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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.
| 101. |
Write balanced equations for the following: (i) `BD_(3)+LiHrarr` (ii) `B_(2)H_(6)+H_(2)Orarr` (iii) `NaH+B_(2)H_(6)rarr` (iv) `H_(3)BO_(3)rarr` (v) `Al+NaOHrarr` (vi) `B_(2)H_(6)+NH_(3)rarr` |
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Answer» (i) `2BF_(3)+6LiHoverset("Heat")rarrB_(2)H_(6)+6LiF` (ii) `B_(2)H_(6)+6H_(2)Orarr2H_(3)BO_(3)+6H_(2)` (iii) `2NaH+B_(2)H_(6)rarr2Na^(+)[BH_(4)]^(-)` (iv) `2H_(3)BO_(3)overset("Heat")rarrB_(2)O_(3)+3H_(2)O` (v) `2Al+2NaOH+2H_(2)Orarr2NaAlO_(20+3H_(2)` (vi) `3B_(2)H_(6)rarr2B_(30N_(3)H_(6)+12H_(2)` |
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| 102. |
Which of the following oxides reacts with both HCl and NaOH?A. CaOB. ZnOC. `N_2O_5`D. `CO_2` |
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Answer» Correct Answer - B ZnO is amphoteric oxide. Therefore reacts with acid as well as base. |
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| 103. |
Which one of the following reacts with conc. `H_2SO_4`?A. AuB. AgC. PtD. Pb |
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Answer» Correct Answer - B `2Ag + 2H_2SO_4 overset"conc."to Ag_2SO_4 + SO_2 + 2H_2O` Au and Pt - inert , `PbSO_4` is insoluble . |
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| 104. |
Which is not the correct statement ?A. The `S_8` ring is not planarB. Oxygen is more electronegative than sulphurC. `SF_(4)` exist but `OF_(4)` does not existD. `SO_(3)^(-) and SO_(3)^(2-)` both have trigonal planar geometry |
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Answer» Correct Answer - D `SO_(3) ` is trigonal planar due to ` sp^(2)`- hybridised sulphur . `SO_(3)^(2-)` is ` sp^(3)` -hybridised , but pyramidal due to the presence of lone pair of electron. |
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| 105. |
Bromine water reacts with `SO_(2)` to formA. HBr and SB. `H_(2)O and HBr `C. `S and H_(2)O`D. `H_(2)SO_(4) and HBr ` |
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Answer» Correct Answer - D When bromine water reacts with `SO_(2)` , it oxidises `SO_(2)` to sulphuric acid and itself get reduced to HBr . ` underset("Bromine water ")(ubrace(Br_(2)+2H_2O))+SO_(2) to 2HBr + H_2SO_(4)` |
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| 106. |
Write balanced equations : (i) `BF_(3) + LiH rarr` (ii) `B_(2)H_(6) + H_(2)O rarr` (iii) `NaH + B_(2)H_(6) rarr` , (iv) `H_(3)BO_(3) overset(Delta)(rarr)` , (v) `Al + NaOH rarr` , (vi) `B_(2)H_(6) + NH_(3) rarr` |
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Answer» (i) `2BF_(3) + 6LiH rarr underset("Diborane")(B_(2)H_(6))+6LiF` (ii) `underset("Diborane")(B_(2)H_(6)) + 6H_(2)Orarrunderset("Orthoboric acid")(2H_(3)BO_(3)) + 6H_(2)` (iii) `2NaH + B_(2)H_(6)rarr underset( " Sod. boronhydride ")(2Na^(+) [BH_(4)]^(-))` , (iv) `underset(" Orthoboric acid ")(H_(3)BO_(3)) overset(Delta)rarr underset(" Metaboric acid ")(HBO_(2)) + H_(2)O` `underset("Metaboric acid ")(4HBO_(2))overset(Delta)underset(-H_(2)O)rarrunderset("Tetraboric acid ")(H_(2)B_(4)O_(7))overset(Delta)rarrunderset("Boron fluoride ")(2B_(2)O_(3))+H_(2)O` (v) `2Al + 2NaOH + 6H_(2)O rarr underset("Sod. tetrahydroxoaluminate (III) ")(2Na^(+)[Al(OH)_(4)]^(-)) + 3H_(2)` (vi) `B_(2)H_(6) + 2NH_(3) rarr underset(" Borane-ammonia complex ")(2BH_(3). NH_(3))` |
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| 107. |
Which of the following is formed by the action of water on sodium peroxide ?A. `H_2`B. `N_2`C. `O_2`D. `CO_2` |
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Answer» Correct Answer - C `2Na_2O_2 + 2H_2O to 2NaOH + O_2` |
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| 108. |
Which one of the following compounds is a peroxide?A. `NO_2`B. `KO_2`C. `BaO_2`D. `MnO_2` |
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Answer» Correct Answer - C `BaO_2` is peroxide. Oxidation state of oxygen is -1 |
