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88801.

A metal complex having the composition or (NH_(3))_(4)Cl_(2)Br has been isolated in two form A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia .The formule of the complex B is

Answer»

`[Cr(NH_(3))_(4)Br]Cl_(2)`
`[Cr(NH_(3))_(4)CLBR]Cl`
`[Cr(NH_(3))_(4)Cl_(2)]Br`
`[Cr(NH_(3))_(4)]BrCl_(2)`

Answer :C
Discussion
88802.

A metal complex having the composition Cr (NH_(3))_(4)Cl_(2) Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia. The spin only value of the magnetic moment of the complex is

Answer»

`sqrt(3)` B.M.
`sqrt(8)` B.M.
`sqrt(15)` B.M.
Zero

Solution :As there are three unpaired ELECTRONS, spin magnetic moment `(MU)`
=`sqrt(N(n+2))` B.M.
`=sqrt(3xx5)` B.M. = `sqrt(15)`
Discussion
88803.

A metal complex of coordination number six having three different types of ligands a, b and c of composition Ma_(2)b_(2)c_(2) can exist in several geometrical isomeric forms, the total number of such isomers is:

Answer»

3
5
7
9

Answer :B
Discussion
88804.

A metal complex having the composition Cr (NH_(3))_(4)Cl_(2) Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia. The formula of the complex A is

Answer»

`[Cr(NH_(3))_(4)Br]Cl_(2)`
`[Cr(NH_(3))_(4)BrCl]Cl`
`[Cr(NH_(3))_(4)Cl_(2)]Br`
Both (a) and (B) are possible

Solution :As form A is readily soluble in dilute AQUEOUS ammonia, it must have `Cl^(-)` outside the coordination sphere, i.e., its formula is `[Cr(NH_(3))_(4)BrCl]^(+)Cl^(-)`.
Discussion
88805.

A metal complex having the composition Cr (NH_(3))_(4)Cl_(2) Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia. The hybridisation of Cr in the complexes A and B respectively is

Answer»

`d^(2) SP^(3) and sp^(3) d^(2)`
`sp^(3) d^(2) and d^(2) sp^(3)`
`sp^(3)d and d sp^(3)`
`d^(2)sp^(3) and d^(2)sp^(3)`

Solution :Both the above FORMS have Cr in +3 oxidation state. Hence, its ELECTRONIC CONFIGURATION is `[Ar] 3d^(3)`, i.e.,

Thus, the hybridisation of Cr in each case is `d^(2)sp^(3)`, i.e., both the complexes are octahedral.
Discussion
88806.

A metal complex having composition Cr(NH_(3))_(4)Cl_(2)Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia, whereas Bgives a yellow precipitate soluble in concentrated ammonia. (i) Write the formulae of A and B. (ii) State hybridisation of chromium in each. (iii) Calculate their magnetic moments for each (spin-only value). (iv) Calculate the EAN for both. (v) Will they conduct electricity or not. (vi) Write the formula of the complexes formed when the precipitates dissolve in aqueous ammonia & the concentrated ammonia respectively.

Answer»

Solution :`(i) [Cr(NH_(3))_(4)CLBR]Cl overset(aq)iff [Cr(NH_(3))_(4)ClBr]^(+)+Cl^(-),Ag^(+)+Cl^(-) to AgCl DARR`(white), soluble in dilute `NH_(3)`.
`[Cr(NH_(3))_(4)Cl_(2)]Br overset(aq)iff [Cr(NH_(3))_(4)Cl_(2)]^(+) +Br^(-),Ag^(+)+Br^(-) to AGBR darr`(yellow), soluble in conc. `NH_(3)`. So , `A=[Cr(NH_(3))_(4)ClBr]Cl and B=[Cr(NH_(3))_(4)Cl_(2)]Br`.
(ii) In both complexes chromium is in +3 oxidation state. Chromium with `3d^(3)` configuration has 3 unpaired electrons with weak field as well as strong field as well as strong field ligand. So, the hybridisation scheme is as follow :

(III) `mu=SQRT(n(n+2))=sqrt(15)`
(iv) EAN=24-3+12=22
(v) Yes, both have two ions per formula unit.
`(vi) AgCl+2NH_(3) iff [Ag(NH_(3))_(2)]Cl,AgBr+2NH_(3)iff[Ag(NH_(3))_(2)]Br`
Discussion
88807.

