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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.
| 851. |
Which of following trihalides of nitrogen behaves as the weakest base?A. `NF_(3)`B. `NCl_(3)`C. `NBr_(3)`D. `Nl_(3)` |
| Answer» Correct Answer - A | |
| 852. |
`Sn+PCl_(5) to` |
| Answer» `Sn+2PCl_(5) to SnCl_(4)+2PCl_(3)` | |
| 853. |
`P_(4)+NaOH+H_(2)O to` |
| Answer» `P_(4)+3NaOH+3H_(2)O to 3NaHPO_(2)+PH_(3)` | |
| 854. |
Write the chemical reactions of `P_(4)O_(6)` with cold and hot water.A. Ortho phosphoric acidB. Meta phosphoric acidC. Pyrophosphoric acidD. Hypophosphorous acid |
| Answer» Correct Answer - A | |
| 855. |
Complete the following chemical equations : (i)` P_(4) (s) +NaOH (aq) +H_(2)O (l) to` `I^(-) (aq) +H_(2)O (l) +O_(3) (g) to` |
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Answer» (i) `P_(4)(s) +3NaOH (aq) +3H_(2)O(l) to underset(" Sod. Hypophosphite.")(3NaH_(2)PO_(2))(aq)+PH_(3)(g)` (ii) `2I^(-) (aq) +H_(2)O(l)+O_(3)(g) to I_(2)(s) +O_(2)(g)+2OH^(-)(aq)` |
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| 856. |
Red phosphorus is less reactive than yellow phosphorus because |
| Answer» Red phosphorus has a polymeric structure in which `P_(4)` molecules are polymerised. It is therefore, less reactive than white phosphorus in which `P_(4)` molecules exist independently. | |
| 857. |
You are given an inorganic mixture with the information that the cations of group II are absent. However, on passing `H_(2)S` gas through the mixture solution acidified with dilute HCl, yellowish turbidity apperas. Explain. |
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Answer» The mixture probably contains some nitrite or nitrate salt which in the acidic medium releases either nitrous acids or nitric acid. The acid oxidises hydrogen sulphide to sulphur which appears as yellow turbidity. `2HNO_(2)+H_(2)S to 2NO+2H_(2)O + underset(("Yellow tubidity"))(S)` `2HNO_(3) +H_(2)S to 2NO_(2)+2H_(2)O + underset(("Yellow turbidity")) (S)` |
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| 858. |
What are the oxidation states of phosphorus in the following: (i)`H_(3)PO_(3)` , (ii)`PCl_(3)` , (iii)`Ca_(3)P_(2)` (iv)`Na_(3)PO_(4)` , (v)`POF_(3)` |
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Answer» Let the oxidation state of p be x (i) `H_(3) PO_(3)` ` 3 + x+ 3(-2) = 0` ` 3 + x - 6 = 0` `x- 3 =0` `x= + 3` (ii) `PCl_(3)` `x + 3 (-1) =0` `x- 3 =0` `x- 3 =0` (iii) `Ca_(3)P_(2)` `3(+2) +2 (x) =0` `6+2x =0` `2x = -6` `x = -3` (iv) `Na_(3)PO_(4)` `3(+1) + x+ 4(-2) =0` `3+ x-8=0` `x-5 = 0` `x = +5` (v) `POF_(3)` `x-5 = 0` `x = +5` |
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| 859. |
In the preparation of `HNO_(3)`, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of `NH_(3)` will be.A. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - A `2NH_(3)+5//2O_(2) to 2NO+3H_(2)O` |
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| 860. |
A brown ring is formed in the ring test for `NO_(3)^(-)` ion. It is due to the formation ofA. `[Fe(H_(2)O)_(5)(NO)]^(2+)`B. `FeSO_(4)cdot NO_(2)`C. `[Fe(H_(2)O)_(4)(NO)_(2)]^(2+)`D. `FeSO_(4)cdot HNO_(3)` |
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Answer» Correct Answer - A `NO_(3)^(-)+3Fe^(2+)+4H^(+)rarrNO+3Fe^(3+)+2H_(2)O` `[Fe(H_(2)O)_(6)]^(2+)+NOrarrunderset("(Brown)")([Fe(H_(2)O)_(5)(NO)])^(2+)+H_(2)O` |
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| 861. |
