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90601.

A complex of platinum, ammonia and chloride produces four ions per molecule in the solution. Thestructure consistent with the observation is :-

Answer»

`[Pt(NH_(3))_(4)]Cl_(4)`
`[Pt(NH_(3))_(2) Cl_(4)`]
`[Pt(NH_(3) )_(5) CL]Cl_(3)`
`[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`

ANSWER :C
Discussion
90602.

A complex of platinum, ammonia and chloride produces four ions per molecule in the solution. The structure consistent with the observation is –

Answer»

`[Pt(NH_(3))_(4)]Cl_(4)`
`[Pt(NH_(3))_(2)Cl_(4)]`
`[Pt(NH_(3))_(5)Cl]Cl_(3)`
`[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`

SOLUTION :`[Pt(NH_(3))_(5)Cl]Cl_(3) to [Pt(NH_(3))_(5)Cl]^(+3) +3CL^(-)`
Discussion
90603.

A complex of platinum, ammonia and Chloride produces four ions per molecule in the solution. The structure consistent with the observation is :

Answer»

`[PT(NH_(3))_(4)]Cl_(4)`
`[Pt(NH_(3))_(2)Cl_(4)]`
`[Pt(NH_(3))_(5)Cl]Cl_(3)`
`[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`

ANSWER :C
Discussion
90604.

A six coordination complex of formula CrCl_(3)*6H_(2)O has green colour. A 0.1 M solution of the complex when treated with excess of AgNO_(3) gave 28.7g of white precipitate. The formula of the complex would be:

Answer»

`[Cr(H_(2)O)_(6)]Cl_(3)`
`[CRCL(H_(2)O)_(5)]Cl_(2).H_(2)O`
`[CrCl_(2)(H_(2)O)_(4)]Cl.2H_(2)O`
`[Cr(H_(2)O)_(3)Cl_(3)]3H_(2)O`

Answer :B
Discussion
90605.

A complex of cobat with ammonia is analyzed for determining its formula,by titrating it against a strandarized acid as follows : Co(NH_(3))_(x)Cl_(y)(aq)+Hclrarr NH_(4)^(+)(aq)+Co^(y+)+(aq)+Cl^(-)(aq) A 1.8 g complex required 20.00 mL 1.54 M HCl to reach the equivalence point. Also, if the reaction mixture at equivalence point is treated with excess of AgNO_(3) solution, 7.735 g of AgCl precipitate was produced. What is the formula of this complex? ["Given : atomic weight of CO = 59 gmol"^(-1)]

Answer»

`CO(NH_(3))_(4)Cl_(3)`
`CO(NH_(4))_(4)Cl_(3)`
`CO(NH_(3))_(3)Cl_(4)`
`CO(NH_(4))_(3)Cl_(4)`

ANSWER :A
Discussion
90606.

A complex of iron and cyanide ions is 100% ionised at 1m (mol). If its elevation in b.p is 2.8^(0), (K_(b)=0.52^(0) "mol"^(-1)Kg), then the complex is :

Answer»

`K_(3)[FE(CN)_(6)]`
`Fe(CN)_(2)`
`K_(4)[Fe(CN)_(6)]`
`Fe(CN)_(4)`

ANSWER :A
Discussion
90607.

A complex of cobalt with ammonia is analyzed for determining its formula, by titrated it against a standardized acid as follows : Co(NH_(3))_(x) Cl_(3) ( aq) + HCl rarr NH_(3)^(+)(aq) + Co^(3+) (aq) + Cl^(-) ( aq)

Answer»

Solution :The balanced chemical REACTION is `:`
`CO(NH_(3))_(X) Cl_(3) + x HCl rarr xNH_(4) ^(+) +o^(3+) + ( x+ 3) Cl^(-)`
`(1.58)/( 165.5+ 17x) " " ( 1.58x)/( 165.5 + 17x) " " (1.58(x+3))/( 165.5 + 17x)`
`rArr ( 1.58x)/( 165.5+ 17x) = ( 23.63 xx 1.5 )/( 1000)`
= moles of HCl , x = 6
mole of AGCL `= ((x+ 3 ) xx 1.58 )/( 165.5 + 17x)`
Discussion
90608.

