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26951.

The number of methoxy groups in a compound can be determined by treating it with :

Answer»

HI and `AgNO_3`
SODIUM carbonate
Sodium HYDROXIDE
ACETIC acid

Answer :A
Discussion
26952.

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +(1)/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) The wave number of electromagnetic radiation emitted during the transition of electron in between two levels of Li ^(2+) ion whose principal quantum numbers sum is 4 and difference is 2 is:

Answer»

`3.5R`
`4R`
8 R
`8//9R`

ANSWER :C
Discussion
26953.

The number of methoxy and ethoxy groups in a compound is determined and estimated by :

Answer»

Zeisel method
Herzig mehtod
HOFMANN method
KOLBE's mehtod.

Solution :NA
Discussion
26954.

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +(1)/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) The difference in the wavelength of the 1 ^(st) line of Lyman series and 2nd line of Balmer series in a hydrogen atom is :

Answer»

`(9)/(2R)`
`4/R`
`(88)/(15R)`
NONE

Answer :D
Discussion
26955.

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +(1)/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) In a collection of H–atom, electrons make transition from 5th excited state to 2nd excited state then maximum number of different types of photons observed are :

Answer»

3
4
6
15

Answer :C
Discussion
26956.

The number of lone pairs on S atom in SF_2, SF_4 and SF_6 are respectively :

Answer»

2, 1 and 0
2, 1 and 1
4, 2 and 0
2,2 and 2

Answer :A
Discussion
26957.

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) The energy photon emitted corresponding to transition n =3 to n =1 is [h=6 xx 10 ^(-34)J-sec]

Answer»

`1.76 XX 10 ^(-18)J`
`1.98 xx 190 ^(-18)J`
`1.76 xx 10 ^(-17)J`
None of these

Answer :A
Discussion
26958.

The number of lone pairs on Chlorine atom in CIO^(-), CIO_(2)^(-), CIO_(3)^(-), ClO_(4)^(-) ions are

Answer»

0,1,2,3
1,2,3,4
4,3,2,1
3,2,1,0

Answer :D
Discussion
26959.

The total number of lone pair present in XeF_(4) is

Answer»

3
4
2
1

Answer :C
Discussion
26960.

The only correct combination which product produces intense blue colour with starch paper will be .

Answer»

(I)(i)(Q)
(II)(i)(R)
Both (A) and (B)
(IV),(i),(S)

SOLUTION :`K_2Cr_2O_7+6KI+14H^(+)to8K_((AQ))^+ +2Cr_((aq))^(+3)+7H_2O+3I_2`
`2KMnO_4+10KI+16H^(+)to12K_((aq))^(+)+2Mn_((aq))^(+2)+5I_2+8H_2O`
Discussion
26961.

The number of lone pairs of electrons on the central atoms of H_(2)O, SnCl_(2), PCl_(3) and XeF_(2) respectively are

Answer»

2,1,1,3
2,2,1,3
3,1,1,2
2,1,2,3

Solution :`:H_(2)underset(..)OVERSET(..)(O)` (O is GROUP 16) `overset(..)(S)nCl_(2)` (Sn is group 14),
`overset(..)(P)Cl_(3)` (P is group 15), `:underset(..)overset(..)(X)eF_(2)` (Xe is group 18).
Discussion
26962.

The number of lone pairs and S-S bonds, in S_(8) molecule, respectively

Answer»

`8&8`
`4&4`
`16&8`
`16&4`

Answer :C
Discussion
26963.

The only correct combination which shows releasing of H_2 gas, will be:

Answer»

(II)(iii)(R)
(III)(iii)(S)
(IV)(iv)(S)
(I)(ii)(Q)

Solution :`CrCl_(2(AQ))+2HCL to2CrCl_(3(aq))+H_(2(g))uarr`
Discussion
26964.

The number of lone pairs and bond pairs present on Xe of XeO_3, molecule

Answer»

1,3
1,6
4,3
6,1

Answer :B
Discussion
26965.

The only correct combination that product gives haloform test :

Answer»

(IV)(III)(P)
(IV)(i)(P)
(IV)(II)(S)
(IV)(iii)(R)

SOLUTION :
Discussion
26966.

The numberof lone pairof electrons on the Xe - atom in XeF_(2),XeF_(4)and XeF_(6) molecules are respectively -

Answer»

3,2,1
4,3,2
2,3,1
3,2,0

Answer :A
Discussion
26967.

The only correct combination that reaction nature is stereospecific and stereoselective.

Answer»

<P>(I)(i)(S)
(II)(i)(P)
(III)(ii)(Q)
(IV)(iv)(P)

SOLUTION :
Discussion
26968.

The number of lattice points per unit cell in B.C.C and end centered lattice respectively

Answer»

6, 6
9, 10
6, 8
6, 10

Answer :B
Discussion
26969.

