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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. |
A gas X is passed through water to form a saturated solution. The aqueous solution on treatement with `AgNO_(3)` gives a white precipitate. The saturated aqueous solution also dissolves Mg ribbon with evolution of colourless gas `Y,X` and Y are respectively:A. `CO_(2),CI_(2)`B. `CI_(2),CO_(2)`C. `CI_(2),H_(2)`D. `H_(2),CI_(2)` |
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Answer» Correct Answer - C `underset((X))(Cl_(2))+H_(2)O rarr HOCl+HCl` `AgNO_(3)+HCl rarr AgCl+HNO_(3)` `Mg+2HCl rarr MgCl_(2)+underset((Y))(H_2)`. |
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| 52. |
The products of the chemical reaction between `Na_(2)S_(2)O_(3)`, `CI_(2)` and `H_(2)O` areA. S,HCI,Na_(2)SO_(4)`B. `S,HCI,Na_(2)S`C. `S,HCI,Na_(2)SO_(4)`D. `S,NaCIO_(3)` |
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Answer» Correct Answer - A `Na_(2)S_(2)O_(3)+H_(2)O+CI_(2)rarr Na_(2)SO_(4)+S+2HCI` |
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| 53. |
The reactions of `CI_(2)` gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two different oxioacids of chlorine, P and Q, respectively. The `CI_(2)` gas reacts with `SO_(2)` gas, in presence of charocal, to give a product R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus. P and Q, respectively, are the sodium salts ofA. Hypochlorous and chloric acidsB. Hypochlorous and chlorous acidsC. Chloric and perchloric acidsD. Chloric and hypochlorus acids. |
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Answer» Correct Answer - A `underset(("Cold+dil".))2NaOH +CI_(2) rarr NaCI + underset(P)NaCIO +H_(2)O` `underset (("Hot+conc."))6NaOH +3CI_(2)rarr 5NaCI+underset(Q)NaCIO_(3) +H_(2)O` |
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| 54. |
Which of the following halides react(s) with `AgNO_(3(aq))` to give a precipitate that dissolves in `Na_(2)S_(2)O_(3(aq))`A. `HCI`B. `HF`C. `HBr`D. `HI` |
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Answer» Correct Answer - A::C::D `AgNO_(3)+HCI rarr AgCI darr` `AgNO_(3)+HBr rarr AgBr darr` `AgNO_(3)+HI rarr AgI darr` All these precipitates will get dissolved in hypo `(Na_(2)S_(2)O_(3))` formaing complex, `Na_(3)[Ag(S_(2)O_(3))_(2)]` |
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| 55. |
The reactions of `CI_(2)` gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two different oxioacids of chlorine, P and Q, respectively. The `CI_(2)` gas reacts with `SO_(2)` gas, in presence of charocal, to give a product R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus. R, S and T, respectively, areA. `SO_(2)CI_(2)`, `PCI_(5)` and `H_(3)PO_(4)`B. `SO_(2)CI_(2)`, `PCI_(3)` and `H_(3)PO_(3)`C. `SOCI_(2)`, `PCI_(3)` and `H_(3)PO_(2)`D. `SOCI_(2)`, `PCI_(5)` and `H_(3)PO_(4)` |
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Answer» Correct Answer - A `SO_(2)+CI_(2)overset(Charcoal)rarr underset(R )SO_(2)CI_(2)` `SO_(2)CI_(2)P_(4)rarr PCI_(5)+SO_(2)` `PCI_(5)+H_(2)O rarr underset(T)H_(3)PO_(4) +HCI` |
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| 56. |
Give balanced chemical reactions for the following: i. Sodium iodate is treated with sodium bisulphite solution. ii. Chlorine is passsed through hot NaOH solution. iii. Chlorine is passed through aqueous potassium hydroxide. iv. Chlorine gas is bubbled through a solution of ferrous bromide. v. Iodine reacts with conc. `HNO_(3)`. vi. Chlorine is passed over slaked lime. vii. Sodium chloride is heated with `K_(2)Cr_(2)O_(7)` and conc. `H_(2)SO_(4)`. viii. Potassium iodide is heated with `MnO_(2)` and conc. `H_(2)SO_(4)`. ix. Chlorine reacts with `Na_(2)SO_(3)` solution. x. Iodine is added to stannous chloride solution. xi. Chlorine is passe through a suspension of idine. xii. `CI_(2)` is passed through a suspension of `CaCO_(3)`. xiii. Chlorine gas is passed through dry and aqueous sulphur dioxide. xiv. Bromine reacts with `Na_(2)CO_(3)` solution. xv. Potassium reacts is added to bleaching powder containing dilute acetic acid. |
