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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.
| 1401. |
The reason for small radius of Ga compared to Al is `"………………."`.A. poor screening effect of d and f orbitalsB. increases in nuclear chargeC. presence of higher orbitalsD. higher atomic number |
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Answer» Correct Answer - A::B Poor shielding of d-electron and increase in nuclear charge. Thus, options (a) and (b) correct. |
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| 1402. |
Which of the following is true for white and red phosphours except that theyA. Are both soluble in `CS_(2`B. Can be oxidised by heating in sirC. Consists of same kind of atomsD. Can be converted into one another |
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Answer» Correct Answer - A White phosphorus is soluble in `CS_(2)` whereas red phosphorus is insoluble in it. |
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| 1403. |
Which of the following is a cyclic phosphate?A. `H_5P_3O_10`B. `H_6P_4O_13`C. `H_5P_5O_15`D. `H_7P_5O_16` |
| Answer» Correct Answer - C | |
| 1404. |
Chile salt petre contains sodium iodate uptoA. `0.02%`B. `0.2%`C. `0.002%`D. `0.5%` |
| Answer» Correct Answer - B | |
| 1405. |
A compound X of boron, reacts with `NH_(3)` on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating `BF_(3)` with Lithium aluminium hydride. The compounds `X` and Y are represented by the formules.A. `B_(2)H_(6), B_(3)N_(3)H_(6)`B. `B_(2)O_(3), B_(3)N_(3)H_(6)`C. `BF_(3),B_(3)N_(3)H_(6)`D. `B_(3)N_(3)H_(6), B_(2)H_(6)` |
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Answer» Correct Answer - A `underset("Diborane (X)")(B_(2)H_(6)) + 6NH_(3) overset(473 K)rarr underset("Inorganic benzene (Y)")(2B_(3)N_(3)H_(6)) + 12H_(2)` `4BF_(3)+3LiAlH_(4)rarr 2B_(2)H_(6) + 3LiF+3AlF_(3)` |
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| 1406. |
Quartz Is extensively used as a piezoelectric material, it contains `"……….."`.A. PbB. SiC. TiD. Sn |
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Answer» Correct Answer - B Quartz is maed up of Si. |
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| 1407. |
Which of the following forces bind together carbon atoms in diamond?A. ionicB. CovalentC. DipolarD. van der waals. |
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Answer» Correct Answer - B Carbon atoms in diamond are linked to each other by covalent bonds. |
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| 1408. |
Assertion: Superoxide ion is paramagnetic whereas peroxide ion is diamagnetic. Reason: Superoxide ion `[O=O]^(-)` has one unpaired electron whereas peroxide ion `[O=O]^(-)` has no unpaired electron.A. If both asseration and reson are true and the reason is the correct explanation of the asseration.B. If both asseration and reason are true ans the reason is the correct explanation of the asseration.C. If asseration is true but reason is false.D. If asseration is false but reason is true. |
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Answer» Correct Answer - A `[O=O]^(-)` has `MO` configuration `sigma1s^(2)sigma**,sigma2s^(2)**2s^(2),sigma2p^(2)pi2p_(x)^(2)pi2p_(y)^(2)pi**2p_(y)^(2)pi**2p_(x)^(2)pi**2p_(y)^(1)` `[O=O]^(2-)` has `MO` configuration `sigma1s^(2)**1s^(2),sigma2s^(2)sigma**2s^(2),sigma2p^(2)pi2p_(x)^(2)pi2p_(y)^(2)pi**2p_(x)^(2)pi**2p_(y)^(2)` |
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| 1409. |
Which of the following leaves no residue on heating ?A. `Pb(NO_3)_2`B. `NH_4NO_3`C. `Cu(NO_3)_2`D. `NaNO_3` |
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Answer» Correct Answer - B `NH_4NO_3` decomposes : `NH_4NO_3 to N_2O + 2H_2O` |
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| 1410. |
A metal `X` on heating in nitrogen gas gives `Y,Y` on treatment with `H_(2)O` gives a colourless gas which when passed through `CuSO_(4)` solution gives a blue colour. `Y` is:A. `Mg (NO_(3))_(2)`B. `Mg_(3)N_(2)`C. `NH_(3)`D. `MgO` |
