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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.
| 51. |
Which of the following is not isostructural with `SiCI_(4)` ?A. `NH_(4)^(+)`B. `SCl_(4)`C. `SO_(4)^(2-)`D. `PO_(4)^(3-)` |
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Answer» Correct Answer - 2 Isostructural means same shape and geometry. `|{:("Species","Steric Number","Geometry","Shape"),(SiCl_(4),4+0,"Tehrahedral","Tetrahedral"),(NH_(4)^(+),4+0,"Tetrahedral","Tetrahedral"),(SCl_(4),4+1,"Trigonal bipyramidal","See-Saw"),(SO_(4)^(2-),4+1,"Tetrahedral","Tetrahedral"),(PO_(4)^(3-),4+1,"Tetrahedral","Tetrahedral"):}|` This question is also creided to chemcial bonding. |
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| 52. |
Which of the following is incorrect about graphite ?A. For electrodes in batteriesB. Crucibles made from graphite are used for its inertness to dilute acids and alkaliesC. For adsorbing poisonous gasesD. Lubricant at high temperature. |
| Answer» Correct Answer - C | |
| 53. |
The interlayer distance in graphite isA. the same as the covalent radius of carbonB. more than twice the covalent radius of carbonC. many times larger than the covalent radius of carbonD. very short, and the layers are tightly packed |
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Answer» Correct Answer - B The distance between the layers is `335` pm. It is appreciably more than twice the covalent radius of `C (2 xx 154 p m = 308 p m)`. |
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| 54. |
On heating with concentrated NaOH solution in an inert atmosphere of `CO_(2)`, white phosphorus gives a gas. Which of the following statements is incorrect about the gas ?A. It is more basic than `NH_(3)`B. Its solution in water decomposes in the presence of light.C. It is less basic than `NH_(3)`D. It is highly poisonous and has smell like rotten fish. |
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Answer» Correct Answer - A `P_(4)+3NaOH+3H_(2)O to 3NaHPO_(2)+PH_(3)` `PH_(3)` is less basic than `NH_(3)` because due to bigger size of phosphorus atom than that of nitrogen, electron density of the lone pair is less in `PH_(3)` than in `NH_(3)`. Therefore, `PH_(3)` is a weaker Lewis base than `NH_(3)`. |
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| 55. |
When chlorine reacts with cold and dilute solution of sodium hydroxide, the products obtained areA. `Cl^(-) " and " ClO^(-)`B. `Cl^(-) " and " ClO_(2)^(-)`C. `ClO^(-) " and " ClO_(3)^(-)`D. `ClO_(2)^(-) " and " ClO_(3)^(-)` |
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Answer» Correct Answer - A `Cl_(2) + 2NaOH overset("Cold") to NaCl+NaOCl+H_(2)O`. |
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| 56. |
When chlorine reacts with cold and dilute solution of sodium hydroxide, the products obtained areA. `Cl^(-) " and " ClO^(-)`B. `Cl^(-) " and " ClO_(2)^(-)`C. `Cl^(-) " and " ClO_(3)^(-)`D. `Cl^(-) " and " ClO_(4)^(-)` |
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Answer» Correct Answer - A `Cl_(2)+2NaOH to NaCl+NaClO+H_(2)O` |
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| 57. |
In which of the following arrangements the given sequence is not strict according to the property indicated against it?A. `HF lt HCI lt HBr lt HI` : increasing acidic strengthB. `H_(2)O lt H_(2)S lt H_(2)Se lt H_(2)Te` : increasing `pK_(a)` valuesC. `NH_(3) lt PH_(3) lt AsH_(3) lt SbH_(3)` : increasing acidic characterD. `CO_(2) lt SiO_(2) lt SnO_(2) lt PbO_(2)` : increasing oxidising power. |
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Answer» Correct Answer - B The acidic strength of hydrides in group 16 elements increases with the increase the size of the element. This means that `pK_(a)` is to be in decreasing order. The increasing `pK_(a)` order is the reverse `H_(2)Te lt H_(2)Se lt H_(2)S lt H_(2)O`. |
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| 58. |
Asseration: A fresh stain of iodine is washed with hypo solution. Reason: Hypo is a bleaching agent and it reduces `I_(2)` to `I^(-)`.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 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 - A Reason is the correct explanation for assertion. |
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| 59. |
