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.
| 88301. |
(A) Nitrogen can not form more than three bonds (R ) Nitrogen has three valence electrons. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 88302. |
A: Nitrobenzene on reduction with Zn//NaOH//Delta gives hydrazobenzene. R: Nitrobenzene on reduction with Sn/HCI gives aniline. |
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Answer» If both ASSERTION & REASON are TRUE and the reason is the correct explanation of the assertion, then MARK (1). |
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| 88303. |
A nitroalkane produces a ketone when it is boiled with HCl. The nitroalkane could be_____________. |
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Answer» `CH_3 CH_2 NO_2` |
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| 88304. |
A nitroalkane reactionwith HNO_(2)to yielda productwhichis insolublein NaOH andgivebluecolouron treatmentwithalkali. The nitroalkanecouldbe |
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Answer» `CH_(3) - CH_(2) - NO_(2)` |
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| 88305. |
A : Nitration of benzoic acid gives ortho and para derivatives of nitrobenzoic acids. R : Carboxyl group is activating group. |
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Answer» If both Assertion & Reason are true and the reason is the CORRECT explanation of the assertion, then mark (1). |
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| 88306. |
(a) Nitric acid is prepared from ammonia in a three-step process. (i) 4NH_3 + 5O_2 to 4NO + 6H_2O (fast)(ii) 2NO + O_2 to 2NO_2 (slow)(iii) 3NO_2 + H_2O to 2HNO_3 + NO (fast)Calculate how much HNO_3 can be produced from 10^5 kg of ammonia assuming 100% efficiency in each of the reactions.(b) If equation (ii) is second-order in NO and first-order in O_2 , calculate the rate of formation of HNO_3when oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M. k = 5.8 xx 10^2 L^6 "mol"^(-2) s^(-1) |
| Answer» Solution :`(a) 2.47 xx 10^3 kg (B) 1.63 xx 10^(-6)` | |
| 88307. |
A nitraite on acid hydrolysis gives compound A, which reacts with thionyl chloride to give compound B. Benzene reacts with compound B in presence of anhydrous AlCl_(3) to give compound C. The compound B reacts with ethyl alcohol to give ester D. Identify the compounds A,B C and D and write the equations. |
Answer» Solution :`underset("Acetonitrile")(CH_(3)-CN)overset(H_(3)O^(+))underset("Acetic ACID")(CH_(3)-COOH) overset(SOCl_(2))underset(Delta)rarrunderset("Acetyl chloride")(CH_(3)-COCl)`  `underset((B))(CH_(3)COCl) + underset("Ethyl alcohol")(C_(2)H_(5)OH)RARR underset("Ethyl acetate")(CH_(3)COOC_(2)H_(5)) +HCl` 
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| 88308. |
(A) Nitration of aniline can be conveniently done by (R) Acetylation increases the electron density in the benzene ring.\ |
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Answer» If both A and R are CORRECT and R is the correct EXPLANATION of A. |
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| 88309. |
Assertion:Ni(CO)_4 complex is tetrahedral in shape Reason:Ni atom undergoes sp^3 hybridization in Ni(CO)_4. |
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Answer» Both A & R are true, R is the correct EXPLANATION of A |
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| 88310. |
A: NiCI, is more stable than PtCl_(2)R: K_(2) PtCl_(6) is more stable than K,NiCI |
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Answer» If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1) |
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| 88311. |
(A) Ni^(+2) salts are pink in colour (R) Ni^(+2) absorb light in IR region |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 88313. |
(A) NH_(3) is a better reducing agent than PH_(3) (R ) NH_(3) is weaker lewis base when compared to PH_(3) |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 88314. |
(A) NH_(3) adsorbs more readily over activated charcoal than CO_(2) (R ) NH_(3)Is non polar |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 88315. |
A new carbon bond is possible in the following reaction reactions: |
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Answer» `C_(6)H_(6)+ CH_(3)CI overset(anhy.AICI_(4))(to)` |