The number of linkage isomers for the compound K_(2)[Cu(CN)_(2)(NCS)_(2)] are

Answer»

4
6
16
9

Solution :With RESPECT to -CN TWO linkageisomers.
With respect to -SCN two LINKAGE isomers.
Discussion
26970.

The only correct combination is :

Answer»

<P>(II)(iv)(P)
(I)(ii)(Q)
(III)(ii)(R)
(IV)(iii)(S)

Solution :(I)First line of Balmer SERIES
`n_1=2 " " n_2=3`
`DeltaP.E=27.2xx(Z)^2[1/n_1^2-1/n_2^2]`
=`27.2xx(1)^2[1/4-1/9]=27.2xx5/36=3.78 ev`
`DeltaE=13.6xx(1)^2[1/4-1/9]=1.88 ev`
`barv=Rxx(1)^2[1/4-1/9]=(5R)/(36)`
`lambda=36/(5R)`
Angular momentum=`(3h)/(2PI)-(2h)/(2pi)=(h)/(2pi)`
(II)Third line of PASCHEN series of `He^+` ion
`n_1=3 " " n_2=6`
`DeltaP.E=27.2xx(2)^2[1/9-1/36]=9.06 ev`

`DeltaE=13.6xx(2)^2[1/9-1/36]=4.54 ev`
`barv=Rxx(2)^2[1/9-1/36]=R/3`
`lambda=3/(R)`
Angular momentum=`(6h)/(2pi)-(3h)/(2pi)=(3h)/(2pi)`
(III)Lyman series limit for`Li^(+2)` ion
`n_1=1 " " n_2=oo`
`DeltaP.E=27.2xx(3)^2[1/1^2-1/oo^2]=27.2xx9=244.8 ev`

`DeltaE=13.6xx(9)[1/1^2-1/oo^2]=122.4 ev`
`barv=Rxx(3)[1/1^2-1/oo^2]=9R`
(IV)`2^(nd)` line of Lyman series for `He^+` ion
`n_1=1 " " n_2=3`
`DeltaP.E=27.2xx(2)^2[1/1-1/9]=96.71 ev`

`DeltaE=13.6xx(2)^2[1/1-1/9]=48.4 ev`
`barv=Rxx(2)^2[1/1-1/9]=(32R)/9,lambda=9/(32R)`
Angular momentum=`(3h)/(2pi)-(h)/(2pi)=(2h)/(2pi)=h/pi`
Discussion
26971.

The number of a bonds in the product formed by passing acetylene through dil.H_2 SO_4 containing mercuric sulphate is

Answer»

zero
one
two
three

Answer :B
Discussion
26972.

The number of isomers possible for the octahedral complex[CoCl_2(en)(NH_3)_2]^+ is ,

Answer»

Two
Three
No isomer
Four ISOMERS

ANSWER :D
Discussion
26973.

The only correct combination that gives two moles of same monosaccharides is :

Answer»

<P>(I)(II)(P)
(II)(III)(P)
(II)(i)(S)
(IV)(i)(P)

SOLUTION :
Discussion
26974.

The only correct combination that gives red ppt. is :

Answer»

(I)(i)(R)
(III)(ii)(R)
(II)(i)(Q)
(I)(iii)(R)

SOLUTION :All reducing sugars GIVE RED PPT with Fehling solution
Sucrose is non reducing sugar
Discussion
26975.

The only correct combination is :

Answer»

(I)(iii)(R)
(II)(i)(S)
(III)(ii)(P)
(IV)(iv)(Q)

Solution :(I)First line of Balmer SERIES
`n_1=2 " " n_2=3`
`DeltaP.E=27.2xx(Z)^2[1/n_1^2-1/n_2^2]`
=`27.2xx(1)^2[1/4-1/9]=27.2xx5/36=3.78 ev`
`DeltaE=13.6xx(1)^2[1/4-1/9]=1.88 ev`
`barv=Rxx(1)^2[1/4-1/9]=(5R)/(36)`
`lambda=36/(5R)`
Angular momentum=`(3h)/(2pi)-(2H)/(2pi)=(h)/(2pi)`
(II)Third line of PASCHEN series of `He^+` ion
`n_1=3 " " n_2=6`
`DeltaP.E=27.2xx(2)^2[1/9-1/36]=9.06 ev`

`DeltaE=13.6xx(2)^2[1/9-1/36]=4.54 ev`
`barv=Rxx(2)^2[1/9-1/36]=R/3`
`lambda=3/(R)`
Angular momentum=`(6h)/(2pi)-(3h)/(2pi)=(3h)/(2pi)`
(III)Lyman series limit for`Li^(+2)` ion
`n_1=1 " " n_2=oo`
`DeltaP.E=27.2xx(3)^2[1/1^2-1/oo^2]=27.2xx9=244.8 ev`

