InterviewSolution
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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.
| 201. |
Assertion (A) : Stability of `NH_(3)` is greater than `PH_(3)` Reason (R) : M-H bond energy increases down the group in the hydries of pnicogens.A. Both A and R are correct and R is the correct explanation of AB. Both A and R are correct and R is not the correct explanation of AC. A is correct R is wrongD. A is wrong R is correct |
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Answer» Correct Answer - C M-H bond energy decreases down the group in the hydrides of pnicogens. |
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| 202. |
The wrong statement about `N_(2)O` is :A. It is called laughing gasB. It is called nitrous oxideC. It is a linear moleculeD. It is a more reactive oxide |
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Answer» Correct Answer - D `N_(2)O` is relatively unreactive |
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| 203. |
Which of the following statement is wrong ?A. Since 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 |
| Answer» Correct Answer - A | |
| 204. |
`E+D darr overset(H_(2)O)larr B overset(P)larr Ca overset(N_(2))rarr A overset(H_(2)O)rarr C darr +E` When gas C is passed through bleaching powder suspension, another gas F comes out, which is not obtained byA. heating `NH_(4)NO_(3)`B. heating `NH_(4)NO_(2)`C. heating `(NH_(4))_(2) Cr_(2)O_(7)`D. heating `Ba(N_(3))_(2)` |
| Answer» Correct Answer - A | |
| 205. |
`E+D darr overset(H_(2)O)larr B overset(P)larr Ca overset(N_(2))rarr A overset(H_(2)O)rarr C darr +E` Which of the following characteristic is same for gases C and D?A. presence of `PH_(3)`B. presence of `P_(2)H_(4)`C. presence of `O_(2)`D. presence of `H_(2)` |
| Answer» Correct Answer - C | |
| 206. |
Assertion (A) : `PH_(3)` is more basic than `NH_(3)` Reason (R) : EN of N is more than that of PA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - D | |
| 207. |
Which of the following statement is wrong ?A. The stablity of hydride increases from `NH_(3)` to `BiH_(3)` in group 15 of the periodic tableB. nitrogen cannot form `d pi - p pi` bondsC. single `N-N` bond is weaker than the single `P-P` bondD. `N_(2)O_(4)` has two resonance structures |
| Answer» Correct Answer - A | |
| 208. |
In Nitrogen family the H-M-H angle in the hydrides `MH_(3)` gradually becomes closer to `90^(@)` on going from N to Sb. This due toA. The basic strength of the hydrides increasesB. Due to the increase in the size of central atom M and increase in its electronegativityC. The bond energies of M-H increaseD. Pure P orbital participating in the bonding |
| Answer» Correct Answer - D | |
| 209. |
The correct order of reducing abilites of hydrides of group 15 elements isA. `NH_(3)ltPH_(3)lt AsH_(3)lt SbH_(3)ltBiH_(3)`B. `NH_(3)gt PH_(3)gtAsH_(3)gt SbH_(3)gt BiH_(3)`C. `NH_(3)lt PH_(3)gt AsH_(3)gt SbH_(3) gt BiH_(3)`D. `SbH_(3)gt BiH_(3)gt AsH_(3)gtNH_(3)gt PH_(3)` |
| Answer» Correct Answer - A | |
| 210. |
Assertion (A) : `NH_(3)` is liquid while the other hydrides V-A group elements are gases at room temp Reason (R) : `NH_(3)` possess inter molecular H-bonds in liquid stateA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - D | |
| 211. |
Sodium hexametaphosphate is known asA. CalgonB. PermutitC. NataliteD. Nitrolim |
| Answer» Correct Answer - A | |
| 212. |
Assertion (A) : The hydrides of VA group elements are good reducing Agents Reason (R) : `NH_(3)` is a weak reducing Agent among the hydrides of VA groupA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - B | |
| 213. |
Which of the following is correct about 15th group Hydrides (from ammonia to Bismuthine)A. Their thermal stability gradually increaseB. Their ease of preparation gradually increaseC. The electron pair donating Nature gradually decreaseD. The bond energies gradually increase |
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Answer» Correct Answer - C From `NH_(3)` to `BiH_(3)` the electron pair donating nature graduall decreases. |
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| 214. |
Assertion (A) : `P_(4)` is more reactive than `N_(2)` Reason (R) : `P-P` bonds are relatively weaker than `N-=N`A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - A | |
