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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.
| 401. |
Which of the following is the correct order of bond dissociation enthalpy?A. `I_(2) lt Br_(2) lt Cl_(2) lt F_(2)`B. `I_(2) lt Cl_(2) lt Br_(2) lt F_(2)`C. `I_(2) lt F_(2) lt Br_(2) lt Cl_(2)`D. `I_(2) lt Br_(2) lt F_(2) lt Cl_(2)` |
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Answer» Correct Answer - 3 All the Group `17` elements from diatomic molecules. It would be expected that the bond dissociation enthalpy would decreases as the atoms become larger because increased size result in less effective overlap of orbitals .`Cl_(2), Br_(2)` and `I_(2)` show the expected trend but the bond dissociation enthalpy of `F_(2)` does not fit the expected trend. The bond dissociation enthalpy of `F_(2)` is abnormally low and this is largely responsible for its very high reactivity. Notice that other element in the second row of the peiodic table also have weaker bonds than the element wich follow in their respective groups . For example ,in group `15`, the `N-N` bond in hydrazine si wealer than `P-P` bond ,similarly in Gropu `16` the `O-O` bond in peroxides is weaKer than `S-S` since `F` atoms are small , the `F-F` disatance is also small `(1.48 Å)` amd hence inter electron -electron replusion is apppreciable. The large electron of the two `F` atoms wealen the bond also .The correct order is: `Cl_(2) gt Br_(2) gt F_(2) gt I_(2)` Thus `F_(2)` and `I_(2)` have low values of bond dissociation enthalpies. |
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| 402. |
Which of the folllowing is the correct order of electron affinity?A. `At lt I lt Br lt F lt Cl`B. `At lt I lt Br lt Cl lt F`C. `At lt I lt Cl lt Br lt F`D. `At lt Br lt I lt Cl lt F` |
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Answer» Correct Answer - 1 Electeon affinities for all the halogens are negative implying that energy is realeased when a halgeon atom gains an electron ,`(X rarr X^(-))`. Thus all halgons form halide ions. Electron affinity of the group `17` elements becomes less and negative from `Cl` onwards. However mthe elctron affinity of `F` is less negative thatn that of chlorine. Due ot small size of `F` atoms, there are strong interelctronic replusion and hence the incoming electrons does not experience much attraction. Thus correct order is `Cl gt F gt Br gt I gt At` |
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| 403. |
Iodine is obtained commercially from Chile saltpetre through the reactionA. `2KI+Cl_(2)rarrI_(2)+2KCl`B. `IO^(-)+5I^(-)+6H^(+)rarr3I_(2)+3H_(2)O`C. `2IO_(3)^(-)+5HSO_(3)^(-)rarr3HSO_(4)^(-)+2SO_(4)^(2-)+H_(2)O+I_(2)`D. `2I^(-)+H_(2)O_(2)rarrI_(2)+2OH^(-)` |
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Answer» Correct Answer - 3 There are two different commericial method of obtaining iodine. The methods used depends on whether the source is chile saptpeter or natural brine. Iodine is manufactured from natural brines (or sea weed) by the oxidation of iodide ion with chlorine. Chile sltpeter is mainly `NaNO_(3)`, but it contain traces of sodium iodate `(NaIO_(3))` and sodium periodate `(NaIO_(4))` . Pure `NaNO_(3)` is obtianed by dissolving saltpeter in water and crystallizing `NaNO_(3)`. Thus the iodate residues accumulate and concentrated in the mother liquor. Eventually, this concentrate is divided into two parts. One parts is reduced with `NaHSO_(3)` to give `I^(-)`. This is mixed with the unreacted part, which is filtered off as a solid and then purified by sublimation: ` 2IO_(3)^(-)+6HSO_(3)^(-)rarr2I^(-)+6SO_(4)^(2-)+6H^(+)` `5I^(-)+IO_(3)^(-)+6H^(+)rarr3I_(2)+3H_(2)O` `NaIO_(3)(aq.)+3NaHSO_(3)(aq.)rarrNal(aq.)+3NaHSO_(4)(aq.)` `NaIO_(3)(aq.)+5NaI(aq.)+3H_(2)SO_(4)(aq.)rarr3I_(2)(s)+3H_(2)O(l)+3Na_(2)SO_(4)(aq.)` |
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| 404. |
