Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Is (--NH-CHR-CO--)_(n) a homopolymer or copolymer ?

Answer»

SOLUTION :Since the unit `(--NH-CHR-CO--)_(N)` OBTAINED from a SINGLE monomer unit, it is HOMOPOLYMER.
Discussion
2.

Is (NHCHR-CO)_n, a homopolymer or copolymer?

Answer»

Solution :It is a homopolymer because the REPEATING STRUCTURAL unit has only one type of monomer unit `(NH_2 - underset(R)underset(|)CH - COOH)` having FUNCTIONALITY two.
Discussion
3.

K_2Cr_2O_7 on strong heating gives :

Answer»

`K_2CrO_4`
`Cr_2O_3`
`O_2`
All

Answer :D
Discussion
4.

Is (NH - CH - CO)_(n) a homopolymer ora copolymer ?

Answer»

Solution :It is a HOMOPOLYMER because the repeating structural unit has only one type of monomer unit, t.e., `NH_(2) – CHR – COOH`. The functionality of this monomer is two.
Discussion
5.

K_(2)Cr_(2)O_(7) is an example of

Answer»

hexagonal
TRICLINIC
cubic
Orthorhombic.

Solution : For `K_(2)Cr_(2)O_(7)`that is, triclinic SYSTEM,`a ne b ne C, ALPHA ne beta ne gamma = 90^(@)`
Discussion
6.

________is necessary to obtain purest form of copper metal.

Answer»

CARBON reduction
hydrogen reduction
Electrolytic process
Thermite process

Answer :C
Discussion
7.

K_(2)Cr_(2)O_(7) + HCl rarr KCl + CrCl_(3) + Cl_(2) + H_(2)O. How many moles of HCl reacts with one mole of K_(2)Cr_(2)O_(7) ?

Answer»


ANSWER :14
Discussion
8.

is mixed and reacted with Br_2//KOH then how many products are obtained .

Answer»


SOLUTION :It is HOFFMAN's Bromamide reaction and in this reaction CROSS over PRODUCTS are not obtained .
Discussion
9.

K_(2)Cr_(2)O_(7) + "conc" H_(2)SO_(4)+ H_(2)O_(2) etherrarrblue precipitate anbydride (in enthereal layer) Blue colour is due to

Answer»

`CrO_(3)`
`H_(2)CrO_(4)`
`H_(2)Cr_(2)O_(3)`
`CrO_(3)`

ANSWER :d
Discussion
10.

K_2Cr_2O_7 acts as a good oxidizing agent in acidic mediumunderset("Orange")(Cr_(2)O_(7)^(2-)) + 14H^(+) + 6e^(-) rarr underset("Green")(2Cr^(3+)) + 7H_2O In alkaline solution, orange colour of Cr_(2)O_(7)^(2-) chages to yellow colour due to formation of Cr_2O_(4)^(2-) and again yellow colour changes to orange colour on changing the solution to acidic medium underset("Orange")(Cr_2O_7^(2-))+2OH^(_) rarrunderset("Yellow")(Cr_2O_7^(2-))+H_2O underset("Yellow")(2CrO_(4)^(2_-)) + 2H^(+) rarr underset("Orange")(Cr_(2)O_(7)^(2-) + H_(2)O) Cr_(4)^(2-) and Cr_(2)O_(7)^(2-) exist in equilibrium at pH =4 and are interconvertible by altering the pH of the solution. When heated with H_2SO_4 and metal chloride K_2Cr_2O_7 gives vapour of chromyl chloride (CrO_2Cl_2) . Chromyl chloride (CrO_2Cl_2)when passed into aqueous NaOH solution, yellow colour solution of CrO_(4)^(2-)is obtained. This on reaction with lead acetate gives yellow ppt. PbCrO_4 . When H_2O_2 is added to an acidified solution of dichromate ion, a complicated reaction occurs. The products obtained depend on the pH and concentration of dichromate. Cr_2O_7^(2-)+2H^(+) + 4H_(2)O_(2) rarr 2Cr(O_2)+5H_2O A deep blue-violet coloured peroxo compound, CrO(O_2)_2,' called chromic peroxide is formed. This decomposes rapidly in aqueous solution into Cr^(3+)and xygen. Which of the following statements is wrong when a mixture of NaCI and K_2Cr_2O_2 is gently walmed with conc. H_2SO_4?

