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26751.

The order of melting point , boiling point and densities of halogens

Answer»

GRADUALLY decreases from `F_(2)` to `I_(2)`
Gradually INCREASES from `F_(2)toI_(2)`
Decreases from `F_(2)`to`Br_(2)` and then increases
Increases from `F_(2)` to `Br_(2)` and then decreases

Answer :B
Discussion
26752.

The number of octahedral voids present in a lattice is _A_. The number of closed packed particles, the number of tetrahedral voids generated is _B_ the number of closed packed particles

Answer»

A-equal, B-half
A-twice, B-equal
A-twice, B-half
A-equal, B-twice

Answer :D
Discussion
26753.

The order of acidic strength of o, m and p-nitrophenol is

Answer»

<P>`o GT m gt p`
`m gt p gt o`
`p gt m gt o`
`p gt o gt m`

ANSWER :B
Discussion
26754.

The number of octahedral void(s) per atom present in a cubic close-packed structure is-

Answer»

2
4
1
3

Answer :B
Discussion
26755.

The order of K_(eq) values for the following keto-enol equilibrium constants is CH_(3)-CHO overset(K_(1))hArrCH_(2)=CH-OH, CH_(3)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)overset(K_(2))hArrCH_(3)-overset(OH)overset(|)(C)=CH-overset(O)overset(||)(C)-CH_(3) {:(""O""OH),("||""|"),(CH_(3)-C-CH_(3)overset(K_(3))hArrCH_(2)=C-CH_(3)):}

Answer»

`K_(1) gt K_(2) gt K_(3)`
`K_(2) gt K_(3) gt K_(1)`
`K_(2) gt K_(1) gt K_(3)`
`K_(1) gt K_(3) gt K_(2)`

Solution :In `CH_(3)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)`, the enol form is stable DUE to INTRAMOLECULAR H-bonding

Hence enol content MAXIMUM
(out of `CH_(3)CHO` and `CH_(3)-underset(O)underset(||)(C)-CH_(3)`)
more the number of hyperconjugating H-atom, more the enol content. Hence enol content in `CH_(3)-underset(O)underset(||)(C)-CH_(3)gt CH_(3)CHO`
`RARR K_(2) gt K_(3) gt K_(1)`
Hence choice (b) is correct while (a), (c) and (d) are incorrect.
Discussion
26756.

The number of octahedral void(s) per atom present in a cubic close -packed structure is :

Answer»

1
3
2
4

Answer :A
Discussion
26757.

The order of ionic character in metal halides is....

Answer»

MF gt MCI gt MBR gt MI
MCI gt MF gt MBr gt MI
MF gt MCI gt MBr lt MI
MF gt MCI lt MBr lt MI

Answer :A
Discussion
26758.

The order of increasing sizes of atomic radii among the elements O, S, Se and As is :

Answer»

`As LT S lt O lt SE `
`Se lt S lt As lt O`
`O lt S lt As lt Se `
`O lt S lt Se lt As `

ANSWER :D
Discussion
26759.

The number of octahedral voids in case of hcp unit cell is

Answer»

`6`
`12`
`4`
`8`

ANSWER :A
Discussion
26760.

The order of increasing reactivity towards HCI of the following compounds will be (1)CH_2=CH_2 (2)(CH_3)_2C=CH_2 (3)CH_3CH=CHCH_3

Answer»

`1 lt 2 lt 3`
`1 lt 3 lt 2`
`3 lt 2 lt 1`
`2 lt 1 lt 3`

ANSWER :C
Discussion
26761.

The number of octahedral sites per sphere in a c.c.p. (f.c.c.) structure is:

Answer»

0
1
2
4

Answer :B
Discussion
26762.

The order of increasing freezing point of C_2H_5OH, Ba_3(PO_4)_2 , Na_2SO_4 , KCl and Li_3PO_4 is

Answer»

`Ba_3(PO_4)_2 lt Na_2SO_4 lt Li_3PO_4 lt C_2H_5OH lt KCl`
`Ba_3(PO_4)_2 lt C_2H_5OHlt Li_3PO_4 lt C_2H_5OH lt KCl`
`C_2H_5OH lt KCl lt Na_2SO_4 lt Ba_3(PO_4)_2 lt Li_3PO_4`
`Ba_3(PO_4)_2 lt Li_3PO_4 lt Na_2SO_4 lt KCl lt C_2H_5OH`

Solution :`DeltaT_f PROP i`
`i` for `C_2H_5OH=1,i` for `Ba_3(PO_4)_2=5,i` for `Na_2SO_4 =3`
`i` for `Li_3PO_4` = 4 , `i` for KCl =2
THUS , depression in FREEZING point will be in order :
`C_2H_5OH lt KCl lt Na_2SO_4 lt Li_3PO_4 lt Ba_3(PO_4)_2`
Thus, freezing point will be in order :
`Ba_3(PO_4)_2 lt Li_3PO_4 lt Na_2SO_4 lt KCl lt C_2H_5OH`
Discussion
26763.