| 215. |
By the reduction of `HNO_(3)` to `NO_(2)` the number of moles of electrons involved per mole of `HNO_(3)` isA. 8B. 6C. 3D. 1 |
| Answer» Correct Answer - D | |
| 216. |
`Zn+conc.HNO_(3) rarr Zn(NO_(3))_(2)+X+H_(2)O` `Zn+dil.HNO_(3) rarr Zn(NO_(3))_(2)+Y+H_(2)O` Compounds X and Y are respectively:A. `NO_(2),NO,N_(2)O,NO`B. `NO,NO,NO_(2),NO_(2)`C. `NO_(2),NO_(2),NO_(2),NO_(2)`D. `NO,NO_(2),N_(2)O,NO_(2)` |
| Answer» Correct Answer - D | |
| 217. |
Assertion (A) : The basic nature of VA group hydrides decreases from ammonia to bismuthine Reason (R) : Availbility of lone pair on the central atom is hydrides decreases down tha group.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - A | |
| 218. |
The number of of P-O bonds and lone pair of electrons present in `P_(4)O_(6)` moleculeA. 12, 16B. 12, 12C. 8, 8D. 12, 4 |
| Answer» Correct Answer - A | |
| 219. |
Which of the following molecules does NOT contain a lone pair of electron ?A. `PCl_(3)`B. `NCl_(3)`C. `AsCl_(3)`D. `PCl_(5)` |
| Answer» Correct Answer - D | |
| 220. |
Nitrogen forms the largest number of oxides as it is capable of forming stable multiple bonds with oxygen. They range of `N_(2)O` (O.S of nitrogen +1) through `NO, N_(2)O_(3),NO_(2),N_(2)O_(4)` "to" `N_(2)O_(5)` (O.S of nitrogen +5). Following points are improtant regarding the study of oxides of nitrogen. (a) All oxides of nitrogen expect `N_(2)O_(5)` are endothermic as a large amount of energy is required to dissociate the stable molecule of oxygen and nitrogen. (b) The small electronegativity difference between oxygen and nitrogen make N-O bond easily breakle to give oxygen and hence oxides of nitrogen are said to be better oxidising agents. (c) Expect `N_(2)O_(5)`, all are gases at ordinary temperature. `N_(2)O_(3)` is stable only at lower temperature (253K). (d) Expect `N_(2)O` and NO which are neutal oxides, all are acidic oxides which dissolve in water forming coresponding oxy acids. (e) They are also good example for illustrating the concept of resonance. The gas which is acidic in nature is :A. `NO`B. `N_(2)O`C. `N_(2)O_(3)`D. both a and c |
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Answer» Correct Answer - C `NO,N_(2)O` - neutral oxides `N_(2)O_(5)` - acidic oxide `N_(2)O_(3) +KOH rarr KNO_(2)+H_(2)O` `NO_(2)` (gas) - paramagnetic `overset((+4))(NO_(2))` : acts both as a reducing agent and oxidising agent Bond angle in `NO_(2)=132^(@)` |
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| 221. |
`4Zn+10HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`. In this reaction one mole of `HNO_(3)` is reduced byA. 32g ZnB. 64g ZnC. 128g ZnD. 256g Zn |
| Answer» Correct Answer - D | |
| 222. |
In `NO_(3)^(-)` ion, the number of bond pair and lone pair of electrons on nitrogen atom are:A. 2,2B. 3,1C. 1,3D. 4,0 |
| Answer» Correct Answer - D | |
| 223. |
`H_(3)PO_(4) overset(Delta)rarr X overset(Delta)rarrY` gives a white precipitate with silver nitrate solution. When ammonium molybdate and conc `HNO_(3)` were added to `H_(3)PO_(4)`, yellow precipitate A is formed. The formula of A is:A. `(NH_(4))_(3)PO_(3).12 MoO_(3)`B. `(NH_(4))_(3)PO_(4).12 MoO_(6)`C. `(NH_(4))_(3)PO_(4).12MoO_(4)`D. `(NH_(4))_(3)PO_(4).12MoO_(3)` |
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Answer» Correct Answer - D `X=H_(4)P_(2)O_(7) , Y = HPO_(3)` A = ammonium 12 - molybdo phosphate `[(NH_(4))_(3)PO_(4).12MoO_(3)]` |
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| 224. |
Among the oxides of nitrogen `N_(2)O, NO and NO_(2)`, molecules with unpaired electrons are:A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true and (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - C | |
| 225. |
The following are aresome statement about oxides of VA group element I) `N_(2)O` molecule is linear II) `NO_(2)` molecule is angular III) `N_(2)O_(5)` molecule is angular The correct combination isA. All are correctB. I & III are correctC. II & III are correctD. I & II are correct |
| Answer» Correct Answer - D | |
| 226. |
When `NH_(4)NO_(3)` is gently heated, an oxide of Nitrogen is formed. What is the oxidation state of Nitrogen in this oxide ?A. `+4`B. `+2`C. `+3`D. `+1` |
| Answer» Correct Answer - D | |
| 227. |