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Answer» Correct Answer - B `underset((X))(3Mg)+N_(2)to underset((Y))(Mg_(3))N_(2)` `Mg_(3)N_(2)+6H_(2)O to 3Mg(OH)_(2)+2NH_(3)` `H_(2)[PtCl_(6)]+2NH_(3) to (NH_(4))_(2)[PtCl_(6)]darr`(yellow) |
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| 1411. |
Which of the following is not correct ?A. Ammonia is used as refrigerant.B. A mixture of `Ca (CN)_(2)` and C is known as nitrolim.C. A mixture of `Ca(H_(2)PO_(4))_(2)` and `CaSO_(4).2H_(2)O` is known as superphosphate of lime.D. Hydrolysis of `NCl_(3)` gives `NH_(3)` and `HOCl`. |
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Answer» Correct Answer - B A mixture of `CaCN_(2)` (calcium cyanamide ) and C is known as nitrolim. |
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| 1412. |
The oxide which does not act as reducing agent is:A. `SO_(2)`B. `NO_(2)`C. `CO_(2)`D. `C iO_(2)` |
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Answer» Correct Answer - C `CO_(2)` fails to act as reducing agent. |
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| 1413. |
An ammoniacal solution of cuprous chloride `(CuCl)` absorbsA. `CO_(2)`B. `SO_(2)`C. `H_(2)SO_(4)`D. `CO`. |
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Answer» Correct Answer - D Ammoniacal solution of cuprous chloride contains `CuNH_(3)Cl` which absorbs CO to form complex. `CuNH_(3)Cl` which absorbs CO to form complex. `CuNH_(3)Cl+Cor rarr[Cu(CO)NH_(3)]Cl` |
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| 1414. |
Name the crystalline forms of carbon. |
| Answer» Correct Answer - Diamond, Graphite | |
| 1415. |
The compound which on strong heating gives oxygen is :A. `AgNO_(3)`B. `BaO_(2)`C. `Pb(NO_(3))_(2)`D. `CaCO_(3)` |
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Answer» Correct Answer - A::B::C `AgNO_(3)overset(Delta)toAg+NO_(2)+1//2 O_(2) , 2BaO_(2)overset(800^(@)C)to2BaO+O_(2)` `Pb(NO_(3))_(2)overset(Delta)toPbO+2NO_(2)+1//2 O_(2) , CaCO_(3) overset(Delta)to CaO+CO_(2)` |
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| 1416. |
A group 16 element exists in monoatomic state in the metallic lattice. It also exists in two crystalline forms. The metal isA. SB. TeC. PoD. Se |
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Answer» Correct Answer - C Polonium - Crystalline from `alpha` from `beta` (both metallic ) |
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| 1417. |
The structure and hybridisation of Si in `Si(CH_(3))_(4)` isA. Bent and spB. Trigonal and `sp^(2)`C. Octahedral and `dsp^(2)`D. Tetrahedral and `sp^(3)` |
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Answer» Correct Answer - D The structure is tetrahedral and hybridisation of silicon is `sp^(3)` |
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| 1418. |
An iorganic white crystalline compound (A) has a rock salt structure. (A) on reaction with conc. `H_(2)SO_(4)` and `MnO_(2)`, evolves a pungent smelling, greenish-yellow gas (B). Compound (A) gives white ppt. of ( C) with `AgNO_(3)` solution. Compounds (A), (B) and (C) will be respectively:A. `NaCl,Cl_(2)AgCl`B. `NaBr,Br_(2),NaBr`C. `NaCl,Cl_(2),Ag_(2)SO_(4)`D. `Na_(2)CO_(3),CO_(2),Ag_(2)CO_(3)` |
| Answer» Correct Answer - A | |
| 1419. |
Nitrogen gas is prepared :A. by heating ammonium nitrateB. by reacting excess chlorine with liquor ammoniaC. by passing `HNO_(3)` vapours on red hot copper.D. by heating lead nitrate |
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Answer» Correct Answer - C (A)`(NH_(4))NO_(3)` gives `N_(2)O`, (B)chlorine (excess) with liquor `NH_(3)` forms `NCl_(3)` and (D) `Pb(NO_(3))_(2)` gives `NO_(2)` not `N_(2)` on heating. (C )`5Cu`(red hot)+`2HNO_(3) to 5CuO + N_(2) uarr +H_(2)O` |
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| 1420. |