A metal complex having compositionhas been isolated in two forms (X) and (Y).The form (X) reacts withto give a white precipitate readily soluble in aqueous ammonia,whereas (Y) gives a pale-yellow precipitate soluble in concentrated ammonia. Choose the correct statement regarding the complex X and Y. (i) Both complexes have same Van’t Hoff factor for same extent of dissociation (ii) Both complexes have exactly same conductivity in H_(2)O (iii) In both the complexesis involved only to secondary valency (iv) In both the complexes electronic configuration oft_(2_(g)) subshell is same. The correct choice is :

Answer»

I,ii and iii
I,ii,iii and iv
ii,iii and iv
I, iii and iv

Solution :`X: [CR(NH_(3))_(4) ClBr] Cl, Y: [Cr(NH_(3))_(4) Cl_(2)]Br`
Both COMPLEXES have same van’t Hoff factor.
Both complex form different ions in water hence conductivity is not exactly same
In both complexes `NH_3` is involved in secondary valency
In both complexes’ electronic configuration ofis same
Hence A, C, D are CORRECT statements.
Discussion
88808.

A metal complex having the composition Cr (NH_(3))_(4)Cl_(2) Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia. The formula of the complex B is

Answer»

`[Cr(NH_(3))_(4)BR]Cl_(2)`
`[Cr(NH_(3))_(4)ClBr]Cl`
`[Cr(NH_(3))_(4)Cl_(2)]Br`
`[Cr(NH_(3))_(4)]BrCl_(2)`

Solution :As form B gives a pale YELLOW PRECIPITATE SOLUBLE in concentrated ammonia, it MUST have `Br^(-)` outside the coordination sphere, i.e., its formula is `[Cr(NH_(3))_(4)Cl_(2)]^(+)Br`.
Discussion
88809.

A metal complex having composition [Cr(NH_(3))_(4)Br_(2)] was isolated in two forms (X) and (Y). Form (X) reacts with AgNO_(3) to gives a plate yellow precipitate which is partially soluble in excess of NH_(4)OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH_(4)OH.

Answer»

X-cls from optically INACTIVE, Y-cls form optically active
X-CIS form optically inactive, Y-trans form optically active
The cis and trans forms of both X and Y are optically active
The cis and trans form of both X and Y are optically inactive.

ANSWER :D
Discussion
88810.

A metal complex having composition [Cr(NH_(3))_(4)Br_(2)] was isolated in two forms (X) and (Y). Form (X) reacts with AgNO_(3) to gives a plate yellow precipitate which is partially soluble in excess of NH_(4)OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH_(4)OH. The hybridization of Cr ini X and Y both are

Answer»

`SP^(3)d^(2)`
`dsp^(3)d`
`d^(2)sp^(3)`
`sp^(3)`

ANSWER :C
Discussion
88811.

A metal complex having composition [Cr(NH_(3))_(4)Br_(2)] was isolated in two forms (X) and (Y). Form (X) reacts with AgNO_(3) to gives a plate yellow precipitate which is partially soluble in excess of NH_(4)OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH_(4)OH. The formula of X and Y are

Answer»

`[Cr(NH_(3))_(4)BR]Br` and `[Cr(NH_(3))_(3)Br_(2)]NH_(3)` respectively.
`[Cr(NH_(3))_(4)IBR` and `[Cr(NH_(3))_(4)Br_(2)]` respectively.
both `[Cr(NH_(3))_(4)I]Brl`
`[Cr(NH_(3))_(2)lBr_(2)](NH_(3))_(2)`

Answer :B
Discussion
88812.

