Explore topic-wise InterviewSolutions in Current Affairs.

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1.

Arrange the following in decreasing order of priority for functional group of organic compound?

Answer»

SOLUTION :
Discussion
2.

Arrange the following in decreasing order of basic strength.

Answer»

Solution :The BASIC character of these hydrides depends upon the STRENGTH of the ADDITIONAL element-hydrogen bond formed when the hydride accepts a proton.

On MOVING down the group, the size of the element increases and the strength of the E-H bond decreases. Consquently, the tendency of the hydride to ACCPET a proton decreases and hence the basic strength decreases down the group, i.e., `NH_(3)gtPH_(3)gtAsH_(3)gtSbH_(3)gtBiH_(3)`.
Discussion
3.

Arrange the following in correct order of nucleophilicity :- CH_(3)-O - CH_(2) - O^(oplus)(I) "" Ph - CH_(2) - O^(oplus)(II) CH_(3) - O^(oplus)(III) "" ChCH_(2) - O^(oplus)(IV)

Answer»

`IV gt III gt II gt I`
`I gt II gt III gt IV`
`III gt IV gt II gt I`
`II gt I gt IV gt III`

SOLUTION :IF DONAR atom is same the nucleophiliticyt is PARALLEL to basicity
Discussion
4.

Arrange the following: (i) LiH, NaH and CsH in order of increasing ionic character. (ii) H-H, D-D and F-F in order of incresing bond dissociaton enthalpy. (iii) NaH, MgH_(2) and H_(2)O in order of incresing reducing powder.

Answer»

Solution :(I) Increasing ioninc CHARACTER: `LiH lt NaH lt CsH`
Reason: The IONISATION ENTHALPHY decreases in the order `LI gt Na gt Cs`. This influences the ionic character adversely which increases as shown
(ii) Increasing bond dissociation enthalphy: `F-F lt H-H lt D-D`.
Reason: The bond dissociation enthalphy of `underset(..)overset(..)( :F)-underset(..)overset(..)(F: )`fluorine is very small `(242-6kJ mol^(-1))` due to the repulsion in the lone paris of electrons PRESENT on the two F atoms. Out of `H_(2) and D_(2)`, the bond dissociation enthalphy of `H-H(435.88 kJ mol^(_1))`is less than of `D-D(443.35kJ mol^(-1))`
(III) Increasing reducing power: `H_(2)O lt MgH_(2) lt NaH`.
Reason: NaH being ioninc in nature is the strongest reducing agent. Both `H_(2)O and MgH_(2)` are covalent in nautre but bond dissociation enthalpy of `H_(2)O` is higher. Therefore, it is weaker reducintg agent than `MgH_(2)`.
Discussion
5.

Arrange the following : (i) -NO_(2), -COOH, -F, -CN, -I, in decreasing order of -I effect. (ii) CH_(3)^(-), D-, (CH_(3))_(3)C-, (CH_(3))_(2)CH-, CH_(3)CH_(2^(-)),in increasing order of +I effect.

Answer»

SOLUTION :(i) `-NO_(2) GT -CN gt -COOH gt -F gt -I`(II) `D - lt CH_(3) - lt CH_(3)CH_(2^(-)) lt (CH_(3))_(2)CH- lt -C(CH_(3))_(3)`.
Discussion
6.

Arrange the following : (i) C_(6)H_(5)overset(*)(C)HCH_(3), C_(6)Hoverset(*)(C)HCH = CH_(2), C_(6)H_(5)CH_(2)overset(*)(C)H_(2), C_(6)H_(5)overset(*)(C)(CH_(3))_(2) in order of increasing stability. (ii) CH_(3)CH_(2)^(+), C_(6)H_(5)CH_(2)^(+), (CH_(3))_(3)C^(+), CH_(2) =CHCH_(2)^(+) in order of decreasing stability. (iii) HC -= C^(-), CH_(2) = CH^(-), CH_(3)CH_(2)^(-), CH_(3)^(-), (CH_(3))_(2)CH^(-), C_(6)H_(5)CH_(2)^(-)in order of increasing stability.

