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InterviewSolution
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.
| 551. |
The complex `K_(4)[Zn(CN)_(4)(O_(2))_(2)]` is oxidised into `K_(2)[Zn(CN)_(4)(O_(2))]`, then whichof the following is//are correct:A. a.Zn(II) is oxidised to Zn(IV)B. b.magnetic moment decreasesC. c.`O-O` bond length decreasesD. d.magnetic moment remains same. |
| Answer» Correct Answer - C::D | |
| 552. |
The `IUPAC` name for `[AI(OH)(H_(2)O)_(5)]^(2+)` is (a) Pentahyroaluminium hydroxide (b) Aquometaaluminate ion (c ) Pentaaquaaluminate(III)hydroxide (d) Pentaaquahydroxoaluminium(III) . |
| Answer» Correct Answer - d | |
| 553. |
`[Pt(NH_(3))_(4)][CuCl_(4)]` and `[Cu(NH_(3))_(4)][PtCl_(4)]` are known asA. ionisation isomersB. coordination isomersC. linkage isomersD. polymerisation isomers. |
| Answer» Correct Answer - B | |
| 554. |
The correct IUPAC name of `K_(2)[Zn(OH)_(4)]` isA. potassium tetrahydroxyzinc (II)B. dipotassium tetrahydroxyzinc (II)C. potassium tetrahydroxyzincate (II)D. potassium tetrahydroxy zincate (III) |
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Answer» Correct Answer - C Potassium tetrahydroxyzincate (II) |
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| 555. |
`[Co(NH_(3))_(5)NO_(2)]Cl_(2) and [Co(NH_(3))_(5)ONO]Cl_(2)` are related to each other as :A. geometrical isomersB. linkage isomersC. coordination isomersD. ionisation isomers |
| Answer» Correct Answer - B | |
| 556. |
Calculate the instability constant for `[Ni(NH_(3))_(6)]^(2+)`, given that `beta_(6)` for this complex is `2 xx 10^(12)` |
| Answer» `K_(("instability")) = (1)/(beta_(6)) = (1)/(2) xx 10^(-2) = 5 x 10^(-13)` | |
| 557. |
Which of the following is wrong statement?A. (a) `Ni(CO)_4` has oxidation number `+4` for `Ni`B. (b) `Ni(CO)_4` has zero oxidation number for `Ni`C. (c) `Ni` is metalD. (d) `CO` is gas |
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Answer» Correct Answer - A `Ni(CO)_4` has O.N. zero for `Ni`. |
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| 558. |
The correct name of the compound `[Cu(NH_(3))_(4)](NO_(3))_(2)`, according to IUPAC system isA. Cuprammonium nitrateB. Tetramminecopper (II) dinitrateC. Tetramminecoper (II ) nitrateD. Tetramminecoper (II ) dinitrite |
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Answer» Correct Answer - C `[Cu(NH_(3))_(4) ](NO_(3))_(2)` tetrammine copper (II) nitrate |
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| 559. |
The correct IUPAC name of `K_(2)[Zn(OH)_(4)]` isA. potassium tetrahydroxyzinc (II)B. dipotassium tetrahydroxyzinc (II)C. potassium tetrahydroxyzincate (II)D. potassium tetrahydroxy zincate (III) |
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Answer» Correct Answer - C Potassium tetrahydroxyzincate (II) |
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| 560. |
The complexes `[Pt(NH_(3))_(4)][PtCl_(6)] and [Pt(NH_(3))_(4)Cl_(2)][PtCl_(4)]` are :A. linkage isomersB. optical isomersC. co-ordination isomersD. ionisation isomers |
| Answer» Correct Answer - C | |
| 561. |
Which of the following metal carbonyls is synthesized by the drect interaction of finely divided metal with CO ?A. `Fe_(3)(CO)_(12)`B. `Fe_(2)(CO)_(9)`C. `Fe(CO)_(5)`D. All of these |
