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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.
| 251. |
The atomic radii of the elements are almost same of which seriesA. `Li - Be - B`B. `Na - K - Rb`C. `F - C1 - Br`D. `Fe - Co - Ni` |
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Answer» Correct Answer - D `Fe - Co - Ni`. With the increase in the d-electrons, screening effect increases, this counter balances the increased nuclear charge due to increase in atomic number. As a result atomic radii remain practically same after chrmium. |
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| 252. |
Manganese show oxidation state from `+ 2` to `+ 7`. The most oxidizing state known in aqueous solution isA. `+ 7`B. `+ 4`C. `+ 3`D. `+ 2` |
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Answer» Correct Answer - A The most oxidizing state in aq. Solution of `Mn` is `+ 7 (KMnO_(4))` |
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| 253. |
Oxygen stablises higher oxidation state becauseA. It is electronegativeB. Of its tendency to form double bondC. Of small sizeD. Of large size |
| Answer» Correct Answer - B | |
| 254. |
Reduction potential of `Mn^(2+)//M` will depend onA. `IE_(1)+IE_(2)`B. `DeltaH_("atomisation")`C. `Hydration energyD. All of these |
| Answer» Correct Answer - D | |
| 255. |
Most powerful oxidizing property of manganese is shown by which of the following oxidation stateA. `Mn(+5)`B. `Mn(+2)`C. `Mn(+4)`D. `Mn(+7)` |
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Answer» Correct Answer - D Manganese is stronger oxidising agent in `+ 7` oxidising state, e.g., `KKnO_(4)` |
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| 256. |
Which of the following have highest magnetic moment ?A. `Fe^(2+)`B. `Mn^(+)`C. `Fe^(3+)`D. `Fe^(+)` |
| Answer» Correct Answer - B | |
| 257. |
Which of the following elements is an essential element ?A. MtB. CoC. RhD. Ir |
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Answer» Correct Answer - B Group 9 elements are cobalt, rhodium, iridium and Meitnerium. Cobalt is a bluish white, hard metal, and like Fe, it is a magnetic (ferromagnetic) material but, is appueous states, is a strong oxidising agent, it can oxidise even water with the evolution oxygen. Vitamin `B_(12)` has `Co(+3)` at the core of the molecule, surrounded by a ring structure similar to the porphyrin ring. Injections of this vitamin are used in the treatment of pernicious anemia. Cobalt is also involved in some enzyme functioning. `Co(+3)` exhibits unique tendency of forming a large number of coordination complexes especially with N-donor ligands. Cobalt compounds find use in ceramic and paint industries and as cataysts. |
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| 258. |
In which of the following the stabiliy of two oxidation states is correctly represented?A. `Fe^(2+) gt Fe^(3+)`B. `Ti^(3+) gt Ti^(4+)`C. `Mn^(2+) gt Mn^(3+)`D. `Cu^(+) gt Cu^(2+)` |
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Answer» Correct Answer - C `Mn^(2 +) (d^(5))` is more stable than `Mn^(3 +) (d^(4))` because `Mn^(2 +)` has half-filled configuration. |
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| 259. |
Which of the following shall have the highest value of magnetic moment?A. `Zn(II)` ionB. `Mn(IV)` ionC. `Fe(II)` ionD. `Ti(III)` ion |
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Answer» Correct Answer - C (a) Valence shell electron configuration of `Zn (II)` is `3d^(10) 4 s^(0)`. So has no upaired electrons. (b) Valence shell electron configuration to `Mn (IV)` is `3d^(3) 4 s^(0)`. So has 3 uparied electrons. (c ) Valence shell electron configuration of `Fe (II)` is `3d^(6) 4 s^(0)`. So has 4 upaired electrons. (d) Valence shell electron configuration of `Ti (III) 3d^(1) 4 s^(0)`. So has one uparied electrons. So, `Fe(II)` has highest value of magnetic moment as given by `B.M` or `B.M` |
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| 260. |
Which of the following is incorrect ?A. The transition metals react with halogens at elevated temperatures to form halidsB. The reactivity of halogens decreases in the order : `I_(2) gt Br_(2) gt CI_(2) gt F_(2)`C. Metals are usually oxidized to their highest oxidation states to form fluoridesD. The lower oxidation states are stabilised in iodides. |
