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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. |
Higher oxidation state of manganese in fluoride is `+4 (MnF_(4))` but highest oxidation state in oxides is `+7(Mn_(2)O_(7))` becauseA. fluorine is more electronegative than oxygenB. fluorine does not possess d-orbitalsC. fluorine stabilises lower oxidation stateD. in covalent compounds, fluorine can form single bond only while oxygen forms double bond. |
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Answer» Correct Answer - D Oxygen can form `p pi-p pi` bonding while fluorine can form only single bonds. |
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| 302. |
Although zirconium belongs to 4d transition series and hafnium to 5d transition series even then they show similar physical and chemical properties because …….. .A. both belong to d-blockB. both have same number of electronsC. both have similar atomic radiusD. both belong to the same group of the periodic table. |
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Answer» Correct Answer - C Due to lanthanoid cantraction Zr and Hf have nearly equal size. |
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| 303. |
Why HCl not used to make the mdeium 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 HCl.D. `KMNO_4` acts as a reducing agent in the presence of HCl. |
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Answer» Correct Answer - B In case HCl is used, it is oxidised to `Cl_2`. `2KMnO_4+16HClto 2KCl +2MnCl_2+8H_2O+5Cl_2` |
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| 304. |
All Cu(II) halides are known except the iodide. The reason for it is thatA. iodide is a bulky ionB. `Cu^(2+)` oxidizes iodide to iodineC. `Cu_((aq))^(2+)` has much negative hydration enthalpyD. `Cu^(2+)` ion has smaller size. |
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Answer» Correct Answer - B `2Cu^(2+)+4I^(-)to 2CuI_2` The `CuI_2` immediately decomposes to liberate `I_2` and insoluble copper (I) iodide. `2CuI_2to 2CuI+I_2` |
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| 305. |
All actinods have high desities exceptA. `Th` and `Pa`B. `Am` and `Cm`C. `Th` and `Am`D. `Pa` and `Cm` |
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Answer» Correct Answer - C Factual statement. |
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| 306. |
Assertion: `MnO` is basic whereas `Mn_2O_7` is acidic. Reason: Higher the oxidation state of a transition metal in its oxide, greater is the acidic character.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. |
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Answer» Correct Answer - A Reason is the correct explanatin is assertion. |
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| 307. |
`MnO_(4)^(2-)` on reduction in acidic medium fromsA. `MnO_(4)`B. `Mn^(+ +)`C. `MnO_(4)^(2-)`D. `Mn` |
| Answer» Correct Answer - B | |
| 308. |
One of the following metals froms a volatile corbony1 compound and this property is taken advantage of its extraction. This metal isA. IronB. NickelC. CobaltD. Tungsten |
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Answer» Correct Answer - B Nickel, `Ni + 4CO rarr [Ni(CO)_(4)] "(volatile")` |
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| 309. |
`Zn` and `Hg` belong to same group, they differ in many of their properties. The property that is shared by both isA. They form oxide readilyB. They react with steam readilyC. They react with out concentrated sulphuric acidD. They react with hot sodium hydroxide |
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Answer» Correct Answer - A They form oxide readily. |
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| 310. |
The main reason for not using a mercury electrolytic cell in `NaOH` manufacture is thatA. `Hg` is toxicB. `Hg` is a liquidC. `Hg` has a high vapour pressrueD. `Hg` is a good conductor of electricity |
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Answer» Correct Answer - D `Hg` is a good conductor of electricity. |
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| 311. |
Mercury is transported in metal containers made ofA. SilverB. IronC. LeadD. Aluminium |
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Answer» Correct Answer - B Iron because mercury does not form amalgam with iron. |
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| 312. |