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Answer» Correct Answer - C Pentacarbonyliron (0), `Fe(CO)_(5)` , is a yellow toxic liquid, which is used for making magnets and iron films. It can be prepared by heating finely divided Fe under CO. `Fe(CO)_(5)` reacts photochemically to give the yellow dimer `Fe_(2)(CO)_(9)` . When heated it forms the black solid, `Fe_(3)(CO)_(12)`. |
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| 562. |
The compound`[CrCl_(2)(NH_(3))_(2)(en)]` can form .A. Geometrical isomersB. Coordination isomersC. Optical isomersD. Linkage isomers |
| Answer» Correct Answer - C | |
| 563. |
The correct `IUPAC` name of `[Mn_(3)(CO)_(12)]` is .A. Dodecacarbonylmanganate`(0)`B. Dodecacarbonylmanganate`(II)`C. Dodecacarbonylmanganate`(0)`D. Manganiododecarbonyl`(0)` |
| Answer» Correct Answer - C | |
| 564. |
What is the coordination number of `Mn` in `[Mn_(2)(CO)_(10)]` ? |
| Answer» Six. Mn form octahedral complex. | |
| 565. |
Pick out the complex compound in which the central metal atom obeys EAN rule strictlyA. (a) `K_4[Fe(CN)_6]`B. (b) `K_3[Fe(CN)_6]`C. (c) `[Cr(H_2O)_6]Cl_3`D. (d) `[Cu(NH_3)_4]SO_4` |
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Answer» Correct Answer - A In complex `K_4[Fe(CN)_6]`, Fe obeys EAN rule strictly. |
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| 566. |
The compounds `[PtCl_(2)(CH_(3))_(4)]Br_(2)` and `[PtBr_(2)(NH_(3))_(4)]Cl_(2)` constitutes a pair ofA. coordination isomersB. linkage isomersC. ionisation isomersD. hydate isomers |
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Answer» Correct Answer - C `[Pt (Cl_(2) (NH_(3))_(4)]Br_(2) and [Pt Br_(2) (NH_(3))_(4)Cl_(2)` shows ionisation isomerism because in ionisation isomerism the difference arises from the position of group within or outside the coordination sphere therefore these isomers give different ions in the solution . |
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| 567. |
The compounds `[PtBr_(2)(NH_(3))_(2)]` can form .A. Geometrical isomersB. Coordination isomersC. Optical isomersD. Linkage isomers |
| Answer» Correct Answer - A | |
| 568. |
The metal ion in complex `underline(A ) ` has EAN identical to the atomic number of krypton . `underline (A) ` is ( Atomic no. Of `Cr =24,Fe=26 ,pd =46) `A. `[Pd(NH_(3))_(6)]Cl_(4)`B. `[Cr(NH_(3))_(5)Cl]SO_(4) `C. `Na_(4)[Fe(CN)_(6)]`D. `K_(3)[Fe(CN)_(6)]` |
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Answer» Correct Answer - C EAN =`Z-(o .N ) +2 (C .N) x` Electrons donated by `CN^(-) ` where ,ON = oxidation number Z = atomic numebr ` [Fe(CN)_(6)]^(4-) :EAN = 26-(2) +2(6) - 2+12=36` |
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| 569. |
Calculate the `EAN` of the central metal in the following complex (a) `[AuCI_(2)]^(Θ)` ,(b) `[AI(C_(2)O_(4))_(3)]^(3-)`,(c ) `[(CdI)_(4))]^(Θ)` . |
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Answer» Oxidation state of Au in `[AuCI_(2)]^(Θ)` is `x + 1 (-1) xx 2 = - 1)` `implies x = +1` EAN of Au in `[AuCI_(2)]^(Θ)` is Atomic number of Au - Oxidation state of Au + number of electrons gained by donation from ligands `=79 -1 + 4 =82` For EAN rule total electrons in the above compound should be 86 next noble gas `Rn(Z = 86)` Hence EAN rule in the above copounds is not valid (b) Oxidation state of Al in `[A1(C_(2)O_(4))_(3)]^(3-)` is `x + (-2) xx 3 = -3` ` x = +3` EAN of A1 in `[A1(C_(2)O_(4))_(3)]^(3-)` Atomic number of A1 - Oxidation state of A1 number of electrons gained by donation