In diboraneA. Four bridged hydrogen atoms and two terminal hydrogen atoms are presentB. Two bridged hydrogen atoms and four terminal hydrogen atoms are presentC. three bridged hydrogen atoms and three terminal hydrogen atoms are presentD. there are no bridged hydrogen atoms in diborane, only hydrogen bonds are present. |
| Answer» Correct Answer - B | |
| 60. |
Noble gases are sparingly soluble in water due toA. London forcesB. Dipole-dipole interactionC. dipole-induced dipole interactionD. lon-dipole interaction |
| Answer» Correct Answer - C | |
| 61. |
Which is correct for the structure of diborane ?A. `[BH_(2)]^(+)[BH_(4)]^(-)`B. `h*underset(H)underset(|)overset(H)overset(|)(B)-underset(H)underset(|)overset(H)overset(|)(B)*H`C. D. |
| Answer» Correct Answer - C | |
| 62. |
Assertion (A) : If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge Reason (R) : Aluminium is trivalent while silicon is tetravalent.A. Both A and R are correct and R is the correct explanation of AB. Both A and R are correct but R is not the correct explanation of AC. Both A and R are not correctD. A is not correct but R is correct |
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Answer» Correct Answer - d Assertion is not correct but reason is correct. Aluminium is trivalent whereas silicon is tetravalent. If aluminium atom replaces a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires neutrally. |
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| 63. |
Dry `SO_(2)` does not bleach dry flowers becauseA. nascent hydrogen responsible for bleaching is produced only in presence of moistureB. water is the actual reducing agent responsible for bleachingC. water is stronger acid than `SO_(2)`D. the `OH^(-)` ions produced by water cause bleaching. |
| Answer» Correct Answer - A | |
| 64. |
Why is sulphur dioxide considered as an air pollutant?A. It increases the temperature of the atmosphere.B. It is used as insecticide which causes air pollution.C. It causes acid rain due to formation of sulphuric acid on combining with `O_(2) and H_(2)O`.D. It is a strong oxidising agent hence oxidises the other components of air. |
| Answer» Correct Answer - C | |
| 65. |
Match the Column I with Column II and mark the approprite choice. `{:("Column I","Column II"),((A)"Galena",(i)"Abrasive"),((B)"Diamond",(ii)"Metal carbonyls"),((C )"Carbon monoxide",(iii) "Hydrides of Si"),((D) "Silanes",(iv) "An ore of lead"):}`A. (A) `to` (iv), (B) `to` (ii), (C ) `to` (i), (D) `to` (iii)B. (A) `to` (iv), (B) `to` (i), (C ) `to` (ii), (D) `to` (iii)C. (A) `to` (ii), (B) `to` (i), (C ) `to` (iii), (D) `to` (iv)D. (A) `to` (i), (B) `to` (ii), (C ) `to` (iii), (D) `to` (iv) |
| Answer» Correct Answer - B | |
| 66. |
Which of the following halogens is the most reactive?A. `Cl_(2)`B. `Br`C. `I_(2)`D. `F_(2)` |
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Answer» Correct Answer - 4 Fluorine is the nost reactive of all the elements in the periodic table. It reacts all the other elements except the lighter noble gars `(He, Ne and Ar)`. It reacts with `Xe` under mild condition to form `Xe` fluorides. Reations of `F_(2)` with many element s are vigorous ,and ofter explosives. In the assive form a few metals such as `Cu, Ni, Fe` and `Al` acquire a protective fluoride coating. However ,if these metals are in powerdered form (with a large suface area) ,or if the reaction mixture is heated ,then the reaction is vigorous. The reactivity of the other halogens decreases in the order `Cl gt Br gt I` .Chlorine and bromine reacts with most of the elements thorugh less vigorously than does fluoride. Iodine is less reacive and does not combine with some elements such as `S` and `Se`. Fluorine and chlorine often oxidize elements further than do bromine and iodine, by this means bringing out ghigher valcances: `PF_(5), PCl_(5), SF_(6)` etc. The great reactivity of fluorine is attributable to two factors: (i) The low bond dissociation enthalpy of the `F-F` bond which results in a low activation energy for the energy. (ii) The very strong bond which are formed. Both of these properties arise form the small size of fluorine. The weak `F-F` bond arises because of repulsion between the lone pair of electons on the two `F` atoms. Strong bonds arise because of the high coordination number `(SF_(6))` and high lattice enthalpy `(CaF_(2))` |