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| 109. |
Which one of the following compounds is a peroxide?A. `NO_2`B. `KO_(2)`C. `BaO_(2)`D. `MnO_(2)` |
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Answer» Correct Answer - C `BaO_(2)` has peroxide linkage `underset("Peroxide linkage")(ubrace([""^(-)O-O^(-)]))Ba^(2+)` |
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| 110. |
Give one example each for a super oxide |
| Answer» `"KO"_(2),"RbO"_(2)` are super oxides. | |
| 111. |
Among the hydrides of chalcogens, which is most acidic and which is most stable ? |
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Answer» `to` Among hydrides of chalcogens, `H_(2)` Te is most acidic. `to` Among hydrides of chalcogens, `H_(2)O` is most stable. |
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| 112. |
Give one example each for a peroxide |
| Answer» `"Na"_(2)O_(2),"BaO"_(2)` are peroxides. | |
| 113. |
Give one example each for a neutral oxide |
| Answer» `"CO,N"_(2)O` are neutral oxides. | |
| 114. |
Give an example of neutral oxide of nitrogen. |
| Answer» Nitrous oxide `(N_(2)O)` and Nitric oxide (NO) are neutral oxides of nitrogen. | |
| 115. |
Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixture ?A. group 2B. groups 2,13and 14C. groups 2 and 13D. groups 2 and 14 |
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Answer» Correct Answer - B Is the correct anwer, Cement also contains (group2),`Al_(2)O_(3)` (group 13) `SiO_(2)` (group 14) |
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| 116. |
`{:(,"Column I",,"Column II"),((A),"An element with +1 stable oxidation state",(p),"lead"),((B),"A neutral oxide",(q),"Dry ice"),((C),"An element with stable +2 oxidation state",(r),"Thallium"),((D),"Solid carbon dioxide",(s),"Carbon monoxides"):}`A. A-p,B-q,C-r,D-sB. A-r,B-s,C-p,D-qC. A-r,B-p,C-q,D-sD. A-q,B-s,C-r,D-p |
| Answer» Correct Answer - B | |
| 117. |
Dry ice isA. Solid `NH_(3)`B. Solid `SO_(2)`C. Solid `CO_(2)`D. Solid `N_(2)` |
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Answer» Correct Answer - C Solid `CO_(2)` is called dry ice. |
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| 118. |
Cement, the important buildig material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixtureA. group 2B. group 2, 13 and 14C. groups 2 and 13D. groups 2 and 14 |
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Answer» Correct Answer - B Cement is mainly a mixture of calcium and aluminium silicates. Thus, it contains elements of group 2 `(Ca)` group `13(Al)` and `Si(14)`. |
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| 119. |
Thermal stability of hydrides of first group elements follows the order :A. `NH_3 gt PH_3 gt AsH_3 gt SbH_3`B. `NH_3 lt PH_3 lt AsH_3 lt SbH_3`C. `PH_3 gt NH_3 gt AsH_3 gt SbH_3`D. `AsH_3 gt NH_3 gt PH_3 gt SbH_3` |
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Answer» Correct Answer - A Stability of hydrides decreases down the group |
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| 120. |
The basic character of hydrides of the `V`-group elements decreases in the orderA. `SbH_(3)gtPH_(3)gtAsH_(3)gtNH_(3)`B. `NH_(3)gtSbH_(3)gtPH_(3)gtAsH+_(3)`C. `NH_(3)gtPH_(3)gtAsH_(3)gtSbH_(3)`D. `SbH_(3)gtAsH_(3)gtPH_(3)gtNH_(3)` |
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Answer» Correct Answer - C `NH_(3)gtPH_(3)gtAsH_(3)gtSbH_(3)` On moving down the group atomic size increases and availability of lone pair decreases. Hence, basic character decreases. |
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| 121. |
The basic character of hydrides of the `V`-group elements decreases in the orderA. `SbH_3 gt PH_3 gt AsH_3 gt NH_3`B. `NH_3 gt SbH_3 gt PH_3 gt AsH_3`C. `NH_3 gt PH_3 gt AsH_3 gt SbH_3`D. `SbH_3 gt AsH_3 gt PH_3 gt NH_3` |