from ligands Note (en) is bidentate ligand and `C_(2)O_(4)^(2-)` is identate ligand EAN of A1 in `[A1(C_(2)O_(4))_(3)]^(3-)` `=13 -3 + 2(3 xx 2) = 22` For EAN rule total electrons should be 36 next noble gas `Kr(Z=36)` So EAN rule is not valid in the above compounds `[(CdI)_(4)]^(Θ)` Oxidation state of Cd in `[(CdI)_(4)]^(Θ)` is ` x + (-1) x 4 = - 1` ` x + = +3` EAN of Cd in `[(CdI)_(4)]^(Θ)` =Atomic number - Oxidation state of Cd+ number of electrons gained by donation from ligands `= 48 -3+8=53` For EAN rule total electrons should be 54 next noble gas Xe with `(Z =54)` So EAN rule is not valid in the above compound . |
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| 570. |
The total number of possibel coordination isomer for the given compounds `[Pt(NH_(3))(4)Br_(2)][PtBr_(4)]` is .A. `2`B. `4`C. `5`D. `3` |
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Answer» Correct Answer - B `[Pt^(4+)CI_(2)(NH_(3))_(4)]^(2+)[Pt^(2+)CI_(4)]^(2-)` (2) `[Pt^(4+)CI_(3)(NH_(3)]^(o+)[Pt^^(2+)CI_(3)(NH_(3))]^(o+)` (3) `[Pt^(2+)CI(NH_(3))_(3)]^(o+)[Pt^(4+)CI_(5)(NH_(3))]^(Θ)` (4) `[Pt^(2+)(NH_(3))_(4)]^(2+)[Pt^(4+)CI_(6)]^(2-)` . |
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| 571. |
Use the `EAN` rule to predict the molecular formula for the simple carbonyls of (a) `Cr(Z = 24)` (b) `Fe (Z =26)` and (c ) `Ni (Z =28)` (Assume that the oxidation state of the metals in the these carbonyls is zero) . |
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Answer» Let molecular formulae for the simple carbony1 of `Cr` Fe and `Ni` are `Cr(CO)_(n) Ni(CO)_(n)` and `Ne(CO)_(n)` respectively EAN of these metals in `Cr(CO)_(n)Ni(CO)_(n)` nad `Ne(CO)_(n)` respectively `EAN` of these metals in `Cr(CO)_(n)Fe(CO)_(n)` and `Ni(CO)_(n)` respectively is equal to the atomic number of `Kr(Z =36)` next nearest noble gas) In `Cr(CO)_(n) Z` of `Cr =24` Oxidation state of `Cr =0` (given) No of electrons donated by n `CO` groups `=2n` `EAN` of Cr in `Cr(CO)_(n) =24 - 0 + 2n = 36` `2n = 36 -24` `n = (36 - 24)/(2) = (12)/(2) =6` Hence molecular formula of `Cr(CO)_(n) = Cr(CO)_(6)` (b) In `Fe(CO)_(n)Z` of `Fe = 26` Oxidation state of `Fe = 0` No of electrons donated by n COgroups `=2n` `EAN` of Fe in `Fe(CO)_(n) = 26 - 0 + 2n = 36` `n = (36 -26)/(2) = 5` Hence molecular formula of Fe `(CO)_(n) = Fe(CO)_(5)` (c) In `Ni(CO)_(n) Z ` of `Ni = 28` Oxidation state of `Ni = 0` No of electrons donated by n `CO` groups `=2n` `EAN` of Ni in `Ni (CO)_(n) = 28 - 0 + 2n = 36` ` n = (36 - 28)/(2) = 4` Hence molecular formula of `Ni(CO)_(n)` =` Ni (CO)_(4)` . |
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| 572. |
Two complexes with empirical formula `Co(NH_(3))_(3)(H_(2)O)_(2) Br_(2)CI` exists in two isomeric forms `(A)` and `(B)` Form A gives two moles of AgBr On treatment with `AgNO_(3)` solution whereas form `B` gives only one mole of AgBr Give the structural formula of both these isomers What are these isomer called . |
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Answer» (A) `Co (NH_(3))_(3)(H_(2)O)_(2)Br_(2)CI` gives two moles of AgBr on treatment with `AgNO_(3)` solution hence two `Br^(Θ)` ions are present as counter ion outside the coordination sphere Hence A is `[Co(NH_(3)_(3)(H_(2)O))_(2)CI]Br_(2)` `Co(NH_(3))_(3)(H_(2)O)_(2)CI]Br_(2)rarr[Co(NH_(3))_(3)(H_(2)O)_(2)CI]^(2+) + 2Br^(Θ)` is present as counter ion i .e `B` is `[Co(NH_(3)_(3)(H_(2)O))BrC1]Br.H_(2)O` 1 mole of `H_(2)O` will be present outside the coordination sphere as coordination number of `Co^(3+)` ion is 6 `[Co(NH_(3))_(3)(H_(2)O)BrC]Br.H_(2)O` `rarr[Co(NH_(3))_(3)(H_(2)O)BrCI]^(o+) + Br^(Θ) + H_(2)O` `Br^(Θ) + AgNo_(3)rarrAgBrdarr+ NO_(3)` This isomers are (A) `[Co(NH_(3)_(3)(H_(2)O)_(2)CI]Br_(2)` and (B)`[Co(NH_(3))_(3)(H_(2)O)BrCI]BrH_(2)O` . |