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| 67. |
Match the uses of the metal aluminium given in column I with its properties given in column II and mark the appropriate choice. `{:("Column I","Column II"),((A)"Transmission cable",(i) "High malleability"),((B)"Aircraft body",(ii)"High electrical conductivity"),((C )"Packing industry",(iii)"High thermal conductivity"),((D) "Utensils",(iv) "Light and tough alloys"):}`A. (A) `to` (ii), (B) `to` (i), (C )`to` (iii), (D) `to` (iv)B. (A) `to` (iv), (B) `to` (iii), (C ) `to` (ii), (D) `to` (i)C. (A) `to` (ii), (B) `to` (iv), (C ) `to` (i), (D) `to` (iii)D. (A) `to` (iii), (B) `to` (iv) , (C ) `to` (i), (D) `to` (ii) |
| Answer» Correct Answer - C | |
| 68. |
Arrange the following in increasing order of the property indicated : (a) `H_(3)PO_(3), H_(3)PO_(4), H_(3PO_(2) ("Reducing Character")` (b) `NH_(3), PH_(3), AsH_(3), SbH_(3), BiH_(3) ("Basic Strength")` |
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Answer» (a) `H_(3)PO_(4) lt H_(3)PO_(3) lt H_(3)PO_(2)` (b) `BiH_(3) lt SbH_(3) lt AsH_(3) lt PH_(3) lt NH_(3)`. |
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| 69. |
An oxide X in the normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens, hydrogen and most of acids and metals even at elevated temperature. It is only attacked by HF and NaOH. The oxide X is.A. `SiO_(2)`B. `CO_(2)`C. `SnO_(2)`D. `PbO_(2)` |
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Answer» Correct Answer - A `SiO_(2)` is attacked by HF and NaOH but not by chemicals under normal conditions |
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| 70. |
Which of the following order are correct as per the properties mentioned against each ?A. `As_(2)O_(3) lt SiO_(2) lt P_(2)O_(3) lt SO_(2)` Acid strength.B. `AsH_(3) lt PH_(3) lt NH_(3)` Enthalpy of vapourisationC. `S lt O lt Cl lt F` More negative electron gain ethalpy.D. `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te` Thermal stability. |
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Answer» Correct Answer - A::D are both correct answers. In general greater the electronegativity of the element more is the acidic strength of the oxide. Thermal stability of the hydrides of oxygen family decreases with increase in bond length. |
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| 71. |
Which of the following increasing order is not correct as mentioned in the property with it?A. `HCIO lt HClO_(2) lt HClO_(3) lt HClO_(4)` (thermal stability)B. `HClO_(4) lt HClO_(3) lt HClO_(2) lt HCIO` (oxidising power)C. `F^(-) lt Cl^(-)lt Br^(-)lt I^(-)` (reducing nature)D. `HIO_(4) lt ICl lt I_(2) lt HI` (oxidation number of iodine) |
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Answer» Correct Answer - D Increasing oxidation number of iodine is in the order: `HIlt I_(2) lt ICl lt HIO_(4)` |
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| 72. |
Match the species given in Column I with the properties mentioned in Column II `{:(,"Column-I",,"Column-II"),("A.","BF"_(4)^(-),1.,"Oxidation state of central atom is +4"),("B.","AlCl"_(3),2.,"Strong oxidising agent"),("C.","SnO",3.,"Lewis acid"),("D.","PbO"_(2),4.,"Can be further oxidised"),(,,5.,"Tetrahedral shape"):}` |
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Answer» A. `BF_(4)^(-)` Tetrahedral shape `sp^(3)` hybridisation regular geometry B. `AlCl_(3^(-))` Octet not complete of Al, act as Lewis acid C. `SnO Sn^(2+)` can show +4 oxidation state D. `PbO_(2)` Oxidation state of Pb in `PbO_(2)` is +4. Due to inert pair effect `Pb^(4+)` is less stable than `Pb^(2+)`, acts as strong oxidising agent. |
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| 73. |
Match the species given in Column I with properties given in Column II `{:(,"Column-I",,"Column-II"),("A.","Diborane",1.,"Used as a flux for soldering metals"),("B.","Gallium",2.,"Crystalline form of silica"),("C.","Borax",3.,"Banana bonds"),("D.","Aluminossilicate",4.,"Low melting, high boiling, useful for measuring high temperatures"),("E.","Quartz",5.,"Used as catalyst in petrochemical industries"):}` |
