Explore topic-wise InterviewSolutions in .

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.

29601.

The isoelectronic pair is

Answer»

`Cl_(2) O, ICl_(2)^(-)`
`Cl_(2)^(-) , ClO_(2)`
`IF_(2)^(+), I_(3)^(-)`
`ClO_(2)^(_) , ClF_(2)^(+)`

ANSWER :D
Discussion
29602.

The isoelectric-points of a colloidally dispersed material is the pH value at which

Answer»

the dispersed PHASE MIGRATE in an electric field
the dispersed phase does not migrate in an electric field
the dispersed phase has PH EQUAL to 7
the dispersed phase has pH equal to zero

Solution :At isoelectric point there is no migration of dispersed phase in an electric field.
Discussion
29603.

The isoelectric (or isoionic) point is the pH where the amino acid is

Answer»

POSITIVELY CHARGED.
NEGATIVELY charged.
neutral.
can't say.

Answer :C
Discussion
29604.

The incrorrect expression amoung the following is

Answer»

`(DeltaG" SYSTEM")/(DeltaS" total")=-T`
In SPONTANEOUS process `W_("reversible")=nRTl"N"(V_(f))/(V_(i))`
`lnK=(DeltaH^(@)-TDeltaS^(@))/(RT)`
`K=e^(DeltaG^(@)//RT)`

Solution :(c ) `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`-RTInK=DeltaH^(@)-TDeltaS^(@)`
`InK=(-DeltaH^(@)-TDeltaS^(@))/(RT)`
Discussion
29605.

The isoelectric point of a protein is :

Answer»

the PHAT which the protein in molecule has no CHARGES on its surface
the pH at which the protien SOLUTION has an EQUAL number of positive and nagtive charge
the electric charge under isothermal CONDITION
none of these

Solution :N//A
Discussion
29606.

The isobars are atoms with the same number of

Answer»

Protons
Neutrons
Protons and neutrons
Nucleons

Answer :D
Discussion
29607.

The indicated atom is not a nuclcophilic site in

Answer»

`underset(UARR )(BH_(4)^(-))`
`underset(uarr )(CH_(3) ) `Mgl
`CH_(3) underset(uarr)(O)H`
`CH_(3) underset(uarr )(N)H_(2)`

Solution :In `BH_(4)^(-)` . there is no lone pair on B-atom. HENCE, t is not a nucleophilic site
Discussion
29608.

The ______ is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.

Answer»


ANSWER :CRYSTAL FIELD MODEL
Discussion
29609.

The islent killer carbon monoxide can be estimated with

Answer»

`I_(2)O_(4)`
`I_(2)O_(5)`
`I_(2)O_(7)`
`BrO_(3)`

Answer :B
Discussion
29610.

The iron salt used in blue prints is :

Answer»

`FeC_(2)O_(4)`
`Fe_(2)(C_(2)O_(4))_(3)`
`K_(4)Fe(CN)_(6)`
`FeSO_(4)`

Solution :`2CrO_(4)^(2-) +2H^(+) to Cr_(2)O_(7)^(2-) +H_(2)O`
`Cr_(2)O_(7)^(2-)+2OH^(-)+2CrO_(7)^(2-)+H_(2)O`
Discussion
29611.

The iron obtained from the blast furnace is called:

Answer»

PIG iron
Cast iron
Wrought iron
Steel

Answer :A
Discussion
29612.

The increasing order of volatitility of hydrides of group 16 element is

Answer»

`H_(2)S LT H_(2)O lt H_(2)SE lt H_(2)Te`
`H_(2)O lt H_(2)Te lt H_(2)Se lt H_(2)S`
`H_(2)O, H_(2)S lt H_(2)Se lt H_(2)Te`
`H_(2)Te lt H_(2)Se lt H_(2)S lt H_(2)O`

Solution :As `H_(2)O` has the HIGHEST boiling point and `H_(2)S` the LOWEST, and boiling point further increases from `H_(2)Se` to `H_(2)TE`, therefore the volatility increases in the order `H_(2)O lt H_(2)Te lt H_(2)Se lt H_(2)S`
Discussion
29613.