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| 573. |
A compound with the empirical formula `Co(NH_(3))_(5)SO_(4)`Br exists in two forms viz red and violet forms Solution of red form gives a precipitate of `AgBr` with `AgNO_(3)` solution The violet form gives no precipitate with `AgNO_(3)` but gives white precipitate with the aquous solution of `BaCI_(2)` From the these observations give the structure of each form . |
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Answer» Red form of Co `(NH_(3))_(5)SO_(4)Br` gives a ppt of AgBr with` AgNO_(3)` solutions this indicates that `Br^(Θ)` solution this indicatest that `Br^(Θ)` is ionisable and hence is present outside the coordination sphere Thus red form is `[Co(NH_(3))_(5)SO_(4)]Br` ` [Co(NH_(3))_(5)SO_(4)]Brrarr[Co(NH_(3))_(5)SO_(4)]^(o+) + Br^(Θ)` `Ag NO_(3) + Br^(Θ) rarr AgBrbarr + NO_(3)^(Θ)` Violet form gives no precipitate with `AgNo_(3)` i. e `Br^(Θ)` is present inside the coordination sphere along with this it gives white ppt with aqueous solution of `BaCI_(2)` this indicates that `SO_(4)^(2-)` is present outside the coordination sphere Thus violet form is `[Co(NH_(3))_(5)Br]SO_(4)` `[Co(NH_(3))_(5)Br]SO_(4)rarr[Co(NH_(3))_(5)Br]^(2+) + SO_(4)^(2-)` `BaCI_(2) + SO_(4)^(2-)rarr BaSO_(4)underset(White ppt)(darr+2CI^(Θ)` . |
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| 574. |
Column-I and Column-II contains four entries each. Entries of Column-I are to be matched with some entries of Column-II One or more than one entries of Column-I may have the matching with the same entries of Column-II. |
| Answer» Correct Answer - D | |
| 575. |
Two complexes with empirical formula `Co(NH_(3))_(3)(H_(2)O)_(2) Br_(2)CI` exists in two isomeric forms `(A)` and `(B)`. Form A gives two moles of AgBr on treatment with `AgNO_(3)` solution whereas form `B` gives only one mole of AgBr. Give the structural formula of both these isomers. What are these isomer called? |
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Answer» For empirical formual `Co(NH_(3))_(2)(H_(2)O)_(2)CI_(2)Br_(2)` one form gives two moles of AgBr on treatmnet with `AgNO_(3)` this idicates that `2Br^(Θ)` ions are present as counter ions and this from is `[Co(NH_(3))_(2)(H_(2)O)_(2)CI_(2)]Br_(2)rarr` `[Co(NH_(3))_(2)(H_(2)O)_(2)CI_(2))]^(2+) + Br^(Θ)` `2Br^(Θ) + 2AgNO_(3)rarr 2AgBrdarr+2NO_(2)^(Θ)` The other form gives one mloe of AgBr this indicates that one `Br^(Θ)` ion is present as counter ion and this form is `[Co(NH_(3))_(2)(H_(2)O))CI_(2)Br]Br.H_(2)O` `(:" Coordination number of" Co^(3+)` is 6, so one `H_(2)O` is present as hydrate moleculte ) `[Co(NH_(3))_(2)(H_(2)O)CI_(2)Br]Br.H_(2)O` `rarr[Co(NH_(3))_(2)(H(2)O)CI_(2)Br]^(o+) + Br^(Θ) + H_(2)O` `AgNO_(3)+ Br^(Θ) rarr AgBrdarr+No_(3)^(Θ)` . |
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| 576. |