Answer»

Deep RED vapour is evolved
The vapour when PASSED through NaOH solution gives a yellow solution of `Na_2CrO_4`
Chlorine GAS is formed
Chromyl chloride is formed

Solution :It is chromyl chloride test `Cr_2O_7^(-2) + Cl^(-) RARR Cl_2 +Cr^(+3)`
Discussion
11.

______is ivolved when an amide is heated with a dilute solution of sodium hydroxide.

Answer»

SOLUTION :AMMONIA.
Discussion
12.

K_2Cr_2O_7 acts as a good oxidizing agent in acidic mediumunderset("Orange")(Cr_(2)O_(7)^(2-)) + 14H^(+) + 6e^(-) rarr underset("Green")(2Cr^(3+)) + 7H_2O In alkaline solution, orange colour of Cr_(2)O_(7)^(2-) chages to yellow colour due to formation of Cr_2O_(4)^(2-) and again yellow colour changes to orange colour on changing the solution to acidic medium underset("Orange")(Cr_2O_7^(2-))+2OH^(_) rarrunderset("Yellow")(Cr_2O_7^(2-))+H_2O underset("Yellow")(2CrO_(4)^(2_-)) + 2H^(+) rarr underset("Orange")(Cr_(2)O_(7)^(2-) + H_(2)O) Cr_(4)^(2-) and Cr_(2)O_(7)^(2-) exist in equilibrium at pH =4 and are interconvertible by altering the pH of the solution. When heated with H_2SO_4 and metal chloride K_2Cr_2O_7 gives vapour of chromyl chloride (CrO_2Cl_2) . Chromyl chloride (CrO_2Cl_2)when passed into aqueous NaOH solution, yellow colour solution of CrO_(4)^(2-)is obtained. This on reaction with lead acetate gives yellow ppt. PbCrO_4 . When H_2O_2 is added to an acidified solution of dichromate ion, a complicated reaction occurs. The products obtained depend on the pH and concentration of dichromate. Cr_2O_7^(2-)+2H^(+) + 4H_(2)O_(2) rarr 2Cr(O_2)+5H_2O A deep blue-violet coloured peroxo compound, CrO(O_2)_2,' called chromic peroxide is formed. This decomposes rapidly in aqueous solution into Cr^(3+)and xygen. CrO_3 on reaction with HCl and the product on reaction with NaOH(aq) give respectively

Answer»

`CrO_(2)Cl,CrO_(4)^(2-)`
`Cr(OH)_2,CrO_(4)^(2-)`
`Cl_(2),Cr_(2)O_(7)^(2-)`
`Cl_(2).Cr(OH)_3`

SOLUTION :`Cl^(+) + CrO_3 RARR CrO_2Cl_2`
Discussion
13.

………………..is known as sweet spirit of nitre.

Answer»


Answer :4% SOLUTION of ETHYL nitrite in alcohol
Discussion
14.

K_2Cr_2O_7 acts as a good oxidizing agent in acidic mediumunderset("Orange")(Cr_(2)O_(7)^(2-)) + 14H^(+) + 6e^(-) rarr underset("Green")(2Cr^(3+)) + 7H_2O In alkaline solution, orange colour of Cr_(2)O_(7)^(2-) chages to yellow colour due to formation of Cr_2O_(4)^(2-) and again yellow colour changes to orange colour on changing the solution to acidic medium underset("Orange")(Cr_2O_7^(2-))+2OH^(_) rarrunderset("Yellow")(Cr_2O_7^(2-))+H_2O underset("Yellow")(2CrO_(4)^(2_-)) + 2H^(+) rarr underset("Orange")(Cr_(2)O_(7)^(2-) + H_(2)O) Cr_(4)^(2-) and Cr_(2)O_(7)^(2-) exist in equilibrium at pH =4 and are interconvertible by altering the pH of the solution. When heated with H_2SO_4 and metal chloride K_2Cr_2O_7 gives vapour of chromyl chloride (CrO_2Cl_2) . Chromyl chloride (CrO_2Cl_2)when passed into aqueous NaOH solution, yellow colour solution of CrO_(4)^(2-)is obtained. This on reaction with lead acetate gives yellow ppt. PbCrO_4 . When H_2O_2 is added to an acidified solution of dichromate ion, a complicated reaction occurs. The products obtained depend on the pH and concentration of dichromate. Cr_2O_7^(2-)+2H^(+) + 4H_(2)O_(2) rarr 2Cr(O_2)+5H_2O A deep blue-violet coloured peroxo compound, CrO(O_2)_2,' called chromic peroxide is formed. This decomposes rapidly in aqueous solution into Cr^(3+)and xygen. What happens when a solution of potassium chromate is treated with an excess of dilute nitric acid?