`DeltaE=13.6xx(9)[1/1^2-1/oo^2]=122.4 ev`
`barv=Rxx(3)[1/1^2-1/oo^2]=9R`
(IV)`2^(nd)` line of Lyman series for `He^+` ion
`n_1=1 " " n_2=3`
`DeltaP.E=27.2xx(2)^2[1/1-1/9]=96.71 ev`

`DeltaE=13.6xx(2)^2[1/1-1/9]=48.4 ev`
`barv=Rxx(2)^2[1/1-1/9]=(32R)/9,lambda=9/(32R)`
Angular momentum=`(3h)/(2pi)-(h)/(2pi)=(2h)/(2pi)=h/pi`
Discussion
26976.

The number of isomers possible for the octahedral complex [CoCI_2 (en)(NH_3)_2]^+is,

Answer»

Two
Three
No isomer
Four isomers

Solution :
Discussion
26977.

The only correct combination is -

Answer»

(I)(iii)(R)
(IV)(i)(Q)
(I)(i)(S)
(IV)(iii)(S)

Solution :`S^(-2)+H_2SO_4(DIL.)toH_2S(uarr)`
`S^(-2)+[FE(CN)_5NO]^(2+)tounderset("voilet")([Fe(CN)_5NOS](darr))`
`S^(-2)+16H^(+)+2MnO_4^(-)to2Mn^(+2)+8H_2O+5S`
Discussion
26978.

The number of isomers possible for square planar complex K_(2)[PdClBr_(2)SCN] are

Answer»


SOLUTION :With RESPECT to cis-2 ISOMERS, with respect to TRANS 2 isomers.
Discussion
26979.

The only correct combination is :

Answer»

(I)(ii)(P)
(IV)(iii)(Q)
(II)(iv)(R)
(III)(i)(S)

Solution :(I)FIRST line of Balmer series
`n_1=2 " " n_2=3`
`DeltaP.E=27.2xx(Z)^2[1/n_1^2-1/n_2^2]`
=`27.2xx(1)^2[1/4-1/9]=27.2xx5/36=3.78 ev`
`DeltaE=13.6xx(1)^2[1/4-1/9]=1.88 ev`
`barv=Rxx(1)^2[1/4-1/9]=(5R)/(36)`
`lambda=36/(5R)`
Angular momentum=`(3h)/(2pi)-(2h)/(2pi)=(h)/(2pi)`
(II)Third line of paschen series of `He^+` ion
`n_1=3 " " n_2=6`
`DeltaP.E=27.2xx(2)^2[1/9-1/36]=9.06 ev`

`DeltaE=13.6xx(2)^2[1/9-1/36]=4.54 ev`
`barv=Rxx(2)^2[1/9-1/36]=R/3`
`lambda=3/(R)`
Angular momentum=`(6H)/(2pi)-(3h)/(2pi)=(3h)/(2pi)`
(III)Lyman series limit for`Li^(+2)` ion
`n_1=1 " " n_2=oo`
`DeltaP.E=27.2xx(3)^2[1/1^2-1/oo^2]=27.2xx9=244.8 ev`

`DeltaE=13.6xx(9)[1/1^2-1/oo^2]=122.4 ev`
`barv=Rxx(3)[1/1^2-1/oo^2]=9R`
(IV)`2^(ND)` line of Lyman series for `He^+` ion
`n_1=1 " " n_2=3`
`DeltaP.E=27.2xx(2)^2[1/1-1/9]=96.71 ev`

`DeltaE=13.6xx(2)^2[1/1-1/9]=48.4 ev`
`barv=Rxx(2)^2[1/1-1/9]=(32R)/9,lambda=9/(32R)`
Angular momentum=`(3h)/(2pi)-(h)/(2pi)=(2h)/(2pi)=h/pi`
Discussion
26980.

The only correct combination is :

Answer»

<P>(I)(iii)(R)
(II)(ii)(P)
(IV)(ii)(S)
(III)(iv)(R)

SOLUTION :NA
Discussion
26981.

The number of isomers possible for square planar complex K_2[PdClBr_2SCN] is:

Answer»

2
3
4
6

Answer :C
Discussion
26982.

The number of isomers of dibromobutane is-

Answer»

9
5
6
8

Solution :9
Discussion
26983.