Select the correct `IUPA` name for `[Pt(C_(5)H_(5)N)_(4)[[PtCI_(4)]` complexA. Tetrapyridineplatinate(II)tetrachloridoplatinate(II)B. Tetrapyridineplatinate(II)tetrachloridoplatinate(II)C. Tetrapyridineplatinate(II)tetrachloridoplatinate(II)D. Tetrapyridineplatinate(II)tetrachloridoplatinate(II) |
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Answer» Correct Answer - D NO `"in" + 1` state is proved by magnetic moment Name of `No^(o+)` = nitrosonium or nitrosylium ion . |
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| 577. |
The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex is not a netural complex. The correct formula and geometry of the first complex is :A. `[Ni(H_(2)O)_(2)(NO_(3))_(2).4NH_(3)`,tetrahedralB. `[Ni(NH_(3))_(4)](CO)_(3))_(2).2H_(2)O`,tetrahedralC. `[Ni(NH_(3))_(4)](NO_(3))_(2).2H_(2)O`,square planarD. `[Ni(NH_(3))_(4)](H_(2)O)_(2)](NO_(3))_(2)`, octahedral |
| Answer» Correct Answer - C | |
| 578. |
The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex is not a netural complex. Which of the following statement are true for the second complex ?A. It has the EAN value of 36B. It can show optical isomerism.C. It cannot show geometrical isomerism.D. It produces three-fold freezing point depression. |
| Answer» Correct Answer - D | |
| 579. |
The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex is not a netural complex. The number of water molecules of crystallization are respectivelyA. zero, twoB. zero, zeroC. two, zeroD. two ,two |
| Answer» Correct Answer - C | |
| 580. |
Assertion (A) Toxic metal ions are removed by the chelating ligands. Reason (R) Chelate complexes tend to be more stable.A. Assertion and reason both are true, reason is correct explanation of assertion.B. Assertion and reason both are true but reason is not the correct explanation of assertion.C. Assertion is true, reason is false.D. Assertion is false, reason is true. |
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Answer» Correct Answer - A Assertion and reason both are correct and reason is the correct explanation of assertion. Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metals ligands chelates the metal ions by formation of stable complex. |
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| 581. |
Consider the following complexes: (I) `K_(2)Ptcl_(6)` (II) `PtCl_(4).2NH_(3)` (III) `PtCl_(4).3NH_(3)` (IV) `PtCl_(4).5NH_(3)` Their electrical conductances in an aqueous solution are:A. 256,0,97,404B. 404,0,97,256C. 256,97,0,404D. 404,97,256,0 |
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Answer» Correct Answer - A Greater the number of ions , greater the conductance 1. `K_(2) [Pt Cl_(6)]hArr ubrace(2K^(+)+[PtCl_(6)]^(2-))_("three ion ")` `2. Pt Cl_(4) .2NH_(3)hArr [Pt(NH_(3))_(2)Cl_(4)]to ` no ion ( least ) `3.[Pt(NH_(3))_(3)Cl hArr ` two ions 4. ` [Pt(NH_(3))_(5)Cl]Cl_(3)hArr ` four ions (maximum ) hence ,(a) is correct |
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| 582. |
Addition of `AgNO_(3)` solution to aqueous solution of each of the `Pt(IV)` amines viz (i) `PtCI_(4).6NH_(3)` (ii) `PtCI_(4).5NH_(3)` (iii) `PtCI_(4)` (iv)` PtCI_(4).3NH_(3)` and (v) `PtCI_(4).2NH_(3)` was found by Werner to give `4,3,2,1` and zero moles of `AgCI` per mole of the amines repectively How did Werner explain these observations? . |