Elements of group 15 form compounds in `+5` oxidatin state. However, bismuth forms only one well characterised compound in `+5` oxidation state. The compound isA. `Bi_(2)O_(5)`B. `BiF_(3)`C. `BiCl_(5)`D. `Bi_(2)S_(5)` |
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Answer» Correct Answer - B The stability of +5 oxidation state decreases down the group due to inert pair effect. The only well characterised Bi(V) compound is `BiF_(5)` as fluorine being most electronegative element is able to unpair ns electrons. |
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| 862. |
On heating ammonium dichromate and barium azide separately we getA. `N_(2)` in both casesB. `N_(2)` with ammonium dichromate and NO with barium azideC. `N_(2)O` with ammonium dichromate and `N_(2)` with barium azideD. `N_(2)O` with ammonium dichromate and `NO_(2)` with barium azide. |
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Answer» Correct Answer - A `(NH_(4))_(2)Cr_(2)O_(7)overset(Delta)rarrN_(2)+4H_(2)O+Cr_(2)O_(3)` `Ba(N_(3))_(2)overset(Delta)rarrBa+3N_(2)` |
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| 863. |
Lithium reacts with hydrogen azide to produceA. `Li_(3)N`B. `LiN_(2)`C. `LiN_(3)`D. `Li_(3)N_(2)` |
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Answer» Correct Answer - 3 Hydrogen azide `HN_(3)` (formerly called hydrazoic acid)dissociates slightly in aqueous solution `(pK_(a)~=5)` .It behaves as a week acid (of similar strength to acetic acid). It reacts with electropositive metals, forming salts called azides, but unlike other acid + metal reactions, no hydrogen is evolved: `6HN_(3)+4Lirarr4LiN_(3)rarrunderset("Lithium azide")(4LiN_(3))+2NH_(3)+2N_(2)` |
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| 864. |
The correct geometery and hybridizationn for `XeF_(4)` areA. Octahedral , `sp^(3)d^(2)`B. Trigonal bipyramidal, `sp^(3)d`C. Planar triangle `sp^(3)d^(3)`D. Square planar, `sp^(3)d^(2)` |
| Answer» Correct Answer - D | |
| 865. |
The correct geometery and hydizatiojn for `XeF_(4)` areA. octahedral `sp^(3) d^(2)`B. trigonal bipyramidel `sp^(3) d`C. planar triangle ,`sp^(3) d^(3)`D. squarer planar `sp^(3)d^(2)` |
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Answer» Correct Answer - 1 `XeF_(4)implies` [steric number =`4` atoms +`2` lone pairs]`implies` geometery is octahedral ,shape is square planar and hybrization id `sp^(3) d^(2)` or Number fo hybrid orbitals `=(1)/(2)[VE+MA-c+a]` `=(1)/(2)[8+4-0+0]=6sp^(3)d^(2)` hydrization |
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| 866. |
(a) Why does `PCl_3` fume in moisture ? (b) Are all the five bonds in `PCl_5` molecule equivalent ? Justify your answer. ( c) How do you account for the reducing behaviour of `H_3 PO_2` on the basic of its structure ? (d) Give the disproportional reaction of `H_3PO_3`. |
| Answer» In `H_(3)PO_(2)`, two H atoms are bonded directly to P atom which imparts reducing character to the acid. | |
| 867. |
Which of the following has the lowest melting point?A. `Pb`B. `Sn`C. `Ge`D. `Si` |
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Answer» Correct Answer - B The overall order of melting points is `C gt Si gt Ge gt Pb gt Sn` The first three elements of group `14` have very high melting points, a characteristic of network covalent bonding, whereas the two metals in the group have low melting points. |
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| 868. |
Which of the following group `14` element has the lowest first ionization enthalpy?A. `Si`B. `Ge`C. `Sn`D. `Pb` |
