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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.
| 1751. |
Which statement is wrong for NO ?A. It is anhydride of nitrous acidB. Its dipole moment is 0.22 DC. It forms dimerD. It is paramagnetic |
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Answer» Correct Answer - A `N_(2)O_(3)` but not NO is the anhydride of nitrous acid `(2HNO_(2)toN_(2)O_(3)+H_(2)O)`. |
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| 1752. |
Which of the following statement is wrong ?A. Single N-N bond is stronger than the single P-P bondB. `PH_(3)` can act as a ligand in the formation of coordination compound with transition elementsC. `NO_(2)` is paramagnetic in natureD. Covalency of nitrogen in `N_(2)O_(5)` is four |
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Answer» Correct Answer - A Due to small size, lone pair of electrons on the two N-atoms repel each other and hence N-N bond is weaker than P-P bond. |
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| 1753. |
Which one of the following compounds is not a protoric acid?A. `SO(OH)_(2)`B. `SO_(2)(OH)_(2)`C. `B(OH)_(2)`D. `PO(OH)_(3)` |
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Answer» Correct Answer - C `Ba(OH)_(3)` is not protonic acid because it does not give proton on ionisation directly. While it acts as Lewis acid due to acceptance of `OH^(-)` form water and forms a hydrated species. `B(OH)_(3)+H_(2)O rarr [B(OH)_(4)]^(-)H^(+)`. |
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| 1754. |
Ammonia , on reaction with hypochlorite anion, can formA. `NO `B. `N_2H_(4)`C. `NH_(4)Cl `D. `HNO_(2)` |
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Answer» Correct Answer - B `3NH_(3)+OCl^(-) to underset("Hydrazine")(NH_(2))-NH_(2)+NH_4Cl+OH^(-)` |
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| 1755. |
Ammonia , on reaction with hypochlorite anion, can formA. `NO`B. `NH_(4)Cl`C. `N_(2)H_(4)`D. `HNO_(2)` |
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Answer» Correct Answer - C `2NH_(3)+NaOCl to NH_(2)Cl+NaOH("fast")`. `2NH_(3)+NH_(2)Cl to NH_(2)NH_(2)+NH_(4)Cl("slow")`. |
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| 1756. |
The thermal stability order for group `14` halides is:A. `GeX_(2) lt SiX_(2) lt SnX_(2) lt PbX_(2)`B. `SiX_(2) lt GeX_(2) lt SnX_(2) lt PbX_(2)`C. `SiX_(2) lt GeX_(2) lt PbX_(2) lt SnX_(2)`D. `PbX_(2) lt SnX_(2) lt GeX_(2) lt SiX_(2)` |
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Answer» Correct Answer - B The stability of group `14` tetrahalides edge or surface bonds like dimonds or grphite. |
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| 1757. |
The correct order of the thermal stability of hydrogen halides `(H-X)` isA. `HlgtHbegtHClgtHF`B. `HFgtHClgtHBrgtHI`C. `HClltHFltHBrltHI`D. `HIgtHClgtHFltHBr` |
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Answer» Correct Answer - B `HFgtHClgtHBrgtHI` (Thermal stability)` |
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| 1758. |
The correct order of thermal stability of hydrogen halides (H-X) is :A. `HI gt HBr gt HCl gt HF`B. `HF gt HCl gt HBr gt Hl`C. `HCl lt HF lt HBr lt HI`D. `HI gt HCl lt HF lt HBr` |
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Answer» Correct Answer - 2 |
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| 1759. |
The thermal stability order for group `14` halides is:A. `GeX_(2) lt SiX_(2) lt SnX_(2) lt PbX_(2)`B. `SiX_(2) lt GeX_(2) lt PbX_(2) lt SnX_(2)`C. `SiX_(2) lt GeX_(2) lt SnX_(2) lt PbX_(2)`D. `PbX_(2) lt SnX_(2) lt GeX_(2) lt SiX_(2)` |
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Answer» Correct Answer - C The stability of group `14` tetrahalides decreases down the group whreas of dihalides increases down the group. |
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| 1760. |
Producer gas is a mixture ofA. `CO` and `N_(2)`B. `CO_(2)` and `H_(2)`C. `CO` and `H_(2)`D. `CO_(2)` and `N_(2)` |
| Answer» Correct Answer - A | |
| 1761. |
Producer gas is a mixture ofA. `CO+N_(2)`B. `CO+H_(2)`C. `N_(2)+CH_(4)`D. `CO+H_(2)+N_(2)` |
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Answer» Correct Answer - A Product gas (a mixture of `CO+N_(2)`) is prepared by incomplete combustion of coal in restrict supply of air. |