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Answer» (i) `PtCI_(4).%NH_(3)` Since the This amine 4moles of AGCI in this amine all the four `CI^(Θ)` ions are ionisable and hence this amine has all four `CI^(Θ)` ions outside the coordination sphere Thus Werner formulated this amine as `Pt(NH_(3))_(6)]CI_(4)` (ii) `PtCI_(4).5NH_(3)` Since the amine gives 3 moles of AgCI in this amine only three `CI^(Θ)` ions are ionisable and hence this amine has only three `Ci^(Θ)` ions outside the coordination sphere Thus Werner formulated this amine as `[Pt(NH_(3))_(5)CI]CI_(3)` (iii) `PtCI_(4).4NH_(3)` Since this amine gives 2moles of AgCI in this amine only two `CI^(Θ)` ions are ionisable and hence this amine has only two `CI^(Θ)` ions outside the coordination sphere Thus Werner formulated this coordination sphere Thus Werner formulated this amine as `[Pt(NH_(3))_(4).C1_(2)]CI_(2)` (iv) `PtCI_(4).3NH_(3)` Since this amine gives 1 mole of agCI in this amine only one `CI^(Θ)` ion is ionisable and hence this amine has only one `CI^(Θ)` ions outside the coordination sphere Thus Werner formulated this amine as `[Pt(NH_(3))_(3)CI_(3)]CI` (iv) `PtCI_(4).2NH_(3)` Since this amine gives no `AgCI` in this amine no `CI^(Θ)` ion is ionisable i.e no `CI^(Θ)` ion is present outside the coordination sphere or all the four `CI^(Θ)` ions are present inside the coordination sphere Thus Werner formulated this amine as `[Pt(NH_(3))_(2).CI_(4)]` (non - electrolyte) . |
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| 583. |
Which of the following complexes exhibits the highest paramagnetic behaviour? where gly=glycine, en=ethylenediamine and bipy =bipyridyl (At. no. `Ti=22, V=23, Fe=26, Co=27`)A. (a) `[Ti(NH_3)_6]^(3+)`B. (b) `[Fe(en)(bpy)(NH_3)_2]^(2+)`C. (c) `[Co(OX)_2(OH)_2]^(-)`D. (d) `[V(gly)_2(OH)_2(NH_3)_2]^(+)` |
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Answer» Correct Answer - A `Fe^(2+)`, `Co^(5+)`, `Ti^(3+)` and `V^(3+)` have 4, 4, 1, 2 unpaired electrons respectively. The pairing leads `Fe^(2+)` with no unpaired electron. |
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| 584. |
The magnetic moment of a complex ion is `2.83BM` The complex ion is `[Cr(H_(2))O)_(6)]^(3+)` (b) `[Cu(CN)_(6)]^(2-)` (c ) `[V(H_(2)O))_(6)]^(3+)` (d) `[MnCI_(4)]^(2-)` . |
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Answer» Correct Answer - c Given `mu =2.83 = sqrt(n(n + 2)` `impliessqrt(n(n+2))~~2sqrt2impliesn=2` `[V(H_(2)O))_(6)]^(3+),V` is in `+3 OS` and its configuration is `3d^(2)i.e, `n=2` . |
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| 585. |
Two research students were instruced intructed to synthesise the complex `[Co(NH_(3))_(5)(NO_(2))]Br_(2)` They synthesised the complexes with identical molecular formula molar mass geometry conductane and spin but they differed in colour Based on the above facts answer the following questions Which of the ligands can show ambidentate property ?A. `NO_(2)^(Θ)`B. `NH_(3)`C. `H_(2)O`D. `CO_(3^(2-)` |
| Answer» Correct Answer - A | |
| 586. |
Sodium nitroprusside is used to test `S^(2-)` ion How many `CN^(Θ)` ion acts as ligand in the compound . |
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Answer» Correct Answer - 5 `[Fe(CN)_(5)(NO_(5))]^(4-)` |
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| 587. |
An octahedral complex with molecular composition `M.5NH_(3).Cl.SO_(4)` has two isomers. A and B. The solution A gives a white precipitation with `AgNO_(3)` solution and the solution of B gives white precipitate with `BaCl_(2)` solution. The type of isomerism exhibited by the complex is :A. linkage isomersB. ionisation isomerismC. coordination isomersD. Geometrical isomerism |