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Answer» Correct Answer - C The first ionization enthalpy decreases on moving down the group from `C` to `Sn` while there is a slight increase the effect of increased nuclear charge `(82-50=32 units)` outweighs the shielding effect of `4f` and `5d` electrons. In fact, all other ionization enthalpies of `Pb` are higher than those of `Sn`. |
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| 869. |
Which of the following compounds has the lowest boiling point ?A. `CaCl_(2)`B. `CaBr_(2)`C. `CaI_(2)`D. `CaF_(2)` |
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Answer» Correct Answer - B `Cal_(2)` has the maximum covalent character and the least boiling point. |
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| 870. |
Which of the following is least likely to behave as Lewis acid?A. `H_(2)O`B. `NH_(3)`C. `BF_(3)`D. `OH^(-)` |
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Answer» Correct Answer - C `BF_(3)` behaves as a Lewis acid and not as a Lewis base. |
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| 871. |
Which is mismatched `,`A. halogens- group 17B. Chalcogents-group 16C. Pnicogens - group 14D. Noble gases - group 18 |
| Answer» Correct Answer - C | |
| 872. |
Which is mismatched regarding the formula?A. Epsom salt `-MgSO_(4).7H_(2)O`B. Bartyte-`BeSO_(4)`C. Copperr pyrite-`CuFeS_(2)`D. Galena`-PbS` |
| Answer» Correct Answer - B | |
| 873. |
The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in `[B(OH_(4))]^(-)` and the geometry of the complex are respectively.A. `sp^(3)`, tetrahedralB. `sp^(3)`, square planarC. `sp^(3)d^(2)`, octahedralD. `dsp^(2)`, square planar |
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Answer» Correct Answer - A Structure of `B(OH)_(4)^(-)` is `{:(" "OH),(" |"),(HO-B^(-)-OH),(" |"),(" "OH),("4 bond pair + 0 lone pair"):}` Hybridisation - `sp^(3)` Geometry - Tetrahedral. |
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| 874. |
Which of the following species has a linear shape?A. `O_(3)`B. `NO_(2)^(-)`C. `SO_(2)`D. `NO_(2)^(+)` |
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Answer» Correct Answer - 4 `|{:("Species","Steric Number","Geometry","Shape"),(O_(3),2+1,"Trigonal planar","V-Shaped"),(NO_(2)^(-),2+1,"Trigonal planar","V-Shaped"),(SO_(2),2+1,"Trigonal planar","V-Shaped"),(NO_(2)^(+),2+1,"Linear","Linear"):}|` This question can also be credited to chemical bonding |
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| 875. |
What is the shape `ClF_(3)` molecule ? |
| Answer» Correct Answer - T-shaped | |
| 876. |
Carbon monoxide cab be prepared in the laboratory by the action of conc. `H_(2)SO_(4)` onA. `NaCN`B. `Fe(CN)_(2)`C. `K_(3)[Fe(CN)_(6)]`D. `K_(4)[Fe(CN)_(6)]` |
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Answer» Correct Answer - D `{:(K_(4)[Fe(CN)_(6)]+3H_(2)SO_(4)to2K_(2)SO_(4)+FesO_(4)+6HCN),([HCN+2H_(2)OtoHCOOH+NH_(3)]xx6),([HCOOH to H_(2)O+CO]xx6),([2NH_(3)+H_(2)SO_(4)to(NH_(4))_(2)SO_(4)]xx3),(bar(K_(4)[Fe(CN)_(6)]+6H_(2)OSO_(4)+6H_(2)Oto 2K_(2)SO_(4)+FeSO_(4)+3(NH_(4))_(2)SO_(4)+6CO)):}` |
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| 877. |
`P_(4)+SO_(2)Cl_(2) to` |
| Answer» `P_(4)+10SO_(2)Cl_(2) to 4PCl_(5)+10SO_(2)` | |
| 878. |
`PH_(3)+HgCl_(2) to` |
| Answer» `2PH_(3)+3HgCl_(2) to underset("Mercuric phosphide")(Hg_(3)P_(2))+6HCl` | |
| 879. |