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| 88316. |
A new carbon-bond formation is possible in |
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Answer» Cannizzaro reaction |
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| 88317. |
A neutral organic compound A (C_(5)H_(8)O_(2)) does not decolorize Bayer.s reagent and on hydrolysis with dilute H_(2)SO_(4) produces B (C_(2)H_(10)O_(3)), which is diastereomeric. B on heating with concentrated H_(2)SO_(4) undergoes dehydration producing C (C_(5)H_(8)O_(2)), which shows geometrical isomerism. Also B on treatment with acidic dichromate solution produced D(C_(5)H_(8)O_(2)), which is enantiomeric and gives a yellow precipitate with NaOI. D on gentle heating produces E (C_(4)H_(8)O),which is non-resolvable. Compound (E)is |
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Answer» |
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| 88318. |
A neutral organic compound A (C_(5)H_(8)O_(2)) does not decolorize Bayer.s reagent and on hydrolysis with dilute H_(2)SO_(4) produces B (C_(2)H_(10)O_(3)), which is diastereomeric. B on heating with concentrated H_(2)SO_(4) undergoes dehydration producing C (C_(5)H_(8)O_(2)), which shows geometrical isomerism. Also B on treatment with acidic dichromate solution produced D(C_(5)H_(8)O_(2)), which is enantiomeric and gives a yellow precipitate with NaOI. D on gentle heating produces E (C_(4)H_(8)O),which is non-resolvable. Compound (C) is |
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Answer» |
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| 88319. |
A neutral organic compound A (C_(5)H_(8)O_(2)) does not decolorize Bayer.s reagent and on hydrolysis with dilute H_(2)SO_(4) produces B (C_(2)H_(10)O_(3)), which is diastereomeric. B on heating with concentrated H_(2)SO_(4) undergoes dehydration producing C (C_(5)H_(8)O_(2)), which shows geometrical isomerism. Also B on treatment with acidic dichromate solution produced D(C_(5)H_(8)O_(2)), which is enantiomeric and gives a yellow precipitate with NaOI. D on gentle heating produces E (C_(4)H_(8)O),which is non-resolvable. Compound (B) is |
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Answer» |
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| 88320. |
A neutral organic compound A (C_(5)H_(8)O_(2)) does not decolorize Bayer.s reagent and on hydrolysis with dilute H_(2)SO_(4) produces B (C_(2)H_(10)O_(3)), which is diastereomeric. B on heating with concentrated H_(2)SO_(4) undergoes dehydration producing C (C_(5)H_(8)O_(2)), which shows geometrical isomerism. Also B on treatment with acidic dichromate solution produced D(C_(5)H_(8)O_(2)), which is enantiomeric and gives a yellow precipitate with NaOI. D on gentle heating produces E (C_(4)H_(8)O),which is non-resolvable. Compound (A) is |
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Answer» |
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| 88321. |
A neutral fertilizer among these compounds is |
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Answer» ammonium NITRATE |
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| 88322. |
A neutralliquid of fromual C_(7)H_(14)O_(2)is hydrolysedto an acid (A)and an alcohol (B). Acid (A) hasneutraliztionequivalentof 84. Alchohol (B)is noteasily oxidizedwith an acid solution of Na_(2)Cr_(2)O_(7) . Whatis thestructure fo the original compound ? |
Answer» SOLUTION :
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| 88323. |
A neutral fertilizer among the following compound is |
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Answer» UREA |
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| 88324. |
A neutral divalent carbon intermediateis called: |
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Answer» FREE radical |
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| 88325. |
A neutral compound with molecular formula C_(3)H_(8)O evolves H_(2) when treated with sodium metal and gives iodoform test. The compound is |
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Answer» `(CH_(3))_(2)CHOH` |
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| 88326. |
A neutral compound gives red colour with ceric ammonium nitrate. This suggests the compound contains a/an |