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Answer» Correct Answer - B The reactivity of halogens decreases in the order, `F_(2) gt CI_(2) gt Br_(2) gt I_(2)` Bonding in fluorides is usually ionic. But the ionic character decreases in chlorides, bromides, and iodides with increasing mass of halogen. For example, `CuF_(2)` is an ionic compound while `CuCI_(2)` and `CuBr_(2)` are covalent forming infinite chains. On account of high active activation energies, high temperature is usually needed to stsrt the reaction between transition metals and halogen but once the reaction starts, the heat of reaction is sufficient to maintain the continuity of the reaction. |
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| 261. |
The atomic number of an element is 22. The highest oxidation state exhibited by it in its compound isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C Highest oxidation state `rarr` no. of `underset (2) S - e^(-) underset (+) +` no . `underset (2) of underset (=4) d - e^(-)` |
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| 262. |
Which is mild oxidising agent?A. `Ag_(2) O`B. `KMnO_(4)`C. `K_(2) Cr_(2) O_(7)`D. `Cl_(2)` |
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Answer» Correct Answer - A `Ag_(2)O` is mild oxidising agent as greater the oxidation number or metal stronger oxidising agent. |
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| 263. |
Silver halides darken in light owing to ------- , a property which is primarily responsible for their use in photography.A. oxidationB. photochemical decompositionC. reaction with `N_(2)`D. photoelectric effect |
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Answer» Correct Answer - B Silver chloried, bromide, and iodide are sensitive to light, and the ready reduction of the silver ion results in a darkening of the solid. (This is why silver compounds and their solutions are stored in dark bottles.) `Ag^(+) (s) + e^(-) rarr Ag (s)` This reaction is the key to the traditional photgraphic procedd. In black - and - white photography, it is simply the impact of light on sensitized silver halide microcrystals that initiartes the production of the negitive image. For color photography, the film is composed of layers, with organic dyes acting as color filters for the AgBr. Thus, only light form a particular spectral region will activate the silver bromide in a specific layer. |
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| 264. |
In photography, sodium, thisulphate is used asA. Complexing agentB. Oxidising agentC. Reducing agentD. None of these |
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Answer» Correct Answer - A In order to make the images permanent, it is necessary to remove the unreduced silver bromide from the surface of the developed film. This operation is called fixing of image. Fixing is done by dipping the developed film or plate in sodium thiosulphate (hypo) solution. The hyposoluiton dissolves the undreduced silver bromide by forming a complex. Thus sodium thiosulphate act as a complexing agent. `AgBr +2Na_(2)S_(2)O_(3)rarr underset(("soluble"))underset("Sodium argentothiosulphate")(Na_(3)[Ag(S_(2)O_(3))_(2)]) +NaBr` |
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| 265. |
Nessleers reagent isA. `K_(2) HgI_(4)`B. `K_(2) HgI_(4) + KOH`C. `K_(2) HgI_(2) + KOH`D. `K_(2) HgI_(4) + Hg` |
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Answer» Correct Answer - B `2 Kl + H_(2)I_(2) rarr underset("Nesslers reagent")underbrace(K_(2) Hgl_(4) + KOH)` |
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| 266. |
Which of the following silver halides is used in photography chiefly for the production of colloidal emulsion plates ?A. AgIB. AgBrC. AgCID. AgF |
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Answer» Correct Answer - A Silver chloried is used in photography chiefly for printing paper for lantern slides, the bromide is used in large quantities for the production of photographic films and plates (light-sensitive materials) and the iodide chiefly for the production of colloidal emulsion plates in photography. A photography plate or film is essentially a layer of an emulsion of a silver halide (most frequntly AgBr) in gelation and water applied to a glass or celluloid sheet. The AgBr with a little iodide is precipitated and the warm mixture is allowed to stand for some time. Teh process is called ripening and making