A complex of cobalt has five ammonia molecules, one nitro group and two chlorine atoms for each cobalt atom. One mole of this compound produces three mole ions in aqueous solution which on treating with excess of AgNO_3 give two mole of AgCl. The formula of the compound is

Answer»

`[CO(NH_3)_4NO_2CI][(NH_3)CI]`
`[Co(NH_3)_5CI][CINO_2]`
`[Co(NH_3)_5NO_2]CI_2`
`[Co(NH_3)_5][(NO_2)_2CI_2]`

ANSWER :C
Discussion
90609.

A complex compound ofCo^(3+) with molecular formula COCl_x. yNH_3 gives a totall of 3 ionswhen dissolved in water. How manyCl^- ions when the primary and secondary valency in this complex ?

Answer»

3
1
4
zero

Answer :B
Discussion
90610.

A complex of certain metal has the magnetic moment of 4.91 BM. whereas another complex of the same metal with same oxidation state has zero magnetic moment. The metal Ion could be:-

Answer»

`CO^(2+)`
`MN^(2+)`
`FE^(2+)`
`Fe^(3+)`

Solution :
Discussion
90611.

A complex of certain metal has the magnetic moment of 4.91 BM whereas another complex of the same metal with same oxidation state has zero magnetic moment. The metal ion could be :

Answer»

`CO^(2+)`
`MN^(2+)`
`FE^(2+)`
`Fe^(3+)`

Solution :
Discussion
90612.

A complex K_(x)[MnF_(6)] has magnetic moment 4.9BM what will be the oxidation state of Mn and value of x :

Answer»

`MN" "(II),x=4`
`Mn" "(III),x=3`
`Mn" "(IV),x=2`
`Mn" "(V),x=1`

Solution :`therefore mu=4.9" "B.M.,` so UNPAIRED electrons should be 4.
Mn (III) :
so, x=3
Discussion
90613.

A complex is represented as CoCl_(3).xNH_(3). Its 0.1 molal solution in water shows Delta T_(f)=0.558k. K_(f) for H_(2)O is 1.86K//"mol". Assuming 100% ionisation of complex and coordination numberof Co in six . The formula of complex is

Answer»

`[Co(NH_(3))_(6)]Cl_(3)`
`[Co(NH_(3))_(5)CL]Cl_(2)`
`[Co(NH_(3))_(4)Cl_(2)]Cl`
`[Co(NH_(3))_(3)Cl_(3)]`

ANSWER :B
Discussion
90614.

A complex MA_(3)X_(3). Where A and X are unidentate ligands may give

Answer»

two geometrical ISOMERS
THREE geometrical isomers
two geometrical isomers which can be resolved into a pair of enantiomers
two geometrical isomers ONE of which can be resolved into a pair of enantiomers .

SOLUTION :`MA_(3)X_(3)` can exist as two geometrical isomers.
Discussion
90615.

A complex is prepared by mixing CoCl_(3) and NH_(3) in the molar ratio of 1 : 4. 0.1 M solution of this complex was found to freeze at -0.372^(@)C. What is the formula of the complex ? Given that molal depression constant of water (K_(f))=1.86^(@)C//m

Answer»

SOLUTION :Theoretical `DeltaT_(f)=K_(f)xxm=1.86xx0.1=0.186^(@)`
Observed `DeltaT_(f)=0.372^(@)` (Given)
As observed `DeltaT_(f)` is double of the theoretical value, this shows that each molecule of the complex dissociates to form two ions. This can be so only if the formula of the complex is `[CO(NH_(3))_(4)Cl_(2)]Cl`.
Discussion
90616.

A complex is prepared by mixing CoCl_3 and Nh_3, 0.1 M solution of the complex was found to freeze at -0.372^(@)C . The formula of the complex is _________. [Molar depression constant of water =1.86^(@)C//m]

Answer»

`[CO(NH_3)_6]Cl_3`
`[Co(NH_3)_5Cl]Cl_2`
`[Co(NH_3)_4Cl_2]CL`
`[Co(NH_3)_3Cl_3]`

Answer :C
Discussion
90617.