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Answer» (i) `2NaIO_(3)+5NaHSO_(3) rarr 3NaHSO_(4)+2Na_(2)SO_(4)+H_(2)O+I_(2)` (ii) `3Cl-(2)+6NaOH(conc.) overset(Hot)(rarr) 5NaCl+NaClO_(3)+3H_(2)O` (iii) `Cl_(2)+2KOH_((aq.)) rarr KCI+KOCl+H_(2)O` (iv) `2FeBr_(2)+3Cl_(2) rarr 2FeCl_(3)+2Br_(3)` (v) `I_(20) +10HNO_(3) rarr 2HIO_(3)+10NO_(2)+4H_(2)O` (vi) `Ca(OH)_(2)+Cl_(2) rarr Ca(OCl)Cl+H_(2)O` (vii) `4NaCl+K_(2)Cr_(2)O_(7)+6H_(2)SO_(4) rarr 2CrO_(2)Cl_(2)+4NaHSO_(4)+2KHSO_(4)+3H_(2)O` (viii) `2KI+MnO_(2)+3H_(2)SO_(4) rarr 2KHSO_(4)+MnSO_(4)+2H_(2)O+I_(2)` (ix) `Cl_(2)+Na_(2)SO_(3)+H_(2)O rarr 2HCl+Na_(2)SO_(4)` (x) `SnCl_(2)+2HCl=I_(2) rarr SnCl_(4)+2HI` (xi) `I_(2)+5Cl_(2)+6H_(2)O rarr 2HIO_(3)+10HCl` (Xii) `2CaCO_(3)+2Cl_(2) rarr CaCl_(2)+Ca(ClO)_(2)+2CO_(2)` (xiii) `ubrace(SO_(2)+Cl_(2))_("Dry") rarr SO_(2)Cl_(2)` `SO_(2)+Cl_(2)+2H_(2)O rarr H_(2)SO_(4)+2HCl` (xiv) `3Br_(2)+underset(("conc. and hot"))(2Na_(2)CO_(3)) rarr 5NaBr+NaBrO_(3)+3CO_(2)` (xv) `CaOCl_(2)+2CH_(3)COOH+2KI rarr (CH_(3)COO)_(2)Ca+2KCl+I_(2)+H_(2)O` . |
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| 57. |
Addition of `Cl_(2)` turns it colourless. Why? |
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Answer» `Cl_(2)` is a stronger oxidising agent than `I_(2)` and hence oxidises `I^(Theta)` to `I_(2)` initialy, which imparts a brown colour to the solution. `Cl_(2)+2KI to 2KCl+underset(("Colourless"))(2HIO_(3))`. |
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| 58. |
When chlorine water is added to an aqueous solution of sodium iodide in the presence of chloroform, a violet colouration is obtained. On adding more of chlorine water and vigorous shakinh, the violet colour disappear. This shows the conversion of …….. into .......A. `I_(2),HIO_(3)`B. `I_(2),HI`C. `HI,HiO_(3)`D. `I_(2),HOI` |
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Answer» Correct Answer - A `3CI_(2)+2NaI rarr 2NaCI+I_(2)` `I_(2)` gives violet colouration in `C CI_(4)`. `5CI_(2)+6H_(2)O+I_(2)rarr underset("Colourless")HIO_(3)+10HCI` |
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| 59. |
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom//atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one `P=O` or `P rarr O` unit and one `P-OH` unit. The `OH` group is ionisable but `H` atom linked directly to `P` is non-ionisabl. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid. Which one has `+ 3` oxidation state ?A. `H_3 PO_4`B. `H_3 PO_3`C. `H_4 P_2 O_7`D. `H_4 P_2 O_6` |
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Answer» Correct Answer - B Let oxidation state of `P` in `H_3 PO_3` is `x(+1) xx 3 + x + (-2) xx 3 = 0, x = + 3`. |
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| 60. |
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom//atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one `P=O` or `P rarr O` unit and one `P-OH` unit. The `OH` group is ionisable but `H` atom linked directly to `P` is non-ionisabl. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid. Which one is monobasic acid ?A. `H_3 PO_2`B. `H_3 PO_3`C. `H_3 PO_4`D. `H_3 PO_5` |
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Answer» Correct Answer - A `H_3 PO_2` contains one `P-OH` group. |
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| 61. |
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom//atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one `P=O` or `P rarr O` unit and one `P-OH` unit. The `OH` group is ionisable but `H` atom linked directly to `P` is non-ionisabl. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid. Which of the following is a cycle oxoacid ?A. `H_4 P_2 O_7`B. `H_4 P_2 O_6`C. `H_3 P_3 O_9`D. `H_5 P_5 O_15` |
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Answer» Correct Answer - C `H_3 P_3 O_9 is (HPO_3)_3`. |
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| 62. |
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom//atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one `P=O` or `P rarr O` unit and one `P-OH` unit. The `OH` group is ionisable but `H` atom linked directly to `P` is non-ionisabl. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid. The acid which forms two series of salts is.A. `H_3 PO_4`B. `H_3 PO_3`C. `HPO_3`D. `H_3 PO_2` |