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Answer» Correct Answer - C Basic character of hydrides decreases down the group |
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| 122. |
Which element from group 15 gives most basic compound with hydrogen?A. NitrogenB. BismuthC. ArsenicD. Phosphorus |
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Answer» Correct Answer - A Nitrogen , Basic character decreases down the group |
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| 123. |
Why is Bi(V) a stroger oxidant than Sb(V) ? Or of Bi (V) and Sb (V) which may be a stronger oxidising agent and why ? |
| Answer» Due to inert effect, +3 oxidation state of Bi is moer stable than its +5 oxidation state while +5 oxidation state of Sb is more stable than its +3 oxidation state. Therefore, Bi(V) can more easily accept a pair of electrons and get reduced to form more stable Bi(III) than Sb(V) can accept a pair of electrons. In other works, Bi(V) is a stronger oxidising agent than Sb(V). | |
| 124. |
`Bi(V)` and `Sb(V)` which may be a stronger oxidizing agent and why ? |
| Answer» Bismuth and antimony both belong to the nitriogen family and exhibit the `+5` oxidation state. However, on moving down the group, i.e. from antimony to bismuth, the stability, the stability of the `+5` oxidation state decreases. This is due to inert pair effect. thus, Bi (V) is a stronger oxidant than Sb (V). | |
| 125. |
Which one of the following elements is most metallic?A. BiB. AsC. SbD. P |
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Answer» Correct Answer - A Bi,Metallic character increases down the group |
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| 126. |
Ionic radii (in `Ã…`) of `As^(3+), Sb^(`(3+)` and `Bi^(3+)` follow the orderA. `As^(3+) gt Sb^(3+) gt Bi^(3+)`B. `Sb^(3+) gt Bi^(3+) gt As^(3+)`C. `Bi^(3+) gt As^(3+) gt Sb^(3+)`D. `Bi^(3+) gt Sb^(3+) gt As^(3+)` |
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Answer» Correct Answer - D Ionic radii increases down the group |
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| 127. |
`XeF_6` on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when `XeF_6` reacts with silica. The compound ‘X’ Is:A. (1)`XeO_3`B. (2)`XeF_4`C. (3)`XeF_2`D. (4)`XeOF_4` |
| Answer» Correct Answer - D | |
| 128. |
`F_(2)` is a stronger oxidising agent than `Cl_(2)` |
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Answer» Correct Answer - The electrode potential of `F_(2)` and `Cl_(2)` are : `F_(2)+2e^(-)to2F^(-) ; E^(@)=+2.87 V` and `Cl_(2)+2e^(-)to2Cl^(-) ; E^(@)=1.36 V` Since `E^(@)` for `F_(2)//F^(-)` electrode is higher than of `Cl_(2)//Cl^(-)` electrode, therefore, `F_(2)` is more easily reduced than `Cl_(2)`.In other words, `F_(2)` is a stronger oxidising agent than `Cl_(2)`.For example, `F_(2)` oxidises `Cl^(-)` ions to `Cl_(2)` but `Cl_(2)` does not oxide `F^(-)` ions to `F_(2)`. `F_(2)+2Cl^(-)to2F^(-) + Cl_(2); Cl_(2) + 2F^(- -) ` |
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| 129. |
Hydrolysis of `XeF_4 and CaCN_2` gives respectively:A. (A) `XeO_3 and CaCO_3`B. (B) `XeO_2 and Ca(OH)_2 `C. (C) `XeOF_3 and Ca(OH)_2 `D. (D) `XeOF_2 and CaCO_3` |
| Answer» Correct Answer - A | |
| 130. |
Electron affinity of chlorine is more than F. Inspite of this `F_(2)` is the better oxidising agent . Why ? |
| Answer» SRP of `F_2` is much higher than that of `Cl_2` on account of smaller bond dissociation energy. | |
| 131. |
Does the hydrolysis of `XeF_4` at `-80^(@)C` lead to a redox reaction ? |
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Answer» `XeF_4+H_2O overset(-80^(@)C)to XeOF_2+2HF` The oxidation states of all the elements in the products remain the same as it was in the reacting state.hence, it is a not redox reaction. |
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| 132. |
What is the utility of the clatherate compunds ? |
| Answer» It can to used to separate mixture of inert gases containing say He and Xe Process is much cheaper than say distillation as a means of separation. | |