Answer»

Solution :`(i) C_(6)H_(5)CH_(2)overset(*)(C)H_(2) lt C_(6)H_(5)overset(*)(C)HCH_(3) lt C_(6)H_(5)overset(*)(C)(CH_(3))_(2) lt C_(6)H_(5)overset(*)(C)H - CH = CH_(2)`.
(ii) `(CH_(3))_(3)C^(+) gt C_(6)H_(5)CH_(2)^(+) gt CH_(2) = CHCH_(2)^(+) gt CH_(3)CH_(2)^(+)`.
(iii) `(CH_(3))_(2)CH^(-) lt CH_(3)CH_(2)^(-) lt CH_(3)^(-) lt CH_(2) = CH^(-) lt C_(6)H_(5)CH_(2)^(-) lt HC -= C^(-)`.
Discussion
7.

Arrange the following, (i) CaH_(2), BeH_(2) and TiH_(2) in order of increasing electrical conductance. (ii) LiH, NaH, CsHis order of increasing ionic character(iii) H-H-D-D and F-F in order of increasing bond dissocation enthalpy. (iv) NaH, MgH_(2) and H_(2)O in order of increasing reducing property.

Answer»

Solution :(i) `BeH_(2)` is a covalenthydride, therefore, it does not conduct electricity at all. `CaH_(2)` conducts electricity in the FUSED statewhile `TiH_(2)` conducts electricity at room temperature. Thus, the order of increasing electrical conductance is : `BeH_(2) LT CaH_(2) lt TiH_(2)`
(ii) Electronegativity decreases down the group from Li to Cs, therefore, the ionic CHARACTER of their hydrides increases in the same order, i.e., `LiH lt NaH lt CsH`
(iii) Due to greater nuclear mass, the bondpair in D-D bond is attracted more strongly than in H-H bond. Therefore, bond dissociation energy of D-D bond is higher than that of H-H bond.
However, due to repulsions between lone pairs of F and the bond pair, F-F bonddissociation enthalpy is the minimum . Thus the bond dissociation enthalpy increases in the order: ` F - F lt H -H lt D -D`
(iv) Ionic hydrides are powerful reducing agents. Both `MgH_(2)` and `H_(2)O` are covalent hydrides but he bond dissociation energy of `H_(2)O` is much higher than that of `MgH_(2)` . Therefore, the reducing character increases in the order : `H_(2) O lt MgH_(2) lt NaH`
Discussion
8.

Arrange the following (i) CaH_2, BeH_2 and TiH_2 in order of increasing electrical conductance. (ii) LiH, NaH and CsH in order of increasing ionic character. (iii) H- H, D- D and F-F in order of increasing bond dissociation enthalpy. (iv) NaH, MgH_2and H_2O in order of increasing reducing property.

Answer»

SOLUTION :(i)`BeH_2 lt CaH_2 lt TiH_2`
(II) LiH `lt` NAH `lt` CSH
(iii) F-F `lt` H-H `lt` D-D
(iv)`H_2O lt MgH_2 lt NaH`
Discussion
9.

Arrange the following hydrogen halides in order of their decreasing reactivity with propene

Answer»

HCl gt HBR gt HI
HBr gt HI gt HCl
HI gt HBr gt HCl
HCl gt HI gt HBr

Solution :Reactivity INCREASES as the STRENGTH of H-X BOND DECREASES, i.e., HI gt HBr gt HCl
Discussion
10.

Arrange the following hydrogen halides in order of their decreasing reactivity with propene.

Answer»

`HCL gt HBr gt HI`
`HBr gt HI gt HCl`
`HI gt HBr gt HCl`
`HCl gt HI gt HBr`

Solution :REACTIVITY of hdydrogen halides depends upon bond dissociation energy and bond strength. As the SIZE of halogen atom increase in hydrogen halides, the bond strength decreases and bond dissociation energy also decreases. Therefore, the CORRECT order of reactivity is `HI gt HBr gt HCl`.
Discussion
11.

Arrange the following halides in order of increasing S_(N)2 reactivity: (CH_(3))_(3)C Cl, CH_(3)Cl, CH_(3)Br, CH_(3)CH_(2)Cl, (CH_(3))_(2)CHCI

Answer»

SOLUTION :`(CH_(3))_(3)Cl lt (CH_(3))_(2)CHCL lt CH_(3)CH_(2)Cl lt CH_(3)Cl lt CH_(3)Br`
Discussion
12.

Arrange the following groups in the increasing order of -M : (i) -NO_(2),-COOR,-CHO,-CN,-COR"(ii) "-overset(O)overset(||)(C)-F,-overset(O)overset(||)(C)-NH_(2),-overset(O)overset(||)(C)-O^(-)

Answer»


ANSWER :`-m:-COORlt-CORltCHOltCNltNO_(2)"(II)"-m:-OVERSET(O)overset(||)C-O^(-)lt-overset(O)overset(||)C-NH_(2)lt-overset(O)overset(||)C-F`
Discussion
13.