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| 1762. |
Graphite is soft solid lubricant extreamly to melt. The reason for this anomalous behaviour is that graphite:A. has molecules of varaible molecular masses like polymersB. has carbon atoms arranged in large plates of ring of strongly bound carbon atoms with weak interpolated bondsC. is a non-crystalline substanceD. is an allotropic form of of diamond |
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Answer» Correct Answer - B A fact about graphite due to `sp^(2)`-hybridation. |
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| 1763. |
How can the following reaction be made to proceed in forward direction ? `B(OH)_(3) + NaOH hArr Na[B(OH)_(4)]`.A. addition of cis-`1-2-diol`B. addition of boraxC. addition of trans`-1,2` dioleD. addition of `Na_(2)HPO_(4)` |
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Answer» Correct Answer - A Due to formation of chelated complex, the reaction proceeds in forwards direction. |
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| 1764. |
How can the following reaction be made to proceed in forward direction ? `B(OH)_(3) + NaOH hArr Na[B(OH)_(4)]`.A. Addition of cis 1,2-diolB. Addition of boraxC. Addition of trans 1,2-diolD. Addition of `Na_(2)HPO_(4)` |
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Answer» Correct Answer - A |
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| 1765. |
How can the following reaction be made to proceed in forward direction ? `B(OH)_(3) + NaOH hArr Na[B(OH)_(4)]`.A. Addtion of cis 1, 2-diolB. Addition of boraxC. Addition of trans 1,2-diolD. Addition of `Na_(2)HPO_(4)` |
| Answer» Correct Answer - A | |
| 1766. |
Consider the compounds, `BCl_(3)` and `"CCl"_(4)` . How will they behave with water ? Justify . |
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Answer» The B atom in `BCl_(3)` has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence `BCl_(3)` undergoes hydrolysis to form boric acid `(H_(3)BO_(3))` and HCl. `BCl_(3) rarr + 3H_(2)O rarr H_(3)BO_(3) + 3HCl` In contrast, C atom is `"CCl"_(4)` has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule and hence neither accepts nor donates a pair of electrons. In other words , it does not accept a pair of electrons from `H_(2)O` molecule and hence `"CCl"_(4)` does not under go hydrolysis in water. |
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| 1767. |
Standard electrode potential values, `E^(@)` at `Al^(3+)//Al` is `-1.66 V` and that of `Tl^(3+)//Tl` is `+1.26 V`. Predict about the formation of `M^(3+)` ions in solution and compare the electropositive character of the two metals. |
| Answer» The negative valueof `E^(@)` for `Al^(3+)//Al` suggests that Al has a strong tendency to form `Al^(3+)` (aq) ions. On the other hand, the positive value of `E^(@)` for `Tl^(3+)//Tl` suggests that Tl does not have high tendency to form `Tl^(3+)` ions. Since Al can form `Al^(3+)` ions more easily than `Tl` does to form `Tl^(3+)` ions, therefore Al is more electronegativity than Tl. | |
| 1768. |
`CO_(2)` is a gas while `SaO_(2)` is a solid. Explain. |
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Answer» Carbon because of its small size and higher electronegativity than silicon forms `ppi-ppi` double bonds with O-atoms to form `CO_(2)` molecule. These molecules of `CO_(2)` are held together by weak van der Waals forces of attracts which can be easily overcome by collision of the molecules at room temperature. Consequently, `CO_(2)` is a gas. Silicon, on the other hand, because of its bigger size and lower eletronegativity than carbon has little tendency to form `ppi-ppi` double bonds with `O`-atoms . Instead, each silicon atom forms four single, covalent bonds with O-atoms which are tetrahedrally arranged around it leading to the formation of a three-dimensional network structure . To break these covalent bonds, a large amount of energy is needed and hence `SiO_(2)` is a high melting solid. |
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| 1769. |
Aluminium trifluoride is insoluble in anhydrous `HF` but dissolves on addition of `NaF`. Aluminium trifluoride precipitates out of the resulting solution when gaseous `BF_(3)` is bubbled through. Give reasons. |