Which of the following statement is false ?A. Superoxides give hydrogen peroxide and oxygen with waterB. `CrO_(3)` is an acidic oxide.C. `SnO_(2)` is an amphoteric oxideD. `KO_(2)` is peroxide which with `H_(2)O` forms hydrogen peroxides only. |
| Answer» Correct Answer - D | |
| 1421. |
Which of the following is only acidic in natureA. `Be(OH)_(2)`B. `Mg(OH)_(2)`C. `B(OH)_(3)`D. `Al(OH)_(3)` |
| Answer» `Mg(OH)_(2)` is only basic `Be(OH)_(2)` and `Al(OH)_(3)` are amphoteric while `B(OH)_(3)` is only acidic. | |
| 1422. |
Why does water carried through lead pipes becomes poisonous ? |
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Answer» Lead slowly dissolves in water containing dissolved air forming lead hydroxide which makes water phoisonous . `2Pb + O_(2) + 2H_(2)O rarr 2Pb(OH)_(2)` This dissolution of lead in water is called plumbosolvency. It increase if water contains nitrates, organic acods and ammonium salts. However, presence of salts like phosphates , carbonate and sulphates in water retards plumbosolvency since they from a coating of the corresponding insoluble lead salts on the surface of the lead pipe which protects it from further action. Since hard water contains chlorides, sulphide and carbonates of `Ca^(2+)` or `Mg^(2+)` which from the corresponding insoluble lead salts, therefore, hard water has no action on lead pipes. |
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| 1423. |
Which of the following compound is tribasic acid?A. `H_(3)PO_(2)`B. `H_(2)PO_(3)`C. `H_(3)PO_(4)`D. `H_(4)P_(2)O_(7)` |
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Answer» Correct Answer - C `HO-overset(O)overset(||)underset(OH)underset(|)(P)-OH`it is ionizes in three steps because three `-OH` group are present. |
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| 1424. |
Complete the equation : `[Fe(H_(2)O)_(6)]^(2+)+NOto` |
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Answer» NO reduces `Fe^(2+)" to "Fe^(+)` and itself gets oxidised to nitrosonium ion `(NO^(+))`. The two species then combine to form a brown complex. `[Fe(H_(2)O)_(6)]^(2+)(aq)+NO(g)tounderset(("Brown complex"))underset("Pentaaquanitrosonium iron (I)")([Fe^(+)(H_(2)O)_(5)NO^(+)](aq)+H_(2)O(I))` |
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| 1425. |
Write the formula of the compound of phosphorus which is obtained when conc. `HNO_(3)` oxidises? |
| Answer» `{:(2HNO_(3)toH_(2)O+2NO_(2)+[O]xx10),(P_(4)+10[O]toP_(4)O_(10)),(P_(4)O_(10)+6H_(2)Oto4H_(3)PO_(4)):}/{:P_(4)+20HNO_(3)tounderset("Orthophosphoric acid")(4H_(3)PO_(4))+20NO_(2)+4H_(2)O:}` | |
| 1426. |
Nitric acid can be obtained from ammonia via the formation of the intermediate compoundsA. nitric oxide and nitrogen dioxideB. nitrogen and nitric oxideC. nitric oxide and dinitrogen pentoxideD. nitrogen and nitrous oxide |
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Answer» Correct Answer - A `underset("Ammonia")(NH_(3))overset([O])tounderset("Nitric oxide")(NO)overset([O])tounderset("Nitrogen dioxide")(NO_(2))overset(H_(2)O)toHNO_(3)+NO` |
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| 1427. |
The existence of lower oxidation state than the group oxidation state of group 13 and 14 elements is explained on the basis of `"………….."`. |
| Answer» Correct Answer - inert pair effect | |
| 1428. |
How to boron obtained from borax ? Give chemical equations with reaction conditions. Write the structure of `B_(2)H_(6)` and its reaction with `HCl`. |
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Answer» `underset("Borax")(Na_(2)B_(4)O_(7).10H_(2)O)+H_(2)SO_(4) ("conc".) rarr Na_(2)SO_(4) + underset(" Tetraboric acid ")(H_(2)B_(4)O_(7)) + 10H_(2)O` `H_(2)B_(4)O_(7) +5H_(2)O rarr underset(" Boric acid ")(4H_(3)BO_(3))` `2H_(3)BO_(3) overset(" Red heat ")rarrunderset(" Boric oxide ")B_(2)O_(3) + 3H_(2)O` `B_(2)O_(3)+3Mgoverset(Delta)rarr2B + 3MgO` For structure of `B_(2)H_(6)`. With dry `HCl, B_(2)H_(6)` evolves `H_(2)` gas `B_(2)H_(6) + HCl ("dry") rarr B_(2)H_(5)Cl + H_(2)` |
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| 1429. |