Arrange the following groups in the increasing order of +M : (i) -I,_Cl,-F,-Br"(ii) "-NH_(2),-OH,-O^(Theta)

Answer»


ANSWER :`+m:-Ilt-Brlt-Cllt-F"(II)"+m:-OHltNH_(2)lt-O^(THETA)`
Discussion
14.

Arrange the following group in increasing order of nitration reaction : C_(6)H_(6), H_(6)H_(5)Cl, C_(6)H_(5)COOH, C_(6)H_(5)NH_(2)

Answer»

Solution :INCREASING ORDER of nitration reaction :
`C_(6)H_(5)COOH LT C_(6)H_(5)Cl lt C_(6)H_(6) lt C_(6)H_(5)NH_(2)`
Discussion
15.

Arrange the following free radicals in order of decreasing stability Methyl (I), Vinyl (II), Allyl (III) Benzyl (IV)

Answer»

`I GT II gt III gt IV`
`III gt II gt I gt IV`
`II gt I gt IV gt III`
`IV gt III gt I gt II`

Solution :BENZYL (IV) `gt` ALLYL (III) `gt` methyl (I) `gt` VINYL (II).
Discussion
16.

The order of metallic character of Si, Na, Mg, P is

Answer»

Solution :Metallic CHARACTER increases down a group and decreases along a PERIOD as we move from LEFT to right. HENCE the order of increasing metallic character is, `Plt Si LT Be lt Mg lt Na. `
Discussion
17.

Arrange the following elements in the increasing order of non - metallic character. B,C,Si,N,F

Answer»

SOLUTION :METALLIC character increases down a group and DECREASES along a period as we MOVE from left to RIGHT. Hence the order of increasing metallic character is, `Plt Si lt Be lt Mg lt Na. `
Discussion
18.

Arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P

Answer»

Solution :METALLIC character increases down a group and decreases along a PERIOD as we MOVE from left to right. HENCE the ORDER of increasing metallic character is, `Plt Si lt Be lt Mg lt Na. `
Discussion
19.

Arrange the followingelements in theincreasing order of metallic character : B ,Al, Mg, K.

Answer»

SOLUTION :B `LT Al lt MG lt K`
Discussion
20.

Arrangethe followingelementsin orderofdecreasingelectrongainenthalpy : B, C,N , O.

Answer»


Solution :N has +ve electron gainenthalpywhile allothershave-ve electrongainenthalpies. Sincesizedecrease inmoving from B `to C to O` thereforeelectron gainenthalpiesbecomemore andmorenegativefrom B `to C to O`. Thusthe OVERALL decreasingorderof electrongainenthalpiesis N ,B ,C, O
Discussion
21.

Arrange the following elements in increasing order of non - metallic character : B, C, Si , N, F .

Answer»

Solution :SI, B,C,N,Fis increasing ORDER of non - metallic CHARACTER.
Discussion
22.

Arrangethe followingelementin theincreasingorderof theirnon - metalscharacter: B, C, SiN , F

Answer»

Solution :Arrangingthe elementsintodifferentgroupsandperiodsinorderof theirincreasingatomic numberwehave .

Sincethe metalliccharacterincreasesdown thegroupand non-metalliccharacterincreasesalong aperiodfrom leftto rightthereforeSi themostmetallicor theleast non-metallicelement . The overall increasingorderof non- metalliccharacteris `: Silt B LT C lt N lt F`
Discussion
23.

Arrange the following compounds in the descending order of their reactivity towards electrophilic substitution. a) Chlorobenzene b) Nitrobenzene c) Benzene d) Phenol e) Toluene

Answer»

`D GT E gt C gt A gt B`
`B gt A gt C gt E gt D`
`D gt E gt C gt B gt A `
`E gt D gt C gt A gt B`

SOLUTION :Phenol (OH) GROUP is strongly ACTIVATING and electron releasing group, so it prefers electrophilic substitution.
Discussion
24.

Arrange the following compound in the descending order of their reactivity towards electrophilic substitution. a) Chlorobenzene b) Nitrobenzene c )Benzene d) Phenol e) Toluene

Answer»

`D GT E gt C gt A gt B`
`B gt A gt C gt E gt D`
`D gt E gt C gt B gt A `
`E gt D gt C gt A gt B`

SOLUTION :PHENOL (OH) group is STRONGLY activating and electron releasing group, so it prefers electrophilic substitution.
Discussion
25.