Sodium nitrate decomposes above `800^(@)C` to give :A. `N_(2)`B. `O_(2)`C. `NO_(2)`D. `Na_(2)O` |
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Answer» Correct Answer - A::B::D are correct answer. `2NaNO_(3) overset(500^(@)C) to 2NaNO_(2)+O_(2)` `2NaNO_(2) overset(800^(@)C) to Na_(2)O +3//2O_(2)+N_(2)` |
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| 880. |
The reduction of an oxide by aluminium is calledA. it highly stable natureB. its highly unstable natureC. its amphoteric natureD. its highly explosive nature. |
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Answer» Correct Answer - A `AI_(2)O_(3)` is highly stable and passive to chemical reactions. |
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| 881. |
How will you prepare `PH_(3)` from (i) Metal phosphides (ii) `H_(3)PO_(2) (iii) H_(3)PO_(3)` ? Draw the structure of `PH_(3)`. |
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Answer» (i) `Ca_(3)P_(2)+6H_(2)O to 3Ca (OH)_(2)+2PH_(3)` (ii) `2H_(3)PO_(2) overset("Heat") to H_(3)PO_(4)+PH_(3)` (iii) `4H_(3)PO_(3) overset("Heat")to 3H_(3)PO_(4)+PH_(3)` |
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| 882. |
How will you prepare fullerene? |
| Answer» By action of laser beam on a sample of graphite in inert atmospere | |
| 883. |
Name the superior quality of coal which burns with non-smoky flame. |
| Answer» Correct Answer - Anthracite. | |
| 884. |
`H_(3)PO_(2)` acts as a monobasic acid. Explain. |
| Answer» Hypophorus acid or phosphinic acid acts as a monobasic acid because the central P atom has only one-OH group attached to it . | |
| 885. |
Why is `N_(2)O_(5)` more acidic than `N_(2)O_(3)` ? |
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Answer» Both these oxides dissovle in water to form acids. `N_(2)O_(3)+H_(2)O overset(+5) to 2HNO_(2)` `N_(2)O_(5)+H_(2)O ot 2HNO_(3)` `N_(2)O_(5)` is more acidic because the oxidation state of nitrogen `(+5)` is more than in `N_(2)O_(3) (+3)`. Greater the oxidation state of the central atom, more is the acidic strength of the oxide. |
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| 886. |
Give reason for the following : (i) Chlorine water loses its colour on standing. (ii) The is more acidic than `H_(2)S`. |
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Answer» (i) Chlorine water (light greenish in colour) onstanding changes to HCl and HClO. Both these products are colourless. `Cl_(2) +H_(2)O to HCl +HClO` (ii) The bond dissociation enthalpy of H-Te bond is less than that of H-S bond. Therefore, `H_(2)Te` is more acidic as compared to `H_(2)S`. |
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| 887. |
Which of the following gases is envolevd when `Al` is boiled with caustic potash ?A. `H_(2)`B. `O_(2)`C. Both `O_(2)` and `H_(2)O`D. `H_(2)O` vapor |
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Answer» Correct Answer - A `2Al + 2KOH + 6H_(2)O overset(Delta)rarr 2NaAlO_(2) . 2H_(2)O + 3H_(2)` |
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| 888. |
Which of the following reactions of `Al` is known as the thermite reaction?A. Reaction with `N_(2)`B. Reaction with `O_(2)`C. Reaction with `NH_(3)`D. Reaction with `NaOH` |
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Answer» Correct Answer - B In thermite reaction, `Al` burns readily in air or oxygen. The rection is strongly exothermic: `2Al(s) + (3)/(2) O_(2)(g) rarr Al_(2)O_(3) + "Energy", Delta_(T)H^(@)` `= -1670 kJ mol^(-1)` It evolves so much heat energy that it can be dangerous as the `Al` becomes white hot and often causes fire. The very strong affinity of `Al` for oxygen is used in the metallurgical extraction of other metals from their oxides: `8Al + 3Mn_(3)O_(4) rarr 4Al_(2)O_(3) + 9Mn` `2Al + Cr_(2)O_(3) rarr Al_(2)O_(3) + 2Cr` |