Which of the following statements is wrong ?A. `Ti^(4+) and Ag^(+)` are repelled by magnetic field.B. `Mn^(2+)` shows maximum magnetic character among the first transition series.C. `Fe^(2+)` is more stavle than `Mn^(2+)` towards oxidation to + state.D. Cr in `Cr_2O_7^(2-)` ion involves `sp^3d^2` hydridisation. |
| Answer» Correct Answer - C | |
| 313. |
Mercury is a liquid metal becauseA. it has a completely filled d-orbital that prevents d-d overlapping of orbitalsB. it has a completely filled d-orbital that causes d-d overlappingC. it has a completely filled s-orbitalD. it has a small atomic size. |
| Answer» Correct Answer - A | |
| 314. |
Compare the chemistry of actinoids with that of the lanthanoids with special reference to: |
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Answer» (i) Electronic configuration The general electronic configuration for lanthanoids is `[Xe]^(54) 4f^(0-14) 5d^(0-1) 6s^(2)` and that for actinoids is `[Rn]^(86) 5f^(1-14) 6d^(0-1) 7s^(2)`. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent. (ii) Oxidation states The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state. (iii) Atomic and lonic sizes Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals. 4. Chemical reactivity In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer). |
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| 315. |
Match the column I with column II and mark the appropriate choice. A. `Ato iv, B to ii,Cto iii,Dtoi, Eto v`B. `Ato ii, B to v,Cto i,Dtoiii, Eto iv`C. `Ato v, B to iii,Cto iv,Dtoii, Eto i`D. `Ato iii, B to iv,Cto ii,Dtov, Eto i` |
| Answer» Correct Answer - D | |
| 316. |
Actinoids in general show more oxidation states than the lanthanoids. The main reason for this isA. higher energy difference between 5f and 6d orbitals than between 4f and 5d orbitalsB. lower energy difference between 5f and 6d orbitals than between 4f and 5d orbitalsC. higher reactivity of actinoids than lanthanoidsD. actinoids are more basic than lanthanoids. |
| Answer» Correct Answer - B | |
| 317. |
Lanthanides and actinides resemble inA. electronic configurationB. oxidation stateC. ionization energyD. fromation of complexes |
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Answer» Correct Answer - A Electric configuration lanthanide `4f^(1-14) 5d^(1) 6 s^(2)` and electronic configuration of actionid `5f^(1-14) 6d^(1), 7s^(2)`. |
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| 318. |
The main reason for larger number of oxidation state exhibited by the actinides than the corresponding lanthanides, isA. more energy difference between `5f` and 6 d-orbitals than between `4f` and `5d`-orbitalsB. lesser energy difference between `5f` and `6d`-robitals than between `4f` and `5d`-orbitalsC. larger atmoic size of actinides than the lanthanidesD. greater reactive nature of the actinides than the lanthanides |
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Answer» Correct Answer - B The main reason for large number of oxidation states exhibited by actinides than corresponding lanthanides is lesser energy difference between `5f` and `6d` orbitals than between `4f` and `5d` orbitals. |
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| 319. |
Lanthanides and actinides resemble inA. electronic configurationB. oxidation stateC. inoization erergyD. formation of complexes |
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Answer» Correct Answer - B The oxidation state in both (lanthanide and actinide) is `+ 3`. The properties of actinide are very similar ot those of lanthanide when both are in ` + 3` state. |
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| 320. |
The main reason for larger number of oxidation state exhibited by the actinides than the corresponding lanthanides, isA. lesser energy difference between `5f` and `6d` orbitals than between `4f` and `5d`-orbitalsB. larger atomic size of actinides than the lanthanidesC. more energy difference between `5f` and `6d` orbitals than between `4f` and `5d`-orbitalsD. greater reactive nature of the actinides than the lanthanides |
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Answer» Correct Answer - A A both `5f` and `4f` orbitals have similar shapes and they resemble in their angular part of the wave functions. But `4f` orbitals is much more penetrated than `5f` orbital towards nucles. |
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| 321. |
Match the column I with column II and mark the appropriate choice. A. `Ato i, B to iii,Cto ii,Dtoiv`B. `Ato ii, B to iii,Cto iv,Dtoi`C. `Ato iv, B to i,Cto iii,Dtoii`D. `Ato iii, B to ii,Cto i,Dtoiv` |