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Answer» (i) Anhydrous `HF` is a covalent compound and is strongly `H`-bonded. Therefore , it does not give `F^(-)` ions and hence `AlF_(3)` does not dissolve in `HF`. In contrast, `NaF` being an ionic compound contains `F^(-)` ions and hence combines with `AlF_(3)` to form the soluble complex. `3NaF + AlF_(3) rarr underset((" soluble complex "))underset(" Sod. hetrafluoroborate (III) ")(3Na [BF_(4)]) + AlF_(3) (s)` (ii) Because of smaller size and higher electronegativity, has high much higher tendency to form complexes than Al, therefore when `BF_(3)` is added to the above solution. `AlF_(3)` gets precipitated. |
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| 1770. |
`AlF_(3)` is insoluble in anhydrous `HF` but dissolves on addition of `NaF. AlF_(3)` precipitates out of the resulting solution when gaseous `BF_(3)` is bubbled through. Give reasons. |
| Answer» Aluminium trifuoride `(AIF_(3))` is insoluble in anhydrous HF because of its covalent nature. However, it forms a complex compound on reacting with `NaF` which bond polarities cancel out. | |
| 1771. |
What is the maximum percentage of available chlorine in a sample of bleaching powder ? |
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Answer» Correct Answer - `55.9%` |
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| 1772. |
Beryllium chloride exists as polymeric chain having fromula `(BeCl_(2))_(n)`. If a compound `K_(x) (Be_(4)Cl_(10))` has finite chain, find the value of x. |
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Answer» Correct Answer - `0002` `K_(x)(Be_(4)Cl_(10)) implies x+8-10=0 implies x=2` |
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| 1773. |
The correct decreasing order of the acidic strength of `HClO, HClO_(2), HClO_(4),HClO_(4)` isA. `HClOgtHClO_(2)gtHClO_(3)gtHClO_(4)`B. `HClO_(4)gtHClO_(3)gtHClO_(2)gtHClO`C. `HClO_(4)gtHClO_(2)gtHClOgtHClO_(3)`D. `HClO_(3)gtHClOgtHClO_(4)gtHClO_(2)` |
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Answer» Correct Answer - B The correct decreasing order of the acidic strength of oxoacids of chlorine is `HClO_(4) gt HClO_(3) gt HClO_(2) gt HClO` |
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| 1774. |
`NCl_(3)` gets readily hydrolysed while `NF_(3)` does not. Why ? |
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Answer» In `NCl_(3)`, Cl has vacant d-orbitals to accept the lone pair of electrons donated by O-atom of `H_(2)O` molecule but in `NF_(3)`, F does not have d-orbitals. Thus, `NCl_(3)` undergoes hydrolysis but `NF_(3)` does not. `NCl_(3)+3H_(2)OtoNH_(3)+HOCl" , "NF_(3)+H_(2)Oto` No reaction |
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| 1775. |
`NCl_(3)` is an endothermic compound while `NF_(3)` is an exothermic compound. |
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Answer» Due to smaller size of F as compared to Cl, the N-F bond is much stronger `(272k kJ mol^(-1))` than N-Cl bond `(201 kJ mol^(-1))` while bond dissociation energy of `F_(2)(158.8 kJ mol^(-1))` is much lower than that of `Cl_(2)(242.6 kJ mol^(-1))`. Therefore, energy released during the formation of `NF_(3)` molecule is more than the energy needed to break `N_(2)(941.4 kJ mol^(-1))` and `F_(2)` molecules into individual atoms. In other words formation of `NF_(3)` is an exothermic reaction or `NF_(3)` is an exothermic compoud. `{:(N_(2)(g)+3F_(2)(g)to2NF_(3)(g),,DeltaH=-214.2Jmol^(-)),(N_(2)(g)+3Cl_(2)(g)to2NCl_(3)(g),,DeltaH=-463.2kJmol^(-1)):}` In contrast, energy released during the formation of `NCl_(3)` molecule is les than the energy needed to break `N_(2)andCl_(2)` molecules into individual atoms. In other words, formation of `NCl_(3)` is an endothermic reaction or `NCl_(3)` is an endothermic compound. |
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| 1776. |
A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). (B) reacts with organic compounds containing -OH group converting them into chloro derivatives. (B) on hydrolysis gives (C) and is finally converted into phosphonc acid. (A), (B) and (C) areA. `P_4, PCl_5 , H_3PO_4`B. `P_4,PCl_5 , H_3PO_3`C. `P_4, PCl_5, POCl_3`D. `P_4, PCl_3 , POCl_3` |