A complex in which the oxidation number of the metal is zero is….. .

Answer»

`K_(4)[Fe(CN)_(6)]`
`[Fe(CN)_(3)(NH_(3))_(3)]`
`[Fe(CO)_(5)]`
both (b) and (c )

Answer :C
Discussion
90618.

A complex in which central atom carries zero oxidation state is

Answer»

`[CO(NH_3)_3Cl_3]`
`PtCl_4 . 2NH_3`
`NI(CO)_4`
`Na_2[(Ni(CN)_4)]`

Answer :C
Discussion
90619.

A complex has empirical formula, PtCl_(2).2 NH_(3). When ground with AgNO_(3), it gives [Pt(NH_(3))_(4)(NO_(3))_(2)] and an insoluble slid Ag_(2)[PtCl_(4)] was also obtained. Name and mention the structure of the complex.

Answer»

Solution :`2 PtCl_(2).2 NH_(3)+2 AgNO_(3)rarr[Pt(NH_(3))_(4)(NO_(3))_(2)]+Ag_(2)[PtCl_(4)]`
On reaction with `AgNO_(3)`, no AgCl is formed as `Cl^(-)` ions are PRESENT as ligands in the COMPLEX `Ag_(2)[PtCl_(4)]`. Hence, the given complex must be
`[PtCl_(2)(NH_(3))_(2)] or [Pt(NH_(3))_(4)][PtCl_(4)]`
In the first CASE, molecular formula is same as empirical formula.
IUPAC name of `[PtCl_(2)(NH_(3))_(2)]` = diammine dichloridoplatinum (II).
In the second case, molecular formula is double of empirical formula.
IUPAC name of `[Pt(NH_(3))_(4)][PtCl_(4)]`= tetraammine PLATINUM (II) tetraachloridoplatinate (II).
Discussion
90620.

A complex has a molecular formula MSO_(4)Cl.6H_(2)O. The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex ?

Answer»

`[M(H_(2)O)_(4)Cl]SO_(4).2H_(2)O`
`[M(H_(2)O)_(6)]SO_(4)`
`[M(H_(2)O)_(5)Cl]SO_(4).H_(2)O`
`[M(H_(2)O)_(3)Cl]SO_(4).3H_(2)O`

ANSWER :B::C::D
Discussion
90621.

A complex containing cobalt(III) ion is associated with two diaminopropane molecules and two chloride ion forming an octahedral geometry. The complex ion will show :

Answer»

LIGAND isomerism
geometrical isomerism
optical isomerism
ionization isomerism

Solution :`[CO("PN")_(2)Cl_(2)]^(+)`
Discussion
90622.

A complex cross-linked polymer (silicone) is formed by

Answer»

HYDROLYSIS of `(CH_(3))_(3)SiCl`
hydrolysis of a mixture of `(CH_(3))_(3)` and `(CH_(3))_(2)SiCl_(2)`
hydrolysis of `CH_(3) SiCl_(3)`
hydrolysis of `SiCl_(4)`

SOLUTION :As CHAIN can grow in three places as
Discussion
90623.

A complex has a composition corresponding to the formula CoBr_(2)Cl.4NH_(3). What is the structural formula if conductance measurements show two ions per formula unit? Silver nitrate solution given an immediate precipitate of AgCl but no AgBr.

Answer»

`[CoBrCl(NH_(3))_(4)]Br`
`[CoCl(NH_(3))_(4)]Br_(2)`
`[CoBr_(2)Cl(NH_(3))_(4)]`
`[CoBr_(2)(NH_(3))_(4)]Cl`

ANSWER :D
Discussion
90624.

A complex compound of one cobalt (III) ion, two 'en' molecules, two chloride ions and one nitrate ion will show:

Answer»

LINKAGE isomerism
ionisation isomerism
geometrical isomerism
optical isomerism

Solution :`[Co("en")_(2)Cl_(2)](NO_(3)) to "Ionisation, geometrical.optical"`.
Discussion
90625.