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| 889. |
Complete the following reactions : (i) `C_(2)H_(4)+O_(2) to` (ii) `Al +O_(2) to` |
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Answer» `(i) C_(2)H_(4)+3O_(2) overset("heat")to 2CO_(2)+2H_(2)O` `(ii) 4Al +3O_(2) overset("heat") to 2Al_(2)O_(3)` |
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| 890. |
After `Al`, the most abundant group `13` elements isA. `Tl`B. `In`C. `Ga`D. `B` |
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Answer» Correct Answer - C Gallium is twice as abundant as boron, but `In` and `Tl` are much less common. |
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| 891. |
Anhydrous alumina `(Al_(2)O_(3))` is known asA. emeraldB. corundumC. sapphireD. ruby |
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Answer» Correct Answer - B It is natural aluminium oxide. It is a crystalline substance nearly as hard as diamond, used as an abrasive. Ruby is a red form of corundum, `Al_(2)O_(3)`, that owes its color to traces of `Cr`. Sapphire is a natural crystalline form of blue, transparent corundum, the color being due to traces of cobalt or other metals. |
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| 892. |
Bleaching of flowers by `Cl_(2)` is permanent while bleaching by `SO_(2)` is temporary, why ? |
| Answer» Bleaching by `Cl_(2)` id due to oxidation. The colour of bleached flower can be restored . The bleaching by `SO_(2)` is due to reduction. The colour of bleached article can be restored on exposure to air or oxygen (oxidation). | |
| 893. |
When `CI_(2)` gas reacts with hot and concentrated sodium hydroxide solution ,the oxidation number of chlorine changes from:A. Zero to `+1` and Zero to `-5`B. Zero to `-1` and Zero to `+5`C. Zero to `-1` and Zero to `+3`D. Zero to `+1` and Zero to `-3` |
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Answer» Correct Answer - 2 Chlorine reacts with hot and conc. Sodium hydroxide solution to yield sodium chloride sodium chlorate: `underset("O.N. of Cl is 0")(3Cl_(2)(g))+6Na^(+)OH^(-) (aq.) rarr underset("O.N. of Cl is -1")(5Na^(+)Cl^(-) (aq.))+underset("O.N. of Cl is +5")(Na^(+) ClO_(3)^(-)(aq.))+3H_(2)O(l)` However, sodium chloride and sodium hypochlorite are formed when chlorine gas reacts with cold and dilute sodium hydroxide solution: `underset("O.N. of Cl is 0")(Cl_(2)(g))+2Na^(+)OH^(-)(aq.) rarr underset("O.N. of Cl is -1")(Na^(+)Cl^(-)(aq.))+underset("O.N. of Cl is +1")(Na^(+) ClO^(-) (aq.))+H_(2)O(l)` |
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| 894. |
Chlorine on reaction with hot and concentrated `NaOH` yieldsA. `NaCl` and `NaCIO_(3)`B. `NaCl` and `NaCIO`C. `NaCl` and `NaCIO_(2)`D. `NaCl` and `NaCIO_(4)` |
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Answer» Correct Answer - 1 Chlorate are formed with hot and concentrated alkali solution : `3Cl_(2)+6NaOH rarr 5NaClO_(3)+3H_(2)O` It is disproportionate reactions as `Cl` from zero oxidation state is changed to `-1`and `+5` oxidation states: `6Cl_(2)+underset("Hot and conc.")(6Ca(OH)_(2))rarr5CaCl_(2)+Ca(ClO_(3))_(2)+2H_(2)O` Hypochlorities are formed with cold and dilute solution of alkali: `Cl_(2)+2NaOH rarr NaCl+NaCIO+H_(2)O` |