Answer»

`Cr^(3+)` and `Cr_(2)O_(7)^(2-)` are FORMED
`Cr_(2)O_7^(2-)` is REDUCED to +3 state of Cr
`CrO_(4)^(2-)` reduced to +3 state of Cr
`CrO_(4)^(2-)` is reduced to 0 state of Cr

Solution :`CrO_4^(-2) overset(H^+) RARR Cr_2O_7^(-2)`
Discussion
15.

K_2Cr_2O_7 cannot be used for

Answer»

PREPARING azo COMPOUNDS
tanning leather
as a laboratory OXIDANT
as a REDUCTANT

Answer :D
Discussion
16.

k_(2)CO_(3) cannot be prepared by solvay process because

Answer»

`KHCO_3` is less soluble than `NaHCO_3`
`KHCO_3` is too soluble to GET SEPARATED from solution
`K_2 O_3` is too soluble to be precipitated by KCL
`K_2CO_3` is less soluble than `Na_2CO_3`

ANSWER :B
Discussion
17.

Is it true that under certain conditions Mg can reduceAl_(2)O_(3) and Al can reduce MgO ?What are those conditions?

Answer»

Solution :Yes, below `1350^(@)C`Mg can REDUCE`Al_(2)O_(3)`and above`1350^(@)C,` Al reduce MgO. This can be inferred from `DeltaG^(THETA)` VS T plots.
Discussion
18.

K_2 Pt Cl_6is well known compound while corresponding Ni compound is not known?

Answer»

Solution :PT in `K_2PtCl_6`is in + 4 oxidation STATE which is more stable than `NI^(4+)`
Discussion
19.

Is it ture under certain conditions Mg can reduce Al_2O_3 and Al can reduce MgO? What are the conditions?

Answer»


Answer :From the ELLINGHAM DIAGRAM it can be seen that below `1350^@C` MG can reduce `Al_2O_3` and above `1350^@`, AL can reduce MgO.
Discussion
20.

k_(1)andk_(2) are the velocity constants of forward and backward reactions. The equilibrium constant K of the reaction is

Answer»

`k_(1)xxk_(2)`
`k_(1)-k_(2)`
`k_(1)//k_(2)`
`(k_(1)+k_(2))/(k_(1)-k_(2))`

SOLUTION :`K=k_(F)//k_(b)=k_(1)//k_(2)`
Discussion
21.

Is ittruethatundercertainconditions, Mg canreduceSiO _ 2andSicanreduceMgO?Whataretheconditions ?

Answer»

Solution :Below 1973 K ,the` Delta _fG^(@) `curvefortheformation of` SiO_ 2` liesabove the`Delta_ fG ^(@) `curvefortheformationofMgO, therefore, attemperaturesbelow1973 K , MG can reduce`SIO _2`tometallic silicon
` ""SiO_2+2Mg OVERSET (lt 1973K)TO2 MgO+Si ,Delta_rG^(@)=- ve`
Above 1973K, the` Delta _ fG^(@) `curvefor theformation of` SiO_2`liesbelowthe CORRESPONDING curvefor theformastionofMgO. Therefore,above1973 K, siliconcan reduceMgOtoMg.
` "" Si +2MgOoverset ( gt 1973K) toSiO _2+2Mg`
Discussion
22.

K for the synthesis of HI is 50. K for dissociation of HI is

Answer»

50
5
`0.2`
`32.0`

Solution :`K_(sigma)"for" H_(2)+I_(2)hArr2HI is 50`
`K_(c_(2))"for"2HIhArrH_(2)+I_(2)`
`K_(c_(2))=(1)/(K_(c_(1)))=1/50=0.02`
Discussion
23.

K for a zero order reaction 2xx10^-2molL^-1sec^-1 If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration must have been:

Answer»

0.5M
1.25M
12.5M
1.0M

Answer :A
Discussion
24.