A metal complex having composition [Cr(NH_(3))_(4)Br_(2)] was isolated in two forms (X) and (Y). Form (X) reacts with AgNO_(3) to gives a plate yellow precipitate which is partially soluble in excess of NH_(4)OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH_(4)OH. Both the (X) form and (Y) form show

Answer»

LINKAGE isomerism
coordination isomerism
ionization isomerism
none of these

ANSWER :C
Discussion
88813.

A metal complex having composition [Cr(NH_(3))_(4)Br_(2)] was isolated in two forms (X) and (Y). Form (X) reacts with AgNO_(3) to gives a plate yellow precipitate which is partially soluble in excess of NH_(4)OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH_(4)OH. The magnetic moment in both is

Answer»

5
3.87
`SQRT(24)`
`sqrt(2)`

ANSWER :B
Discussion
88814.

A metal complex Cr(NH_(3))_(4)Cl_(2) having molar conductivity corresponding to 1:1 electrolyte have been isolated in two Forms (X) and (Y), (X) reacts with AgNO_(3) giving white precipitate which turns yellow on reaction with sodium arsenite whereas (Y) gives a yellow precipitate soluble in concentrate NH_(4)OH Which of the following statements is true for the complex (Y) ?

Answer»

Its trans form is non-planar
Its IUPAC name is tetraaminedichloridochromate(III) bromide.
Its 'spin only' magnetic moment is 3.83 BM.
All of these

Solution :`underset(d^(3))(X) to underset(d^(2)sp^(3))[CR(NH_(3))_(4)ClBr]Cl,(Y) to underset(d^(2)sp^(3),mu=3.8 B.M.)[Cr(NH_(3))_(4)Cl_(2)]Br`
`"EAN" to 24-3+12=33`
Both X and Y can show G.I.
Discussion
88815.

A metal complex having composition Cr(NH_(3))_(4) Cl_(2)Br has been isolated in two forms (A) and (B). The form (A) reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia, whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write the formula of(A) and (B) and state the hybridization of chromium in each Calculate their magnetic moments (spin--only value)

Answer»

SOLUTION : `Cr(NH_(3)_(4),Cl_(2)Br` have two forms A and B. `[Cr(NH_(3))_(4)BrCl]Cl, (d^(2)sp^(3)), [Cr(NH_(3))_(4)CI_(2)]`Br.`d^(2)sp^(3))`, 3.87 BM, 3.87 BMForm .A. gives white ppt. wth `AgNO_(3)` hence it must have chloride ion in form of noncomplex ion i.e. outside the complex sphere as<
`Cr(NH_(3))_(4)CIBr]Cl`
`UNDERSET(From A)[Cr(NH_(3))_(4)CIBr]Cl+AğNO_(3)tounderset(White ppt)(AgCI)downarrow(+)[Cr(NH_(3))_(4)CIBr]+NO_(3)`
These precipitates of AgCl are soluble in `NH_(4) OH` due to formation of complex salt
`underset(whiteppt)(AgCI)+2NH_(4)OH tounderset(complex salt)([Ag(NH_(4))_(2))CI]+2H_(2)O`
Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in `NH_(4)OH` Form .B. is `[Cr(NH_(3))_(4)CI_(2)]Br`
`AgNO_(3)+[Cr(NH_(3))_(4)C1_(2)]Br tounderset(pale yellow ppt) (AgBr)downarrow(+)+[Cr(NH_(3))_(4)Cl_(2)]^(+)+ NO_(3)^(-)`
`underset(pale yellow ppt)(AgBr)+2NH_(4)OHto [Ag(NH_(3))_(2)Br]+2H_(2)O`
In both complexes, chromium is present as central ion and its oxidation number is +3 So in these
`_(24) Cr = 1s^(2), 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5), 4s^(1)`
`Cr6(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
Number of ligands are six and `Cr^(3+)` shows `d^(2)sp^(3)` hybridization in both complexes A and B
Hence in it, number of unpaired electrons are 3. So MAGNETIC moment `(mu) = sqrt(n(n + 2)).= sqrt(3(3+2)) = sqrt(15)` = 3.872 BM`
Discussion
88816.