Bromine is commercially produced by the oxidation of bromide ions in natural brine withA. `CI_(2)`B. `F_(2)`C. `I_(2)`D. `O_(2)` |
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Answer» Correct Answer - 1 Bromine is obtained from sea water and brine lakes. Sea water contains about `65p p m Br^(-)`. Thsu `15` tonnes of sea water contain about `1` kg of bromine. Bromine is extracted from sea water but it is more economical to use more concrntrated brine sources (such as the Dead sea) which contain `2000-5000p p m` of `Br^(-)`. First `H_(2)SO_(4)` is added to adjust the`pH` about `3.5`. Then `Cl_(2)(g)` is passed throught the solution to oxidize the`Br^(-)` to`Br_(2)` (an example of dispacement of one elementby another higher in the electrochemical series): `Cl_(2)+2Br^(-)rarr2Cl^(-)+Br^(2)` The `Br_(2)` is removed by a stream of air, because `Br_(2)` is quite volatile. The gas is passed through a solution of `Na_(2)CO_(3)` when the `Br_(2)` is absorbed, forming a mixture of `NaBr` and `NaBrO_(3)` . Finally the solution is acidified and disstilled to giv epure bromine. `3Br_(2)+3NaCO_(3)rarr5NaBr+NaBrO_(3)+3CO_(2)` `5NaBr+NaBrO_(3)+3H_(2)SO_(4)rarr5HBr+HBrO_(3)+3Na_(2)SO_(4)` `5HBr+HBrO_(3)rarr3Br_(2)+3H_(2)O` |
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| 405. |
The compound that does not produce nitrogen gas upon thermal decomposition is :A. `(NH_(2))_(2)SO_(4)`B. `Ba(N_(3))_(2)`C. `(NH_(4))_(2)Cr_(2)O_(7)`D. `NH_(4)NO_(2)` |
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Answer» Correct Answer - A Except for `(NH_(4))_(2)SO_(4)` which evolves `NH_(3)` upon strong heating, all other compounds decompose the produce `N_(2)` gas `(NH_(4))_(2)SO_(4) overset("heat")to 2NH_(3)+H_(2)SO_(4)` `(NH_(4))_(2)Cr_(2)O_(7) overset("heat") to N_(2)+4H_(2)O + Cr_(2)O_(3)` `(NH_(4))_(2)NO_(2) overset("heat")to N_(2)+2H_(2)O` `Ba(N_(3))_(2) overset("heat") to Ba+3N_(2)` |
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| 406. |
Dinitrogen is commercially prepared byA. the action of alkaline hypobromite solution on ureaB. the oxidation of `NH_(3)`by passing it over heated copper oxide.C. fractional distillation of liquefied airD. heating barium aside |
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Answer» Correct Answer - 3 `N_(2)` is obtained commercially by condensing air to the liquid state and then fractionally distilling the liquid air.`N_(2)` has a lower boiling point than `O_(2)` ( the boiling point of liquids `N_(2)` and liquid `O_(2)` are `-196^(@)C`amd `-183^(@)` respectively)and therefore molecular nitrogen will boil off before molecular oxygen does during the fractional distillation of liquid air. Six industrial gases are obtained in this way: `N_(2), O_(2), Ne, Ar, kr` and `Xe`. |
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| 407. |
Which of the following statements is wrong ?A. Stability of hydrides increases from `NH_(3) " to " BiH_(3)` in group 15 of the periodic tableB. Nitrogen cannot form `d pi - p pi` bond.C. N-N bond is weaker than P-P bondD. `N_(2)O_(4)` has two resonating strutures. |
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Answer» Correct Answer - A The stability of hydrides decreases down the group 15 elements. |
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| 408. |
Which of the following statements is wrong ?A. The stability of hydrides increases from `NH_(3) " to " BiH_(3)` in group 15 of the periodic table.B. Nitrogen cannot form `d pi - p pi` bond.C. Single N - N bond is weaker than single P-P bondD. `N_(2)O_(4)` has two resonating strutures. |
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Answer» Correct Answer - A The stability of hydrides decreases from `NH_(3) " to " BiH_(3)` in group 15 because the bond length of M - H bond decreases. |
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| 409. |