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Answer» A. `BH_(3)` is unstable forms diborane `B_(2)H_(6)` by 3 centre-2 electron bond show banana bond B. Gallium with low melting point and high boiling point makes it useful to measure high temperatures C. Borax is used as a flux for soldering metals for heat, scratch resistant coating in earthenwares D. Alumino silicate used as catalyst in petrochemical industries E. Quartz, is a crystalline form of silica. |
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| 74. |
Which of these is not a monomer for a high-molecular mass silicone polymer?A. `Me_(2)SiCl_(2)`B. `Me_(3)SiCl`C. `PhSiCl_(3)`D. `MeSiCl_(3)` |
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Answer» Correct Answer - B The hydrolysis of `(CH_(3))_(3)SiCl` gives `(CH_(3))_(3)SiOH` which condenses to give a dimer: `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-OH+HO-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-CH_(3)overset(-H_(2)O)rarr` `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-CH_(3)` |
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| 75. |
`NH_(3)+NaOClto NH_(2)-NH_(2)+NH_(4)Cl` To obtain this product in large amount, we should useA. Tap water as mediumB. Glue or gelatine in mediumC. Heavy metal ion in solutionD. by takin NaOCl in excess amount |
| Answer» Correct Answer - B | |
| 76. |
Statement-1: `R_(2)SiCl_(2)` is the starting material for linear polymer of silicone. Statement-2: `R_(2)SiCl_(2)` is the starting material for 3-D polymer of silicone. Statement-3: `R_(3)SiCl` is the starging material for 3-D polymerr of silicone. |
| Answer» Correct Answer - A | |
| 77. |
Which of the following oxides of carbon is known as dry ice?A. Solid `CO`B. Solid `C_(3)O_(2)`C. Solid `CO_(2)`D. Solid `C_(5)O_(2)` |
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Answer» Correct Answer - C Solid `CO_(2)` is called dry ice or cardice because it sublimes directly to the vapor state (without going through the liquid state) at `-78^(@)C` under atmospheric pressure. As a result, unlike ordinary ice `[H_(2)O (s)]`, it does not wet the surface on which it is kept. |
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| 78. |
Choose the correct regarding bond lengthA. `CO^(+) lt CO`B. `NO^(+) lt NO`C. `N lt N_(2)^(+)`D. `All of these |
| Answer» Correct Answer - D | |
| 79. |
Dry ice isA. solid `NH_(3)`B. solid `sO_(2)`C. solid `CO_(2)`D. solid `N_(2+)` |
| Answer» Correct Answer - C | |
| 80. |
Select the correct option regarding the properties of dioxygen?A. Dioxygen never reacts with metals.B. Dioxygen is diamagnetic in nature.C. Combination of dioxygen with other elements is highly exothermic process.D. Dioxygen liquefies at 55 K and freezes at 90 K. |
| Answer» Correct Answer - C | |
| 81. |
A pale blue gas having a characteristic sharp smell restores the colour of blackened lead paintings.The gas isA. `O_(3)`B. `O_(2)`C. `SO_(2)`D. `CI_(2)` |
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Answer» Correct Answer - 1 Black lead sulphide is oxidized to while lead sulphate `{:(O_(3)rarr O_(2) +O"]"xx4),(PbS+4O rarr PbSO_(4)),(bar(PbS+4O_(3) rarr PbSO_(4)+4O_(2))):}` |
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| 82. |
A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with `NH_(3) an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from . . . . . . . . . .A. `-3" to "+3 `B. `-3" to" 0`C. `- 3" to "+ 5`D. `0" to "- 3` |
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Answer» Correct Answer - A `underset("(black)")(MnO_(2))+4HClrarrMnCl_(2)+underset("(greenish yellow gas)")(Cl_(2)uarr)+2H_(2)O` `overset("(-3)")(NH_(3))+underset("(excess)")(3Cl_(2))rarroverset(+3)(NCl_(3))+3HCl` |
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| 83. |
`Cl_(2)+underset((Excess))(F_(2)) overset(573K)to(A)` Shape of compound (A) isA. LinearB. TetrahedralC. Bent T shapeD. Trigonal bipyramidal |
| Answer» Correct Answer - C | |
| 84. |
Statement-1: Molten state of `IF_(5)` is slightly conducting. Statement-2:`IC l` is more reactive than `Cl_(2)`. Statement-3: In molten state of `ClF_(3)`, chlorine may possess `sp^(3)d^(2)` hybridisation.A. FTFB. TFTC. TTTD. FFF |
| Answer» Correct Answer - C | |
| 85. |
Which of the following halogens can oxidise `H_(2)S` to S?A. `Cl_(2)`B. `F_(2)`C. `Br_(2)`D. All of these |
| Answer» Correct Answer - D | |
| 86. |
Oxygen difluoride`(OF)`, a pale yellow gas ,is prepared by passing `F_(2)` intoA. concentrated `NH_(3)` solutionB. concentrated `NaOH` solutionC. dilute `NaOH_(4)` solutionsD. dilute `NaOH` solution |
