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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.
| 51. |
The co-ordination number of Cr in `[Cr(NH_(3))_(3)(H_(2)O)_(3)]Cl_(3)` isA. 3B. 4C. 6D. 2 |
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Answer» Correct Answer - C C.N.of Cr in the give complex is 6. |
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| 52. |
In the coordination compound, `K_4[Ni(CN)_4]` oxidation state of nickel isA. `-1`B. 0C. `+1`D. `+2` |
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Answer» Correct Answer - B `K_(3)[Ni(CN)_(4)]` of Ni = `4 xx 1-4 xx 1` = 0 |
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| 53. |
Explain how two complexes of nickel, `[Ni(CN)_4]^(2-)` and `Ni(CO)_4` have different structures but do not differ in magnetic behaviour. |
| Answer» The complex `[Ni(CN)_(4)]^(2-)` and `[Ni(CO)_(4)]` is square planar while `[Ni(CO)_(4)]` is tetrahedral in nature. The differnence in nature. The difference arises because of the oxidation states of the metal atom/ion in the two complexes. But since all the electron orbitals of hte central metal atom/ion in these complexes are paired, both are therefore, diamagnetic in nature. | |
| 54. |
The IUPAC name of the `K_(2)[Ni(CN)_(4)]` is :A. potassium tetracyanonickelate (II)B. potassium tetracynonickelate (III)C. potassium tetracyanotonickelate (II)D. Potassium tetracyanonickel (III) |
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Answer» Correct Answer - A a) it is the correct answer. |
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| 55. |
(a) Write the IUPC names of the following co-ordination compounds (i) `[Ce(NH_(3))_(4) (H_(2)O)_(2)]Cl_(3)` (ii) `[PtCl_(2)(NH_(3))_(4)] [PtCl_(4)]` (b) State the hybridisation and magnetic property of `[Fe(CN)_(6)]^(3-)` ion according to the valence bond theory . |
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Answer» (a) (i) tetraamminediaquachromium (III) chloride (ii) tetramminedichloridoplatinum (IV) tetrachloridoplatinate (II) (b) `d^(2)sp^(3)` hybridisation, inner orbital complex and weakly paramagnetic . For details , consult section 10. |
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| 56. |
The oxidation state of nickel in `K_(4)Ni(CN)_(4)` is: |
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Answer» Correct Answer - A `K_(4)[Ni(CN)_(4)]` `x+4(-1)=-4` `x=-4+4`= zero |
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| 57. |
`[NiCl_(4)]^(2-)` is paramagnetic while `[Ni(CO)_(4)]` is diamagnetic though both are tetrahedral. Why? |
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Answer» The `Cl^(-)` is a weak field ligand and cannot cause the electron pairing . On the other hand , CO is a strong field ligand and can cause the electron pairing . That is why , `(NiCl_(4))^(2-)` is paramagnetic while `[Ni(CO)_(4)]` is diamagnetic nature . For details , consult section 10 . |
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| 58. |
`Ag^(+)+hArr|Ag(NH_(3))_(2)|^(+), k_(1) = 6.8 xx 10^(-3)` `[Ag(NH_(3))]^(+) + NH_(3)hArr |Ag(NH_(3))_(2)|, k_(2) = 1.6xx10^(-3)` Then the formation constant or `|Ag(NH_(3))_(2)|^(+)` isA. `1.08 xx 10^(-7)`B. `1.08 xx 10^(-5)`C. `1.08 xx 10^(-9)`D. none of these |
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Answer» Correct Answer - B As the reaction takes place in two stages `k = k_(1)xx k_(2) = (6.8 xx 10^(-3)) xx (1.6xx 10^(-3))` = `10.8 xx 10^(-6) = 1.08xx 10^(-5)` |
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| 59. |
From the following given co-ordination compounds find the odd one out `[Ag(NH_(3))_(2)]^(+)`, `[AgCl_(2)]^(-)`, `[Ag(CN_(2)]^(-)`, `[Ag(CN)_(4)]^(2-)`A. `[Ag(NH_(3))_(2)]^(+)`B. `[AgCl_(2)]^(-)`C. `[Ag(CN)_(2)]^(-)`D. `[Ag(CN)_(4)]^(2-)` |
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Answer» Correct Answer - D Silver has shown a coordination number of four |
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| 60. |
Calculate the oxidation state of metal ion in: i) `[Ag(CN)_(2)]^(-)` ii) `[Ni(CO)_(4)]` iii) `[Fe(CN)_(6)]^(3-)` |
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Answer» i) `[Ag(CN)_(2)]^(-)] : x+2(-)=-1 or x=+1` ii) `[Ag(CN)_(2)]^(-): x+4(0)=0` or x=0 iii) `[Fe(CN)_(6)]^(3-)` : `x+6(-1)=-3` or `x=+3`. |
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| 61. |
In the silver plating of copper, `K[Ag(CN)_(2)]` is used instead of `AgNO_(3)` . The reason isA. A thin layer of Ag is formed on CuB. More voltage is requiredC. `Ag^(+)` ions are completely removed form solutionD. Less availability of `Ag^(+)` ions, as Cu cannot displace Ag from `[Ag(CN)_(2)]^(-)` ion. |