A complex compound of cobalt contains five NH_(3) molecules. One nitro group and two chloride ions for one Co atom. One mole of this compound produces three mole ions in aq. solution. On reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated. EAN, of central atom of the complex is

Answer»

35
36
37
54

Solution :EAN of CO `=27-3+6xx2=36`
Discussion
90626.

A complex compound of Co^(3+) with molecular formula CoCl_(x).yNH_(3) gives a total of 3 ions on dissovling it in water.To satisfy both primary and secondary valencies in this complex, the number of Cl^(-) ions required is

Answer»

one
four
three
zero

Solution :As cobalt is PRESENT as `Co^(3+)` as coordination NUMBER of cobalt is 6, the molecular formula of compound should be `CoCl_(3). yNH_(3)`. Now, as it gives a total of three ion when dissoolved in water, its structural formula must be `[CoCl(NH_(3))_(5)]Cl_(2)`.
`[CoCl(NH_(3))_(5)]Cl_(2)hArr[CoCl(NH_(3))_(5)]^(2+)+2Cl^(-)`
THUS only one `Cl^(-)` ion is satisfying both primary and secondary VALENCY of `Co^(3+)` in this compound.
Discussion
90627.

A complex compound of cobalt contains five NH_(3) molecules. One nitro group and two chloride ions for one Co atom. One mole of this compound produces three mole ions in aq. solution. On reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated. The formula of the complex compound is

Answer»

`[CO(NH_(3))_(4)NO_(2)Cl](NH_(3))Cl`
`[Co(NH_(3))_(5)Cl]Cl(NO_(2))`
`[Co(NH_(3))_(5)(NO_(2))Cl_(2)`
`[Co(NH_(3))_(5)][(NO)_(2)Cl_(2)]`

Solution :To give 2 mole of AGCL ppt `2Cl^(-)` ions mustpresent outside the []
Discussion
90628.

A complex compound of cobalt contains five NH_(3) molecules. One nitro group and two chloride ions for one Co atom. One mole of this compound produces three mole ions in aq. solution. On reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated. IUPAC name of the complex compound is

Answer»

pentaammine nitrito-N-COBALT (III) chloride
pentaaminenitrito-N-cobalt (III) chloride
pentaamminenitrito-N-cobalt (III) chloride
pentamine chlorido cobalt (II) nitrite

Solution :
Discussion
90629.

A complex compound of chromium chromium contains five NH_(3) molecules, one nitro group and two chloride ions for one Cr^(3+) cation. One molecule of this compound produces three ions in aq. Solution, on reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated Q. Magnetic moment off complex compound is:

Answer»

0 BM
`sqrt(24)` BM
`sqrt(15)` BM
`sqrt(3)` BM

Answer :C
Discussion
90630.

A complex compound of chromium chromium contains five NH_(3) molecules, one nitro group and two chloride ions for one Cr^(3+) cation. One molecule of this compound produces three ions in aq. Solution, on reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated Q. The types of isomerism shown by the complex compound is:

Answer»

GEOMETRICAL, ionization
ionization, linkage
linkage, optical
geometrical, optical

Answer :B
Discussion
90631.

A complex compound of chromium chromium contains five NH_(3) molecules, one nitro group and two chloride ions for one Cr^(3+) cation. One molecule of this compound produces three ions in aq. Solution, on reacting with excess of AgNO_(3) solution, two moles of AgCl get precipitated Q. The formula of the complex compound is:

Answer»

`[CRCL(NO_2)(NH_(3))_(4)]NH_(3)*Cl`
`[CrCl(NH_(3))_(5)]Cl*NO_(2)`
`[CR(NO_(2))(NH_(3))_(5)]Cl_(2)`
`[Cr(NH_(3))_(5)]NO_(2)*Cl_(2)`

Answer :C
Discussion
90632.

A complex cationis formed by Pt( in some oxidation state ) with ligands ( in proper number so that coordination number of Pt becomes six ). Which of the following can be its correct IUPAC name ?

