Cr ^(+3)` are colouredB. all are colouredC. all are colourlessD. only `Cr^(+3)` is coloured, `Cu^(+1) and Fe ^(+2)` are colourless |
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Answer» Correct Answer - A `Cu^(+) ` has `d ^(10)` configuration. Hence it has no d-d tranition hence `Cu^(+)` is colourless `Fe ^(2+)and Cr ^(3+)` have `d^(6)and d^(3)` configuration respectively. They contains unpaired electrons in d-orbitals. Hence these are coloured. |
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| 338. |
Amongst `[TiE_(6)]^(2-), [CoF_(6)]^(3-), Cu_(2)Cl_(2)` and `[NiCl_(4)]^(2-)` [Atomic no. `Ti = 22, Co = 27, Cu = 29, Ni = 28`] the colourless species are : (A) `[TiF_(6)]^(2-)` and `[Cu_(2)Cl_(2)]` (B) `Cu_(2)Cl_(2)` and `[NiCl_(4)]^(2-)` (C) `[TiF_(6)]^(2-)` and `[CoF_(6)]^(3-)` (D) `[CoF_(6)]^(3-)` and `[NiCl_(4)]^(2-)`A. `COF_(6)^(3-) and NiCl_(4)^(2-)`B. `TiF_(6)^(2-) and COF_(6)^(3-)`C. `Cu_(2)Cl_(2) and NiCl_(4)^(2-)`D. `TiF_(6)^(2-) and Cu_(2)Cl(2)` |
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Answer» Correct Answer - D Transition metal ions having unpaired electrons are coloured while those which those which have no unpaired electron are colourless In `TiF_(6)^(2-),Ti^(4+):[Ar]3d^(0),0` unpaired electron, colourless In `Cu_(2)Cl_(2),Cu^(+):[Ar]3d^(10),0` unpaired electrons, colourless In`COf_(6)^(3-),CO^(3+):[Ar]3d^(6),4` In` NiCl_(4)^(2-), Ni^(2+):[Ar]3d^(8),2` unpaired electrons, coloured |
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| 339. |
Maximum oxidation state shown by Mn (At. no. 25) isA. `+7`B. `+8`C. `+6`D. `+4` |
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Answer» Correct Answer - A Maximum oxidation state of Mn is +7 as the electronic configuration is `3d^5 4s^2`. All 3d and 4s can be involve in forming bond, thereby giving oxidation state of +7 |
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| 340. |
The metal ion which does not form coloured compound isA. CrB. FeC. ZnD. Mn |
| Answer» Correct Answer - C | |
| 341. |
Which of the following ion is coloured ?A. `Sc^(3+)`B. `Zn^(3+)`C. `Cu^(1+)`D. `Cr^(3+)` |
| Answer» Correct Answer - D | |
| 342. |
Which of the following ion of Ianthan ides is / are coloured?A. `Ce^(+3)`B. `Tb^(+3)`C. `Er^(+3)`D. Both (b) and (c ) |
| Answer» Correct Answer - D | |
| 343. |
Which of the following transition metal ion is not coloured?A. `Cu^(+)`B. `V^(3+)`C. `Co^(2+)`D. `Ni^(2+)` |
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Answer» Correct Answer - A `Cu^(+)` ion (`3d^(10)` configuration) can exhibit seven (7) variable oxidation states. |
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| 344. |
Amongst `Cu^(+1), Fe ^(+2)and Cr^(+3)` (At. No. `Cu =29, Fe =26, Cr= 24`)A. `Cu^(+1)` is colourless, `Fe^(+2)` and `Cr^(+3)` are colouredB. all are colouredC. all are colourlessD. only `Cr^(+3)` is coloured, `Cu^(+1)` and `Fe^(+2)` are colourless |
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Answer» Correct Answer - A `Cu^(+1)` is colourless as it has no unpaired electron `(3d^10 4s^0) ,Fe^(+2)` is coloured as it has one unpaired electron `(3d^4 4s^0), Cr^(+3)` is coloured as it has three unpaired electron `(3d^3 4s^0)` |
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| 345. |
Which of the following is wrong?A. `K_(2)Cr_(2)O_(7)rarr`OrangeB. `CuSO_(4).5H_(2)Orarr` BlueC. `MnSO_(4)rarr`YellowD. `Cr_(2)(SO_(4))_(3)rarr`Purple |
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Answer» Correct Answer - C `MnSO_(4)` is light pink. |
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| 346. |
The aqueous solutions of the following salts will be coloured in case ofA. `Zn (NO_(3))_(2)`B. `Co(NO_(3))_(2)`C. `CrCl_(3)`D. both `Co(NO_(3))_(2)and CrCl_(3)` |
| Answer» Correct Answer - D | |
| 347. |
Which ion will give colour in the aqueous solution ?A. `Zn^(+2)`B. `Cu^(+1)`C. `Ti^(+4)`D. `Cu^(2+)` |
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Answer» Correct Answer - D `Cu^(2+)` , since all ions contain paired electrons . `Zn^(+2) 3d^10 4s^0` `Cu^(+1) 3d^10 3s^0 ` `Ti^(+4 ) 3d^0 4s^0` |
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| 348. |
Coloured ion among the following isA. `SO_4^(-2)`B. `I^-`C. `Cu^(+2)`D. `Cu^(+1)` |
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Answer» Correct Answer - C Coloured ion is `Cu^(+2)` as it contain one unpaired electron. (Configuration is `3d^9 4s^0`) |
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| 349. |
The aqueous solution of the following salts will be coloured in the case ofA. `Zn (NO_(3))_(2)`B. `Co(NO_(3))_(2)`C. `Cu(NO_(3))_(2)`D. both b and c |
| Answer» Correct Answer - D | |
| 350. |
The highest oxidation state of `Mn` is shown byA. `KMnO_4`B. `MnO_2`C. `Mn_2O_3`D. `K_2MnO_4` |
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Answer» Correct Answer - A Maximum oxidation state of Mn is in `KMnO_4` (+7) `{:(+1 +7-8,+4-4,+4" "-6,+2" "+6""-8),(K " " M" " O_4, Mn " " O_2 , Mn_2 " " O_3 , K_2 " " Mn" "O_2),(+1 " "x -2 , x " "-2 , 2x " "-2 , +1 " "x " "-2):}` |
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