The increasing order of + I effect shown by H, CH_3, C_2H_5 and C_3H_7 is:

Answer»

`HgtCH_3ltC_2H_5ltC_3H_7`
`HltCH_3ltC_2H_5gtC_3H_7`
`HLT C_2H_5 LT CH_3 lt C_3H_7`
NONE of these

Answer :A
Discussion
29614.

The iron obtained from blast funkce is

Answer»

PIG IRON
Silver
Soft iron
Steel

Answer :A
Discussion
29615.

The iron catalyst used in the synthesis of ammonia in Haber's process is poisoned by

Answer»

`As_2O_3`
`V_2O_5`
`H_2S`
Glycerine

Answer :C
Discussion
29616.

The increasing order of the reactivity of the following halides for the Sn^(1) reaction is underset((I))(underset(Cl)underset(|)(CH_(3)CH)CH_(2)CH_(3))""underset((II)(CH_(3)CH_(2)CH_(2)Cl)""underset((III)(pH_(3)CO-C_(6)H_(4)-CH_(2)Cl)

Answer»

`(II) lt (I) lt (III)`
`(I) lt (III) lt (II)`
`(II) lt (III) lt (I)`
`(III) lt (II) lt (I)`

Solution :Reactivity towards `SN^(1)` reaction `PROP` stability of CARBOCATION IIltIltIII

[Stability ORDER of carbocation]
Discussion
29617.

The iron obtained from blast funace is

Answer»

PIG IRON
Silver
SOFT iron
Steel

Answer :A
Discussion
29618.

The increasing order of the reactivity of the following halides for the S_(N)1 reaction is …… underset((I))(CH_(3)underset("Cl ")underset("| ")("C ")HCH_(2)CH_(3)) "" underset((II))(CH_(3)CH_(2)CH_(2)Cl) "" underset((III))(p-H_(3)CO-C_(6)H_(4)-CH_(2)Cl)

Answer»

`(iii)lt(ii)lt(i)`
`(II)lt(I)lt(III)`
`(I)lt(III)lt(II)`
`(II)lt(III)lt(I)`

Solution :REACTIVITY of `S_(N)1` Reaction `PROP` STABILITY of carbocation.
Discussion
29619.

The IR stretchin frequencies of free CO, and CO [V(CO)_(6)]^(-),[Cr(CO)_(6)]^(-) and [Mn(CO)_(6)]^(-) are 2143cm^(-1),1860cm^-1),2000cm^(-1) and 2090cm^(-1), respectively. Then correct statemet about metal carbonyl is:

Answer»

C-O' bond is STRONGEST in the cation and WEAKEST in the anion.
C-O' bond is weakest in the cation and strongest in the anion
C-O' bond is longer in the cation than in the anion
M-C' pi bonding is HIGHER in the cation

Solution :In metal carbonyls:
Higher the negative oxidation state of CENTRAL metal: `prop `Bond LENGTH of C-O
`:prop(1)/("Bond order of CO")`
: Bond order of M-C bond.
Discussion
29620.

The increasing order of the strength of hydrogen bond in the following mentioned linkages is : (i)O-H-S(ii)S-H-O (iii)F-H-F(iv)F-H-O

Answer»

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

Solution :Strength of H-bond depends on following FACTORS
(i)ELECTRONEGATIVITY of ELEMENT covalently bonded to hydrogen atom
(ii)Size of ELECTRONEGATIVE element.
(iii)Ease of donation of lone pair of electrons by electronegative element.
Discussion
29621.

The IP of H is 13.6 eV. It is exposed to electromagnetic waves of 1028Å and given out induced radiatiions. Find the wavelength of these induced radiations.

Answer»


ANSWER :`6558Å`
Discussion
29622.

The ions that can be precipitated by both HCl and H_(2)S are

Answer»

`PB^(+2)`
`Cu^(+2)`
`Ag^(+)`
`Sn^(+2)`

Solution :`Sn^(+2)` can be PRECIPITATES only by `H_(2)S` but not by HCI
Discussion
29623.