Answer»

Diammineethylenediaminedithicocyanato-S-PLATINUM (II) ion
Diammineethylenediaminedithicocyanato-S-platinate (IV) ion
Diammineethylenediaminedithicocyanato-S-platinum (IV) ion
Diamminebis (ethylenediamine ) dithiocyanate -S- platinum (IV) ion

Answer :C
Discussion
90633.

A complex compound in which the oxidation number of a metal is zero

Answer»

`[Ni(CO)_(4)]`
`[Pt(NH_(3))_(4)]Cl_(2)`
`K_(4)[Fe(CN)_(6)]`
`K_(3)[Fe(CN)_(6)]`

Solution :Co is a neutral ligand , HENCE it has ZERO oxidation state in the cpmplex `[Ni(CO)_(4)]`.
Discussion
90634.

A complex cation is formed by Pt (in some oxidation state) with ligands (in proper number so that coordination number of Pt becomes six). Which of the following can be correct IUPAC name?

Answer»

Diammineethylenediaminefithiocyanto-S-platinum (II) ion
Diammineethylenediaminefithiocyanto-S-platinate (IV) ion
Diammineethylenediaminefithiocyanto-S-platinum (IV) ion
Diamminebis (ethylenediamine) dithiocyanate-S-platinum (IV) ion

Solution :`[Pt(NH_(3))_(2)(en)(SCN)_(2)]^(2+`
Discussion
90635.

A complex cation is formed by ppt in some oxidation state) with ligands (in proper number so that CN of ppt becomes six). Which of the following can be its correct IUPAC name?

Answer»

Diammine ETHYLENE diamine dithiocyanato-s-platinum (II)
Diammine ethylene diamine dithiocyanato-s-platinum (IV) ion
Diammine ethylene diamine dithiocyanato-s-platinum (III) ion
Diamminebis (ethylene diamine) dithiocyanato-s-platinum (IV) ion

Solution :Cyanid (Turn bull.s blue) turn bull.s blue `[Pt(NH_(3))_(2)(EN)(SCN)_(2)]^(2-)`
Discussion
90636.

A complex anion is formed by Osmium (in some oxidation state) with ligands (in proper number so that coordination number of osmium becomes six). Which of the following can be its correct IUPAC name ?

Answer»

pentachloridonitridoosmium(VI)
pentachloridonitridoosmate(VI)
azidopentachloridoosmate(VI)
NONE of these

Solution :`[overset(+6)(Qs)overset(-5)(Cl)_(5)overset(-3)N]^(2)`
Discussion
90637.

a completely filled d orbital (d^(10))

Answer»

SPHERICALLY SYMMETRICAL
Has OCTAHEDRAL symmetrical
Has octahedral symmetry
Depends on the atom

Answer :B
Discussion
90638.

(a) Complete the following reactions: (i) CH-=CH+H_(2)O underset(HgSO_(4), 333K)overset(Dil. H_(2)SO_(4))to ? (ii) CH_(3)CHO+HCN to ? (iii) HCHO+NaOH to ? (b) How will you prepare RCOOH from? (i) Alcohol (ii) Nitrile.

Answer»

Solution :(a) (i) `CH-=CH+H_(2)O underset(HgSO_(4), 333K)overset(Dil. H_(2)SO_(4))to CH_(3)-CHO`
(ii)
(iii) `2HCHO+underset((50%))(NaOH) to CH_(3)OH +HCOONa`
(b) (i) `R-CH_(2)OH underset(KMnO_(4) // OH^(-))overset((O))to RCHO underset(KMnO_(4) // OH^(-))overset((O))to RCOOH`
(ii) `R-CN+2H_(2)O underset(Delta)overset(Dil. HCl)to RCOOH+NH_(3)`
Discussion
90639.

(a) Complete the following reaction : (i) NH_(3) + 3Cl_(2) rarr overset(?)(………) +3HCl (ii) Cl_(2)+F_(2) overset(473 K)rarr overset(?)(……) (b) Write the structure of perchloric acid (HClO_(4)).

Answer»

SOLUTION :(a) (i) `NCl_(3)`
(II) 2CIF
(B)
Discussion
90640.