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Answer» Correct Answer - A `underset"(A)"(P_)4 + 10Cl_2 to underset"(B)"(4PCl_5)` `PCl_5 + C_2H_5OH to C_2H_5Cl + POCl_3 + HCl` `PCl_5 + H_2O to underset"(C )"POCl_3 + 2HCl` `POCl_3 + 3H_2O to H_3PO_4 + 3HCl` |
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| 1777. |
Statement-I : `PCl_(5) "and" PbCl_(4)` are thermally unstable. Statement-II : They produce same gas on thermal decompositionA. If both statement -I & Statement-II are true & the Statement-II is correct explaination of the statement-I.B. If both Statement-I & Statement-II are true but Statement-II is not a correct explainatin of Statement-I.C. If Statement-I is True but the Statement-II is FalseD. If the Statement-I is False but the Statement-II is True |
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Answer» Correct Answer - B `PCl_(5)overset(triangle)toPCl_(3)+Cl_(2)` `PCl_(5) "decomposes into" PCl_(3) "&" Cl_(2)` as in its structure two `P-Cl` axial bonds are longer than other three `p-Cl` equatorial bonds. `PbCl_(4)overset(triangle)toPbCl_(2)+Cl_(2)` Due to inert pair effect, `Pb^(4+)` is unstable therefore it changes into more stable `Pb^(++)` |
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| 1778. |
Thermal decomposition of ammonium dichromate givesA. `N_(2),H_(2)OandCr_(2)O_(3)`B. `N_(2),NH_(3)andCrO`C. `(NH_(4))_(2)CrO_(4)andH_(2)O`D. `N_(2),H_(2)OandCrO` |
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Answer» Correct Answer - A `(NH_(4))_(2)Cr_(2)O_(7)overset(Delta)toN_(2)+Cr_(2)O_(3)+4H_(2)O` |
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| 1779. |
Very pure `N_(2)` can be obtained byA. thermal decomposition of ammonium dichromateB. treating aqueous solution of `NH_(4)ClandNaNO_(2)`C. liquefaction and fractional distillation of liquid airD. thermal decomposition of sodium azide |
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Answer» Correct Answer - D Pure `N_(2)` is obtained by thermal decomposition of sodium azide `underset("Sod.azide")(2NaN_(3))to2Na+N_(2)` |
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| 1780. |
`{:(,"Column I",,"Column II"),((A),"Inorganic benzene",(p),"An allotrope of carbon"),((B),"Fulerene",(q),"Orthosilicate"),((C),"Phenacite",(r),"An ore of boron"),((D),"Colemanite",(s),"Borazine" (B_(3)N_(3)H_(6))):}`A. A-s,B-p,C-r,D-qB. A-p,B-s,C-q,D-rC. A-q,B-r,C-s,D-pD. A-s,B-p,C-q,D-r |
| Answer» Correct Answer - D | |
| 1781. |
Why `BBr_(3)` is a stronger Lewis acid as compared to `BF_(3)` through florine is more electronegative than bromine ? |
| Answer» The B atom in `BF_(3)`or `BBr_(3)` has only six electron inits valence shell and hence can accept a pair of electrons to complete its octet. Therefore, both `BF_(3)` and `BBr_(3)` at as Lewis acids. But in `BF_(3)`, the sizes of empty `2p-`orbital of B and the 2p-orbital of F containing the lone pair of electrons are almost identical and hence effective `ppi - ppi` bounding occurs. As a result, the lone pair of F is donate to B atom and hence the electrons deficiency of boron decreases . In contrast, in `BBr_(3)`. the size of 4p-orbital ov Br containing the lone pair of electrons is much bigger than the empty 2p-orbital of B and hence donation of lone pair of electrons of Br to B does not occur to an signification extent. As a result , the electron deficiency of B is much higher to `BBr_(3)` than in `BF_(3)`, and hence `BBr_(3)` i a stronger Lewis and than `BF_(3)`. | |
| 1782. |
Amongst the following, the maximum number of compounds which do not behave as Lewis acids are : `SnCl_(2),H_(3)BO_(3),AlCl_(3),CF_(4),SiF_(4),"CCl"_(4),BF_(3),SnCl_(4)` |
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Answer» Correct Answer - 3 Three , `CF_(4), SiCl_(4), "CCl"_(4)`. |
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| 1783. |
The tendency of `bF_(3), BCl_(3)` and `BBr_(3)` behave as Lewis acid decreases in the sequneceA. `BCl_(3) gt BF_(3) gt BBr_(3)`B. `BBr_(3) gt BCl_(3) gt BF_(3)`C. `BBr_(3) gt BF_(3) gt BCl_(3)`D. `BF_(3) gt BCl_(3) gt BBr_(3)` |