| Answer» Correct Answer - B | |
| 322. |
Lanthanides and Actinides generally differ inA. Oxoion formationB. Radioactive natureC. Tendency towards complex formationD. All of these |
| Answer» Correct Answer - D | |
| 323. |
Lanthanides and actinides resemble inA. electroic configurationB. oxidation stateC. ionization energyD. formation of complexes |
| Answer» Correct Answer - A | |
| 324. |
The second and third row elements of transition maetals resemle each other much more than they resemble the first row because ofA. lanthanoid contraction which results in almost same radii of second and third row metalsB. diagonal relationship between second and third row elementsC. similar ionisation elethalphy of second and third row elementsD. similar oxidation states of second and third row metals. |
| Answer» Correct Answer - A | |
| 325. |
Actinoid contraction is greater from element to element than lanthanoid contraction Why? |
| Answer» This is due to poor screening of 5f orbital . | |
| 326. |
The most common oxidation states of cerium areA. `+2` and `+3`B. `+3` and `+4`C. `+2` and `+4`D. `+3` and `+5` |
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Answer» Correct Answer - B The formation of `Ce(+4)` is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common `+3` state. Thus, ceric sulphate, `Ce(SO_(4))_(2)` is most widely used as an oxidizing agent in the volumetric analysis. On heating in oxygen, lanthanides burn easily to give `M_(2)O_(3)` xcept for cerium which form `CO_(2)` as `Ce_(2)O_(3)` is oxidized by `O_(2)` . |
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| 327. |
Which of the following properties form Ce so Lu ?A. Hardness of elementsB. Melting points of elementsC. Boiling points elementD. Basic strength of Ln `(OH)_(3)` |
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Answer» Correct Answer - D The atomic and ionic radii (`Ln^(3+)` ions) of lanthanides decrease with increasing atomic number (lanthanoidcontraction). It occurs because the poor shielding effect of 4f electrons is coupled with increased attraction between the nucleus and the added electrons. The hardness, meltin point and boiling point of elements increase from Ce to Lu because the attraction between the atoms increases as the size decreases. The basic strength of `Ln(OH)_(3)` decreases on account of increasing charge density of `Ln^(3+)` ion due to lanthanide contraction. Thus `Ce(OH)_(3)` is the most basic while `Lu(OH)_(3)` is the least basic. |
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| 328. |
Out of following which is more than basic ` Lu(OH)_(3) and Ce(OH)_(3)` . |
| Answer» Correct Answer - `Ce(OH)_(3)` | |
| 329. |
Why ` Ce(OH)_(3)` is more basic than `Lu(OH)_(3)` ? |
| Answer» Due to small size of `Lu^(3+)Lu(OH)_(3)` is less basic | |
| 330. |
Write the disproportionate reaction of `MnO_(4)^(2-)` ? |
| Answer» `3MnO_(4)^(2-) to 2MnO_(2) +MnO_(4)^(-)+2H_(2)O` | |
| 331. |
When `(NH_(4))_(2)Cr_(2)O_(7)` is heated, the gas evolved isA. `NO_(2)`B. `N_(2)`C. `O_(2)`D. `N_(2)O` |
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Answer» Correct Answer - B `(NH_(4))_(2) Cr_(2)O_(7)` `overset (Delta) rarr` `Cr_(2)O_(3) + 4H_(2)O + N_(2)` |
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| 332. |
ActinidesA. are synthetic elementsB. include element 104C. have short lived isotopesD. exhibit variable valencies |
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Answer» Correct Answer - D All the actinides after U (Z = 92) are synthetic elements. Unlike lanthanides, actides exhibit variable oxidation states. |
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| 333. |
Give the general electronic configuration of actinides.A. `(6-2)f^(1-14)(6-1)d^(1-10)6s^(2)`B. `(6-2)f^(1-14)(6-1)d^(0-1)6s^(2)`C. `(7-2)f^(1-14)(7-1)d^(1-10)7s^(2)`D. `(7-2)f^(1-14)(7-1)d^(0-1)7s^(2)` |
| Answer» Correct Answer - D | |
| 334. |
Lanthanoid which is radioactive isA. PromethiumB. EuropiumC. PlutoniumD. Neptunium |
| Answer» Correct Answer - A | |
| 335. |
Size of lanthanoid decrease becaue of poor screening ofA. 4fB. 3dC. 5fD. 4d |
| Answer» Correct Answer - A | |
| 336. |