Why does not boron form `B^(3+)` ions ? |
| Answer» Boron has three electrons in the valence shell. Because of its small size and high sum of the first ionization enthalphis (i.e, `Delta_(i) H_(1) + + Delta_(i) H_(2) + Delta_(i) H_(3))`, boron does not lose all its valence electrons to form `B^(3+)` ions. | |
| 1430. |
Which of the following oxidation states are the most characteristic for lead and tin respectively :A. `+2,+2`B. `+4,+2`C. `+2,+4`D. `+4,+4` |
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Answer» Correct Answer - C Due to inert pair effect , +2 state of Pb is more stable while +4 state of tin is more stable, i.e., option (c ) is correct. |
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| 1431. |
Which one of the following pentafluorides cannot be formed?A. `PF_5`B. `AsF_(5)`C. `SbF_(5)`D. `NF_(5)` |
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Answer» Correct Answer - D The +5 oxidation state of N is unstable due to inert pair effect . Thus , `NF_5` cannot be formed . |
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| 1432. |
Aluminium metal is frequently used as a reducing agent for the extraction of metals like Cr,Mn,Fe etc. Explain. |
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Answer» Aluminium has a great affinity for oxygen since it is a reduc ing agent. As aresult, if removes oxygen from the oxdies of less electropositive elements listed above and is. Therefore, used for their extraction. `Cr_(2)O_(3)[email protected]+Al_(2)O_(3)` |
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| 1433. |
Aluminium is a highly electropositive metal but it find extensive use as a structural material. Why? |
| Answer» Aluminium has a tend ency to combine with other metals to form useful alloys. For example the alloy duralumin contain `Al(95%)`, Cu(4%), Mg(0.5%) and Mn (0.5%). It is used in making parts of aeroplanes and automobiles. Moreover, the metal has very high tensile strength wich is necessary in making structure. | |
| 1434. |
Total number of `CN^(-)` ion that are present in ICN(liq.) |
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Answer» Correct Answer - `0003` `ICN(l) rarr(I_(2)(CN)^(+))+(I(CN)_(2))^(-)` |
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| 1435. |
In diborane, the banana bond is formed betweenA. `2` electrons, `3` atomsB. `2` electrons, `1` atomsC. `1` electrons, `2` atomsD. `2` electrons, `2` atoms |
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Answer» Correct Answer - A `3` centred, `2bar(e)` (banana bond) |
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| 1436. |
Inorganic graphite isA. `B_(3)N_(3)H_(6)`B. `B_(3)N_(3)`C. `SiC`D. `Fe(CO)_(5)` |
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Answer» Correct Answer - B Structure of `B_(3) N_(3)` is similar to that of graphite i.e. it has a layer structure consisting of hexagon rings. |
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| 1437. |
The number of chlorine to oxygen bonds in `Cl_2O_7` isA. 7B. 8C. 6D. 10 |
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Answer» Correct Answer - B `Cl_2O_7` is `O=undersetunderset(O)(||)oversetoverset(O)(||)Cl-O-undersetunderset(O)(||)oversetoverset(O)(||)C=O` Thus 8-chlorine links to oxygen |
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| 1438. |
Among the oxo-acid of chlorine the correct order of increasing acid strength isA. `HClO_4 lt HClO lt HCIO_2 lt HClO_3 `B. `HClO_3 lt HClO_2 lt HClO_4 lt HClO`C. `HClO_4 gt HClO_3 gt HClO_2 gt HClO`D. `HClO_4 lt HClO_2 lt HClO lt HClO_3` |
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Answer» Correct Answer - C More oxygen atom more acidic character |
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| 1439. |
Among hypohalous acid the acid with highest Ka value isA. HIOB. HFOC. HClOD. BrOH |
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Answer» Correct Answer - C Greater electronegativity HFO do not exist |