The order of increasing ionic radius of the following is:

Answer»

`K^(+) lt LI^(+) lt MG^(2+) lt Al^(3+)`
`K^(+) lt Mg^(2+) lt Li^(+) lt Al^(3+)`
`Li^(+) lt K^(+) Mg^(2+) lt Al^(3+)`
`Al^(3+) lt Mg^(2+) lt Li^(+) lt K^(+)`

ANSWER :D
Discussion
26764.

The number of octahedral sites in a cubical close pack array of N spheres is :

Answer»

N/2
2N
4N
N

Answer :D
Discussion
26765.

The order of increasing basic strength among m-toluidine (I), p-toluidine (II) and o-toluidine (III) is

Answer»

`III lt II lt I`
`II lt III lt I`
`III lt I lt II`
`II lt I lt III`.

Solution :p-Toluidine `(K_(b) =12 xx 10^(-10))`
m-Toluidine `(K_(b)= 5 xx 10^(-10))`
o-Toluidine `(K_(b) = 2.6 xx 10^(-10))`
`:.` Correct order is
`III lt I lt II`
Discussion
26766.

The number of octahedral sites for a lattice consisting of N-atoms is :

Answer»

N
2N
`N//2`
6N

Solution :For each ATOM there is ONE OCTAHEDRAL SITE.
Discussion
26767.

The order of heat of hydrogenation in following compounds :-

Answer»

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

Solution :Heat of hydrogenation `ALPHA(1)/("stability of ALKENE")`.
Discussion
26768.

The order of following reaction is …………… H_(2(g))+Cl_((g))overset(hv)to2HCl_((g))

Answer»

2
1
1.5
0

Answer :D
Discussion
26769.

The number of O_3molecules in 16 g of ozoneis approximately.

Answer»

`2 XX 10^23`
`3 xx 10^23`
`4 xx 10^23 `
`6 xx 10^23`

SOLUTION :48 gof `O_3` CONTAIN molecules = `6.02 xx 10^23`
16 g of `O_3` contain molecules
` = (6.02 xx 10^23)/(48) xx 16= 2 xx 10^23`
Discussion
26770.

The number of O-U-O linkages in Uranyl nitrate dihydrate UO_(2)(NO_(3))_(2).2H_(2)O is .......... .

Answer»


SOLUTION :
Discussion
26771.

The order of esterification of alcohols is ….

Answer»

TERTIARY `GT`PRIMARY `gt` SECONDARY
Tertiary `gt`Secondary`gt`Primary
Primary`gt`Secondary`gt`Tertiary
Secondary`gt`Primary `gt`Tertiary

Solution :Primary`gt`Secondary`gt`Tertiary
Discussion
26772.

The order of equivalent conductance at infinite dilution for LiCl, NaCland KCl is

Answer»

`LiCl gt NaCl gt KCl`
`KCl gt NaCl gt LiCl`
`NaCl gt KCl gt LiCl`
`LiCl gt KCl gt NaCl`

SOLUTION :`Li^(+)` ,due to its high polarising POWER gets HYDRATED and its ionic mobility is reduced. Smallar the cation, higher is the HYDRATION enthalpy and heavier becomes the cation. Mobility and CONDUCTANCE decrease.
Discussion
26773.

The number of o bonds, t bonds and lone pair of electrons present in acetic acid are

Answer»

7`sigma`-BONDS, 2 `pi`-bonds, 2 lone pair of `e^(-)`
6`sigma`-bonds, 1 `pi`-bonds, 4 lone pair of `e^(-)`
7`sigma`-bonds, 1 `pi`-bonds, 4 lone pair of `e^(-)`
NONE of these

Answer :C
Discussion
26774.

The number of o and pi -bonds between two carbon atoms in calcium carbide are

Answer»

ONE `sigma`, one `pi`
one `sigma`, TWO `pi`
two `sigma`, one `pi`
one `sigma, 1 (1)/(2) pi `

Answer :B
Discussion
26775.