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Answer» Correct Answer - 4 `2F_(2)+2NaOH(2%) rarr OF_(2)(g)+2NaF+H_(2)O` On the hand ,dioxygen difluoride `(O_(2)F_(2))` an unstable orange -yellow solid, is prepared by passing an electric discharge though a mixture of `F_(2)` and `O_(2)` at a very low pressure and liquids-air temperature. In `O_(2)F_(2)` the `O-O` bond is shorter than the `O-O` bond in `H_(2)O_(2)` |
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| 87. |
Choose the correct regarding heating of S.A. At `160^(@)`, the `S_(8)` rings break and the diradicals so formed polymerise forming long chains of up to a million atomsB. Due to heating viscosity decreases regularlyC. Due to heating viscosity increases regularlyD. at `200^(@)C` all `S_8` molecule break up to `S_(2)`. |
| Answer» Correct Answer - A | |
| 88. |
Choose the correct statement regarding ice.A. Hexagonal rings are present.B. In ice, one oxygen is surrounded by four hydrogen atomsC. In ice, movement of `H^(+)` is faster than waterD. All of these |
| Answer» Correct Answer - D | |
| 89. |
(a) Write the reaction of the thermal decomposition of sodium azide. (b) Why does `NH_3` act as a Lewis base ? |
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Answer» Thermal decomposition of sodium azide gives dinitrogen gas `2NaN_(3) to 2Na + 3N_(2)` |
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| 90. |
Pure oxygen is colourless gas but liquid and solid `O_(2)` is pale blue or blue coloured, this is becauseA. Due to electronic transition triplet to singlet formation is not favourable in gaseous stateB. Density of gas is less than density of liquidC. Singlet state is diamagneticD. `O_(2)` cannot absorb photon in gaseous state. |
| Answer» Correct Answer - A | |
| 91. |
Statement-1: All halogens are coloured. Statement-2: Halogens exist in diatomic gaseous state. Statement-3: `F_(2)` is more reactive than `Cl_(2)`A. TTTB. TFTC. FTTD. TFF |
| Answer» Correct Answer - B | |
| 92. |
Statement-1: `O_(2)F_(2)` is an unstable orange yellow coloured solid. Statement-2: In `O_(2)F_(2)`, bond order of O-F is less than 1.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 correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - B | |
| 93. |
What is the maximum oxidation state of sulphur in its compounds ? |
| Answer» Correct Answer - `+6` | |
| 94. |
`PH_(3)` has lower boiling point that `NH_(3)`. Why ? |
| Answer» Unlike `NH_(3),PH_(3)` is not associated through hydrogen bonding in liquid state. | |
| 95. |
Why are the elements of group 17 generally coloured ? |
| Answer» Because of the excitation of non-bonded electrons in halogen molecules by absorbing radiations from visible region of light. | |
| 96. |
Bond angle in `NH_(3)` is more than in `PH_(3).` Explain. |
| Answer» Bond angle in `PH_(3)(93.6^(@))` is less than in `NH_(3)(107.8^(@))` because nitrogen is more electronegative (3.0) than phosphorus (2.1). As a result, there is a greater repulsion in the shared electron pairs around nitrogen than around phosphorus. | |
| 97. |
Why does `PH_(3)` have lower boiling point than `NH_(3)` ? |
| Answer» Due to the absence of intermolecular hydrogen bonding which is present in `NH_(3)` molecules. | |
| 98. |
Among the following statements which one is incorrect ?A. Nitrogen has the ability to form `p pi - p pi` bonds with itself.B. Bismuth forms metallic bonds in elemental state.C. Catenation tendency is higher in nitrogen when compared with other elements of the same group.D. Nitrogen has higher first ionization enthalpy when compared with other elements of the same group. |
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Answer» Correct Answer - C Phosphorus has the highest catenation tendency as compared to the other members of the group 15. |
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| 99. |
Why is the bond angle in `PH_(3)` molecule lesser than that in `NH_(3)` molecule? |
| Answer» Due to greater repulsion in the shared pairs of electrons around nitrogen than around phosphorus atom. | |
| 100. |
All bonds in the molecule of `PCl_(5)` are not equivalent. Explain. |
| Answer» `PCl_(5)` has trigonal bipyramidal structure and has both axial and equivalent bonds which are not equivalent. | |