A metal complex Cr(NH_(3))_(4)Cl_(2) having molar conductivity corresponding to 1:1 electrolyte have been isolated in two Forms (X) and (Y), (X) reacts with AgNO_(3) giving white precipitate which turns yellow on reaction with sodium arsenite whereas (Y) gives a yellow precipitate soluble in concentrate NH_(4)OH The forms (X) and (Y) are example of :

Answer»

linkage iosmers
ligand isomers
ionisation isomers
COORDINATION position isomers

Solution :`underset(d^(3))(X) to underset(d^(2)sp^(3))[Cr(NH_(3))_(4)ClBr]CL,(Y) to underset(d^(2)sp^(3),mu=3.8 B.M.)[Cr(NH_(3))_(4)Cl_(2)]Br`
`"EAN" to 24-3+12=33`
Both X and Y can show G.I.
Discussion
88817.

A metal complex Cr(NH_(3))_(4)Cl_(2) having molar conductivity corresponding to 1:1 electrolyte have been isolated in two Forms (X) and (Y), (X) reacts with AgNO_(3) giving white precipitate which turns yellow on reaction with sodium arsenite whereas (Y) gives a yellow precipitate soluble in concentrate NH_(4)OH Which of the following statements is true for the complex (X) and (Y) ?

Answer»

The EFFECTIVE atomic number of chromium in both the COMPLEXES is 33.
Both can exist as diastereomers.
Both CIS and TRANS forms of (X) have same dipole moment.
All of these

Solution :`underset(d^(3))(X) to underset(d^(2)sp^(3))[Cr(NH_(3))_(4)ClBr]Cl,(Y) to underset(d^(2)sp^(3),mu=3.8 B.M.)[Cr(NH_(3))_(4)Cl_(2)]Br`
`"EAN" to 24-3+12=33`
Both X and Y can show G.I.
Discussion
88818.

A metal complex having composition Co(en)_(2)Cl_(2)Br has been isolated in two forms Aand B. (B) reacted with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide . Whereas A gives a pale yellow precipitate. Write the formula of A and B, state the hybridization of Co in each and calculate their spin only magnetic moment.

Answer»

Solution :A metal complex having composition `Co(en)_(2)Cl_(2)Br` has been isolated in two forms `A` and `B`.
`(i) (B)` reacts with silver nitrate to give a white PRECIPITATE readily SOLUBLE in ammonium hydroxide . This shows (B) has `Cl^(OPLUS)` counter ion. Hence B is `[Co(en)_(2)ClBr]Cl` form.
`(ii)` (A) reacts with silver nitrate to give a pale yellow precipitate. This shows (A) has `Br^(oplus)` counter ion. Hence A is `[Co(en)_(2)Cl_(2)]Br`.
`(iii)` Formula of A and B
`A-[Co(en)_(2)Cl_(2)]Br`
`B-[Co(en)_(2)ClBr]Cl`
`(a) A-[Co(en)_(2)Cl_(2)]Br`
Electronic configuration of Co atom
Electronic configuration of `Co^(3+)` atom
Hybridisation and formation of
`[Co(en)_(2)Cl_(2)]Br` complex
`d^(2)sp^(3)` hybridisation
`en` is strong FIELD ligand
No unpaired electrons, hence it is diamagnetic
MAGNETIC moment,
`mu_(s)=sqrt(n(n+2))`
`n=0`
`mu_(s)=0`

`(b) A-[Co(en)_(2)ClBr]Cl`
Electronic configuration of Co atom
Electronic configuration of `Co^(3+)` atom
Hybridisation and formation of
`[Co(en)_(2)ClBr]Cl` complex
`d^(2)sp^(3)` hybridisation
`en` is strong field ligand
No unpaired electrons, hence it is diamagnetic
Magnetic moment,
`mu_(s)=0 :. {n=0`
Discussion
88819.