| Answer» Correct Answer - B | |
| 588. |
Magnetic moments `2.84 B.M` is given by : (At. nos. ni = 28, Ti = 22, Cr = 24, Co = 27).A. `Ni^(2+)`B. `Ti^(3+)`C. `Cr^(3+)`D. `Co^(2+)` |
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Answer» Correct Answer - A Magnetic moment `mu = sqrt( n(n+2)) BM` where `therefore 2.84= sqrt(n(n+2)) BM` `(2.84)^(2) = n(n+2)` `8=n^(2)+2n` `n^(2)+2n -8=0` `n^(2)+4n-2n-8=0` `n^(2) +4n-2n-8=0` `n(n+4)-2(n+4)=0` `n=2` ` Ni^(2+) =[Ar] 3d^(8)4s^(0)`(two uppaired electrons ) ` Ti^(3+)=[Ar] 3d^(1) 4s^(0) ` (one unpaired electron ) ` Cr^(3+) = [Ar] 3d^(3)` (three unpaired electrons ) ` Co^(2+) =[Ar] 3d^(7)4s^(0)` ( three unpaired electrons ) SO , only `Ni ^(2+)` has 2 unpaired elecrons |
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| 589. |
Which of the following complexes exhibits the highest paramagnetic behaviour? where gly=glycine, en=ethylenediamine and bipy =bipyridyl (At. no. `Ti=22, V=23, Fe=26, Co=27`)A. (a) `[Fe(en)(bipy)(NH_3)_2]^(2+)`B. (b) `[Co(OX)_2(OH)_2]^(-)`C. (c) `[Ti(NH_3)_6]^(3+)`D. (d) `[V(gly)_2(OH)_2(NH_3)_2]^(+)` |
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Answer» Correct Answer - C `Fe^(2+), Co^(5+), Ti^(3+)`, and `V^(3+)` have 4, 4, 1, 2 unpaired electron respectively. The pairing leads `Fe^(2+)` with no unpaired electron. |
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| 590. |
Among `(A)` `TiF_(6)^(2-),(B) CoF_(6)^(3-), (C ) Cu_(2)CI_(2)` and` (D) NiCI_(4)^(2)` (atomic number of `Ti=22,Co=27,Cu=29,Ni=28)` the colourless species are (a) `(B)` and `(D)` (b) `(A)` and `(B)` `C ` and (D)` `(A)` and `(C )` . |
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Answer» Correct Answer - d Completely filled d-orbital system or completely unfilled d-orbital system are colourless Thus colour of a complex is explained on the basis of d-transition `{:(In,Cu_2Cl_2,Cu,is i n,+1OS),(i.e.,3d^(10),system,and,),(i n,TiF_6^(2-),Ti,is i n,+4OS),(i.e.,3d^(0),system,,):}}"Colourless"` . |
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| 591. |
What type of isomerism exists in the following pairs of complexes ? (i) `[Co(NH_(3))_(5)NO_(3)]SO_(4)` and `[Co(NH_(3))_(5) SO_(4) ] NO_(3)` (ii) `[Co(en)(H_(2)O)_(2) Cl_(2)] Cl` and `[Co(en) (H_(2)O) Cl_(3) ] H_(2)O`A. (i) Ionisation (ii) HydrateB. (i) Linkage (ii) HydrateC. (i) Ionisation (ii) LinkageD. (i) Linkage (ii) Coordination |
| Answer» Correct Answer - A | |
| 592. |
Ammonia acts as a very good ligand but ammonium ion does not form complexes becauseA. `NH_(3)` is a gas while `NH_(4)^(+)` is in liquid formB. `NH_(3)` undergoes `sp^(3)` hybridisation while `NH_(4)^(+)` undergoes `sp^3` d hybridisationC. `NH_(4)^(+)` ion does not have any lone pair of electronsD. `NH_(4)^+` ion has one unpaired electron while `NH_(3)` has two unpaired electrons . |
| Answer» Correct Answer - C | |
| 593. |
Among `TiF_(6)^(2-), CoF_(6)^(3-), Cu_(2)C1_(2)` and `NiC1_(4)^(2-)` (At. No. `Ti = 22, Co = 27, Cu = 29, Ni = 28)`, the colourless species are -A. (a) `CoF_6^(3-)` and `NiCl_4^(2-)`B. (b) `TiF_6^(2-)` and `CoF_6^(3-)`C. (c) `Cu_2Cl_2` and `NiCl_4^(2-)`D. (d) `TiF_6^(2-)` and `Cu_2Cl_2` |
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Answer» Correct Answer - D `Ti^(4+): 3d^(0)` and `Cu^(+): 3d^(10)` cannot show d-d transition and thus colourless. |