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Answer» Correct Answer - B As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, `BF_(3)` is the weakest Lewis acid. This is because of the `ppi-ppi` back bonding between the fully filled unutilised 2p-orbitals of F and vacant 2p-orbitals of boron which makes `BF_(3)` less electron dificients. Such back donation is not possible in case of `BCl_(3)`of `BBr_(2)` due to larger energy difference between their orbitals. Thus, these are more electrons deficient. Since on moving down the group the energy difference increases, the Lewis acid characters also increases. Thus, the tendency to behave as Lewis acid follows the order `BBr_(3) gt BCl_(3) gt BF_(3)` |
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| 1784. |
`{:(,"Column I",,"Column II"),((A),B_(2)H_(6),(p),"Borax"),((B),BF_(3),(q),"Lewis acid"),((C ),AlCl_(3),(r ),ppi-ppi " back bonding "),((D),H_(3)BO_(3),(s),NaBH_(4)):}` |
| Answer» Correct Answer - (A-s; B-q,r; C-q ; D-p,q) | |
| 1785. |
Consider the following statements I. `BBr_(3)` is stronger acid than `BF_(3)`. II. `ppi-ppi` back bonding occurs in the haldies of aluminium. III. Borazine is less reactive than boron. IV. Al is unstable in air and water. The set of incorrect statement isA. I and IIB. II and IIIC. III and IVD. None of the above |
| Answer» (c ) Borazine is more reactive than boron. Almunium is stable in air and water. | |
| 1786. |
Which one of the following arrangements does not give the correct picture of the trends indicated against it ?A. `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)` Oxidising powerB. `F_(2), Cl_(2) gt Br_(2) gt I_(2)` Electron gain enthaplyC. `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`D. `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)` Electronegtivity |
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Answer» Correct Answer - C Generally as the size of the atom increases bond dissociation energy decreases, so in halogens `I_(2)` have lowest bond dissociation energy, but the bond dissocitation energy of chlorine is higher than that of fluorine because in fluorine there is a greater repulsion between non-bonding electrons `(2p)`. Hence, the order of bond dissociation energy is `{:(,Cl_(2),gt,F_(2),gt,Br_(2),gt,I_(2)),(underset("energy" (kJ//mol))("Bond dissociation"),243,,153,,193,,151):}` |
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| 1787. |
Which of the following elements can be involved in `ppi-dpi` bonding?A. CarbonB. NitrogenC. PhosphorusD. Boron |
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Answer» Correct Answer - C Due to availacility of d-orbitals . |
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| 1788. |
Which of the following has `ppi-dpi` bonding?A. `NO_(3)^(-)`B. `BO_(3)^(3-)`C. `SO_(3)^(2-)`D. `CO_(3)^(2-)` |
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Answer» Correct Answer - C `SO_(3)^(2-)` is `bar(O) -underset(ul(O))underset(|)=O`, havind d (of S ) and p (of O) `pi`-bonds. |
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| 1789. |
Statement-1. `BF_(3)` is a weaker Lewis acid than `BCl_(3)`. Statement-2. The `ppi-ppi` back bonding is stronger in `BF_(3)` than in `BCl_(3)`.A. Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for statement-1.B. Statement-1 is True, Statement-2 is True , Statement-2 is not a correct explanation for statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 1790. |
Which of following elements can be involved in `ppi-dpi` bonding ?A. CarbonB. NitrogenC. PhosphorusD. Boron |
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Answer» Correct Answer - C Phosphorus belongs to 3rd period and hence contains d-orbitals which can which form `ppi-dpi` bonds. |
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| 1791. |
Statement-1. `(SiH_(3))_(H)N` is a weaker base than `(CH_(3))_(3)N`. Statement-2. Due to `ppi-dpi` back bonding the availability of electrons on the N atom in `(SiH_(3))+(3)N` decreases.A. Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for statement-1.B. Statement-1 is True, Statement-2 is True , Statement-2 is not a correct explanation for statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 1792. |
Although thallium (Z = 81) has only slightly higher atomic radius (170 pm) than that of indium (167 pm) but its ionization enthalpy `(589 kJ "mol"^(-1))` is much higher than that of indium `(558 kJ "mol"^(-1))` . Explain why / |