A metal complex Cr(NH_(3))_(4)Cl_(2) having molar conductivity corresponding to 1:1 electrolyte have been isolated in two Forms (X) and (Y), (X) reacts with AgNO_(3) giving white precipitate which turns yellow on reaction with sodium arsenite whereas (Y) gives a yellow precipitate soluble in concentrate NH_(4)OH The hybridization of chromium in complexes (X) and (Y) are :

Answer»

both `SP^(3)d^(2)`
both `d^(2)sp^(3)`
`d^(2)sp^(3)" and "sp^(3)d^(2)` respectively
`sp^(3)d^(2)`

Solution :`underset(d^(3))(X) to underset(d^(2)sp^(3))[Cr(NH_(3))_(4)ClBr]Cl,(Y) to underset(d^(2)sp^(3),mu=3.8 B.M.)[Cr(NH_(3))_(4)Cl_(2)]Br`
`"EAN" to 24-3+12=33`
Both X and Y can show G.I.
Discussion
88820.

A metal complex Cr(NH_(3))_(4)Cl_(2) having molar conductivity corresponding to 1:1 electrolyte have been isolated in two Forms (X) and (Y), (X) reacts with AgNO_(3) giving white precipitate which turns yellow on reaction with sodium arsenite whereas (Y) gives a yellow precipitate soluble in concentrate NH_(4)OH The correct formula and the IUPAC name of the complex (X) is :

Answer»

`[Cr(NH_(3))_(4)Cl_(2)]Br` and tetraaminedichloridochromate(II) BROMIDE.
`[Cr(NH_(3))_(4)ClBr]CL` and tetraamminebromidochloridochromate(II) chloride.
`[Cr(NH_(3))_(4)ClBr]Cl` and tetramminebromidochloridochromium(III) chloride.
`[Cr(NH_(3))_(4)ClBr]Cl` and tetraamminebromidochloridochromium(III) chloride.

Solution :`underset(d^(3))(X) to underset(d^(2)sp^(3))[Cr(NH_(3))_(4)ClBr]Cl,(Y) to underset(d^(2)sp^(3),mu=3.8 B.M.)[Cr(NH_(3))_(4)Cl_(2)]Br`
`"EAN" to 24-3+12=33`
Both X and Y can show G.I.
Discussion
88821.

A metal comples having composition Cr(NH_(3))_(4)CI_(2)Br has been isolated in two forms A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia, where as B gives a plae yellow precipitate soluble in concentrated ammonia. The hybridisation of Cr in A and magnetic momentum in B are respectively

Answer»

`d^(2)sp^(3),2.8BM`
`sp^(3)d^(2),3.8MB`
`d^(2)s^(3),3.8BM`
`sp^(3)d^(2),2.8BM`

Solution :(A) : `CL^(-1)` ION must be out of co-ordination sphere to react with `AgNO_(3)`.
`[Crunderset((A))((NH_(3))_(4)BRCL](sp^(3)d^(2)),[Crunderset((B))((NH_(3))_(4)Cl_(2))]Br(3.8Bm)`
Discussion
88822.

A metal chloride (X) shows the following reactions : (i) When H_(2)S is passed in acidified aqueous solution of (X) a black ppt. is obtained. (ii) The ppt. obtained in step (i) is not soluble in yellow ammonium sulphide. (iii) When a solution of stannous chloride is added to an aqueous solution of X, a white ppt. is obtained which turns grey on addition of more stannous chloride. (iv) When an aqueous solution of KI is added to an aqueous solution of (X) a red ppt. is obtained which dissolved on addition of excess of KI. Identify X and write down equation for the reactions at step(i), (ii), (iii) & (iv).