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Answer» ALCOHOLIC group |
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| 88327. |
A neutral compound gives red colour with ceric ammonium nitrate. It suggests that the compound has : |
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Answer» ALCOHOL gp. |
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| 88328. |
Aneutral compound C_(4) H_(8)O_(2), reduce Fehling'ssolution , andliberate H_(2) gas when treated with sodium metal and give positive iodoform test. The compound is |
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Answer» `CH_(3) - CHOHCH _(2) - CHO` |
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| 88329. |
A neutral atom will have the lowest ionization potential when electronic configuration is : |
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Answer» `1s^1` |
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| 88330. |
(A): Neon is used to fill advertisement glass bulbs (R): Neon gives coloured discharge at low pressure |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 88331. |
(A) Neon bulbs are used in botanical gardens(R) Neon is lighter than argon |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 88332. |
A negatively charged sol can be formed by peptizing a solution of |
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Answer» `Agl with AgNO_(3)` |
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| 88333. |
A negatively charged suspenstion of clay in water needs for precipitation the minimum amount of |
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Answer» `AlCl_3` |
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| 88334. |
A negative catalyst will |
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Answer» raise the energy of activation for a given reaction |
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| 88335. |
a ne bc and alpha ne beta ne gamma follow |
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Answer» Triclinic |
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| 88336. |
(a) NCl_(3) gets readily hydrolysed while NF_(3) does not. Why ? (b) What kind of molecules show disproportionation reactions ? Give one example of a compound each of nitrogen and phosphorus which show disproportiionation reactions. Write chemical equation in each case. |
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Answer» Solution :(a) 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) + 3 H_(2)O rarr NH_(3) + 3HOCl"," ""NF_(3) + H_(2)O rarr No` reaction (b) Compounds in which one of elements exists in three DIFFERENT oxidation states can SHOW disproportionation reactions. For example, `3 Hoverset(+3)(NO)_(2) rarr overset(+5)(HNO_(3)) + 2overset(+2)(NO)+H_(2)O","""4H_(3)overset(+3)(PO_(3))overset(Delta)rarr 3H_(3)overset(+5)(PO_(4)) + overset(-3)(PH_(3))` |
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| 88337. |
(A) NCl_(3) is known, but not NCl_(5) (R ) Nitrogen can not expand its valency beyond four. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 88338. |
A nauseating smell in the carbylamine test for primary aminesis due to the formation of |
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Answer» ISOCYANIDE |
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| 88339. |
(a) NCl_(3) gets hydrolysed to form NH_(3) and HOCl while PCl_(3) on hydrolysis gives H_(3)PO_(3) and HCl. Explain why ? (b) SOCl_(2) can act as a weak acid as well as a weak base. Explain. |
Answer» Solution :(a) N does not have d-orbitals to accommodate the electrons donated by O of `H_(2)O`, therefore, ATTACK of `H_(2)O` occurs on the CL atom which has d-orbitals to accommodate the extra electrons donated by `H_(2)O`. Consequently Cl-O bond is fomed leading to the formation of HOCl and `NH_(3)` as shown below :  In contrast, both P and Cl have d-orbitals to accommodate electrons donated by `H_(2)O`. But P-O bond is much stronger than Cl-O bond. As a result, attack of `H_(2)O` molecules occurs preferentially on P of `PCl_(3)` to form `H_(3)PO_(3)` and HCl.  (b) The basic character of `SOCl_(2)` is due to the presence of a lone pair of electrons on the S-atom. Futher, in `SOCl_(2)`, S ALSO has empty d-orbitals which can be used to accept electrons. Therefore, `SOCl_(2)` also behaves as a LEWIS acid. 