the emulsion more sensitive to light. The emulsion is allowed to cool and set and then washed with water to removed soluble material. It is then melted and applied to glass or celluloid sheet and dried. After the plate or film is ready for use, the following process are undertaken for photography in stages: (1) Exposure: Teh exposure of photographic plate or film to the object results in the photochemical decomposition of Ag halides coated on the film: `Ag^(+) + e^(-) rarr Ag` Consequently submicroscopic particles of Ag are formed whilst the halogen combines with the gelatin (A complex protein formed by the hydrolysis of collagen in animal cartillages and bones, by boiling with water. Soluble in water, the solution has the property of setting to a jelly). (2) Developing: The film is now developed by adding an organic reducing agent when more AgBr is reduced, the rate of reduction depending upon the intensity of illumination during the exposure period. Thus, parts of the film which were most strongly illuminated become the darkest. It is this intensification of the latent image first formed that permits the use of short exposure times and causes Ag halide to occupy their unique position in photography. (3) Fixing: The film is now washed with aqueous solution of sodium thiosulphate which removes the nuchanged Ag halide as a complex `[Ag(S_(2)O_(3))_(2)]^(3-)` . The negative is now ready and can be handled in day-light. In order to make print form the negative, light is allowed to pass through it on the Ag Br containing photographic paper which is then treated in a similar way as the negative. The negative image is thus reversed. |
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| 267. |
Which of the following compound is not coloured ?A. `Na_(2) CuCl_(4)`B. `Na_(2) CdCl_(4)`C. `K_(4) Fe(CN)_(6)`D. `K_(3) Fe(CN)_(6)` |
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Answer» Correct Answer - B `Na_(2) CdCl_(4) -` no uparied electrons. |
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| 268. |
Which of the following silver halides is insoluble in liquor ammonia ?A. AgFB. AgCIC. AgBrD. AgI |
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Answer» Correct Answer - D When we add liqour ammonia, the `Ag^(+)` ions (if present in sufficient concentration) may react with `NH_(3)` molecules to give the liner diammine silver (I) ion: `Ag^(+) (aq) + 2NH_(3) (aq) rarr [Ag(NH_(3))_(2)]^(+)` This tends to shift the following equilibrium `Ag X (s)hArr(aq) + X^(-) (aq)` towads right making AgX soluble. The precipitate of AgCI is very readily soluble, silver bromide is only slightly soluble, and silver iodide is insoluble in dilute ammonia. This happens because AgCI has the highest `K_(sp)`, thereby providing large enough concentration of `Ag^(+)` ions to form complex with `NH_(3)` whilst AgI has the lowest `K_(sp)`, thereby providing a very low conc. of `Ag^(+)` ions which is unable to form complex with `NH_(3)` |
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| 269. |
Assertion : `CuSO_(4).5H_(2)O` on heating to `250^(@)C` losses all the five `H_(2) O` molecules and becomes anhydrous. Reason : All five `H_(2) O` molecules are coordinated to the central `Cu^(2+)` ion.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are ture and reason is not the correct explanationof the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C Assertion : `CuSO_(4). 5H_(2)O ("blue") overset(250^(@)C)rarr CuSO_(4)` (white) `+ 5 H_(2) O uarr`. Anhydrous copper sulphate becomes white because there is no d-d transition of electron in absence of ligands according to crystal field theory. Reason : Four water molecules are coordinated to `Cu^(2+)` and fifth one is hydrogen bonded to two water moleucles and anion, `SO_(4)^(2-)` |
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| 270. |
On adding excess of `NH_(3)` solution to `CuSO_(4)` solution, the dark blue colour is due toA. `[Cu(CH_(3))_(4)]^(++)`B. `[Cu(NH_(3))_(2)]^(++)`C. `[Cu(NH_(3))]^(+)`D. None of the above |
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Answer» Correct Answer - A `CuSO_(4) + 4NH_(3) = [Cu(NH_(3))_(4)]^(+ +) SO_(4)^(- -)` |
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| 271. |