(a) Complete the following reaction and write its name 2C_(6)H_(5)CI + 2Na overset(“Dry Ether”) to ? + 2NaCI(b) Explain Zaitsev rule with an example. (c) A Haloalkane when boiled with aqueous KOH which gives an alcohol having inversed configuration. Name the mechanism involved in this reaction.

Answer»

Solution :
(a) Name of the reaction : FITTIG Reaction
(b)In dehydrohalogenation reaction the preferred product is that ALKENE which has GREATER number of alkyl GROUPS attached to the double bonded carbon atom.
(C ) `SN_(2)`
Discussion
90641.

(a) Complete the following equations:(i)RCH_(2)OH overset(1. Alk. KMnO_(4))underset(2. H_(2)O)(to) (ii) CH_(3)COONa overset(CaO)underset(NaOH)(to)(b) Explain Aldol condensation reaction for acetaldehyde. Write equation.(c) What is Formalin solution?

Answer»

Solution :(a) (i) R - COOH Carboxylic acid.
(ii) `CH_(4) + NO_(2) CO_(3)`
(B) Aldehydes and ketones having atleast one hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form b-hydroxyaldehydes (aldol) or b-hydroxy ketones (KETOL).The aldol and ketol, on heating readily lose WATER to give, p, b-unsaturated carbonyl compounds. This reaction is CALLED aldol condensation. Example-1 `underset("Ethanal")(2CH_(3)CHO) hArr underset("Aldol")(CH_(3)-underset(OH)underset(|)(C)H-CH_(2))-CHOunderset(-H_(2)O)overset(Delta)to underset("Aldol condensation product")(CH_(3)-CH=CH-CHO)`
Example 2: `2CH_(3)COCH_(3) overset(Ba(OH)_(2))hArr underset("Ketol")(CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)(C)-CH_(@))COCH_(3) underset(-H_(2)O)overset(Delta)to underset("Aldol condensation product")(CH_(3)-overset(CH_(3))overset(|)(C)=CH-CO-CH_(3))`
(c) 40% aqueous solution of formaldehyde.
Discussion
90642.

a) Complete the following equations: i) PbS_((s))+4O_(3(g))rarr ii) 2NaOH+SO_(2)rarr

Answer»

Solution :a) i) `PbS_((s))+40_(3(s))rarrPbSO_(4(s))+40_(2(s))`
ii) `2NaOH+SO_(2)rarrNa_(2)SO_(3)+H_(2)O`
Discussion
90643.

(a) Complete the following equations: (i) Cr_2O_(7)^(2-) + 2OH^(-) to (ii)MnO_(4)^(-) + 4H^(+) + 3e^(-) (b) Account for the following: (i) Zn is not considered as a transitionelement. (ii) Transition metals form a large number of complexes. (iii) The E^(0) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple.

Answer»

Solution :(a)(i) `Cr_2O_(7)^(2-) + 2OH^(-) to CrO_(4)^(2-) + H_2O`
(ii) `MnO_(4)^(-) + 4H^(+) + 3e^(-) to MnO_(2) + 2H_2O`
(b) (i) `Zn//Zn^(2+)` has fully FILLED d-orbitals.
(ii) This is due to smaller IONIC sizes/higher ionic charge and availability of d-orbitals.
(iii) Because `MN^(2+)` is more stable `(3d^5)` than `Mn^(3+) (3d^4)`.`Cr^(+3)` is more stable due to `t_(2g)^(3)//d^(3)`, configuration.
Discussion
90644.

(a) Complete the following equations : (i) Cr_(2)O_(7)^(2-)+2OH^(-)to (ii) MnO_(4)^(-)+14H^(+)+3e^(-)to (b) Account for the following: (i) Zn is not considered a transition element. (ii) Transition metals form a large number of complexes. (iii) The E^(@) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple.