| 133. |
Why `BiH_3` is strongest reducing agent amongst group `15` hydrides ? |
| Answer» `BiH_(3)` is stronger reducing agent because it has low bond dissociation energy than `SbH_(3)` due to longer bond length. | |
| 134. |
Although chlorine and oxygen have nearly same electronegativity yet only oxygen from hydrogen bond explain it ?A. It is lessB.C.D. |
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Answer» Correct Answer - Oxygen has smaller size than size, it cannot act as central atom in higher oxidation state. |
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| 135. |
For making good quality mirrors, the plates of flint glass are used. These are obtained by floating molten glass over a liquid metal which does not solidify before glass. The metal used can beA. sodiumB. magnesiumC. mercuryD. tin |
| Answer» Correct Answer - C | |
| 136. |
Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphiteA. has molecules of variable molecular masses like polymersB. has carbon atoms arranged in large plate or rings of strongly bonded carbon atoms with weak interplate bondsC. is a non-crystalline substanceD. is an allotropic form of diamond |
| Answer» Correct Answer - B | |
| 137. |
Among `XeO_(2),XeO_(2)F_(2)" and "XeF_(6)`, the molecules having same number of lone pairs on Xe areA. `XeF_(6)" and "XeO_(2)F_(2)`B. `XeO_(3)" and "XeO_(2)F_(2)`C. `XeO_(3)" and "XeF_(6)`D. `XeO_(3), XeO_(2)F_(2)" and "XeF_(6)` |
| Answer» Correct Answer - B | |
| 138. |
Give reasons : (i)Xenon does not form fluorides such as `XeF_(3) and XeF_(5)`.(ii)Out of noble gases, only xenon is known to form established chemicals compounds. |
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Answer» Correct Answer - (i)All the filled orbitals of Xe have paired electrons.The promotions of one, two or three electrons from the 5p-filled orbitals to the 5 d-vacant orbitals will give rise to two, four and six half-filled orbitals respectively.So Xe can combine with even but not odd number of F atoms.Hence, it cannot form `XeF_(3)` and `XeF_(5)`. (ii)Except radon which is radioactive, Xe has least ionization enthalpy among noble gases and hence it readily forms chemical compounds particularly with `O_(2)` and `F_(2)`. |
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| 139. |
Nitrogen and oxygen exist as diatomic but their congeners are `P_(4)` and `S_(8)` respectively because `:`A. phosphorus and sulphur are solids.B. phosphorus and sulphur catenate due to the existance of `d-` orbitals and from strainless structures.C. phosphorus and sulphur polymerise as soon as they are formedD. catenation tendency of P and S is stronger because of the high `P-P` and `S-S` bond energies as compared to `N-N` and `O-O` bond energies |
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Answer» Correct Answer - 4 N and O have ability to form `p pi - p pi` multiple obnds with it self on account of smaller size of atoms. `N-N` and `O-O` bond energies are less on account of repulsion between non`-` bonded pairs of electrons due to smaller size of atoms. `S-S` bond energy `(265 kJ mol^(-1))` is next to `C-C`. |
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| 140. |
Nitrogen can be purified from the impurities of oxides of nitrogen and ammonia by passing throughA. conc. HCIB. alkaline solution of pyrogallolC. a solution of `K_2Cr_2O_7` acidified with `H_2SO_4 `D. a solution of KOH |
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Answer» Correct Answer - D Preparation and purification of `N_2` |
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| 141. |
When `HNO_(3)` is dropped into the palm and washed with water, it turns into yellow. It shows the presence ofA. `NO_(2)`B. `N_(O)`C. NOD. `N_(2)O_(5)` |