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| 88340. |
A biological catalyst is |
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Answer» An AMINO ACID |
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| 88341. |
(A) : Natural rubber and gutta percha are examples of cis-trans isomers. (R) : Cis-trans isomerism arises due to the difference of geometrical arrangement of two different groups on the double bonded carbon atoms. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 88342. |
A natural linear polymer of 2-methyl-1, 3-butadiene becomes hard on treatment with sulphur between 373 to 415 K and -S-S- bonds are formed between chains. Write the structure of the product of this treatment ? |
Answer» Solution :These process is KNOWN as VULCANIZATION of rubber and the PRODUCT is known as valcanized rubber. Its STRUCTURE is :  
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| 88343. |
A natural linear polymer of 2-methyl -1,3-butadiene becomes hard on treatment with sulphur between 373 to 415 and -S-S- bonds are formed between chains. Write the structure of the product of this treatment ? |
Answer» SOLUTION :The NATURAL linear polymer of 2-methyl -1,3- butadiene is CIS -polyisoprene (natural rubber) . On heating with sulphur between `373-415 K`, vulcanization occurs as a result of which `-S-S-` bridges or CROSS- links are introduced between polymer chains through their reactive allylic positions as shown below : 
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| 88344. |
A natural linear polymer of 2 methyl-1, 3-butadiene becomes hard on treatment with sulphur between 373 to 415K and -S-S-bonds are formed between chains. Write the structure of the product of this treatment? |
Answer» SOLUTION :
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| 88345. |
A natural gas may be assumed to be a mixture of CH_(4) and C_(2)H_(6) only. On complete combustion of 10 L of the gas at STP, the heat evolved was 476.6 kJ. Assuming DeltaH_(CH_(4))=-894 kJ/mol and Delta_(C_(2)H_(6))=-1560 kJ/mole. Calculate the % of volume of each gas in the mixture. |
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| 88346. |
A natural gas samplecontains 84% (by volume) of CH_(4) , 10% of C_(2)H_(6) , 3% of C_(3)H_(8)and3% of N_(2) . Ifa seriesof catalyticreactions couldbe used forconvertingall thecarbon atoms of the gas intobutadiene, C_(4)H_(6), whit100% efficiency , howmuchbutadiene couldbe prepared from 100 g of the naturalgas ? |
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| 88347. |
A natural gas may be assumed to be a mixture of CH_4 and C_2H_6 only. On complete combustion of 10 L of the gas at STP, the heat evolved was 474.6 kJ. Assuming DeltaHc(CH_4) = – 894 kJ//mol and ""DeltaHC(C_2H_6) = - 1560 kJ//"mole". Calculate the % by volume of each gas in the mixture |
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Answer» Solution :LET, V be the VOLUME of `CH_4` in 10L MIXTURE. We have Amount of methane = `v/(22.4)`, Amount of ethane = `(10- v)/(22.4)` The expression of heat evolved, we will get `V/(22.4) xx 894 + (10 - V)/(22.4) xx 1560 = 474.6` V = 7.45 litre % of `CH_4 = 74.5%` % of `C_2H_6 = 25.5%`. |
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| 88348. |
A natural gas may be assumed to be a mixture of CH_(4) and C_(2)H_(6) only. On complete combination of 10 L of the gas at STP, the heat evolved was 474.6 kJ. Assuming DeltaH_(c)(CH_(4)(g)) = -894 kJ/mol and DeltaH_(c)(C_(2)H_(6)(g)) = -1560 kJ//mol. Calculatethe % of volume of each gas in the mixture. |
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Answer» Solution :`K_(s) to K(G), DeltaH_(5) = 89.8 KJ mol^(-1)` We have to find `DELTAH` of the followingequation. |
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| 88349. |
A nation's industrial strength can be judged by the quantity of ____________ its produces and consumes. |
| Answer» SOLUTION :SULPHURIC ACID | |
| 88350. |
A nation's industrial strength can be judged by the quantity of a chemical produced and consumed by it. What is that? |
| Answer» SOLUTION :SULPHURIC ACID | |