Group 13 elements show +1 and +3 oxidation states. Relative statibility of +3 oxidation may be given asA. `TI^(+) gt In^(3+) gt Ga^(3+) gt AI^(3+) gt B^(3+)`B. `B^(3+) gt AI^(3+) gt Ga^(3+) gt In^(3+) gt TI^(3+)`C. `AI^(3+) gt Ga^(3+) gt TI^(3+) gt In^(3+) gt B^(3+)`D. `AI^(3+) gt B^(3+) gt Ga^(3+) gt TI^(3+) gt In^(3+)` |
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Answer» Correct Answer - B Stability of +3 oxidation state decreases from AI to TI. B always shows +3 oxidation between state in all of its compounds. |
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| 410. |
`SiF_(6)^(2-)` exist but not `CF_(6)^(2-)` explain why ? |
| Answer» The main reasons are : (i) six large chloride ions cannot be accommodated around `Si^(4+)` due to limitation of its size. (ii) interaction between lone pair of chloride ion and `Si^(4+)` is not very strong. | |
| 411. |
a. Select the member `s` of group `14` that (i) forms the most acidic iodide (ii) is commonly found in `+2` oxidation state and (iii) used as semiconductor . (Solved NCERT Problem `11.5`) b. `|SiF_5^(2-) ` is known, whereas `|SiCl_5|^(2-)` not. Glve possible preasons . (Solved NCEGT Pronlem `11.6`) c . Diamond is covalent , yet covelnt , yet it has high melting point . Why ? (Solved NCEGT Pronlem `11.7`) . |
| Answer» (i) carbon (ii) lead (iii) silicon and germanium | |
| 412. |
Anomalous behaviour of oxygen is due to |
| Answer» Oxygen cannot expand its octet, it can form `p pi- p pi` bond. | |
| 413. |
Given three reasons for anomalous behaviour of carbon. |
| Answer» Small size, high electronegativity, unavailability of vacant d-orbitals. | |
| 414. |
The aqueous solution of `AICI_(3)` is acidic due toA. `AI(OH)_(3) + HCI`B. `[AI(H_(2) O)_(6)]^(3+) + 3CI^(-)`C. `AICI_(3). 2H_(2)O`D. `AI_(2)O_(3) + HCI` |
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Answer» Correct Answer - B `AICI_(3)` in acidified aqueous solution forms `[AI(H_(2)O)_(6)]^(3+)` ion. `AICI_(3) + 9 H_(2) O to [AI(h_(2)O)_(6)] (Oh)_(3) + 3H^(+) + 3CI^(-)` |
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| 415. |
Anhydrous aluminium chloride is prepared byA. reaction of HCI and AI metalB. reaction of dry HCI gas and metal AI metalC. passing Conc `HNO_(3)` gas over heated AI metalD. reaction of `AI_(2) O_(3)` with dil. HCI |
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Answer» Correct Answer - B `2AI + 6HCI_((g)) overset("Heat")(to) 2AICI_(3) + 3H_(2)` |
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| 416. |
Name the compound of N with oxidation state `-3 " and " +2`. |
| Answer» Correct Answer - `NH_(3), NO` | |
| 417. |
Among the following oxioacids, the correct decreasing order of acid strength isA. `HClO_(2) gt HClO_(4) gt HClO_(3) gt HOCl`B. `HOCl gt HClO_(2) gt lt HClO_(3) gt HClO_(4)`C. `HClO_(4) gt HOCl gt HClO_(2) gt HClO_(3)`D. `HClO_(4) gt HClO_(3) gt HClO_(2) gt HOCl` |
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Answer» Correct Answer - D It is the correct decreasing order of acidic strength. |
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| 418. |
The halogen which is most easily reduced is :A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
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Answer» Correct Answer - A `F_(2)` is most easily reduced because HF is the most stable acid among the halogen acid. |
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| 419. |
When copper is heated with conc. `HNO_(3)` it produces?A. `Cu(NO_(3))_(2), NO " and " NO_(2)`B. `Cu(NO_(3))_(2) " and " N_(2)O`C. `Cu(NO_(3))_(2) " and " NO_(2)`D. `Cu(NO_(3))_(2) " and " NO`. |
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Answer» Correct Answer - C `Cu+4HNO_(3)("conc.") to Cu(NO_(3))_(2) +2NO_(2)+2H_(2)O` |
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| 420. |
Which of the following properties is not shown by `NO` ?A. It is diamagnetic in the gaseous stateB. It is a neutrla oxideC. It combines with oxygen to form nitrogen dioxideD. Its bond order is 2.5. |