Arrange the following compounds in the increasing order of their boiling points:(i) CH_3CH = CHCH_3(cis) (ii) CH_3CH = CHCH_3 (trans) (iii) CH_3CH = CHCl (cis) (iv) CH_3CH = CHCl (trans)

Answer»

(ii) lt (i) lt (iii) lt (iv)
(iv) lt (ii) lt (i) lt (iii)
(ii) lt (i) lt (iv) lt (iii)
(iii) lt (i) lt (ii) lt (iv)

Solution :BOILING points DEPENDS up on polarity ORDER = cis > TRANS
Discussion
26.

Arrange the following compounds in the increasing order of bond length of O-O bond O_(2), O_(2) [AsF_(6)], KO_(2) Explain on the basis of ground state electronic of dioxygen in there molecules .

Answer»

Solution :`O_(2) [AsF_(6)]` has ` O_(2)^(-) ` ion while `KO_(2) ` has ` O_(2)^(-)` ion .
E.C. of ` O_(2) = KK sigma _(2S)^(2) sigma _(2s)^(**2) sigma _(2p_(z))^(2) pi_(2p_(X))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**1)pi_(2p_(y))^(**1) ""` Bond order ` (1)/(2) (8-4) = 2 `
E.C. of ` O_(2)^(+) = KK sigma _(2s)^(2) sigma _(2s)^(**2) sigma _(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**1)pi_(2p_(y))^(**0) ""` Bond order ` = (1)/(2) (8-3) = 2.5`
E.C. of ` O_(2)^(-) =KK sigma _(2s)^(2) sigma _(2s)^(**2) sigma _(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**2)pi_(2p_(y))^(**1) ""` Bondorder `= (1)/(2) (8-5) = 1.5 `
Higher the bond order , smaller is the bond length . Hence , order of O-O bond length is :
` O_(2) [AgF_(6)] lt O_(2) lt KO_(2)`.
Discussion
27.

Arrange the following compounds in the descending order of their boiling points a) n-pentane b) iso pentane c) neopentane

Answer»

`cgtbgta`
`AGTBGTC`
`bgtcgta`
`cgtagtb`

ANSWER :B
Discussion
28.

Arrange the following compounds in order of their decreasing reactivity with an electrophile, E^+ (A)chlorobenzene , (B)2,4-dinitrochlorobenzene , (C )p-nitrochlorobenzene

Answer»

C gt B gt A
B gt C gt A
A gt C gt B
A gt B gt C

Solution :Electron-withdrawing groups in the benzene ring decrease the reactivity towards electrophilic substitution. Since Cl is a weaker electron-withdrawing GROUP that `NO_2`, therefore , chlorobenzene (A) is more reactive than p-nitrochlorobenzene (C ) which , in turn, is more reactive than 2,4-dinitrochlorobenzene (B), i.e., option (c) is correct.
Discussion
29.

Arrange the following compounds in order ofltbr. Increasing dipole moment Toluene (I), m-dichlorobenzene (II), o-dichloro-benzene (III) , p-dichlorobenzene (IV)

Answer»

`IltIVltIIltIII`
`IV ltI lt II lt III`
`IV lt I III lt II`
`IV lt II lt I lt III`

SOLUTION :
Discussion
30.

Arrange the following compounds in order of decreasing acidity :

Answer»

`II gt IV gt I gt III`
`I gt II gt III gt IV`
`III gt I gt II gt IV`
`IV gt III gt I gt II`

Answer :C
Discussion
31.

Arrange the following compoundsin order of decreasing reactivity in the elimination (bimolecular ) reaction with C_2H_5ONa

Answer»

II GT I gt III gt IV
IV gt III gt I gt II
III gt I gt II gt IV
I gt III gt IV gt II

Answer :A
Discussion
32.

Arrange the following compounds in order of decreasing acidity:

Answer»

`III GT I gt II gt IV`
`IV gt III gt I gt II`
`II gt IV gt I gt III`
`I gt II gt III gt IV`

Solution :Electron withdrawing groups will increase the acidic character while electron releasing group will decrease the acidic character. Further R- effect is stronger than I-effect.
Here, `-Cl` has `-I` effect and `+R` effect,
`-CH_(3)` has weak `+I` effect,
`-NO_(2)` has `-I` and `-R` effect and
`-OCH_(3)` has `-I` and `+R` effect.
THUS CORRECT order is :
`III gt I gt I gt IV`.
Discussion
33.