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Answer» Correct Answer - B Since `H_3 PO_3` is dibasic, it forms two series of salts. |
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| 63. |
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom//atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one `P=O` or `P rarr O` unit and one `P-OH` unit. The `OH` group is ionisable but `H` atom linked directly to `P` is non-ionisabl. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid. The number of or `P = O` and `P - O- H` bonds in `H_3 PO_4` are.A. `3, 1`B. `2, 2`C. `1, 2`D. `1, 3` |
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Answer» Correct Answer - D Number of `P = 0 bonds = 1` Number of `P - OH bonds = 3`. |
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| 64. |
Fluorine does not form oxyacid, but other halogens do. Why? (b) Both `NO` and `ClO_(2)` are odd electron species. `NO` dimeries but `ClO_(2)` does not . Why? |
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Answer» (a) In oxy-acids the central elements always exhibits positive oxidation states. Fluorine , being the most electronegative , never shows positive oxidation states, hence `F` does not form oxy-acids. Other halogens have the tendency to show positive oxidation states and hence form oxy-acids. (b) In `NO` the size of nitrogen atom is small and the odd electrons is attracted by only oxygen atom while in `ClO_(2)` the size of chlorine atom is comparatively large and an odd electron is attracted by two oxygen atoms.As a result, the odd electron on `N` in `NO` is localised while the odd electron on chlorine in `ClO_(2)` is delocalised. Thus, `NO` has a tendency to dimerise but `ClO_(2)` does not. |
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| 65. |
(a) Halogens have maximum negative gain enthalpy in the respective periods of the periodic table. Why? (b) Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine . Why? Fluorine exhibits only-1 oxidation state, wherease other hlaogens exhibit `+1,+3,+5, and +7` oxidation states also. explain. |
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Answer» Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence, they readily accept one electron to acquire noble gas configuration. (b) Fluorine is a stronger oxidising agent than chlorine, due to : (i) Low enthalpy of dissociation of `F-F` bond. (ii) High hydration enthalpy of fluorine. (iii) Fluorine is the most electronegative element and cannot exhibit positive oxidation state. Other halogens have vacant d-orbitals and therefore, can expand their octests and show `+1,+3,+5 and +7` oxidation states also. |
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| 66. |
The negative value of electron gain enthalpy is less for fluorine than for chlorine . Why? |
| Answer» Atomic size of fluorine is less than chlorine. Hence , the interelectronic repulsion, i.e., the electrons already preset in `F` and the electron to be added is much more than the `Ci` atom. Hence, the incoming electron is accepted with diffculty in `F`. The energy released during formation of `F_((g))^(Theta)` from `F_((g))` is less than that of formation of `Cl^(Theta)` from `Cl_((g))`. Thus, negative gain enthalpy of `F` is lower than that of `Cl`. | |
| 67. |
`PCl_5` is kept in well stopered bottles becauseA. It is highly volatileB. It reacts readily with moistureC. It reacts with oxygenD. It is explosive |
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Answer» Correct Answer - B `PCl_5 + 4H_2 O rarr H_3 PO_4 + 5 HCl`. |
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| 68. |
The number of steps, in which orthophosphoric acid is ionised, areA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C (i) `H_3 PO_4 rarr H^(oplus) + H_2 PO_4^(Ө)` (ii) `H_2 PO_4 ^(Ө) rarr H^(oplus) + HPO_4 ^(2-)` (iii) `HPO_4^(2-) rarr H^(oplus) + PO_4^(3 -)`. |
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| 69. |