The order of energies of following combination (a)2HHe (b)H_2+He_2 (c )He_2+2H and (d)H_2+2He is

Answer»

`d LT a ltbltc`
`d ltbltaltc`
`cltaltblt d`
`C lt B ltaltd`

ANSWER :A
Discussion
26776.

The number of Nucleotide pairs present in one turn of DNA helix

Answer»

10
9
8
4

Answer :A
Discussion
26777.

The order of electron gain enthalpy of VI A group elements is

Answer»

`S GT SE gt TE gt PO gt O`
`S gt Se gt Te gt O gt Po`
`O gt Se gt S gt Te gt Po`
`O gt Te gt Se gt S gt Po`

ANSWER :A
Discussion
26778.

The number of nonbonding electron pairs in O_2molecule is

Answer»

2
6
4
8

Answer :C
Discussion
26779.

The order of E.A of F, Cl, Br, and I is :

Answer»

F LT Clgt BRGT I<BR>F gt Clgt Brgt I
F lt CL lt Br lt I
F gt Cl lt Br lt I

Answer :A
Discussion
26780.

The number of nodel planes present in sigma^(*)santibonding orbitals is

Answer»

0
3
1
2

Solution :The molecular orbital `SIGMA^(**)s` formed by the SUBRACTION of overlapping of TWO s- orbitals.
Discussion
26781.

The order of density of the nucleus is

Answer»

`10^(8) KG// C c`
`10^(11) kg// c c`
`10^(15) kg// c c`
`10^(18) kg// c c`

ANSWER :B
Discussion
26782.

The number of nodal planes is greatest for the orbital:

Answer»

1s
2p
3d
3p

Answer :C
Discussion
26783.

The order of decrese in atomic radii for Be, Na, and Mg :

Answer»

NA GT MG gt Be
Mg gt Na gt Be
Be gt Na gt Mg
Be gt Mg gt Na

Answer :A
Discussion
26784.

The number of nitrogen atoms present in reduced product obtained on reducing nitrobenzene using LiAlH_(4) followed by aqueous work.

Answer»


ANSWER :2
Discussion
26785.

The order of decreasing stability of the carbanions (1) (CH_(3))_(3)bar(overset(..)(C))(2) (CH_(3))_(2)bar(overset(..)(C))H (3) CH_(3)bar(overset(..)(C))H_(2)(4) C_(6)H_(5)bar(overset(..)(C))H_(2) is

Answer»

`1 gt 2 gt 3 gt 4`
`4 gt 3 gt 2 gt 1`
`4 gt 1 gt 2 GT3`
`1 gt 2 gt 4 gt 3`

Solution :`UNDERSET("Benzyl carbanion")(C_(6)H_(5)-CH_(2)) gt underset("Ethyl carbanion")(CH_(3)CH_(2))gt`
`underset("Isoproyl carbanion")((CH_(3))_(2)overset(-)overset(..)(C)H) gt underset(underset("Cabanion")("Tert-butyl"))((CH_(3))_(3)overset(-)overset(..)(C))`
Discussion
26786.

The number of neutrons in the parent nucleus which gives ""^14Non beta emission is

Answer»

14
7
8
6

Answer :C
Discussion
26787.

The number of neutrons in the parent nucleus which gives N^14 on beta-emission and the parent nucleus is

Answer»

`8, C^(14)`
`6, C^(12)`
`4, C^(13)`
None of these

Solution :`._(6)C^(14) rarr ._(7)N^(14) + ._(+1)e^(0)`
No. of neutrons in `C^(14) = 14 - 6 = 8`
Discussion
26788.

The order of decreasing stability of the carbanions (CH_(3))_(3) C^(-), (CH_(3))_(2) CH^(-) (II) , CH_(3) CH_(2)^(-) (III) , C_(6) H_(5) CH_(2)(IV) is

Answer»

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

Solution :`C_(6) H_(5) CH_(2) (IV) gt CH_(3) CH_(2) (III) gt (CH_(3))_(2) CH^(-) (II) gt (CH_(3))_(3) C^(-) (I)`
Discussion
26789.

The order of decreasing reactivity towards electrophilic reagent, for the following compounds : (a) Benzene (b) Toluene (c) Chlorobenzene (d) Phenol Would be :

Answer»

`B GT d gt a gt C`
`d gt c gt b gt a`
`d gt b gt a gt c`
`a gt b gt c gt d`

SOLUTION :
Discussion
26790.

The numberof neutrons in deuterium is

Answer»

2
3
1
0

Answer :C
Discussion
26791.