The increasing order of the reactivity of the following halides for the S_(N^(1)) reaction is: underset(I)(CH_(3)underset(Cl)underset(|)(C)HCH_(2)CH_(3))" "underset(II)(CH_(3)CH_(2)CH_(2)Cl)" "underset(III)(p-H_(3)COC_(6)H_(4)CH_(2)Cl).

Answer»

`(I)lt(III)lt(II)`
`(II)lt(III)lt(I)`
`(III)lt(II)lt(I)`
`(II)lt(I)lt(III)`

Solution :In `S_(N^(1))` MECHANISM, the reactivity ORDER is linked with the relative stabilities of the carbocations in the slow STEP. In the present case, the alkyl halides ionise as follows:
.
The CORRECT order `(II) lt (I)lt (III)`.
Discussion
29624.

The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^(-), small molecules such as H_(2)O or NH_(3), larger molecules such as H_(2)NCH_(2)CH_(2)NH_(2) or N(CH_(2)CH_(2)NH_(2))_(3) or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl^(-), H_(2)O or NH_(3), the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H_(2)NCH_(2)CH_(2)NH_(2) (ethane-1, 2-diamine) or C_(2)O_(4)^(2-) (oxalate), the ligand is said to be didentate and when several donor atoms are present in single ligand as in N(CH_(2)CH_(2)NH_(2))_(3), the ligand is said to be polydentate Ethylenediamineteraacetate ion [EDTA^(4-)] is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. How many nitrogen atoms and oxygen atoms in ethylenediaminetetraacetate are bonded to the metal in the complex compounds ?

Answer»

SOLUTION :TWO NITROGEN atoms and FOUR oxygen atoms in ethylenediaminetetraacetate ligand are BOUND to the central atom.
Discussion
29625.

The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^(-), small molecules such as H_(2)O or NH_(3), larger molecules such as H_(2)NCH_(2)CH_(2)NH_(2) or N(CH_(2)CH_(2)NH_(2))_(3) or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl^(-), H_(2)O or NH_(3), the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H_(2)NCH_(2)CH_(2)NH_(2) (ethane-1, 2-diamine) or C_(2)O_(4)^(2-) (oxalate), the ligand is said to be didentate and when several donor atoms are present in single ligand as in N(CH_(2)CH_(2)NH_(2))_(3), the ligand is said to be polydentate Ethylenediamineteraacetate ion [EDTA^(4-)] is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. What type of ligands are H_(2)O and NH_(3) ?

Answer»

Solution :`H_(2)O` and `NH_(3)` are unidentate LIGANDS.
Discussion
29626.

The increasing order of the reactivity of the following halides for the S_(N)^(1) reaction is : CH_(3)underset(CI)(CHCH_(2) CH_(3)) CH_(3) CH_(2) CH_(2)Cl p-H_(3)CO–C_(6)H–CH_(2)CI

Answer»

`(II) LT (I) lt (III)`
`(I) lt (III) lt (II)`
`(II) lt (III) lt (I)`
`(III) lt (II) lt (I)`

SOLUTION :
Discussion
29627.

The ion(s) that act/s as oxidizing agent in solution is/are

Answer»

`TL^(+)` and `Al^(3+)`
`B^(3+)`and `Al^(3+)`
`Tl^(3+)` only
`B^(3+)` only

Solution :`Tl^(+)` ions are more stable than `Tl^(3+)` ions and thus, `Tl^(3+)` ions change to `Tl^(+)` ion thereby acting as oxidising AGENT.
`underset("(LESS stable oxidising agent)")(Tl^(3)" Compounds ")+2e^(-) RARR Tl^(+)underset(("(More stable"),("reducing agent)"))("Compounds")`
Discussion
29628.