The noble gases found dissolved in spring water areA. HeB. NeC. KrD. Ar |
| Answer» Correct Answer - a,d | |
| 70. |
Which of the following noble gases do not reasct with functionA. KrB. XeC. HeD. Ne |
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Answer» Correct Answer - a,d Due to very high ionisation enthalpies of He and Na |
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| 71. |
The noble gases which do not form any clatherateA. HeB. NeC. AsD. Kr |
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Answer» Correct Answer - a,b Due to very small size of He and Na, they do not fit into the cavities formed by the bost molucule |
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| 72. |
Noble gases have compleately filled valance shall i.e. `m^(2)sp^(2)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`, `XeF_(4)` and `XeF_(4)` are expected to beA. ReductingB. OxidisingC. InertD. Basic |
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Answer» Correct Answer - b Xemon flacrides are strongly oxidising as `Xe` more stable |
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| 73. |
Noble gases have compleately filled valance shall i.e. `m^(2)sp^(2)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`, Argon is used in are welting due to itsA. FlammabilityB. zeroC. Law reactivityy with metalD. Lower the melting with metal |
| Answer» Correct Answer - c | |
| 74. |
Noble gases have compleately filled valance shall i.e. `m^(2)sp^(2)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`, Structure of `XeF_(4)` isA. LinearB. Square platerC. TetrahedralD. Pyramidel |
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Answer» Correct Answer - b `Xe` in `XeF_(4)` is `sp^(3)d^(2)` hybridised with `2lps` |
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| 75. |
Noble gases have compleately filled valance shall i.e. `m^(2)sp^(2)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`, Oxidetion state of `Xe` in `XeF_(2)` isA. `+2`B. `+4`C. `+6`D. `+8` |
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Answer» Correct Answer - a `XeF_(2) x + (-1) xx 2 = 0 rArr s = +2` oxidation state of `Xe` in `XeF_(2)` is `+ 2` |
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| 76. |
If phosphorous acid is allowed to react with sufficient quantity of `KOH`, the product obtained is.A. `K_3 PO_3`B. `KH_2 PO_3`C. `K_2 HPO_3`D. `KHPO_3` |
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Answer» Correct Answer - C `H_3 PO_3 + 2KOH rarr K_2 HPO_3 + 2H_2 O`. |
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| 77. |
Despite the fact than fluorine is more electronegative than iodine, yet `HF` is less acidic as compared to `HI` . Explain. |
| Answer» Fluorine atom is much smaller in size than iodine, hence `H-F` bond length is less than `H-I` and the bond dissociation energy of `H-F` is very high as compared to `H-I`. As a result, `H-I` bond breaks more readily than `H-F` and `HI` is a stronger acid than `HF`. | |
| 78. |
Assertion (A): Salts of `CIO_(3)^(Theta)` and `CIO_(4)^(Theta)` are well known but those of `FO_(3)^(Theta)` are unknown. Reason (R ): F is more electronegative than O, while CI is less electronegative than O. |
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Answer» Correct Answer - B Correct Reason: F cannot show positive oxidation state of `+5` and `+7`, due to absence of vacant d-orbital in its valence shell, hence `FO_(3)^(Theta)` and `FO_(4)^(Theta)` are unknown. |
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| 79. |
Assertion (A): ICl on hydrolysis gives HI and HOCl. Reason (R ): Water can attack iodine more readily. |
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Answer» Correct Answer - D Correct Reason: ICI on hydrolysis gives HCI and HOI. `overset(delta-delta+)ICI+H_(2)O rarr HOI+HCI` |
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| 80. |
Assertion (A): Bond energy of CI-CI bond is more than F-F bond. Reason (R ): Shorter the bond length, stronger the bond, more is the bond energy. |