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| 1440. |
The halogen-halogen bond length is longest forA. `F_2`B. `Cl_2`C. `Br_2`D. `I_2` |
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Answer» Correct Answer - D Bond length increases down the group from `F_2` to `I_2` |
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| 1441. |
Which element would readily replace oxygen from an oxide ?A. FluorineB. ChlorineC. NitrogenD. Sulphur |
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Answer» Correct Answer - A Fluorine being most electronegative will readily replace oxygen from an oxide. |
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| 1442. |
Which one of the following is a pseudohalide ?A. `SCN^-`B. `I Cl_2^-`C. `OSN^-`D. `CN^-` |
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Answer» Correct Answer - B `I Cl_2^-` is not a pseudo halide. |
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| 1443. |
Which of the following halic acid and maximum oxidising strength?A. `HClO_3`B. `HC lO`C. `HClO_4`D. `HClO_2` |
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Answer» Correct Answer - C Cl in `HClO_4` has highest oxidation state |
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| 1444. |
Which of the following is the correct statement for red lead?A. It is an active form of leadB. Its molecular formula is `Pb_(2)O_(3)`C. It decomposes into `PbO` and `CO_(2)`D. It decomposes into `PbO` and `O_(2)` |
| Answer» Correct Answer - D | |
| 1445. |
Which statement is not true about `CO`?A. It is colourless gasB. It is an odourless gasC. it is highly soluble in waterD. it is poisnous gas. |
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Answer» Correct Answer - C Statement (c) is false because `CO` has low solubility in water. |
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| 1446. |
Assertion: All clatharate compound of noble gas are the compounds in which the molecules of noble gases are trapped in cavities in te crystal lattice of other compounds. Reason: He and Ne having smaller size do not form clatrate compound molecules are small because are small because their ehough to escape from cavities.A. If both asseration and reson are true and the reason is the correct explanation of the asseration.B. If both asseration and reason are true ans the reason is the correct explanation of the asseration.C. If asseration is true but reason is false.D. If asseration is false but reason is true. |
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Answer» Correct Answer - A Both facts and explanation is correct reason for statement. |
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| 1447. |
Group 17 elements are known asA. HalogenB. Noble gasC. Transition elementD. Metalloids |
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Answer» Correct Answer - A 17-group is called halogens |
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| 1448. |
The number of np-electrons in the outer shell of elements of group 17 isA. FiveB. ThreeC. FourD. Seven |
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Answer» Correct Answer - A The general formula for group 17 is `ns^2 np^5` `therefore np^5`-Five |
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| 1449. |
The electronic structure of four elements `A,B, C, D` are (a) `1s^(2)` (b) `1s^(2), 2s^(2), 2p^(2)` ( c) `1s^(2), 2s^(2), 2p^(5)` (d) `1s^(2), 2s^(2) 2p^(6)` The tendency to form electrovalent bond is largest inA. DB. CC. BD. A |
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Answer» Correct Answer - B `because` It is an halogen with valency -1 |
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| 1450. |
Which of the following sets of element is known as halogens ?A. Li, Na, KB. F, Cl, BrC. He, Ne, ArD. Cu, Ag, Au |
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Answer» Correct Answer - B The element of group 17 are F,Cl, Br , I , At. |
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