The increasing order of the reactivity of the following halides for the S_(N)1 reaction is - CH_(3)underset((I))underset(Cl)underset(|)CHCH_(2)CH_(3)CH_(3)CH_(2)underset(II)CH_(2)Cl p-H_(3)CO-C_(6)H_(4)-CH_(2)Cl (III)

Answer»

`(II) LT (I) lt (III)`
`(I) lt (III) lt (II) `
`(II) lt (III) lt (I)`
`(III) lt (II) lt (I)`

ANSWER :A
Discussion
29629.

The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^(-), small molecules such as H_(2)O or NH_(3), larger molecules such as H_(2)NCH_(2)CH_(2)NH_(2) or N(CH_(2)CH_(2)NH_(2))_(3) or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl^(-), H_(2)O or NH_(3), the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H_(2)NCH_(2)CH_(2)NH_(2) (ethane-1, 2-diamine) or C_(2)O_(4)^(2-) (oxalate), the ligand is said to be didentate and when several donor atoms are present in single ligand as in N(CH_(2)CH_(2)NH_(2))_(3), the ligand is said to be polydentate Ethylenediamineteraacetate ion [EDTA^(4-)] is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. What is meant by didentate ligand ? Give an example other than ethane-1,2-diamine.

Answer»

Solution :A LIGAND that binds to the central metal at TWO PLACES is CALLED didentate ligand.
Discussion
29630.

The increasing order of the rate of HCN addition to compounds, A-D is A. HCHO B. CH_(3)COCH_(3) C. PhCOCH_(3) D. PhCOPh

Answer»

`A LT B lt C lt D`
`D lt B lt C lt A`
`D lt C lt B lt A`
`C lt D lt B lt A`

ANSWER :C
Discussion
29631.

The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^(-), small molecules such as H_(2)O or NH_(3), larger molecules such as H_(2)NCH_(2)CH_(2)NH_(2) or N(CH_(2)CH_(2)NH_(2))_(3) or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl^(-), H_(2)O or NH_(3), the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H_(2)NCH_(2)CH_(2)NH_(2) (ethane-1, 2-diamine) or C_(2)O_(4)^(2-) (oxalate), the ligand is said to be didentate and when several donor atoms are present in single ligand as in N(CH_(2)CH_(2)NH_(2))_(3), the ligand is said to be polydentate Ethylenediamineteraacetate ion [EDTA^(4-)] is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. Write the formula of ethylenediamine.

Answer»

SOLUTION :`H_(2)NCH_(2)CH_(2)NH_(2)` is the formula of ETHYLENEDIAMINE.
Discussion
29632.

The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^(-), small molecules such as H_(2)O or NH_(3), larger molecules such as H_(2)NCH_(2)CH_(2)NH_(2) or N(CH_(2)CH_(2)NH_(2))_(3) or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl^(-), H_(2)O or NH_(3), the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H_(2)NCH_(2)CH_(2)NH_(2) (ethane-1, 2-diamine) or C_(2)O_(4)^(2-) (oxalate), the ligand is said to be didentate and when several donor atoms are present in single ligand as in N(CH_(2)CH_(2)NH_(2)_(3), the ligand is said to be polydentate Ethylenediamineteraacetate ion [EDTA^(4-)] is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. What is meant by the term ligand ?

Answer»

Solution :The IONS or molecules bound to the CENTRAL atom/ion in the coordination COMPOUNDS are CALLED ligands.
Discussion
29633.

The ions O^(2-) , Na^+ , F^- , Mg^(2+) and Al^(3+)are isoelectronic,Their ionic radii show

Answer»

a significant decrease from `O^(2-) to Al^(3+)`
an increase from `O^(2-) to F^-` and then decrease from `Na^+ to Al^(3+)`
a decrease from `O^(2-)` to `F^-` and then increase from `Na^+` to `Al^(3+)`

Answer :A
Discussion
29634.