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCI, gave a precipitate (P) and a filtrate (Q). The precipitate (P) was found to dissolved in hot water. The filtrate (Q) remained unchanged, when treated with `H_(2)S` in a dilute mineral acid medium. However, it gave a precipitatew (R) with `H_(2)S` in an ammoniacal medium. The precipitate (R) gave a coloured solution (S), when treated with `H_(2)O_(2)` in an aqueous NaOH medium. The precipitate (P) contains ------ while the colored solution (S) containsA. `Pb^(2+)` , `Na_(2)CrO_(4)`B. `Hg_(2)^(2+)` , `CuSO_(4)`C. `Hg^(2+)` , `ZnSO_(4)`D. `Ag^(+)` , `Fe_(2)(SO_(4))_(3)` |
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Answer» Correct Answer - A `Pb^(2+)+2HCl toPbCl_(2)darroverset(Hot)underset(Water)(to)"soluble"` `Cr^(3+)underset("ammoniacal")overset(H_(2)S)toCr(OH_(3))darr` `Cr(OH)_(3)overset(NaOH)underset(H_(2)O_(2))tounderset("Yellow solution") (Na_(2)CrO_(4))` |
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| 337. |
Four successive members of first row transition element are listed belw. Which one of them is expected to have highest `E_((M^(3+))/(M^(2+))^(ɵ)` value?A. `Mn (Z = 25)`B. `Fe (Z = 26)`C. `Co (Z = 27)`D. `Cr (Z = 24)` |
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Answer» Correct Answer - C Simple compounds containing `Co(+3)` are oxidizing and are relatively unstable. However, `Co(+3)` is stable and is very impoertant in complexes. `E_(Cr^(3+)//Cr^(2+))^(@) (-0.41 V)`, `E_(Mn^(3+)//Mn^(2+))^(@) (+1.57 V)`, `E_(Fe^(3+)//Fe^(2+))^(@) (+0.77 V)`, `E_(Co^(3+)//Co^(2+))^(@) (+1.97 V)` |
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| 338. |
Unlike s-block metals, have very high melting and boiling points theyA. are more meltallicB. are more denseC. utilize (n-1) d as well as ns electrons for boilingD. have higher ionization enthaplies |
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Answer» Correct Answer - C The high melting and boiling points of d-block elements are attributed to the stronger forces that bind their atoms together. Besides metallic bonding through the use of ns electrons, d-block metals engage themselves in convalent bonding through the use of their (n-1) d electrons. Therefore, the presence of one or more unpaired (n-1) d electrons contributes to higher interatomic force on account of covalent bonding and therefore, to higher melting and boiling points. Thus, with increasing availability of the unparied d electrons, the strength of the interatomic bonding increases, resulting in higher melting points. Consequently, when we move across any series in the d-block, the melting point increases, rises to a maximum value near the middle of each series and then it decreases on account of pairing of (n-1) d electrons. However, Mn and Tc have abnormally low melting points on account of their copmplex crystal structures. Among the d-block metals, the melting points of Zn, Cd and Hg are abnormally low due to non availability of unpaired (n-1) d electrons for covalent bonding. |
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| 339. |
Which of the following has the highest melting point in the first row transition elements ?A. VB. MoC. WD. Cr |
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Answer» Correct Answer - D This is associated with the maximum participation of d-electrons in meltallic bonding. The maximum melting point in the second and third row transition occurs with Mo and W respectively. |
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| 340. |
The melting point of copper is higher than that of zinc becauseA. the s, p as well as d- electrons of copper are involved in metallic bonding.B. the atomic volume of copper is higherC. the d - electrons of copper are involved in metallic bondingD. the s as well as d- electrons of copper are involved in metallic bonding. |
| Answer» Correct Answer - C | |
| 341. |
Identify the transition elementA. `1 s^(2), 2 s^(2) , 2 p^(6), 3 s^(6), 3 p^(6), 4 s^(2)`B. `1 s^(2), 2 s^(2), 2P^(6), 3s^(2), 3p^(6), 4 d^(10), 4 s^(2) 4 p^(2)`C. `1 s^(2), 2 s^(2), 2 p^(6), 3 s^(2), 3p^(6), 4 d^(2), 4 s^(2)`D. `1 s^(2), 2 s^(2), 2p^(6), 3 s^(2), 3P^(6), 3p^(6), 3 d^(10), 4 s^(2) 4 p^(1)` |
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Answer» Correct Answer - C We know that transition element are those element which have partially filled d-subshell in their elementary form. Therefore, the genral electronic configuration or d-block element is `(n - 1) d^(1 - 10) ns^(1 - 2)` |
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| 342. |
Assertion : `AgNO_(3)` produces a black stain on the skin. Reason : `AgNO_(3)` is a dye.A. If both the assertion and reason are ture but the reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of the assertionC. If the assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - C `AgNO_(3)` reduces to `Ag` on coming in contact with protein. |