Is it true that under certain conditions, Mg can reduce Al_(2)O_(3)" and "Al can reduce MgO? What are those conditions?

Answer»

Solution :Looking at the Ellingham diagram, we see that the curve of oxidation of Mg lies below the curve for oxidation af AL upto about `1270""^(@)C`. After that TWO CURVES CHANGE their positions. That is, the curve for Mg lies above the curve for Al. Therefore before the temperature `1270""^(@)C`, Mg can reduce `Al_(2)O_(3)` but above this temperature, Al can reduce MGO.
Discussion
25.

K, Ca and Li metals may be arranged in the decreasing order of their standard electrode potential as

Answer»

K, CA, LI
Ca, K, Li
Li, Ca, K
Ca, Li, K

Solution :ACCORDING to electrochemical series.
Discussion
26.

K, A & E_a of a process at 25^@Crespectively are 5 xx 10^(-4) s^(-1) , 6 xx 10^14 s^(-1)&108 kJ/mol. Then the value of rate constant as timeto oo will be

Answer»

`1.2 xx 10^18`
0
`6 xx 10^14`
`5 xx 10^(-4)`

Answer :D
Discussion
27.

Is it true that under certain conditions, Mg can reduce Al_2O_3 and Al can reduce MgO ? What are those conditions ?

Answer»

Solution :Yes, below `1350^@C`, Mg can reduce `Al_2O_3` and above `1350^@C`, Al can reduce MGO. This can be INFERRED from ELLINGHAM DIAGRAM.
Discussion
28.

Jute is

Answer»

NATURAL fibre
POLYMER fibre
SYNTHETIC fibre
regenerated fibre

ANSWER :A
Discussion
29.

Justify the presence of kefo group inC_(2) in fructose.

Answer»

Solution :On oxidation with nitric ACID, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in FRUCTOSE. This SHOWS that a KETO GROUP is present in C-2.
Discussion
30.

Justify the position of lanthanides and actinides in the periodic table.

Answer»

Solution :In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar CHEMICAL properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. Thisposition can be justified as follows.
(a) Lanthanoids have GENERAL electronic configuration [Xe] `4f^(2-14) 5d^(0-1) 6s^(2)`
(b) The common oxidation state of lanthanoids is +3
(C) All these elements have similar physical and chemical properties.
(II) Similarly the fourteen elements following actinium resemble in their physical and chemical properties.
(iii) If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.
(iv) HENCE a separate position is provided to the inner transition elements at the bottom of the periodic table.
Discussion
31.

Is it true that certain conditions Mg can reduce SiO_(2) and Si can reduce MgO? What are those conditions?

Answer»

Solution :Free energy for the OXIDATION of an element changes with temperature and sometimes there is an abrupt change in its value due to phase change `(s to l)" or "(l to v)`.
AMORPHOUS SILICON is obtained by the reduction os `SiO_(2)` with burning Mg wire
`SiO_(2) +2Mg to Si+ 2MgO`
Si, when HEATED with MgO at very high temperature say `1000""^(@)C" form "SiO_(2)` and Mg metal is obtained.
`2MgO+Si overset(1000""^(@)C)(to) 2Mg+SiO_(2)`
We can ascertain from Ellingham DIAGRAM which metal oxide can be reduced by which metal and at which temperature.
Discussion
32.

Justify thepositionof lanthanidesand actinidesin theperiodic table .

Answer»

Solution :(i) Theactualpositionof Lanthanoidsin theperiodictableis atgroupnumber 3and periodnumber6 .Howeverin the sixthperiodafterlanthanum , theelectrons arepreferentiallyfilledin inner4fsubshellandthesefourteenelements followinglanthanumshowsimilarchemicalproperties.Therefore THESEELEMENTS are groupedtogether andplacedat thebottomof theperiodictable . Thispositioncan bejustifiedas follows.
(a)Lanthanoids have generalelectronicconfiguratration`[Xe] 4f^(gamma -14) 5D^(0-1) 6s^(2)`
( b)The commonoxidationstateoflanthanoides is +3
( C )All theseelementshave similar(ii) Simiilarlythe fourtheenelementsfollowingactiniumresemble in theirphysicaland CHEMICALPROPERTIES .
(iii)Similarto lanthanoids activitiesare alsoplacedin a separaterowat the bottonof theperiodictable.
Discussion
33.

Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation, state and hydride formation.