Arrange the following compounds in increasing order of their density. (A) C Cl_4(B) CHCl_3(C ) CH_2Cl_2(D) CH_3Cl

Answer»

D LT C lt B lt A
C GT B gt A gt D
A lt B lt C lt D
C gt A gt B gt D

SOLUTION :D < C < B < A
Discussion
34.

Arrange the following compounds in increasing order of their density? (A) C Cl_(4) (B) CHCl_(3) (C) CH_(2)Cl_(2) (D) CH_(3)Cl

Answer»

`D LT C lt B lt A`
`C gt B gtA gt D`
`A lt B lt C lt D`
`C gt A gt B gt D`

SOLUTION :`D lt C lt B lt A`
Discussion
35.

Arrange the following compounds in increasing order of their density. (A) C CI_4 ""(B) CHCI_3 "" (C) CH_2 CI_2 "" (D) CH_3 CI,

Answer»

`D ltClt B lt A `
` C gtB GT A gt D`
`A lt B LTC lt D`
`C gt Agt B gt D`

Solution :`D ltClt B lt A `
Discussion
36.

Arrange the following compounds in increasing order of their acidic strength.

Answer»

`I LT II lt III lt IV lt V`
`IV lt I lt III lt II lt V`
`III lt II lt I lt IV lt V`
`II lt I lt III lt IV lt V`

ANSWER :A
Discussion
37.

Arrange the following compounds in increasing order of acidity. (i)CH_(3) CH (Br) CH_(2) COOH (ii)CH_(3) CH_(2) CH (Br) COOH (iii)CH_(3) CH_(2)C (Br)_(2) COOH (iv)CH_(2) Br CH_(2) CH_(2) COOH

Answer»

SOLUTION :` square ` Br group causes polarity .
` square ` As the distance between Br and COOH increases ,
` square ` Order of acidity : ` III gt II gt I gt IV `
[Hint:INDUCTIVE effect DECREASES due to polarity ] .
Discussion
38.

Arrange the following compounds in decreasing order of their reactivity order with E^(+) : C_(6)H_(5)NH_(2), C_(6)H_(5)NHCOCH_(3), C_(6)H_(6), C_(6)H_(5)SO_(3)H

Answer»

Solution :Decreasing order of their REACTIVITY order with `E^(+)` is as FOLLOWS :
`C_(6)H_(5)NH_(2) gtC_(6)H_(5)NHCOCH_(3) gt C_(6)H_(6) gt C_(6)H_(5)SO_(3)H`
Discussion
39.

Arrange the following compounds in decreasing order of acidic charater I. Butene II. But1-1-ene III. But-1-yne IV. Pt//H_(2)

Answer»

IgtIIgtIIIgtIV
IVgtIIgtIIIgtI
IVgtIIIgtIIgtI
IgtIIIgtIIgtIV

Answer :C
Discussion
40.

Arrange the following compounds in decreasing order of acidic character of CH. CH_(2)=CH_(2), CH_(3)-CH_(3), C_(6)H_(6), CH-=CH

Answer»

Solution :`CH-= CH rarr CH_(2)=CH_(2) rarr C_(6)H_(6) rarr CH_(3)-CH_(3)` ACCORDINGLY the ACIDIC character of C-H decreases.
Discussion
41.

Arrange the following: CH_(3)CH_(2)CH_(2)Cl (I), CH_(3)CH_(2)-CHCI-CH_(3) (II), (CH_(3))_(2)CHCH_(2)CI (III) and (CH_(3))_(3) C-CI (IV) in order of decreasing tendency towards S_(N)2 reactions

Answer»

`I gt III gt II gt IV`
`III gt IV gt II gt I`
`II gt I gt III gt IV`
`IV gt III gt II gt I`

Solution :The decreasing ORDER of reactivity towards `S_(N)2` REACTION is
`underset((I))(CH_(3)CH_(2)CH_(2)Cl) gt underset((III))((CH_(3))_(2)CHCH_(2)Cl) gt underset((II))(CH_(3)CH_(2)CHClCH_(3)) gt underset((IV))((CH_(3))_(3)C Cl)`
This is because of increasing steric hindrance.
Discussion
42.