The increasing order of the rate of HCN addition to compound A-D is (A)HCHO , (B)CH_3COCH_3 , (C ) PhCOCH_3 , (D)PhCOPh

Answer»

D lt Clt Blt A
C lt D lt B lt A
A lt Blt C lt D
D lt B lt C lt A

Solution :Addition of HCN to CARBONYL compounds is nucleophilic addition reaction. The order of reactivity of carbonyl compounds is Aldehydes (smaller to higher), ketones (smaller to higher), Then `HCHO gt CH_3COCH_3 gt PhCOCH_3 gt PhCOPh`
The lower reactivity of ketones is due to presence of two ALKYL group which shows +I effect. The reactivity of ketones decreases as the size of alkyl group increases.
Discussion
29635.

The ions having 4f ^(0), 5d^(0)6s^(0) stable electronic configuration are

Answer»

`LA ^(3+) and CE ^(4+)`
`La ^(3+)and Ce ^(3+)`
only `Ce ^(4+)`
only `La ^(3+)`

ANSWER :A
Discussion
29636.

The increasing order of the pK_(b)of the following compound is:

Answer»

`(B) lt (D) lt (A) lt (C)`
`(B) lt (D) lt (C) lt (A)`
`(A) lt (C) lt (D) lt (B)`
`(C) lt (A) lt(D) lt (B)`

SOLUTION :`pK_(a)` value is directly PROPORTIONAL to the –I effect.
Discussion
29637.

The ions O^(2-), F^(-),Na^(+),Mg^(2+)and Al^(3+) are isoelectric. Their radii show

Answer»

an increase from `O^(2-)` to `F^(-)` and then decrease from `Na^(+)` to `Al^(3+)`
a decrease from `O^(2-)` to `F^(-)` and then increase from `Na^(+)` to `Al^(3+)`
A significant increase from `O^(2-)` to `Al^(3+)`
A significant decrease from `O^(2-)` to `Al^(3+)`

ANSWER :D
Discussion
29638.

The increasing order of the ionic radii of the given isoelectronic species is

Answer»

`K^(+),S^(2-),Ca^(2+),Cl^(-)`
`Cl^(-),Ca^(2+),K^(+),S^(2-)`
`S^(2-),Cl^(-),Ca^(2+),K^(+)`
`Ca^(2+),K^(+),Cl^(-),S^(2-)`

Solution :Correct order of increasing ionic radii is
`Ca^(2+)ltK^(+)LTCL^(-)LTS^(2-)`
With increase in nuclear CHARGE. SIZE decrease in isoelecronic species.
`Ca^(2+)(+20)ltK^(+)(+19)ltCl^(-)(+17)ltS^(2-)(+16)`
Discussion
29639.

The ions discharged at anode by the electrolysis of very dilute H_2SO_4solution are:

Answer»

`H_3O^+`
`OH^-`
`HSO_4^_`
`SO_4^(2-)`

ANSWER :B
Discussion
29640.

The increasing order of the crystal field splitting power of some common ligands is

Answer»

`H_(2)OltOH^(-)ltCl^(-)LTF^(-)ltCN^(-)`
`CN^(-)ltH_(2)OltOH^(-)ltF^(-)ltCl^(-)`
`F^(-)ltCN^(-)ltOH^(-)ltCl^(-)ltH_(2)O`
`Cl^(-)ltF^(-)ltOH^(-)ltH_(2)OltCN^(-)`

Answer :D
Discussion
29641.

The increasing order of the density of alkali metals

Answer»

`LiltKltNaltRbltCs`
`LiltNaltKltRbltCs`
`CsltRbltNaltKltLi`
`CsltRbltKltNaltLi`

ANSWER :A
Discussion
29642.

The ionization potential of hydrogen atom is 13.6 eV. The energy required to remove an electron from the n=2 state of hydrogen atom is:

Answer»

27.2 eV
13.6 eV
6.8 eV
3.4 eV

Answer :D
Discussion
29643.

The ionization potential for the electron in the ground state of hydrogen atom is 13.6eV. what would be the ionization potential for the electron in the first excited state of Li^(2+)?

Answer»

`54.4`EV
`5.4` eV
`30.6`eV
`84.4e`V

Answer :C
Discussion
29644.