The number of isomers of C_(6)H_(14) is

Answer»

4
5
6
7

Solution :(i) `CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(3)`
(ii) `{:(CH_(3)-CH-CH_(2)-CH_(2)-CH_(3)),("|"),(""CH_(3)):}`
(iii)`{:(CH_(3)-CH-CH-CH_(3)),("|""|"),(""CH_(3)" "CH_(3)):}`
(iv) `{:(""CH_(3)),("|"),(CH_(3)-C-CH_(2)-CH_(3)),("|"),(""CH_(3)):}`
(V) `{:(CH_(3)-CH_(2)-CH-CH_(2)-CH_(3)),("|"),(""CH_(3)):}`
Discussion
26984.

The only correct combination is :

Answer»

<P>(II)(i)(S)
(IV)(i)(Q)
(III)(iv)(P)
(I)(ii)(Q)

SOLUTION :NA
Discussion
26985.

The number of isomers obtained on monochlorination of propane

Answer»

4
3
2
1

Solution :Propane has two types of hydrogen ATOM i.,e `1^(@)` hydrogen and `2^(@)` hydrogen. Hence it GIVES two isomeric product on chlorination i.e., CHLOROPROPANE and 2-chloropropane.
Discussion
26986.

The number of isomers including stereoisomers possible for dibromobutane is

Answer»

six
eight
four
ten

Solution :
Discussion
26987.

The only correct combination in which one compound has only one carbon in its parent chain ?

Answer»

(I)(i)(Q)
(II)(ii)(R)
(III)(ii)(R)
(IV)(i)(P)

Solution :IUPAC name of `Ph-CH_2-NH-CH=O` is N-Phenylmethylmethanamide [Parent chain with one carbon atom only ]
`Ph-NH-CH_2-CH=O overset("Tautomerism")HARR Ph-NH-CH=CH-OH`(enol)
Discussion
26988.

The number of isomers for thecompounds with molecularformula C_(2)BrClFI is:

Answer»

`3`
`4`
`5`
`6`

SOLUTION :
Discussion
26989.

The only CORRECT combination in which polymerization of hydrolysed product gives condensation polymer.

Answer»

(I)(ii)(S)
(II)(i)(R)
(III)(IV)(Q)
(IV)(iv)(Q)

SOLUTION :
Discussion
26990.

The number of isomers (including stereoisomers) of C_5H_(10) are

Answer»

10
11
12
13

Answer :D
Discussion
26991.

The only correct combination in which one compound has optical active diastereomers ?

Answer»

<P>(I)(i)(P)
(II)(ii)(Q)
(III)(iii)(S)
(IV)(iv)(P)

Solution :
It has TWO CHIRAL atoms and total 4-stereoisomers including (cis-trans) PAIR of diastereomers.
Discussion
26992.

The number of isomers for the compound with molecule formula C_2HDFCl is :

Answer»


SOLUTION :
Discussion
26993.

The number of isomers for the compound with molecular formula C_2BrClFI is:

Answer»

3
4
5
6

Answer :D
Discussion
26994.

The only correct combination in which isocyanide is formed

Answer»

<P>(I)(i)(S)
(III)(IV)(P)
(III)(iv)(Q)
(I)(iv)(P)

SOLUTION :
Discussion
26995.

The only correct combination in which carboxylic acid is formed.

Answer»

<P>(II)(i)(R)
(II)(iii)(P)
(IV)(iii)R
(III)(i)(P)

Solution :
Discussion
26996.

The number of isomers for the compound with molecular formula C_(2)H_(3)Cl_(3) is :

Answer»

TWO
Three
Four
Five

Answer :A
Discussion
26997.

The number of isomers for the compound with molecular formula C_(2)BrCIFI are:

Answer»


ANSWER :6
Discussion
26998.

The only correct combination in which both compounds can have two optical active stereoisomers ?

Answer»

<P>(I)(i)(P)
(II)(ii)(Q)
(IV)(iii)(P)
(III)(iv)(R)

SOLUTION :
Discussion
26999.

The only correct combination in which Azo dye is formed.

Answer»

(I)(II)(Q)
(III)(ii)(Q)
(I)(i)(S)
(II)(iii(R)

SOLUTION :
Discussion
27000.

The only cations present in a slightly acidic solution are Fe^(3+),Zn^(2+) and Cu^(2+). The reagent that when added in excess to this solution would identify the separate Fe^(3+) in one step is

Answer»

2M HCl
6M `NH_(3)`
6 NAOH
`H_(2)S` gas

Solution :`Fe^(3+)` is a THIRD group radical, whose reagent is `NH_(4)OH` in presence of `NH_(4)CL`. So if 6M `NH_(3)` is added in the slightly ACIDIC (HCl) solution of IONS it will lead to the
`Fe^(3+)+Zn^(2+)+Cu^(2+) overset("6 M"NH_(3))to underset("Brown ppt.")(Fe(OH)_(3))+underset("Soluble")([Zn(NH_(3))_(4)]^(2+))+underset("Soluble")([Cu(NH_(3))_(4)]^(2+))`
Discussion