The number of neutrons emitted when._(92)^(235)U undergoes controlled nuclear fission to ._(54)^(142)Xe and ._(38)^(90)Sr is

Answer»


SOLUTION :`._(92)U^(235) rarr ._(54)Xe^(142) + ._(38)Sr^(90) + 3 ._(0)N^(1)`
Discussion
26792.

The order of decreasing ease of abstraction of hydrogen atoms in the following molecule is

Answer»

`H_(a) gt H_(b) gt H_(c)`
`H_(a) gt H_(c) gt H_(b)`
`H_(b) gt H_(a) gt H_(c)`
`H_(c) gt H_(b) gt H_(a) `

Solution :A more stable PRODUCT formation will give EASY abstraction.

`2^(@)` free radical but not resonance STABILIZED.
Thus, STABILITY order of free radicals : `(I) gt (II) gt (III) `
`therefore` Abstraction order will be : `H_(a) gt H_(c) gt H_(b)`
Discussion
26793.

The number of neutrons emitteed when ._(92)^(235)U undergoes controlled nulclear fission to ._(54)^(142)Xe and ._(38)^(90)Sr is:

Answer»


Solution :`._(92)^(235)U+._(0)n^(1) rarr ._(54)^(142) Xe+._(38)^(142)Xe+._(38)^(90)Sr+x ._(0)n^(1)`
Applying MASS number balance
`235+1=142+90+x`
`x=4`
Discussion
26794.

The order of decreasing ease of reaction withammonia is

Answer»

ANHYDRIDES, ESTERS, ETHERS
Anhydrides, ethers, esters
Ethers, anhydrides, esters
Esters, ethers, anhydrides

Solution :`:NH_3` being nucleophile will attack acid anhydride and esters readily but will not attack ANOTHER nucleophile (Lewis base ) ether. Therefore order will be Anhydrides gt Esters gt Esters
Discussion
26795.

The number of neutrons emitted when ._(92)^(235) U undergoes controlled nuclear fission to ._(54)^(142)Xe and ._(38)^(90)Sr is :

Answer»


ANSWER :4
Discussion
26796.

The order of decreasing ease of abstraction of hydrogen atoms in the following molecule

Answer»

`H_(a) gt H_(B) gt H_(C)`
`H_(a) gt H_(c) gt H_(b)`
`H_(b) gt H_(a) gt H_(c)`
`H_(c) gt H_(b) gt gtH_(a)`

SOLUTION :N//A
Discussion
26797.

The number of neutrons emitted when ""_(92)^(235) U undergoes controlled nuclear fission to ""_(54)^(142) Xe and ""_(38)^(90) Sr is

Answer»


Solution :`""_(92) U^(235) + ""_(0) n^(1) to ""_(54) Xe^(142) + ""_(38) Sr^(90) + y(""_(1) n^(0)) , (235 + 1) = (142 + 90) + y`
`y = (236- 232) = 4` NET no. of .n. emitted = `(4 - 1)` = 3
Discussion
26798.

The number of neutrons accompanyng the formation of _(54)^(139)Xe and _(38)^(139)Xe from the absorption of a slow neutron by _(92)^(235)U followed by nuclear fision is:

Answer»


Solution :SINCE the charge in mass is only due to the emmission of `alpha`-particle, we have
Number of `alpha`-particle emitted `=(234-206)/(4)=7`
Now the associated decrease in atomic number would be `14 (=2xx7)` and thus the atomic number of the daughter atom would be `76(=90-14)`. But the actual atomic number of lead is `82 E`. the atomic number is six more than EXPECTED. This is because of the emission `beta`-particle. Since there is an INCREASE of one in atomic number due to the emission of one `beta`-particle, we have
Number of `beta`-particles emitted `=(82-76)/(1)=6`
Hence, number of `alpha`-particles emitted `=7` ltbr,. Number of `beta`-particles emittes `=6`
ANSWER is `6+7=13`
Discussion
26799.

The order of decomposition of acetaldehyde is ......

Answer»

1
1.5
2
`5//2`

ANSWER :B
Discussion
26800.

The number of neutrons accompanying the formationof ""_54^139Xe and ""_38^94Sr from the absorption of slow neutrons by ""_92^235U followed by nuclear fission is

Answer»

0
2
1
3

Solution :`""_92^235U + ""_0^1n to ""_54^139Xe + ""_38^94 SR+ x ""_0^1n`
Equating mass NUMBER on both sides
235+1= 139+94+(x x 1)
x=3
Thus, 3 neutrons released in the PROCESS.
Discussion