`H_(3)PO_(4) overset(Delta)rarr X overset(Delta)rarrY` gives a white precipitate with silver nitrate solution. X is aA. `PH_(3)`B. `H_(4)P_(2)O_(7)`C. `H_(3)PO_(4)`D. `P_(4)O_(10)` |
| Answer» Correct Answer - C::D | |
| 228. |
`H_(3)PO_(4) overset(Delta)rarr X overset(Delta)rarrY` gives a white precipitate with silver nitrate solution. Y can form polyanions. The one that represents the polyanion is:A. `P_(3)O_(9)^(3-)`B. `P_(2)O_(7)^(2-)`C. `P_(4)O_(13)^(6-)`D. `P_(3)O_(10)^(5-)` |
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Answer» Correct Answer - A `X=H_(4)P_(2)O_(7) , Y = HPO_(3)` A = ammonium 12 - molybdo phosphate `[(NH_(4))_(3)PO_(4).12MoO_(3)]` |
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| 229. |
`H_(3)PO_(4) overset(Delta)rarr X overset(Delta)rarrY` gives a white precipitate with silver nitrate solution. X is aA. di basic acidB. tri basic acidC. mono basic acidD. tetra basic acid |
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Answer» Correct Answer - D `X=H_(4)P_(2)O_(7) , Y = HPO_(3)` A = ammonium 12 - molybdo phosphate `[(NH_(4))_(3)PO_(4).12MoO_(3)]` |
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| 230. |
VA group elements form tri oxides and pentoxides of the type `M_(2)O_(3)` and `M_(2)O_(5)` respectively (M-VA group element). The trioxides and pentoxides of nitrogen are monomers and that of P,As,Sb are dimers. Nitrogen forms various oxides ranging from `NO` to `N_(2)O_(5)`. Oxides of phosphorous have cage like stuctures. Number of P-O-P bonds in `P_(4)O_(10)`: |
| Answer» Number of P-O-bonds : 6 | |
| 231. |
The correct order of the oxidation states of nitrogen in `NO, N_(2)O, NO_(2)` and `N_(2)O_(3)` is : |
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Answer» Correct Answer - C `N_(2)O_(5)`: does not contain N- N linkage |
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| 232. |
`P_(4)("excess")+Cl_(2)rarrX` `P_(4)+Cl_(2)("excess")rarrY` Then the correct statement isA. Hydrolysis of X and Y forms oxyacids of phoshorus having equal reducing natureB. Hydrolysis of X and Y forms oxyacids of phosphorus contains equal number of `pi` bondsC. Hydrolysos of X and Y form oxyacids of phosphorus which have equal basicityD. Hydrolysis of X and Y form oxyacids of phosphorus of X and Y form oxyacids of phosphorus which have equal oxidation state of phosphorus atoms. |
| Answer» Correct Answer - B | |
| 233. |
Which of the following statements are correct about the reaction between the copper metal and dilute `HNO_(3)`?A. The principal reducing product is `NO` gasB. Cu metal is oxidised to `Cu^(2+)` (aq) ion which is blue in colourC. NO is paramagnetic and has one unpaired electron in antibonding molecular orbitalD. NO reacts with `O_(2)` to produce `NO_(2)` which is linear in shape |
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Answer» Correct Answer - A::C `NO_(2)-` bent molecule |
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| 234. |
Which of the following statements regarding `N_(2)O_(4)` is/are correct?A. It is planar moleculeB. It is used as non-aq solventC. It involves N-N bond which is shorter than the N-N bond in hydrazineD. It is dimer of `NO_(2)` |
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Answer» Correct Answer - A::B::D `N -N` bond length in `N_(2)H_(4)` is shorter than the `N-N` bond in hydrazine |
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| 235. |
`P_(4)("white")+X rarrA+SO_(2)+S_(2)Cl_(2)` `P_(4)("white")+Yrarr B+SO_(2)` Then identify Y and A ?A. `Y = SOCl_(2) , A =` colourless oily liquidB. `Y = SO_(2) Cl_(2) , A` = colourless oily liquidC. `Y = SOCl_(2) , A =` Yellowish white powderD. `Y = SO_(2)Cl_(2) , A = Y` Yellow white powder |
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Answer» Correct Answer - B `P_(4) + SOCI_(2) rarr PCI_(3) +SO_(2) +S_(2)CI_(2)` `P_(4)+SO_(2)CI_(2) rarr PCI_(5) +SO_(2)` `PCI_(3)-` colourless liquid, `PCI_(5) -` yellow |
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| 236. |
`P_(4)+SO_(2)CI_(2) rarr X+SO_(2)`, then X =_____A. `PCI_(3)`B. `PCI_(5)`C. `SO_(2)`D. `SCI_(2)` |
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Answer» Correct Answer - C Orthophosphoric acid: `H_(3)PO_(4)` Hypophosphoric acid: `H_(4)P_(2)O_(6)` |
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| 237. |