Transition metals combine with halogens at high temperature to form compounds called halides. On account of high activation energy , the reactions require high temperature to start, but once the the reaction is started , the heat of reaction is sufficient to maintain the continuity . Metals in higher oxidation state form flourides as it is the most electronegative element . Flourides are ionic in nature . The chlorides , bromides and iodides have ionic as well as covalent character . Halides of metals is higher oxidation states are relatively unstable and hydrolysed very easily . `DeltaH_(f)` is negative forA. FlouridesB. BromidesC. IodidesD. All of these |
| Answer» Correct Answer - D | |
| 272. |
The highest oxidation state is exhibited by the transition metals with configuration:A. `(n - 1) d^(3) ns^(2)`B. `(n - 1) d^(5) ns^(1)`C. `(n - 1) d^(5) ns^(2)`D. `(n - 1) d^(8) ns^(2)` |
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Answer» Correct Answer - C it show maximum oxidation state equal to `+ 7` because the energy of `(n - 1)` d and ns orbitals are nearly same and thus seven electrons can participate in bonding. |
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| 273. |
If excess of `NH_(4)OH` is added to `CuSO_(4)` solution, it forms blue coloured complex which isA. `Cu (NH_(3))_(4) SO_(4)`B. `Cu(NH_(3))_(2)SO_(4)`C. `Cu (NH_(4))_(4) SO_(4)`D. `Cu (NH_(4))_(2)SO_(4)` |
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Answer» Correct Answer - A `CuSO_(4) + 4NH_(4)OH rarr underset(Blue)([Cu(NH_(3))_(4)]SO) + 4H_(2)O` |
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| 274. |
Transition metals are related to which blockA. s-blockB. d-blockC. p-blockD. None of these |
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Answer» Correct Answer - B d-block elements are known as transition elements. These show variable valency due to their incomplete d-subshell. |
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| 275. |
When metallic copper comes in contact with mositure, a green powdery/pasty coating can be seen over it. This is chemically known asA. copper sulphide-copper carbonateB. copper carbonate-copper sulphateC. copper carbonate-copper hydroxideD. copper Sulphate-copper sulphide |
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Answer» Correct Answer - C `Cu + O_(2) + CO_(2) + H_(2) O rarr Cu (OH)_(2). CuCO_(3)` |
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| 276. |
Which of the following elements is not an actinide?A. CuriumB. CaliforniumC. UraniumD. Terbium |
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Answer» Correct Answer - D Curium, californium, and uranium are actinide elements. |
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| 277. |
The actinides showing `+7` oxidation state are:A. `U, Np`B. `Pu,Am`C. `Np, Pu`D. `Am,Cm` |
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Answer» Correct Answer - C Np and Pu show maximum oxidation number `+ 7`? |
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| 278. |
Which of the following elements is not an actinide?A. TerbiumB. CaliforniumC. UraniumD. Cruium |
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Answer» Correct Answer - A Terbium is lanthanide as it belongs to `4f`-series having configuration `[Xe] 4f^(9) s^(2)`. However the remaining members belong to `5f`-series (actinides). |
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| 279. |
Which of the following is not an actinide?A. CaliforniumB. UraniumC. CuriumD. Rutherfordium |
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Answer» Correct Answer - D Actinoids are from thorium `(Z = 90)` to Lawrencium `(Z = 103)` in the periodic table. So rutherfordium `(Z = 104)` is not an actionoid. |
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| 280. |
Assertion : Copper metal is turned green when exposed to atmospheric `CO_(2)` and moisture. Reason: Copper gets covered with a green layer of basic copper carbonate.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 281. |
Which of the following lanthanide is commonly used?A. LanthanumB. NobeliumC. ThoriumD. Cerium |
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Answer» Correct Answer - D Ce is most commonly used lanthanide, nobelium (No) and Th (thorium) are actinides. |
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| 282. |