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Answer» Correct Answer - B Correct Reason: Lower bond energy of `F_(2)`, denspite shorter bond length is due to high 1p-1p repulsion in `F_(2)` molecule. |
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| 81. |
HBr and HI can reduce sulphurie acid, HCI can reduced `KMnO_(4)` and HF can reduce…………….A. `H_(2)SO_(4)`B. `KMnO_(4)`C. `K_(2)Cr_(2)O_(7)`D. None of these |
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Answer» Correct Answer - D HF does not behave as a reducing agent. Therefore, it cannot be oxidised by any of these. |
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| 82. |
Assertion (A): HBr is weaker acid than HI. Reason (R ): HBr is more polar than HI. |
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Answer» Correct Answer - C Correct Reason: Due to low bond energy of H-I. |
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| 83. |
HBr is stronger acid than HI because of hydrogen bonding. |
| Answer» H-I bond is weaker than H-Br, HI is a stronger acid than HBr. | |
| 84. |
Electron affinity for a noble gas is appoximately equal toA. that of halogensB. zeroC. that of oxygen familyD. that of nitrogen family |
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Answer» Correct Answer - b Noble gases have completely filled valence sheel and thus have no scope for for addition of extra electron. |
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| 85. |
Reducing property of dioxides decreases from `SO_2` to `TeO_2`. Why ? |
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Answer» `SO_2` undergoes oxidation to `SO_4^(2 -)` and thus is reducing. `overset(4+) SO_2 rarr underset("More stable state") (overset(6+) SO_4^(2-) + 2e^(Ө)` (ii) `TeO_2` undergoes reduction to `TeO` and thus is oxidising or less reducting. This is due to the inert pair effect. Lower `O.S` is more stable down the group. `2e^(Ө) + overset (4+) (TeO_2) rarr overset (2+) (TeO)`. |
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| 86. |
How will you obtain the following from sulphuric acid ? (a) `SO_2` (b) `SO_2` (c) `SO_2 Cl_2`. |
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Answer» (a) `SO_2` is obtained by heating copper with `conc. H_2SO_4`. `Cu + 2H_2 SO_4 rarr Cu SO_4 + SO_2 + 2 H_2 O` It can also be obtained by boiling sulphur with `conc. H_2 SO_4`. `S + 2H_2 SO_4 rarr 3SO_2 + 2H_2 O`. (b) `H_2 SO_4` when titrated with `P_2 O_5` loses water and forms `SO_3`. `H_2 SO_4 + P_2 O_5 rarr 2HPO_3 + SO_3` (c) `SO_2 Cl_2` is formed when `conc. H_2 SO_4` is treated with excess of `PCl_5`. `H_2 SO_4 + 2 PCl_5 rarr SO_2 Cl_2 + 2POCl_3 + 2 HCl`. |
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| 87. |
How will you obtain the following from sulphuric acid ? (a) `SO_2` (b) `SO_2` (c) `SO_2 Cl_2`. |
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Answer» (a) `SO_2` is obtained by heating copper with `conc. H_2SO_4`. `Cu + 2H_2 SO_4 rarr Cu SO_4 + SO_2 + 2 H_2 O` It can also be obtained by boiling sulphur with `conc. H_2 SO_4`. `S + 2H_2 SO_4 rarr 3SO_2 + 2H_2 O`. (b) `H_2 SO_4` when titrated with `P_2 O_5` loses water and forms `SO_3`. `H_2 SO_4 + P_2 O_5 rarr 2HPO_3 + SO_3` (c) `SO_2 Cl_2` is formed when `conc. H_2 SO_4` is treated with excess of `PCl_5`. `H_2 SO_4 + 2 PCl_5 rarr SO_2 Cl_2 + 2POCl_3 + 2 HCl`. |
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| 88. |
Give reason for the following : (a) `Conc. H_2 SO_4` cannot be used for drying `H_2`. (b) `KMnO_4` should not be dissolved in `conc. H_2 SO_4`. |
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Answer» (a) Hydrogen sometimes ignites with the large amount of heat released when water is absorbed by `conc. H_2 SO_4`. (b) `Mn_2 O_7` is formed when `KMnO_4` is dissolved in `conc. H_2 SO_4. Mn_2 O_7` formed is highly unstable and explosive in nature. `2 KMnO_4 + 2 H_2 SO_4 rarr K_2 SO_4 + (MnO_3)_2 SO_4 + 2H_2 O` `(MnO_3)_2 SO_4 + H_2 O rarr Mn_2 O_7 + H_2 SO_4`. |
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| 89. |
Why in the manufacture of `H_2 SO_4` by contact process, sulphur trioxide is not directly dissolved in water ? |