Answer»

Solution :(a) (i) `Cr_(2)O_(7)^(2-)+2OH^(-)to 2CrO_(4)^(2-)+H_(2)O`
(ii) `MnO_(4)^(-)+4H^(+)+3e^(-)to MnO_(2)+2H_(2)O`
(b) (i) ZN is not considered as a transition element.
Zn ATOM has completely filled d-orbitals `(3d^(10))` in its ground as well as in its oxidised state.
Therefore, it is not regarded as a transition element.
(ii) Transition metals form a large number of complexes because of the FOLLOWING:
They have comparatively smaller sizes of the metal ions.
They have high ionic charges.
d-orbitals are available for bond formation.
(III) `E^(@)` value for `Mn^(3+)//Mn^(2+)` couple is much more positive i.e., this REDUCTION takes place with greater tendency because `Mn^(2+)` has `d^(5)`, half-filled stable configuration. In the `Cr^(3+)//Cr^(2+)` couple the change is from `d^(3)` to `d^(4)` which takes place with smaller tendency, hence `E^(@)` value is less positive (in fact, it is negative).
Discussion
90645.

Complete the following equations:

Answer»

SOLUTION :
(II) `C_6H_5COOH`
(III) `CH_4+Na_2CO_3` .
Discussion
90646.

(a) Complete the following chemical reaction equations : (i) MnO_(4(aq))^(-) + C_(2)O_(4(aq))^(2-) + H_((aq))^(+) to (ii) Cr_(2) O_(7 (aq))^(2-) + Fe_((aq))^(2+) + H_((aq))^(+) to (b) Explain the following observations about the transition/inner transition elements : (i) There is in general an increase in density of element from titanium (Z = 22) to copper ( Z= 29). (ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3^(rd) series). (iii) The members in the actinoid series exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.

Answer»

Solution :(i) `5 C_(2)O_(4)^(2-) + 2Mn O_(4)^(-) + 16 H^(+) to 2Mn^(2+) + 8H_(2)O + 10 CO_(2)`
(ii) `Cr_(2)O_(7)^(2-) + 6FE^(2+) + 14 H^(+) to 2Cr^(3+) + 6 Fe^(3+) + 7H_(2)O`
(b) (i) As we move along a transition series from left to right (i.e. Ti to Cu) , the atomic radii decrease due to increase in nuclear charge . Hence the atomic volume decreases. At the same time , atomic mass increases. Hence the density from TITANIUM (Ti) to Copper (Cu) increases.
(ii) Since `3^(rd)` series of transition elements show the variable oxidation state .
(iii) Actinoids series show a large number of oxidation state due to very small energy gap between 5f , 6 d and 7s subsells.
Discussion
90647.

(a) Complete the following chemical reaction equations : (i) Fe_((aq))^(2+) + MnO_(4 (aq))^(-) + H_((aq))^(+) to (ii) Cr_(2)O_(7 (aq))^(2-) + I_((aq))^(-) + H_((aq))^(+) to (b) Explain the following observations : (i) Transition elements are known to form many interstitial compounds . (ii) With the same d^(4) d-orbital configuration Cr^(2+) ion is reducing while Mn^(3+) ion is oxidising . (iii) The enthalpies of atomisation of the transition elements are quite high.

Answer»

Solution :`5Fe_((aq))^(2+) + MnO_(4 (aq))^(-) + 8H_((aq))^(+) to Mn_((aq))^(2+) + 4H_(2)O(l) + 5Fe_((aq))^(3+)`
(ii) `Cr_(2)O_(7 (aq))^(2-) + 6I_((aq))^(-) + 14H_((aq))^(+) to 2 Cr_((aq))^(3+) + 7H_(2)O(l) + 3I_(2)(g)`
b (i) The transition metals form a larger number of interstitial compounds in which small ATOMS such a hydrogen , carbon , boron and nitrogen occupy the empty spaces (interstitial sites ) in their lattices .
(III) DUE to the presence of large number of unpaired electrons in their atom, they form strong interatomic metallic bonds . Hence they have high ENTHALPIES of atomisation .
Discussion
90648.