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| 594. |
Which of the following complexes will mostly likely abosorb visible light ? (At nos. Sc=21,Ti=22,V=23,Zn=30)A. `[Sc(H_(2)O)_(6)]^(3+)`B. `[Ti(NH_(3))_(6)]^(4+)`C. `[V(NH_(3))_(6)]^(3+)`D. `[Zn(NH_(3))_(6)]^(2+)` |
| Answer» Correct Answer - C | |
| 595. |
Discovery of complexes was made byA. B. M. TassaertB. C.K. JorgensonC. Alfred WernerD. Kasimiv Fajan |
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Answer» Correct Answer - A In 1798 B.M. Tassaert found that a solution of cobalt (II) chloride in aqueous ammonia exposed to air (an oxidizing agent) deposits orange yellow crystals. He assigned the formula `CoCI_(3)` . `6NH_(3)` to these crystals. Later, a similar compound of platinum was assigned the formula `PtCI_(4)` . `6NH_(3)` . Because these formulas suggested that the substance were somehow composed of two stble compounds platinum (IV) chloride and ammoina in the latter case-they were called complex compounds or simply complexes. The vasic explanation for the structure of these complexes was given by the swiss Alfred Werner in 1893. |
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| 596. |
The sum of coordination number and oxidation number of the metal M in the complex `[M(en)_(2) (C_(2)O_(4))]CI` (where en is ethylenediamine) is:A. 7B. 8C. 9D. 6 |
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Answer» Correct Answer - C Ligand en is neutral while ligand `C_(2)O_(4)` is `2-` . Let the oxidation number of M be x: `(x) + 2(0) + 1(-2) = + 1` `:. X = +3` Both en and `C_(2)O_(4)` are bidentate ligands. Thus, coordination number of M is 6. Sum of coordination number and oxidation number is `6 + 3 = 9` |
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| 597. |
The magnitude of stability constant gives an indication of the stability of ______in____ . |
| Answer» Correct Answer - Complexes,solution | |
| 598. |
`d_(xy),d_(yz)` and `d_(zx)` orbitals have________ energies and are collectively termed_________ or __________orbitals . |
| Answer» Correct Answer - Equal,`dgama,e_(g)`,Axial | |
| 599. |
In case of octahedral compex, if the `e_(g)` orbitals `(d_(x^(2) - y^(2)) " and " d_(z^(2)))` are asymmetricaaly filled, their degeneracy is destroyed and the ligands approaching along `+Z` and `-Z` directions experiences different amount of repulsions than the ligands approaching along the `+X, -X, +Y " and " -Y` directions. As a result, the symmetrical nature of such complexes is lost and either elongation or compression along Z-axis taken place. Answer the following three questions based on the above situation. Which of the following is incorrect regarding `K_(4) [Cr F_(6)]` ?A. It has two long and four short `Cr-F` bondsB. It has four long and two short `Cr-F` bondsC. spin only magnetic moment of the complex is approximately `4.9 Bm`D. If `Cr` is replaced by `Cu`, similar types of deformation in the regular octahedral geometry are observed |
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Answer» Correct Answer - B |
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| 600. |
`d_(x2),d_(yz)` and `d_(zx)` orbitals have________ energies and are collectively termed_________ or __________or___________orbitals . |
| Answer» Correct Answer - Equal` din, t_(2)g,` None axial | |