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Answer» Sulphur trioxide `(SO_(3))` reacts the water to form `H_(2) SO_(4)` which is not easily condensed and excapes in the atmosphere. Hence, `SP_(3)` is absorbed in `conc. H_(2) SO_(4)` to form oleum, which is then diluted to form `H_(2) SO_(4)`. `H_(2) SO_(4) + SO_(3) rarr H_(2) S_(2) O_(7)` `H_(2) S_(2) O_(7) + H_(2) O rarr 2 H_(2) SO_(4)`. |
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| 90. |
Which shows the least chemical reactivity?A. AmmoniaB. MethaneC. ArgonD. Hydrogen sulphide |
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Answer» Correct Answer - c Ar due to its completely filled valance shell electronic configuration is inert and shown least chemical reactivity |
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| 91. |
Which statement regarding He is incorrect?A. a.it is used in gas cooled nuclear reactorB. it is used as a cryogenic agent for carrying out experiment at low temperatureC. b.it is used to produce and sustain powerful superconducting magnetsD. c.it is used to fill gas balloons instead of `H_(2)` because it is lighter and non-combustible |
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Answer» Correct Answer - d Option (a),(b)and (c ) are uses of He. Helium is heavier than `H_(2)` |
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| 92. |
Which is called stranger gas ?A. KrB. XeC. HeD. Ne |
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Answer» Correct Answer - b Xenon, Greek xenos means stranger |
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| 93. |
Which compound is prepared by the following reaction `Xe + 2F_(2) overset(Ni vestel)underset(673 K, 5-6 atm)(to)` `(1:5 volume ratio)`A. `XeF_(2)`B. `XeF_(6)`C. `XeF_(4)`D. `XeOF_(2)` |
| Answer» Correct Answer - c | |
| 94. |
Which of the following cannot he formed ?A. `He^(2+)`B. `He^(oplus)`C. `He`D. `He_(2)` |
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Answer» Correct Answer - d Bond order for `He_(2)` is zero, this molecule does not exist. |
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| 95. |
(a)Why are the elements of group `18` known we moble gases? (b)Noble gases have very how bulting points why? (c ) Does the hydrogen of `XeF_(2)` lend in a radus reaction? |
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Answer» The elements present in group `18` have their valence shelf orbatin comlentely filled and therefore react with a few element only under certain condition .Therefore , they qare known as nible gases. (b) Noble gases being monocatemic have no interatonic forces except weak displersion forces and therefore they are liquefied at very low temperature .Hence they have low holding points. (c ) No, the produces of hydrogen are `XaOF_(4) and XeO_(2)F_(2)` where the oxidetaion state of all the elements remain the same as it was in the reacting state. `overset(+6)(XeF_(6)) + H_(2)O to overset(+6)(XeOF_(4)) + 2HF` `overset(+6)(XeF_(6)) + 2H_(2)O to overset(+6)(XeO_(2)F_(4)) + 4HF` |
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| 96. |
Which of the following compound cannot be prepared by direct betyween the consituent element?A. `XeF`B. `XeO_(3)`C. `XeF_(4)`D. `XeO_(2)F_(2)` |
| Answer» Correct Answer - b,d | |
| 97. |
Which of the following mases are used for the group `18` elements?A. Zero group elementsB. AerogensC. Noble gasesD. Chalcogens |
| Answer» Correct Answer - a,b,c | |
| 98. |
Which amongs the following statement are correct?A. `XeF_(4) and SbF_(3)` combine to form salfB. `He and Na` do not form clabrancesC. He has highest bolling poin in the groupD. He diffuses through rubber |
| Answer» Correct Answer - a,b,d | |
| 99. |
Which of the following does not react with florime?A. KrB. AtC. XeD. All of these |
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Answer» Correct Answer - b Compound of Ar with fluorine are not know due to very high ionisation enthalpy of Ar |
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| 100. |
Xenon directly combines withA. oxygenB. rubidiumC. flaorineD. chlorine |
| Answer» Correct Answer - c | |