Which of the following statements concering lanthanide elements is false ?A. All lanthanides are highly dense metals.B. More characteristic oxidation state of lanthanide elements is +3.C. Lanthanides are separated from one another by ion exchange method.D. Ionic radii of tricalent lanthanides stradily increases with increases in the atomic number. |
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Answer» Correct Answer - D Ionic radii of tracalent lanthanides decrease with increase in atomic number. |
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| 283. |
Which of the following statements is not correct?A. `La(OH)_(3)` is less basic than `Lu(OH)_(3)`B. In lanthanide series, ionic radius of `Ln^(3+)` ions decreasesC. `La` is actually an elements of transition series rather than lanthanide seriesD. Atomic radii of `Zr` and `Hf` are same because of lanthanide contraction |
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Answer» Correct Answer - A `La(OH)_(3)` is more basic than `Lu(OH)_(3)`. |
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| 284. |
Consider the following statement, (I) The size of the lanthanide `M^(3+)` ions decreases s the atomic number o `M`increases. (II) Electoronic spectra of lanthanide show very broad bands. (III) As wihth transition metal, coordination number 6 is very common in lanthanide complexes.A. I onlyB. I and IIC. I and IIID. III only |
| Answer» Correct Answer - B | |
| 285. |
Which of the following elements shows maximum number of different oxidation states in its compounds ?A. `Gd`B. `Eu`C. `Am`D. `La` |
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Answer» Correct Answer - C Americuium have `+ 3, + 4, + 5` and `+ 6` oxidation states. |
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| 286. |
Reason of lanthanide contraction isA. negligble sereening effect of f-orbitalB. increasing unuclear chargeC. decreasing nuclear chargeD. decreasing screening effect |
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Answer» Correct Answer - A On account of poor shielding effect of `f`-electrons, nucleus exerts a strong attraction as atomic number increase and thus size of atom decrease. |
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| 287. |
Which of the following elements shows maximum number of different oxidation states in its compounds ?A. AmB. GdC. LaD. Eu |
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Answer» Correct Answer - A Element La shows only one oxidation state `(+3)`, elements Eu exihibits two oxidation states `(+2` and `+3)`, element Gd exihibits only one oxidation state `(+3)` while the element Am exhibits a wide range of oxidation states `(+2, +3, +4, +5` and `+6)` of which `+3` is the most stable one. In general, unlike lanthanoids, actiods exhiit a greater range of oxidation states. |
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| 288. |
What are the characteristics of th transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements? |
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Answer» Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell. |
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| 289. |
Addition of `K_(4)[Fe(CN)_(6)]` solution to `FeCl_(3)` solution givesA. Ferro-ferrocyanideB. Ferro-ferricyanideC. Ferri-ferrocyanideD. Ferri-ferricyanide |
| Answer» Correct Answer - C | |
| 290. |
Which one of the following may be regarded as an organometallic compound ? (1) Nickel tetracarbonly (0) (2) Chlorophyll (3) `K_(3)[Fe (C_(2)O_(4))_(3)]` (4) `[Co (en)_(3)]CI_(3)`A. (i) and (iii)B. (i), (ii), (iii) and (iv)C. (i), (ii), and (iii)D. (i) and (ii) |
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Answer» Correct Answer - D Organometallic compounds are those in which there is a direct bonding between metal and carbon. |
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| 291. |
Many lanthanides are used to prepareA. water softenersB. ceramic materialsC. superconducting materialsD. enzyme catalyst. |
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Answer» Correct Answer - C Lanthanide elements are present in warm superconductors such as `La(2 - x)` `Ba_(x)CiuO(4 - Y)`, and `Yba_(2)Cu_(3)O_(7-x)` and others (Sm, Eu, Nb, Dy and Yb). |
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| 292. |
Cinnabar is an ore ofA. ZnB. CuC. HgD. Cr |
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Answer» Correct Answer - C Mercury ores are rare. It occurs in small quantities in the native state. Cinnabar, HgS, is the most important ore of mercury. |