Answer»

SOLUTION :`X` is `HgCl_(2)`
Discussion
88823.

A metal chloride solution on mixing with K_2 CrO_4 solution gives a yellow precipitate, insoluble in acetic acid. The metal may be

Answer»

Mercury
Zinc
silver
Lead

Solution :`PbCl_(2)+K_(2)CrO_(4) to underset("Yellow ppt.")(PbCrO_(4) DARR)+2KCl`
Discussion
88824.

A metal chloride solution on mixing with K_(2) Cr O_(4) solution gives a yellow ppt, which are insoluble in water. The metal may be

Answer»

mercury
zinc
silver
lead

Solution :`PbCl_2 + K_2 CrO_4 to UNDERSET("Yellow PPT.")(PbCrO_(4)) + 2 KCl`
Discussion
88825.

A metal chloride onheatingwith K_(2)Cr_(2)O_(7) gives ayellow ppt insolublein aceticacid .The metalmay be

Answer»

Hg
Zn
Pb
Ag

Answer :C
Discussion
88826.

A metal carbonate is sparingly soluble in water and evolves CO_2 on heating. The metal is:

Answer»

An alkali metal
A NOBLE metal
An alkaline EARTH metal
None of the above

Answer :C
Discussion
88827.

A metal carbonate decomposes according to the following reaction M_(2)CO_(3) (s) to M_(2)O(s) + CO_(2)(g) Percentage loss in mass on complete decomposition of M_(2)CO_(3)(s) (Atomic mass of M= 102 )

Answer»

`100/3 %`
`50/3 %`
`25/3 %`
`15%`

ANSWER :B
Discussion
88828.

A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g cm^(-3). Find the volume of this unit cell.

Answer»


Solution :`M = 50 G MOL^(-1), d = 5.96 g cm^(-3)`, Z = 2 for bcc unit cell , `d =(ZM)/(N_A xx a^3)`
`a^3=(Z xx M)/(N_A xx d)=(2 xx 50)/(6.023 xx 10^(23) xx 5.96) cm^3 = 2.78 xx 10^(-23) cm^3`
Discussion
88829.

A metal (A) brun in dinitrogen to give an ionic compund (B). (B)reactswith waterto give (C) and(D). When CO_(2)gas in passedthrough thethatsolution , the solutionbecomestransparent agin. Whenthe gas (D)is passedthrough CuSO_(4)soluitonthe solutionbecomesdeep bluein colour. Identify (A), (B) , ( C) and (D) givethe reactionsinvloved .

Answer»

Solution :A : `CA, B: Ca_(3)N_(2), Ca(OH)_(2),D : NH_(3)`
Discussion
88830.

(A) Meso tartaric acid is optically inactive. (R) Meso tataric acid has plane of symmetry.

Answer»


ANSWER :A
Discussion
88831.

(A) Mercury is liquid at room temperature (R) In mercury, there is no unpaired d-electron and thus metallic bonding is weakest.

Answer»

Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Solution :Both (A) and (R) are true and (R) is the correct explanation of (A)
Discussion
88832.

(a) Mention two applications of adsorption.(b) What are emulsions? Give an example for O //W emulsion.(c) What is the cause for Brownian movement?

Answer»

Solution :(a) (i) In gas masks(II) CONTROL of humidity
(b) Liquid-liquid colloidal system are called emulsions
eg. Milk, vanishing cream.
(c) Due to the collision of the dispersed PHASE PARTICLES with the molecules of the DISPERSION medium.
Discussion
88833.

Write the IUPAC name of (CH_(3))_(2)N-CH_(2)-CH_(3).

Answer»

Solution :N, N-dimethylmethanamine
(b)
(a) IUPAC name of `(CH_(3))_(2)-N-CH_(3)` is N, N-dimethyl methanamine.
(b) ANILINE from nitrobenzene :

(c) EQUATION for conversion of aniline to 4-bromo aniline :
Discussion
88834.