Answer»

Solution :(i) Electronic configuration : All these elements have the common `ns^(2) np^(4)` (n = 2 to 6) valence shell electronic configuration.
`""_8O=[He] 2s^(2) 2p^4 , ""_(16)S = [Ne]3s^(2) 3p^4 , ""_16S= [ Ne ] 3s^(2) 3p^4 , ""_(34) Se = [Ar] 3d^(10) 4s^(2) 4p^(4)`
`""_(52) Te = [Kr] 4d^(10)5s^(2) 5p^(4) and ""_(84) Po = [Xe] 4f^(14) 5d^(10) 6s^(2) 6p^(4)`
Hence, it is justified to place them in Group 16 of the periodic table.
(II) Oxidation states : They need two more electrons to form dinegative ions and acquire the nearest inert gas configuration. Thus, the minimum oxidation state of these elements should be -2. Oxygen and sulphur being electronegative show an oxidation state of -2. Other elements of this group, being more electropositive than O and S, do not show negative oxidation states. Since these elements have six electrons in the valence shell, THEREFORE, at the maximum they can show an oxidation state +6. Other positive oxidation states SHOWN by these elements are +2 and +4. However, due to the absence of d-orbtials, oxygen does not show oxidation states of +4 and +6. Thus, on the basis of minimum and maximum oxidation states, these elements are justified to be placed in the same group.
(iii) Formation of hydrides : All the elements complete their respective octets by SHARING two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula `EH_2` i.e., `H_2O, H_2S , H_2 Se , H_2 Te and H_2Po`. Thus, on the basis of formation of hydrides of the general formula EH,, these elements are justified to be placed in Group 16 of the periodic table.
Discussion
34.

Justify the placement of O,S,Se,Te and Po in the same group of the periodic table interms of electronic configuration, oxidation state and hydride formation.

Answer»

Solution :(i). Electronic configuration: O, S, Se and Te have the same `ns^2np^4` valence shell electron configuration and hence are justified to be placed in group 16 of the periodic table.
`O(8)[He]2s^22p^4`
`S(16)[Ne]3s^23p^4`
`Se(34)[Ar]3d^(10)4s^24p^4`
`Te(52)[Kr]4d^(10)5s^25p^4`
(ii). Oxidation states: O, S, Se and Te need twio more electrons to form dinegative ion and acquire oxidation state of these elements shoud be -2. OXYGEN predominantly and sulphur to some extent SHOWS `O.S.-2`. Other elements of group 16, being less electronegative, have less stable `O.S.-2`. Since group 16 elements have 6 electrons in their valence shell, at maximum they can show `+6` O.S. other positive oxidation sattes shown by these elements is `+4` and `+6`. Thus on the basis of oxidation states shown by these elements they are justified to be placed in the same group i.e., group 16 of the periodic table.
(iii). Formation of hydrides: All the elements i.e., O,S Se and Te complete their respective octets by sharing of their valence electrons with 2 s-orbital of hydrogen to form hydrides of the type `EH_2` i.e., `H_2O,H_2S,H_2Se` and `H_2Te`. Thus on the basis of formation of hydrides of the general FORMULA `EH_2` these elements are justified to be placed in the group IE., group 16 of the periodic table.
Discussion
35.

Is it possible to store copper sulphate in an iron vessel for a long time ? Given : E_(Cu^(2+)|Cu)^(@)=0.34V and E_(Fe^(2+)|Fe)^(@)=+0.44V

Answer»

Solution :`E_("CELL")^(@)=E_("OX")^(@)+E_("red")^(@)=0.44V+0.34V=0.78V`
These `+veE_("cell")^(@)` values shows that iorn will OXIDISE and copper will GET reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.
Discussion
36.

Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer»

Solution :(i) Electronic configuration. All these elements have the same `ns^(2)np^(4)` (n = 2 to 6) valence shell electronic configuration and HENCE are justified to be placed in group 16 of the periodic table :
`{:(._(8)O = [He] 2s^(2)2p^(4) "," ._(16)S = [Ne] 3s^(2)3P^(4)","._(34)Se = [Ar] 3d^(10)4s^(2)4p^(4)),(._(52)Te=[Kr]4d^(10)5s^(2)5p^(4) "and"._(84)Po = [Xe] 4f^(14)5d^(10) 6S^(2)6p^(4)):}`
(ii) Oxidation states. They need two more electrons to form dinegative ions by acquiring the nearest inert gas configuration. Thus, the minimum oxidation state of these elements should be -2. Oxygen predominatly and sulphur to some extent being electronegative show an oxidation state of -2. Other elements of this group, being less electropositive than O and S, do not show negative oxidation states. Since these elements have six electrons in the valence shell, THEREFORE, at the maximum they can show an oxidation state of +6. Other positive oxidation states shown by these elements are +2 and +4. However, oxygen due to the absence of d-orbitals does not show oxidation states of +4 and +6. Thus, on the basis of minimum ans maximum oxidation states, these elements are justified to be placed in the same group, i.e., gorup 16 of the periodic table.
(iii) Formation of hydrides. All the elements complete their respective octets by sharing two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula `EH_(2)`, i.e., `H_(2)O, H_(2)S, H_(2)Se, H_(2)Te and H_(2)PO`. Thus, on the basis of formation of hydrides of the general formula `EH_(2)`, these elements are justified to be placed in the same group, i.e., group 16 of the periodic table.
Discussion
37.

Is it possible to store copper sulphate in an iron vessel for a long time? Given : E_(Cu^(2+)|Cu)^(@)=0.34V " and " E_(Fe^(2+)|Fe)=-0.44V

Answer»

Solution :`(E_("ox")^(@))_(Fe|Fe^(2+))=-0.44V " and " (E_("red")^(@))_(Cu^(2+)|Cu)=0.34V`
These +ve emf values SHOWS that iron will OXIDISE and COPPER will get reduced i.e., the vessel will dissolve. HENCE it is not possible to store copper SULPHATE in an iron vessel.
Discussion
38.

Justifythe followijngstatement . " Elementsof the firsttransitionseriespossesmanypropertiesdifferent fromthose of heaviertransitionelements "

Answer»

Solution :Theheavier transitionelements belongto fourth(4d ) , fifth (5d )and SIXTH (6d)transitionseries.
Theirpropertiesare expectedto be differentformthe elementsbelonging to the first(3D)seriesdue tofollowingreasons.
(i) ATOMIC :radii: SIZEOF thetransitionthoseof the correspondingelementof thefirsttransitionseriesthoughthoseof 4dand 4d seriesare verycloseto each other.
(ii) Ionisationenthalpyof 5dseriesare higherthan thecorrespondingelement of 3d and4d series .
(iii) A tomisationhigherthan the correspondingelements of tehfirstseries.
(iv)the elementof the firsttransitionseriesgenerally formlow orhighspincomplexes. dependinguponhte higherof ligandfiled.
Howeverthe HEAVIER transitionelementsform lowspincomplexesirrespectiveof the strength ofligandfiled.
Discussion
39.

Is it possible to oxidise t-butyl alcohol using acidified dichromate to form a carbonyl compound.

Answer»

SOLUTION :`3^(@)`-alcohols do not undergo oxidation reaction under normal condition, but at elevated temperature, under strong oxidising agent CLEAVAGE of C - C bond takes place to give a mixture of carboxylic acid.
Yes, it is possible. `t` - butyl alcohol is readily oxidising in ACIDIC solution `(K_(2)Cr_(2)O_(7)//H_(2)SO_(4))` to a mixture of a KETONE and an acid each containing lesser number of carbon atoms than the original alcohol. The oxidation presumably occur via alkenes formed through dehydration of alcohols uder acidic conditions.
Discussion
40.

Is it possible to oxidise t - butyl alcohol using acidified dichromate to form a carbonyle compound.

Answer»

Solution :`3^@` alcohols do not undergo oxidation reaction under normal condition, but at elevated temperature, under strong oxidising agent cleavage of C - C bond takes place to give a mixture of carboxylic acid.
Yes, it is possible. t - butyl ALCOHOL is READILY oxidsing in ACIDIC solution `(K_2Cr_2O_7//H_2SO_4)` to a mixture of a ketone and an acid each containing lesser number of carbon atom: original alcohol. The oxidation presumably occur VIA alkenes formed through dehydration of than the alcohols under acidic conditions.
Discussion
41.

Is it possible to know the size and shape of colloidal particles by using ultramicroscope ?

Answer»

SOLUTION :With the unltramicroscope we can see only the light scattered by colloidal PARTICLES, but not ACTUAL colloidal particle.
Thus, ultramicrope does not provide any information about the SIZE and SHAPE of colloidal particles.
Discussion
42.