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Answer» Correct Answer - A NO is paramagnetic in the gaseous state. The total no. of electrons is `15(7+8)`. There is one unpaired electron present on the oxygen atom. |
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| 421. |
Which among the following is the most reactive gtA. `I_(2)`B. IclC. `Cl_(2)`D. `Br_(2)` |
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Answer» Correct Answer - B Interhalogen compounds are more reactive than halogens. |
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| 422. |
Which of the following solutions will turn violet when a drop of lime juice is added to it ?A. A solution of NaIB. A solution mixture of KI and `NaIO_(3)`C. A solution mixture of NaI and KID. A solution mixture of `KIO_(3) " and " NaIO_(3)` |
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Answer» Correct Answer - B In acidic medium, `H^(+)` ions will oxidise both `l^(-)` ions (Kl) and `lO^(-) " ions" (NalO_(3))` to liberate iodine accompanied by violet fumes. |
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| 423. |
`CaF_(2)+H_(2)SO_(4) to` |
| Answer» `CaF_(2)+H_(2)SO_(4) to CaSO_(4) +2HF` | |
| 424. |
Which is incorrectly given according to order incicated?A. `F_(2) gt Cl_(2) gt Br_(2) gt l_(2),`Oxidising powerB. `Hl gt HBr gt HCl gt HF`: Acidic strengthC. `F_(2) gt Cl_(2) gt Br_(2) gt l_(2)`, Bond dissociation enthalpyD. `HF gt Hl gt HBr gt HCl`, Boiling point |
| Answer» Correct Answer - C | |
| 425. |
Which has maximum bond angle?A. `NH_(3)`B. `PH_(3)`C. `AsH_(3)`D. `SbH_(3)` |
| Answer» Correct Answer - A | |
| 426. |
The colour shown by halogen is incorrectly given byA. `F_(2)`=yellowB. `Cl_(2)`=colourlessC. `Br_(2)`=RedD. `I_(2)`=Violet |
| Answer» Correct Answer - B | |
| 427. |
`4 As+Cl_(2) to ` |
| Answer» `4As+6Cl_(2) to 4 AsCl_(3)` | |
| 428. |
`(NH_(4))_(2)SO_(4)+NaOH to` |
| Answer» `(NH_(4))_(2)SO_(4)+2NaOH overset("heat")to 2NH_(3)+Na_(2)SO_(4)+2H_(2)O` | |
| 429. |
`FeSO_(4)+NO+H_(2)O to` |
| Answer» `FeSO_(4)+NO+5H_(2)O to [Fe(NO)(H_(2)O)_(5)]^(2+)SO_(4)^(2-)` | |
| 430. |
`S-S` bond is present inA. `H_(2)S_(2)O_(7)`B. `H_(2)S_(2)O_(8)`C. `H_(2)S_(2)O_(6)`D. `H_(2)SO_(3)` |
| Answer» Correct Answer - C | |
| 431. |
Radioactive element of group 17 isA. Polonium.B. FranciumC. AstatineD. Radium |
| Answer» Correct Answer - C | |
| 432. |
`Xe+O_(2)F_(2) overset(155 K) to` |
| Answer» `Xe+O_(2)F_(2) overset(155 K) to XeF_(2)+O_(2)` | |
| 433. |
In brown ring test for nitrate ions, brown ring is formed due to complexA. `[Fe(H_(2)O)_(6)}^(2+)`B. `[Fe(H_(2)O)_(5)NO]^(2+)`C. `[Fe(H_(2)O)_(5)NO]^(3+)`D. `[Fe(H_(2)O)_(5)NO_(2)]^(2+)` |
| Answer» Correct Answer - B | |
| 434. |
Which of the following metals does not dissolve in concentrated `HNO_(3)` ?A. CrB. ZnC. AlD. Both 1 and 3 |
| Answer» Correct Answer - D | |
| 435. |
`XeOF_(4) +SiO_(2) to` |
| Answer» `underset("oxytetrfluoride")underset("Xenon")(2XeOF_(4)) + SiO_(2) to underset("dioxydifluoride")underset("Xenon")(2XeO_(2)F_(2))+SiF_(4)` | |
| 436. |
Name the product when iodine is heated with concentrated `HNO_(3)`. |
| Answer» Correct Answer - Iodic acid | |
| 437. |
Among the following, the correct order of acidity is:A. `HClO_(2) lt HClO lt HClO_(3) lt HClO_(4)`B. `HClO_(4) lt HClO_(2) lt HClO lt HClO_(3)`C. `HClO_(3) lt HClO_(4) lt HClO_(2) lt HClO`D. `HClO lt HClO_(2) lt HClO_(3) lt HClO_(4)` |
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Answer» Correct Answer - A It is the correct order of increasing acidic strength. |
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| 438. |
Select the correct order of acidity :A. `HI gt HBr gt HCl gt HF`B. `HClO_(4) gt HBrO_(4) gt HIO_(4)`C. `HClO lt HBrO lt HIO`D. `HClO_(4) gt HClO_(3) gt HClO_(2) gt HClO` |