In which of the following reactions, the products shown are incorrect?A. `2PCI_(3)+O_(2) rarr 2POCI_(3)`B. `NaNH_(2)+N_(2)O overset("moderate heating")rarr NaOH +N_(2)`C. `2H_(3)PO_(4) overset(220^(@)C)rarr H_(4)P_(2)O_(7)+H_(2)O`D. `NaOCI +NH_(3) overset(OH^(-))underset("gelatin")rarr N_(2)H_(4)+NaCI +H_(2)O` |
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Answer» Correct Answer - B `2NaNH_(2)+N_(2)O rarr NaN_(3)+NH_(3)+NaOH` |
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| 238. |
Au and Pt dissolves in aqua regia forming the soluble compounds X and Y respectively. The oxidation states of Au and Pt in X and Y are:A. `+1,+2`B. `+2,+4`C. `+3,+2`D. `+3,+4` |
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Answer» Correct Answer - D `X =H[AuCI_(4)], Y = H_(2) [PtCI_(6)]` |
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| 239. |
The nitrate which when heated gives off a gas (or) a mixture of gases which cannot relight a glowing splinter isA. Sodium nitrateB. Ammonium nitrateC. Lead nitrateD. Potassium nitrate |
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Answer» Correct Answer - B `NH_(4)NO_(3)overset(Delta)(rarr)N_(2)O+2H_(2)O.N_(2)O` is non-combustable |
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| 240. |
When compared to `CN^(-),NO^(+) and CO,N_(2)` is chemically inert. Explain. |
| Answer» Nitrogen is inert because of non-polar bond nature and high bond dissociation energy `(N -=N)`. The rest all are polar molecules which can easily dissociate to participate in chemical reactions. | |
| 241. |
A colourless paramagnetic gas among the followingA. Nitric OxideB. Nitrous OxideC. Nitrogen dioxideD. Dinitrogen trioxide |
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Answer» Correct Answer - A `NO=5+6=11` paramagnetic, colourless oxide |
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| 242. |
Describe being odd electron molecule, `NO` is colourless. Explain. |
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Answer» In NO molecule, the odd electron is involved in bonding between two bonded atoms, and the excitation of it is difficult (possible only in uv region) In `NO_(2)` the unpaired electron can be excited easily by absorption of vissible light Thus NO is colourless and `NO_(2)` is coloured. |
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| 243. |
(a) Why does `NO_2` dimerise ? (b) In what way can it be proved that `PH_3` is basic in nature ? |
| Answer» `NO_(2)` contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to a stable `N_(2)O_(4)` molecule, with even number of electrons. | |
| 244. |
The shape and bind angle of white phosphorou molecule isA. Linear and `180^(@)`B. Trigonal planar and`120^(@)`C. Tetraheral and `109^(@) 28^(1)`D. Tetrahedral and `60^(@)` |
| Answer» Correct Answer - D | |
| 245. |
Phosphine reacts with copper sulphate solution to formA. Copper phosphide `(Cu_(3)P_(2))`B. Coper phosphateC. Copper phosphiteD. Copper pyrophosphate |
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Answer» Correct Answer - A `3CuSO_(4)+ 2PH+(3) rarr Cu_(3)P_(2)+3H_(2)SO_(4)` |
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| 246. |
When white phosphorous is reacted with caustic soda.A. `PH_(3)` and `NaH_(2)PO_(2)` are formedB. `P_(2)O_(5) and Na_(2)HPO_(3)` are formedC. This reaction is an example of oxidation and reductionD. This reaction is an example of neutralisation |
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Answer» Correct Answer - A::C `overset(0)(P_(4))+ NaOH +H_(2)O rarr overset(-3)(PH_(3))+ NaH_(2)overset(+1)(PO_(2))` |
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| 247. |
Aqueous NaOH reacts with white Phosphorous to form Phosphine andA. `NaH_(2)PO_(2)`B. `P_(2)O_(5)`C. `Na_(3)PO_(3)`D. `P_(2)O_(3)` |
| Answer» Correct Answer - A | |
| 248. |
White phosphorous reacts with caustic soda to give phosphine and sodium hypophosphite In this reaction phosphorous undergoesA. OxidationB. ReductionC. BothD. None of these |
| Answer» Correct Answer - C | |
| 249. |
Which of the following elements combines directly with nitrogen to form its nitride ?A. MgB. AlC. LiD. Fe |
| Answer» Correct Answer - A::B::C | |
| 250. |
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 |
| Answer» Correct Answer - A | |