Justify that peroxymono and peroxydisulphuric acids have a peroxy linkage. How are they structurally diffferent?

Answer»

Solution :The peroxymono and peroxydisulphuric acid may be considered to have been DERIVED from `H_(2)O_(2)` by REPLACING one H and both H by `HSO_(3)` groups respectively.
`H_(2)SO_(5)` the monoacid and`H_(2)S_(2)O_(8)` the peroxydisulphuric acid, both on hydrolysis YIELD `H_(2)O_(2)`.
Both permono and perdi sulphuric acids contain `-O-O-` linkage in their molecular structure.
Discussion
43.

Justify and arrange the following compounds of each set in increasing order of reactivity towards the asked displacement : 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-Methylpropane (S_(N)2 reaction)

Answer»

SOLUTION :2-Bromo-2-Methylpropane `lt` 2-Bromobutane `lt` 1-Bromobutane
The nucleophile faces the MAXIMUM hindrance from the BULKY GROUPS in 2-Bromo-2Methylpeopane followed by 2-Bromobutane and 1-Bromobutane. Hence the above order.
Discussion
44.

Is it possible to get original egg after boiling?

Answer»

SOLUTION : When an egg is boiled, the protein loses its biological activity due to disruption of hydrogen bonds among them LEADING to INVERSIBLE DENATURATION. So renaturation is impossible.
Discussion
45.

Justify and arrange the following compounds of each set in increasing order of reactivity towards the asked displacement : 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-Methylpropane (S_(N)1 reaction)

Answer»

Solution :1-Bromobutane `lt` 2-Bromobutane `lt` 2-Bromo-2-Methylpropane
2-Bromo-2-Methylpropane FORMS the CARBOCATION
`CH_(3)-UNDERSET(CH_(3))underset(|)(""^(+)C)-CH_(3)`
which can stabilise to the MAXIMUM EXTENT. This is followed by 2-Bromobutane and 1-Bromobutane.
Discussion
46.

Is it possible to extract Al by the electrolysis of an aqueous solution of aluminium sulphate? Explain.

Answer»

Solution :The tendency of `Al^(3+)` ION to be reduced is less tha that of `H^(+)` ion, Therefore `H_(2)` gas is LIBERATED at the CATHODE INSTEAD of aluminium,
Discussion
47.

Justification of putting H in VII A group is

Answer»

H is GAS
H is non METAL
It form NAH like salt
It has ortho and para allotropes

Answer :C
Discussion
48.

Is it possible to determine Lambda_(oo) of propionic acid experimentally. Give reason.

Answer»

SOLUTION :No, since propionic ACID is weak acid conductance at LOW concentration cannot be DETERMINED.
Discussion
49.

Just like human beings, plants also need various nutrients for their healthy growth. Iron is one of these. The deficiency of iron results in disorder known as iron chlorosis. It appears in the form of yellow leaves. It adversely affect the yield of fruits from citrus trees. (i) In which oxidation state is iron generally present in the soil ? (ii) Why is iron hydroxide not assimilated in the soil ? (iii) Which complex of iron is readily absorbed by soil ? (iv) What is the value associated with the use of this complex ?

Answer»

Solution :(i) Iron is present in + 3 oxidation state as Fe (III)
(ii) Iron hydroxide i.e., `Fe(OH)_3` is insoluble in water and is therefore, not assimilated in the soil.
(iii) Fe(III)-EDTA complex is READILY absorbed by the soil as it is water soluble.
(iv) The addition of this complex compound in proper AMOUNT MAKES plants healthy and the yield of fruits GETS increased.
Discussion
50.

Is it necessary to acidify a solution before group II cations are precipitated with H_(2)S?

Answer»

Solution :Yes, because for the precipitation of cations of group II only small CONCENTRATION of SULPHIDE ions `(S^(2-))` is required this condition is ACHIEVED by passing `H_(2)S` gas in a solution acidified with `HCL` (dil.) due to common ion effect concentration of `S^(2-)` ions decreases as presence of hydrochloric acid suppresses the ionisation of `H_(2)S` resulting only in the precipitation of sulphides of group II metals.
Other metal sulphides remain in solution as they required higher concentration of `S^(2-)` ions for their precipitation.
Discussion