(a) Mention the I.U.P.A.C. name of (CH_(3)CH_(2))_(2)-N-CH_(3). (b) How is Aniline prepared from nitrobenzene ? (c ) Give the equation for the conversion of aniline to 4-bromo aniline.

Answer»

Solution :N, N-dimethylmethanamine
(b)
(a) IUPAC NAME of `(CH_(3))_(2)-N-CH_(3)` is N, N-dimethyl METHANAMINE.
(b) ANILINE from nitrobenzene :

(C) Equation for conversion of aniline to 4-bromo aniline :
Discussion
88835.

Write any two postulates of Werner's theory of co-ordination compounds.

Answer»

Solution :(a) 1. In coordination compounds, metal posses two types of valences.
(i) The primary valence.
(ii) The secondary valence
2. The primary valence is ionisable wDhereas, the secondary valence is non-ionisable.
3. The primary valence is variable whereas, the secondary valence is fixed.
4. Primary valence are satisfied by negative ions whereas secondary valence are satisfied bynegative ions or neutral molecules (ligands)
5. The secondary valences are directed TOWARDS fixed position in space around the central ION.
6. The primary valence corresponds to oxidation state of central metal atom and secondary valence corresponds to coordination NUMBER.
(b) Linkage Isomerism
(b) TYPE of isomerism in `[Co(NH_(3))_(5)SCN]Cl_(2)` and `[Co(NH_(3))5NCS]Cl_(2)` complex compounds.
These show linkage isomerism as central metal atom .cobalt. is bonded through S-atom of SCN while with N-atom of NCS in other case.
In other words .SCN. and .NCS. are ambidentate ligands. These have two donor atoms (S-taoms andN-atom respectively).
Discussion
88836.

Indicate the type of Isomerism in the following set of complex compounds. [Co(NH_3)_5SCN]Cl_2 and [ Co(NH_3)5NCS]Cl_2

Answer»

Solution :(a) 1. In coordination compounds, metal posses two types of valences.
(i) The primary valence.
(ii) The secondary valence
2. The primary valence is ionisable wDhereas, the secondary valence is non-ionisable.
3. The primary valence is variable WHEREAS, the secondary valence is fixed.
4. Primary valence are satisfied by negative ions whereas secondary valence are satisfied bynegative ions or neutral molecules (ligands)
5. The secondary valences are directed towards fixed position in space around the central ION.
6. The primary valence corresponds to oxidation state of central metal atom and secondary valence corresponds to coordination number.
(b) Linkage Isomerism
(b) TYPE of isomerism in `[Co(NH_(3))_(5)SCN]Cl_(2)` and `[Co(NH_(3))5NCS]Cl_(2)` complex compounds.
These show linkage isomerism as central metal atom .cobalt. is bonded through S-atom of SCN while with N-atom of NCS in other case.
In other words .SCN. and .NCS. are ambidentate ligands. These have two donor atoms (S-taoms andN-atom respectively).
Discussion
88837.

A mentalis in comnined state as sulphide. Identify thesteps (A), (B) ,(C)

Answer»


ANSWER :N//A
Discussion
88838.

(A) Melting point of carboxylic acids with even number of carbon atoms is higher than the next higher or lower carboxylic acid with odd number of carbon atoms. (R) Carboxylic acid with even number of carbon atoms fit closely in the crystal lattice as compared to acids with odd number of carbon atoms.

Answer»

Both (A) and (R) are true and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :A
Discussion
88839.

(A)Meltingpointfo neopentane is greater thanthatofn- pentane butthe boilingpointof n-pentaneis morethan thatof neo pentane . (R )Meltingpoint depends uponpackingincrystal litticewhereasbooiling point depends upon surface area of themolecule .

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if both(A) and (R ) are CORRECT and ( R) is thecorrect EXPLANATIONOF the(A)
ifboth(A) and (R ) are correct but ( R) is not thecorrectexplanationof the(A)
if (A) is correct but ( R) is incorrect
is (A)in incorrect but( R)is correct

ANSWER :A
Discussion
88840.