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Answer» Correct Answer - A::B::D are the correct order of increasing acidity. |
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| 439. |
Carbonic acid is aA. Weak tribasic acidB. Weak dibasic acidC. Strong tribasic acidD. Strong dibasic acid |
| Answer» Correct Answer - B | |
| 440. |
Which one of the following is properties of CO gas?A. It is a colourless gasB. it is an odourless gasC. it is a neutral oxideD. All of these |
| Answer» Correct Answer - D | |
| 441. |
`H_(3)PO_(3) overset("Heat") to` |
| Answer» `underset("Phosphorus acid")(4H_(3)PO_(3)) overset("Heat") to underset("Phosphoric acid")(3H_(3)PO_(4)+PH_(3))` | |
| 442. |
`(NH_(4))_(2)Cr_(2)O_(7)` on heating gives a gas which is also given by :A. Heating `NH_(4)NO_(2)`B. Heating `NH_(4)NO_(3)`C. `Mg_(3)N_(2)+H_(2)O`D. `NH_(4)Cl` |
| Answer» Correct Answer - A | |
| 443. |
Name the gas evolved when ammonium dichromate `(NH_(4))_(2)Cr_(2)O_(7)` is heated. |
| Answer» Correct Answer - `N_(2)` | |
| 444. |
On heating, `(NH_(3))_(2)Cr_(2)O_(7)` gives rise to a gas which on treatment with Mg ribbon gives a white solid. On dissolving white solid in water another gas (X) is evolved. It gives white fumes when a rod dipped in conc. HCl is brought in its contact. Identify the gas X. |
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Answer» The available information suggests that the gas (X) is ammonia. The chemical reactions involved are as follows : `(NH_(4))_(2)Cr_(2)O_(7) overset("heat")to Cr_(2)O_(3)+N_(2)4H_(2)O` `3Mg+N_(2)to Mg_(3)N_(2)` ` Mg_(3)N_(2)+6H_(2)O to 3Mg(OH)_(2)+underset("White fumes with HCl")underset((X))(2NH_(3))` |
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| 445. |
In graphite, the bond isA. IonicB. CovalentC. Co-ordinateD. Metallic |
| Answer» Correct Answer - B | |
| 446. |
FullerenesA. Interlocking of hexagonal carbon ringsB. Interlocking of pentagonal carbon ringsC. Interlocking of hexagonal and pentagonal carbon ringsD. Interlocking of hexagonal and heptagonal carbon rings |
| Answer» Correct Answer - C | |
| 447. |
Graphite hasA. 2-D sheet structureB. Van der waals forces between different layersC. `sp^(2)` hybridised carbon linked with other three carbon atoms in hexagonal planar structureD. All of these |
| Answer» Correct Answer - D | |
| 448. |
The carbon atoms in diamond areA. `sp^(3)` -hybridized and are connected by double bondsB. `sp^(2)` -hybridized and are connected by single bondsC. `sp^(3)` -hybridized and are connected by single bondsD. `sp^(2)` -hybridized and are connected by double bonds |
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Answer» Correct Answer - C In diamond, each `C` atom is `sp^(3)` -hydridized and is tetrahedrally linked to four neighboring `C` atoms through four strong `C - C (sp^(3) - sp^(3) , sigma)` bonds. |
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| 449. |
`XeF_(6)+H_(2)O to` |
| Answer» `XeF_(6)+3H_(2)O to XeO_(3)+6HF` | |
| 450. |
Which one of the following statement is correct when `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution?A. Green `Cr_(2)(SO_(4))_(3)` is formedB. The solution truns blueC. The solution is decolourizedD. `SO_(2)` is reduced |
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Answer» Correct Answer - 1 In presence of moisture `SO_(2)` acts as a good reducting agent .It reducing character is due to the evolution of nascent hydrogen: `SO_(2)+2H_(2)OrarrH_(2)SO_(4)2[H]` Sulphur dioxide truns orange coloured potassium dichromatic solution green : `{:(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4) rarr H_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O+3[O]),(SO_(2)+H_(2)O+[O] rarr H_(2)SO_(4)"]"xx3),(bar(underset(("orange"))(K_(2)Cr_(2)O_(7))+H_(2)SO_(4)+3SO_(2) rarr K_(2)SO_(4))underset(("Green"))(+)Cr_(2)(SO_(4))_(3)+H_(2)O):}` |
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