The increasing order of the crystal field splitting power of some common ligands is :

Answer»

`H_(2)O lt OH^(-) lt Cl^(-) lt F^(-) lt CN^(-)`
`H_(2)O lt Cl^(-) lt OH^(-) lt CN^(-) lt F^(-)`
`CN^(-) lt H_(2)O lt OH^(-) lt F^(-) lt Cl^(-)`
`F^(-) lt CN^(-) lt OH^(-) lt CL^(-) lt H_(2)O`

SOLUTION :(E) is the correct ANSWER
Discussion
29645.

The ionization of hydrogen atom would give rise to :

Answer»

HYDRIDE ion
Hydronium ion
Proton
hydroxyl ion

Answer :C
Discussion
29646.

The ionization of hydrogen atom would give rise to

Answer»

HYDROXYL ION
HYDRONIUM ion
Proton
Hydride ion

ANSWER :C
Discussion
29647.

The increasing order of the boiling points for the partition compounds is :underset((I))(C_(2)H_(5)OH)"" underset((II))(C_(2)H_(5)Cl) ""underset((III))(C_(2)H_(5)CH_(3))"" underset ((IV))(C_(2)H_(5)OCH_(3))

Answer»

`(III)LT(IV)lt(II)lt(I)`
`(IV)lt(III)lt(I)lt(II)`
`(II)lt(III)lt(IV)lt(I)`
`(III)lt(II)lt(I)lt(IV)`

Solution :(I) `C_(2)H_(5)OH`Hydrogen bonding Hydrogen bonding `GT` Dipole - Dipole
(II)`C_(2)H_(5)CL`Dipole - Dipole `C_(2)H_(5)Cl` is more polar than`C_(2)H_(5)OCH_(3)` so ,
Dipole - Dipole `C_(2)H_(5)Clgt` Dipole - Dipole
`C_(2)H_(5)OCH_(3)`
(III)`C_(2)H_(5)CH_(3)`No hydrogen
bonding and dipole -
dipole interaction
(IV) `C_(2)H_(5)OCH_(3)`Dipole - Dipole
Boiling POINTS order is -
`(II)lt(IV)lt(II)lt(I)`.
Discussion
29648.

The ionization of H_(2)SO_(4) is completed in 2 steps .(i)H_(2)SO_(4)toH^(+)+HSO_(4)^(-1) k_(a)(1)=1.4xx10^(-2) mol L^(-1) s^(-1) (ii)HSO_(4)^(-)=3.5xx10^(-2) mol L^(-1)S^(-1) Then which of the following equation given below is true for rate?

Answer»

Rate=`K_(a)(1)[H_(2)SO_(4)]`
Rate =`K_(a)(2)[HSO_(4)^(-)]`
Rate =`K_(a)(1)[H_(2)SO_(4)]`
Rate=`K_(a)(2)[HSO_(4)^(-)]`

SOLUTION :The rate of the second reaction is very les.So the rate can be taken same as other LESS reaction.
(ii)`HSO_(4)^(-1)TOH^(+)+SO_(4)^(2-)`Ka (2) `3.5xx10^(-2)`
`therefore` Rate will be =`K_(a)(2)HSO_(4)^(-1)`
So,option (B) is correct.
Discussion
29649.

The ionization isomer of [Cr(H_2O)_4Cl(NO_2)]Clis

Answer»

`[Cr(H_2O)_4(NO_2)]Cl_2`
`[Cr(H_2O)_4Cl_2]NO_2`
`[Cr(H_2O)_4Cl(ONO)]CL`
`[Cr(H_2O)_3Cl_2(NO_2)]H_2O`

Answer :B
Discussion
29650.

The increasing order of reactivity of the following halides for the S_(N)^(1) reaction is (i) CH_(3) - CH - underset(CI)underset(|)CH_(3)(ii) CH_(3) - CH_(2) - CH_(2) - CI (iii) P - H_(3) CO - C_(6)H_(4) - CH_(4) - CI

Answer»

iii lt ii lt i
ii lt I lt iii
I lt iii lt ii
ii lt iii lt i

Answer :B
Discussion