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Answer» Correct Answer - A Nitric acid turns the skin yellow because it reacts with protein giving a yellow compound called xanthoprotein. It is because of this reason that nitric acid turns the skin as well as wool yellow. |
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| 142. |
Dinitrogen can be purified from the impurities of `NO` and `NH_(3)` by passing through :A. concentrated `HCl`B. alkaline solution of phrogallol.C. an acidified solutin of potassium dichromate.D. an aqueous solution of `KOH`. |
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Answer» Correct Answer - C `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+Cr(SO_(4))_(3)+H_(2)O+[O],NO+[O]+H_(2)O toHNO_(3)(l)2NH_(3)+H_(2)SO_(4) to (NH_(4))_(2)SO_(4)` |
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| 143. |
Nitrogen can be purified from the impurities of oxides of nitrogen and ammonia by passing throughA. conc. `HCl`B. alkaline sulution of pyrogallolC. a solution of `K_(2)CrO_(7)` acidified with `H_(2)SO_(4)`D. a solution of `KOH(aq.)` |
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Answer» Correct Answer - D Oxides of nitrogen are acidic and are dissolved in `KOH` (alkali). |
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| 144. |
Which of the following phosphorus is most stable?A. RedB. WhiteC. BlackD. All stable |
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Answer» Correct Answer - A Due to the less reactivity. |
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| 145. |
Calcium carbide an heating with dinitrogen at `1100^(@)C`givesA. Calcium cyanideB. calcium cyanamideC. calcium carbonateD. calcium nitrade |
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Answer» Correct Answer - B `CaC_(2)+N_(2)overset(500-600^(@)C)underset(6-8atm)(rarr)CaCN_(2)+Ca` |
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| 146. |
In `NH_(3)` and `PH_(3)`, the common isA. odourB. combustibilityC. basic natureD. none of these |
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Answer» Correct Answer - C `ddot(N)H_(3)` and `ddot(P)H_(3)` both are basic because of the presence of lone pait of electrons. |
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| 147. |
the correct structural formula of hypophosphorous acid isA. B. C. D. |
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Answer» Correct Answer - A `H_(3)PO_(2)` is hypophosphorus acid |
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| 148. |
Which one has the highest percentage of nitrogen?A. UreaB. Ammonium sulphateC. Ammonium nitrateD. Calcium nitrate |
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Answer» Correct Answer - A `NH_(2)CONH_(2):` `%"of"N=("Mass of " N)/("Mass of compound")xx100` `=(28)/(60)xx100=46%` |
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| 149. |
The number of P-P-P bridges in the structure of phosphorus pentoxide and phosphorus trioxide are respectivelyA. 5,5B. 5,6C. 5,6D. 6,6 |
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Answer» Correct Answer - D `P_(2)O_(5) , i.e., P_(4)O_(10) implies "Six" P-O-P` bridges `P_(2)O_(3) , i.e., P_(4)O_(6) implies "Six" P-O-P ` bridges |
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| 150. |
The crystalline form of borax hasA. tetrabnuclear `[B_(4)O_(5)(OH)_(4)]^(2-)` unitB. all boron atoms in the same planeC. equal number of `sp^(2)` and `sp^(3)` hybridized boron atomsD. one terminal hydroxide per boron atom |
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Answer» Correct Answer - A::C::D It contains tetranuclear `[B_(4)O_(5)(OH)_(4)]^(2-)` units having two `sp^(2)` and two `sp^(3)`-hybridized boron atoms. It also contains one terminal hydroxide per boron unit. Thus, option (b) which states that all boron atoms lie in the same plane is wrong since all the boron atoms are not `sp^(2)`-hybridized . Thus, all the remaining (a,c,d) options are correct. |
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