Arrange the following carboxylic acids in the decreasing order of reactivites (1) CH_(3)COOH (2) ClCH_(2)COOH (3) Cl_(2)CHCOOH (4) Cl_(3)C COOH

Answer»

`2 GT 4 gt 1 gt 3`
`2 gt 1 gt 3 gt 4`
`1 gt 2 gt 3 gt 4`
`4 gt 3 gt 2 gt 1`

ANSWER :D
Discussion
43.

Arrange the following carbonions in order of their decreasing stability. (A)H_3C-C -=C^- , (B)H-C-=C^- , (C )H_3C-CH_2^-

Answer»

A gt B gt C
B gt A gtC
C gt BGT A
C gt A gt B

Solution :sp-Hybridized carbon is more electronegative than a `sp^3`-hybridized carbon and hence can accommodate the negative charge better. THEREFORE, (A) and (B) are more stable than (C ). Since `CH_3` group has +I -effect, therefore , it intensifies the negative charge and hence destabilizes A relative to B. Thus, the overall stability DECREASES in the ORDER : B gt A gt C
Discussion
44.

Arrange the following carbanions in order of their decreasing stability. (A) H_(3)C-C-=C^(-) (B) H-C-=C^(-) (C) H_(3)C-bar(C)H_(2)

Answer»

`A gt B gt C`
`B gt A gt C`
`C gt B gt A`
`C gt A gt B`

Solution :`+I` effect decreases the stability of CARBON ANION. SINCE `CH_(3)` group has +I effect. In B, there is no +I effect, in A , +I effect is there but distance is more from negatvely charged carbon and in C, +I effect `(CH_(3))` direct to NEGATIVELY charged carbon.
Discussion
45.

Arrange the following carbanions in order of their decreasing stability. (I). H_(3)C-C-=C^(-) (II). H-C-=C^(-) (III). H_(3)C-CH_(2)^(-)

Answer»

IgtIIgtIII
IIgtIgtIII
IIIgtIIgtI
IIIgtIgtII

Answer :B
Discussion
46.

Arrange the following carbocations in decreasing order of their stability:

Answer»

<P>

ANSWER :`(P) I gt III gt IV gt II ; (Q) IV gt III gt II gt I ; (R) III gt II gt IV gt I`
Discussion
47.

Arrange the following carbanions in order of their decreasing stability. (I) H_(3)C-C-=C^(-) (II) H-C-=C^(-) (III) H_(3)C-CH_(2)^(-)

Answer»

IgtIIgtIII
IIgtIgtIII
IIIgtIIgtI
IIIgtIgtII

Solution :The order of decreasing stability of carbons is:
`underset((I))(H-C-=C^(-)) gt underset((I))(CH_(3)-C-=C^(-)) gt underset((III))(CH_(3)-CH_(2)^(-))`
sp-hybridised carbon ATOM is more electronegative than `sp^(3)`-hybridised carbon atom annd hence, can accomodate the negative charge more effectively. `-CH_(3)` group has +I effect, therefore, it intensifies the negative charge and hence, destabilises the carbanion .
Discussion
48.

Arrange the following carbocations in order of increasing stability I. (CH_(3))_(3)C CH_(2)^(+)II. (CH_(3))_(3)C^(+) III. CH_(3)CH_(2)CH_(2)^(+)IV. CH_(3)-overset(+)(C)H - CH_(3)

Answer»

`IV lt III lt II lt I`
`III lt IV lt I lt II`
`I lt III lt IV lt II`
`II lt IV lt III lt I`

Solution :`(CH_(3))_(3)OVERSET(+)(C)` (II) is stabilized by nine, `CH_(3)overset(+)(C)HCH_(3)(IV)` is stabilized by SIX, `CH_(3)CH_(2)CH_(2)^(+)` (III) is stabilized by twowhile `(CH_(3))_(3)C CH_(2)^(+)` (I) has no hyperconjugation structure. Therefore, II is the most stable, followed IV, then by III while I is the least stable. Thus, option (c) is CORRECT.
Discussion
49.

Arrange the following bonds in the order of increasing ionic character : C-H,F-H,Br,Na-I , K-F and Li-Cl

Answer»

Solution :`CO_(2), 180^(@)` (DUE to linear structure)
Discussion
50.

Arrange the following bonds in order of increasing polarity: P-H,H-O,C-Cl

Answer»

SOLUTION :ORDER of POLARITY is : `P-H LT C-Cl lt H-O`
Discussion