A medicine which promotes the secretion of urine is called

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Uretic
Monouretic
Diuretic
Triuretic

Answer :C
Discussion
88841.

A measured temperature on Fahrenheit scale in 200^(@)F. What will this reading be on Celsius scale?

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`40^(@)C`
`94^(@)C`
`93.3^(@)C`
`30^(@)C`

Solution :`.^(@)C=(5)/(9)(.^(@)F-32)=(5)/(9)(200-32)=93.3.`
Discussion
88842.

[A](M)""[B](M)"""Initial rate"(Ms^(-1)) 0.4""0.3""2xx10^(-3) 0.8""0.3""0.8xx10^(-2) 1.2""0.9""0.54xx10^(-1) From the above dat athe rate law for the equation A+Bto products is equal to

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`"K[A][B]"`
`"K[A]"^(2)"[B]"^(2)`
`"K[A]"^(2)"[B]"`
`"K[A][B]"^(2)`

ANSWER :C
Discussion
88843.

A maxima or minima is obtained in the temperature. The composition curve of a mixture of two liquids does not indicate

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That the LIQUIDS are immiscible with ONE another
That the liquids are PARTIALLY miscible at the MAXIMUM or minimum.
An azeotropic mixture.
A eutectic formation.

Solution :Greater is the intermolecular FORCE on mixing, more negative will be the devaition `p lt P_(A)^(@)chi_(A) + p lt P_(B)^(@)chi_(B)`.
Experimental vapour pressure will be less than calculated vapour pressure.
Discussion
88844.

A maxima or minima obtained in the temperature, composition curve of a mixture of two liquids indicates:

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An azeotropic mixture
An EUTECTIC formation
That the liquids are immiscible with ONE another
That the liquids are partially MISCIBLE at the maximum or minimum

Answer :A
Discussion
88845.

A match box exhibits:

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CUBIC geometry
Monoclinic geometry
Orthorhombic geometry
Tetragonal geometry

Answer :C
Discussion
88846.

(A) Mass spectrometer is used for determination of atomic mass of isotopes. (R) Isotopes are the atoms of same element having same atomic number but different mass numbers.

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If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A).
If both (A) and (R) are correct but (R) is not the correct explanation of (A).
If (A) is correct but (R) is WRONG.
If (A) is wrong but (R) is correct.

Answer :A
Discussion
88847.

(A) Many copper(I) compounds are unstable in aqueous solution. (R) Copper(I) compounds undergo only reduction in aqueous solution

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Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Solution : (A) is true but (R) is false
Discussion
88848.

A manometer attached to a flask contains NH_(3) gas have no difference in mercury level initially as shown in diagram.After the sparking into the flask, it have difference of19 cm in mercury level in two columns.Calculate % dissciation of ammonia

Answer»


ANSWER :0.25
Discussion
88849.

A man weighs 175 1b. Express has weight in kg. Given that 1 kg = 2.205 1b.

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Solution :`"1 kg = 2.205 1b"therefore1=(2.205)/(1kg)=(1kg)/(2.2051b)`
Hence, `"175 1b"=175 cancel(1b)=(1kg)/(2.205cancel(1b))=79.4kg.`
The UNITS '1b' cancel out from the numerator and denominator and the answer is in REQUIRED units, i.e., kg.
Note. If the conversion factor were not correctly USED, the answer WOULD have absured units, e.g., if we WRITE `1751b=1751bxx(2.2051b)/(1kg)`,the answer will have the units `1b^(2)kg^(-1).`
Discussion
88850.

A major component of Borsch reagent is obtained by reacting hydrazine hydrate with which ofthe following?

Answer»




SOLUTION :The MAJOR COMPONENT of BORSCH reagent is 2,4-dinitrophenyl hydrazine which can be obtained by REACTION of 2,4-dinitrochloro benzene and hydrazine
Discussion