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| 895. |
When on excess of chlorine is treated with ammonia ,the products formed areA. `N_(2)` and `NH__(4)Cl`B. `NCl_(3)` and `HCl`C. `N_(2)` and `HCl`D. `NH_(3)Cl` and `NH_(4)Cl` |
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Answer» Correct Answer - 2 The products are nitrogen and ammonium chloride when ammonia is in excess: `{:(2NH_(3)+3Cl_(2) rarr N_(2)+6HCl),(NH_(3)+HCl rarr NH_(4)Cl"]"xx6),(bar(8NH_(3)+3Cl_(2) rarr N_(2)+6NH_(4)Cl)):}` The produced are nitrogen trichloride and hydrogen chloride when chloride is in excess: `NH_(3)+3Cl_(2)rarrNCl_(3)+3HCl` |
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| 896. |
From `B_(2)H_(6)`, all the following can be prepared exceptA. `H_(3)BO_(3)`B. `B_(2)(CH_(3))_(4)H_(2)`C. `B_(2)(CH_(3))_(6)`D. `NaBH_(4)` |
| Answer» Correct Answer - C | |
| 897. |
Which of the following statements is not correct?A. Helium has the lowest boiling point among the noble gasesB. Argon is used in electric bulbsC. Krypton is obtained during radioactive disintegrationD. `Xe" forms "XeF_(6).` |
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Answer» Correct Answer - C Kr is not obtained during radioactive disintegration. |
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| 898. |
Which of the following statements are incorrect ?A. `SO_(3)` is a stronger oxidising agent and more acidic than `SO_(2)`.B. Selenium forms only two oxoacids i.e., selenous acid `(H_(2)SeO_(3))` and selenic acid `(H_(2)SeO_(4))`.C. The acidic strength and oxidising power of oxoacids is greater in +6 oxidation state than in +4 oxidation state.D. The thermal stability of oxides of group 16 elements decreases in the order : `SO_(2) gt SeO_(2) gt TeO_(2) gt PoO_(2)` |
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Answer» Correct Answer - D Thermal stability of the oxides of group 16 elements decreases in the order: `SO_(2) gt TeO_(2) gt SeO_(2) gt PoO_(2)` and not in the order as mentioned. `TeO_(2)` decomposes at temperature `gt 733^(@)C" while "SeO_(2)" has " ~ 600^(@)C` as the decomposition temperature. `PoO_(2)` decomposes at `500^(@)C` |
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| 899. |
White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calcualte the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.A. 200 gB. 400 gC. 219 gD. 100 g |
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Answer» Correct Answer - C Phosphorus reacts with limited `Cl_(2)` as: `P_(4)+6Cl_(2)rarr4PCl_(3)" …(i)"` `PCl_(3)` is hydrolysed as : `(PCl_(3)+3H_(2)OrarrH_(3)PO_(3)+3HClxx4)/(P_(4)+6Cl_(2)+12H_(2)Orarr4H_(3)PO_(3)+12HCl)" ....(ii)"` ltbr Moles of white `P=(62)/(124)=0.5` mol 1 mol of white `P_(4)` produces = 12 mol of HCl 0.5 mol of white `P_(4)` will produce `12 xx 0.5 = 6` mol of HCl Mass of `HCl = 6 xx 36.5 = 219 g` |
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| 900. |
Calculate the weight of HI obtained by the reaction of 62 g of red phosphorus with iodine in presence of water. |
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Answer» `P_(4)+6I_(2) to 4PI_(3)` `PI_(3)+3H_(2)O to H_(3)PO_(3)+3HI]xx4` `P_(4)+6I_(2)+12H_(2)O to 4H_(3)PO_(3)+12HI` `31xx4 " " 12 (1+127)` `=124 g " " =1536 g` 124 g or red phosphorus form HI =1536 g 62 g of red phosphorus form HI`=((1536g))/((124 g))xx(62g)=768g` |
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