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| 293. |
Which one of following statements is incorrect ?A. Metal iron is corroded readily in noist airB. Argentite is an ore of silverC. oxidation state of Cr in `K_(2)Cr_(2)O_(7)` is `+12`D. Photographic plates and films have an essential ingredient of silver bromide. |
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Answer» Correct Answer - C Oxidation state of Cr in `K_(2)Cr_(2)O_(7)` is `+6`. Argentic or silver glance `(Ag_(2)S)` is an ore of Ag. The photographic plate or film consists of a glass plate or thin strip of celluloid, which is coated with the thin layer of an emulsion of Ag Br dispersed in galatin. |
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| 294. |
Nitriding is a process of hardening steel by treating it inan atmosphere ofA. `NH_(3)`B. `H_(2)S`C. `N_(2)`D. `O_(3)` |
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Answer» Correct Answer - A The process of producing a hard coating of iron nitride on surface of steel is called nitrding. To achieve this, steel is heated in the atmosphere of dry `NH_(3)` at `500 - 600^(@)C` for about 3 to 4 days when a hard coating of iron nitride is produced on the surface. |
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| 295. |
Which method of parification is represented by the following equation ? `Ti(s) + 2I_*(2)(g) rarr TiI_(4)(g) rarr Ti(s) + 2I_(2)(g)`A. CupellationB. van Arkel processC. Electrolytic refiningD. Zone refining |
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Answer» Correct Answer - B This scheme take place in van Arkel process by this process ultrapure metal is prepared, the impure metal is first converted into a volatile stable compound generally iodide leaving behind the impurities which is then decomposed at a higher temperature to give the pure metal. Metals like titanium, zirconium are purified by this method. |
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| 296. |
Which of the following sulphides when heated strongly in air gives the corresponding metal?A. `Cu_(2)S`B. `HgS`C. `Fe_(2) S_(3)`D. `FeS` |
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Answer» Correct Answer - B `HgS` on strong heating gives `Hg`. |
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| 297. |
Which one of the following does not correctly represent the correct order of the property indicated against it ?A. `TI lt V lt Mn lt Cr` : increasing 2 nd ionization enthlpyB. `Ti lt V lt cr lt Mn` : increasing number of oxidation statesC. `Ti^(3+) lt V^(3+) lt Cr^(3+) lt Mn^(3+)` : increasing magnetic momentD. `Ti lt V lt Cr lt Mn` : increasing melting points |
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Answer» Correct Answer - D The correct order of melting points is `Mn lt Ti lt V lt Cr` |
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| 298. |
The first ionisation energies of the elements of the transition series.A. increases as the atomic number increase.B. decrease as the atomic number increase.C. do not show any change as the addition of electrons takes place in the inner `(n-1)` d-orbitals.D. increase from `Ti` to `Mn` and then decrease from `Mn` to `Cu`. |
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Answer» Correct Answer - A First ionisation energies of `3d`-series (first transition series) increase with increase in atomic number due to `(A)` increase in nuclear charge `(B)` decrease in atomic size. |
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| 299. |
How would you account for the irregular of ionisation enthalpies (first) in the first series of the transition elements? |
| Answer» Irregular variation of ionisation enthalpies is mainly attributed to varyingdegree of stability of different 3d-configurations (e.g., `d^(0),d^(5),d^(10)` are exceptionally stable). | |
| 300. |
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 invreasesC. do not show any change as the addition of electrons takes place in the inner (n- 1) d orbitalsD. increases from Ti to Mn and then decreases from Mn to Cu. |
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Answer» Correct Answer - A First ionisation energy of the elements of first transition series `(Ti to Cu)` increases as the atomic number increases due to increases of nuclear charge. The increase however is not regular. |
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