| Answer» Althrough on moving from `In (Z = 49)` to `Tl (Z = 81)`, the nuclear increases by 32 units `(81-49)` yet due to poor shielding effect on intervening 4f- and 5d-electrons, effective nuclear charge action on Tl is much higher than that on In and hence `Delta_(i)H_(1)` of `Tl` is much higher than that of `In`. | |
| 1793. |
The electronic configuration of an element is `1s^2 1s^2 2p^6 3s^2 3p^6 3d^(10) 4s^2 4p^3`. Its properties would be similar to which of the following elements?A. BoronB. OxygenC. NitrogenD. Chlorine |
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Answer» Correct Answer - C Outer electronic configuration of N is `2s^2 2p^3` similar to `4s^2 4p^3` of given element. |
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| 1794. |
Can `PCl_(5)` act as an oxidising as well as reducing agent ? Justify. |
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Answer» The oxidation state of P in `PCl_(5)" is "+5`. Since P has five electrons in its valence shell, therefore, it cannot increase its oxidation state beyond +5 by donating electrons, therefore, `PCl_(5)` cannot act as a reducing agent. However, it can decrease its oxidation number from +5 ot +3 or some lower value, therefore, `PCl_(5)` acts as an oxidisting agent. For example, it oxidises Ag to AgCl, Sn to `SnCl_(4)andH_(2)` to HCl. `2overset(0)Ag+overset(+5)PCl_(5)to2overset(+1)AgCl+overset(+3)PCl_(3),overset(0)Sn+2overset(+5)PCl_(5)tooverset(+4)SnCl_(4)+2overset(+3)PCl_(3)` `Poverset(+5)Cl_(5)+H_(2)tooverset(+3)PCl_(3)+2overset(+1)HCl` In These reacitons, oxidation number of P decreases form +5 in `PCl_(5)` to +3 in `PCl_(3)` and that of Ag and H increases from 0 to +1 and that of Sn from 0 to +4. Thus, `PCl_(5)` acts as an oxidising agent. |
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| 1795. |
What happens when `PCl_(5)` is heated ? |
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Answer» `PCl_(5)` has three equatorial (202 pm) and two axial (240 pm) bonds (Fig. 11.35, page 11/134). Since axial bonds are weaker (repelled by three bond pairs) than equatorial bonds (repelled only by two bond pairs), therefore, when `PCl_(5)` is heated, the less stable axial bonds break to form `PCl_(3)`. `PCl_(5)overset("Heat")toPCl_(3)+Cl_(2)`. |
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| 1796. |
In `PCl_(5)`, phosphorus is in `sp^(3)` d hybridised state but all its five bonds are not equivalent. Justify your answer with reason. |
| Answer» `PCl_(5)` has trigonal bipyramid geometry. (Fig. 11.35, page 11/134). It has three equatiorial and two axial bonds. Since each axial P-Cl bond is repelle by three bond pairs and each erquatorial P-Cl bond is repelled by only two bond pairs, therefore, axial bands are longer are longer (240 pm) than equatroial bonds (202 pm). Thus, all the five P-Cl bonds in `PCl_(5)` are not equivalent. | |
| 1797. |
Write a balanced equation for the hydrolytic reaction of `PCl_(5)` with heavy water. |
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Answer» It reacts with heavy water of form phosphorus oxychloride `(POCl_(3))` and deuterium chloride (DCl). `{:(PCl_(5)+D_(2)OtoPOCl_(3)+2DCl),(POCl_(3)+3D_(2)OtoD_(3)PO_(4)+3DCl))/(PCl_(5)+4D_(2)OtoD_(3)PO_(4)+underset("Deuterium chloride")(5DCl))` |
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| 1798. |
Are all the five bonds in `PCl_(5)` equivalent ? Justify. |
| Answer» `PCl_(5)` has trional bipyramid structure (Fig. 11.35, page 11//134). It has three equatorial bonds inclined at angle of `120^(@)` and two axial bonds inclined at an angle of `90^(@)`. Since two axial P-Cl bonds are repelled by three bond pairs each while three equatorial P-Cl bonds are repelled by two bond pairs each, therefore aixad bonds are longer (240 pm) and hence weaker as compared to more stable equatorial bonds (202 pm). Thus, all the five P-Cl bonds in `PCl_(5)` are not equivalent. | |
| 1799. |
Why is `N_(2)` less reactive at room temperature ? |
| Answer» Dinitrogen is inert of less reactive because the bond enthalpy of `N-=N` bond is very high. | |
| 1800. |
Chlorine is liberated, when we heatA. `KMnO_4+NaCl`B. `K_2Cr_2O_7+MnO_2`C. `Pb(NO_(3))_(2)+MnO_(2)`D. `K_2Cr_2O_7+HCl` |
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Answer» Correct Answer - D |
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