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| 343. |
On strong heating, `AgNO_(3)` produces the gasesA. `NO_(2)` and NOB. NO and `O_(2)`C. `N_(2)O` and NOD. `NO_(2)` and `O_(2)` |
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Answer» Correct Answer - D Silver nitrate (Lunar caustic) is teachnically the mopst important of Ag compounds. It is prepared by dissolving metallic silver in dilute nitric acid: `3Ag + 4HNO_(3)` `overset (Delta) rarr` `3AgNO_(3) + NO + 2H_(2)O` (Dilute) From solution it crystallizes as larger rhombic plates, mp. 482K. It is not hygroscopic and is very soluble is water. The solubility are susceptible to decomposition by light. Thus Ag `NO_(3)` is usually kept in amber-colored battles. When heated it decomposes in two stages: On heating above its melting points, it decomposes to silver nitrite and oxygen: `2 AgNO_(3)` `overset (723K) rarr` `2AgNO_(2) + O_(2)` when heated at red heat, it further decomposes to metallic silver. `2AgNO_(3)` `overset (980K) rarr` `2 Ag + 2NO_(2) + O_(2)` Sliver nitratre is also decomposed by organic matter, such as glucose, paper, skin, and crok. It has also a caustic and destructive effect on organic tissues. In contact with oranic substances it blackens due to decomposition into metallic silver. Thus, it leaves black stains when comes in contant with skin and clothes. It produces burning sensation like the moon luna on skin and thus, clled Lunar caustic. Largequantites of `AgNO_(3)` are used in the production of light sensitive plates, films and papers. in the laboratories it is extensively used as a group reagent as a medicine in nervous diseases. A solution of `AgNO_(3)` is used in marking lines. It is also used in silvering of mirror. |
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| 344. |
The melting points of Cu, Ag and Au follow the orderA. `Cu gt Ag gt Au`B. Cu gt Au gt AgC. Au gt Ag gt CuD. Ag gt Au gt Cu |
| Answer» Correct Answer - B | |
| 345. |
Which of the following is the correct squence of atomic weights of given elements?A. `Fe gt Cogt Ni`B. `Co gt Nigt Fe`C. `Ni gt Co gt Fe`D. `Fe gt Ni gt Co` |
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Answer» Correct Answer - B The atomic weights of `Co, Ni` and Fe are 58.90,58.60,55.85 respectively. Therefore `Co gt Ni gt Fe` is the correct sequence of atoic weight. |
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| 346. |
For making `Ag` from `AgNO_(3)`, Which of the following is used?A. `PH_(3)`B. `AsH_(3)`C. `NA_(2)CO_(3)`D. `NH_(3)` |
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Answer» Correct Answer - A `2AgNO_(3) overset(PH_(3))rarr 2Ag + 2NO_(2) + O_(2)` |
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| 347. |
Correct statement about calomel isA. Ionises as `Hg_(2)^(2+) and 2Cl^(-)` ionsB. Cation is diamagneticC. Used in medicine as purgativeD. With aqueous ammonia it turns black |
| Answer» Correct Answer - A::B::C::D | |
| 348. |
Which of the following species is expected ot show the highest magnetic moment? (At.Nos.: `Cr = 24, Mn = 25, Co = 27, Ni = 28, Cu = 29`)A. `Cr^(2+)`B. `Cu^(2+)`C. `Mn^(2+)`D. `Co^(2+)` |
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Answer» Correct Answer - C Highest magnetic moment depends upon number of uparied electron since `Cr^(2) = 3 d^(4) 4 s^(0), Mn^(2 +) = 3 d^(5) 4 s^(0)` `Cu^(2 +) = 3 d^(9) 4 s^(0), Co^(2 +) = 3 d^(7) 4 s^(0), Ni^(2 +) = 3d^(6) 4 s^(0)` So `Mn^(2 +)` contain maximum number of unparied electron i.e., 5. |
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| 349. |
STATEMENT-1 : Common oxidation states of iron and +2 and +3 in its compound . and STATEMENT-2: Iron can have only +2 and +3 oxidation states in its compounds .A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. 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 FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - C | |
| 350. |
Assertion : Pure iron is not used for making tools and machines. Reason : Pure iron is hard.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. |
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Answer» Correct Answer - C Pure iron is not used for making tools and machines as it is soft. Therefore, cannot be used for this purpose. Assertion is ture but reason is false. |
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