(a) Complete the following chemical reaction equation :(i)MnO_(4(aq))^(-)+C_(2)O_(4(aq))^(2-)+H_(aq)^(+) to (ii) Cr_(2)O_(7(aq))^(2-)+Fe_((aq))+H_((aq))^+to (b) Explain the following observation about the transition /inner transition elements: (i) There is in general an increase in density of element from titanium (Z=22) to copper (Z=29) (ii) there occurs much more frequent metal bonding in compounds of heavy transition elements (3^(rd) series) (iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series (a) Complete the following chemical equation for reaction: (i) 8MnO_(4)^(-)+H_(2)O+3S_(2)O_(3)^(2-) to (ii) Cr_(2)O_(7)^(2-)+3H_(2)S+8H^(+) to (b) Give an explanation for each of the following observations: (i) The gradual decrease in size (actinoid contractions) from element to element is greater among the actinoids than among the lanthanoids (lanthanoid contraction). (ii) The greatest number of oxidation states are exhibited by the members in the middle of a transition series. (iii) With the same d-orbital configuration (d^(4) Cr^(2+)) ion is a reducing agent but Mn^(3+) ion is an oxidising agent.

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SOLUTION :N//A
Discussion
90649.

(a) Complete the following chemical equations : (i) MnO_(4)^(-)(aq)+S_(2)O_(3)^(2-)(aq)+H_(2)O(l)to (ii) Cr_(2)O_(7)^(2-)(aq)+Fe^(2+)(aq)+H^(+)(aq)to (b) Explain the following observations: (i) La^(3+) (Z = 57) and Lu^(3+) (Z = 71) do not show any colour in solutions. (ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism. (iii) Cu^(+) ion is not known in aqueous solutions.

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Solution :(a) (i) `8MnO_(4)^(-)(aq)+3S_(2)O_(3)^(2-)(aq)+H_(2)O(l)to 8MnO_(2) (s)+6SO_(4)^(2-)(aq)+2OH^(-)(aq)`
(ii) `Cr_(2)O_(7)^(2-)(aq)+6Fe^(2+)(aq)+14H^(+)(aq)to2Cr^(3+)(aq)+6Fe^(3+)(aq)+7H_(2)O(l)`
(b) (i) `La^(3+)` (Z = 57) and `Lu^(3+)`. (Z = 71) do not show any COLOUR in solutions. There is no f-electrons in `La^(3+)` and `Lu^(3+)`. HENCE, there is no possibility of f-f transition. Therefore, they show no colour in solution.
(ii) `Mn^(2+)` has the `3d^(5)` CONFIGURATION. Thus, it has maximum (5) unpaired electrons. Therefore, it exhibits maximum paramagnetism.
(iii) `Cu^(+)` ion undergoes disproportionation in AQUEOUS solution as under:
`2Cu^(+)toCu^(2+)+Cu`
Discussion
90650.

(a) Complete the following chemical equations : (i) Cr_(2)O_(7)^(2-)(aq)+H_(2)S(g)+H^(+)(aq)to (ii) Cu^(2+)(aq)+I^(-)(aq)to (b) How would you account for the following: (i) The oxidising power of oxoanions are in the order VO_(2)^(+)lt Cr_(2)O_(7)^(2-)lt MnO_(4)^(-). (ii) The third ionisation enthalpy of manganese (Z = 25) is exceptionally high. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+).

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SOLUTION :(a) (i)
(ii) `2Cu^(2+)(aq)+4I^(-)(aq)toCu_(2)I_(2)(s)+I_(2)(s)`
(b) (i) Oxidising power of oxoanions is in the order :
`VO_(2)^(+)lt Cr_(2)O_(7)^(2-)lt MnO_(4)^(-)`
This is due to increasing stability of the lower species to which they are reduced.
(ii) This is due to exceptional stability of `Mn^(2+)(3d^(5))` which is due to highly stable half-filled configuration.
(III) `Cr^(2+)` is a stronger reducing agent than `FE^(2+)`. This is because after LOSING one electron `Cr^(2+)(d^(4))` changes to `Cr^(3+)(d^(3))` and these electrons are accommodated in lower `t_(2g)` orbitals which gives `Cr^(3+)` greater stability. On the other hand, `Fe^(2+)(d^(6))` on losing one electron change to `Fe^(3+)(d^(5))`. Although half-filled, the configuration is comparative less stable than `Cr^(3+)`.
Discussion