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Answer» Correct Answer - D In the silver plating of copper, `K[Ag(CN)_(2)]` is used instead of `AgNO_(3)`. This is because if `AgNO_(3)` is used , Cu will displace `Ag^(+)` from `AgNO_(3)` . The deposit so obtained is black , soft , flaky and non-adhering . To get a good shining deposit `[Ag(CN)_(2)]^(-)` ions are used . As it is a stable complex, the concentration of `Ag^(+)` ions is very small in the solution. As such no displacent of `Ag^(+)` ions with Cu is possible. |
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| 62. |
`Ag^(+) + NH_(3) ltimplies [Ag(NH_(3))]^(+), k_(1)=6.8 xx 10^(-5)` `[Ag(NH_(3))]^(+) + NH_(3) ltimplies [Ag(NH_(3))_(2)]^(+)`, `k_(2) = 1.6xx10^(-3)` The formation constant of `[Ag(NH_(3))_(2)]^(+)` is :A. `1.08 xx 10^(-7)`B. `1.08 xx 10^(-5)`C. `1.08 xx 10^(-9)`D. none of these |
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Answer» Correct Answer - a k = `k_1 xx K_2` = `(6.8 xx 10^(-5)) xx (1.6 xx 10^(-3))` = `1.08 xx 10^(-7)`. |
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| 63. |
The magnetic moment of salt containing `Zn^(2+)` ion is : |
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Answer» Correct Answer - a a) it is the correct answer. |
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| 64. |
Why do compounds having similar geometry have different magnetic moment ? |
| Answer» This is because of the nature of the ligands present in the complex. If the crystal field stabilising energy (CFSE) i.e. `Delta_(0)` for the complex is high, the complex will have less number of unpaired electrons and less value of magnetic moment. In case `Delta_(0)` is low, the complex will have more number of unpaired electrons and high value of magnetic moment. For exmaple, both `[CoF_(6)]^(3-)` and `[Co(NH_(3))_(6)]^(3+)` are octahedral complexes. Whereas former is paramagnetic, the latter is of diamagnetic nature. This is due to the reason the `F^(-)` ion has lesser `Delta_(0)` value as compared to `NH_(3)` molecule. | |
| 65. |
a) State the hybridisation & magnetic behaviour of `[Cr(CO)_(6)]` b) What are the various factors affecting crystal field splitting energy? c) Which of the two is more stable and why? `K_(4)[Fe(CN)_(6)]` or `[K_(3)[Fe(CN)_(6)]` |
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Answer» a) Hybridisation state of Cr in the complex is `d^(2)sp^(3)`. It is octahedral and is diamagnetic. c)`K_(3)[Fe(CN)_(6)]` is more stable than `K_(4)[Fe(CN)_(6)]` because Fe is in `+3` oxidation state while in the othe complex, it is in `+2` oxidation state. Moreover, the size of `Fe^(3+)` ion is smaller than that of `Fe^(2+)` ion. As a result, the ligands are held more closely by the `Fe^(3+)` ion than by `Fe^(2+)` ion. |
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| 66. |
Arrange following complex ions in increasing order of crystal field splitting energy `(Delta_(0))`, `Cr(Cl)_(6)]^(3-)`, `[Cr(CN)_(3)]^(3-)`, `[Cr(NH_(3))_(6)]^(3+)` |
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Answer» The order of increasing `Delta_(0)` value for the complex ions is: `[CrCl_(6)]^(3-) lt [Cr(NH_(3))_(6)]^(3+) lt [Cr(CN)_(4)]^(3-)` This is because of increasing order of the `Delta_(0)` valuesof the ligands in the spectrochemical series. |
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| 67. |
An arrangement of various ligands in decreasing order of crystal field splitting tendency is know asA. Nephelauxetic seriesB. Spectrochemical seriesC. Electrochemical seriesD. Macrocylic effect. |
| Answer» See Note Worthy Points . | |
| 68. |
Cobalt is present inA. Vitamin `B_(1)`B. Vitamin `B_(2)`C. Vitamin `B_(6)`D. Vitamin `B_(12)` |
| Answer» The central metal atom in vitamin `B_(12)` is cobalt (Co). | |
| 69. |
Phthalocyanin blue, the blue pigment present in feathers of a peacock is a complex compound ofA. CopperB. CobaltC. NickelD. None of these. |
| Answer» Phthalocyanin is a complex compound of copper. | |
| 70. |
Arrange the following complexes in the increasing order of conductivity of their solution `[Co(NH_(3))_(3)Cl_(3)], [Co(NH_(3))_(4)Cl_(2)]Cl, [Co(NH_(3))_(6)]Cl_(3), [Cr(NH_(3))_(5)Cl] Cl_(2)` |
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Answer» All the complexes are of octahedral nature. The increasing order of conductivity is related to the number of ions that are available in solution. The correct order is: `[Co(NH_(3))_(3)Cl_(3)] lt [Cr(NH_(3))_(4)Cl_(2)]Cl lt Co(NH_(3))_(5)Cl_(2) lt [Co(NH_(3))_(6)]Cl_(3)` |
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| 71. |
The green pigment present in plants , chlorophyll contains the metalA. AlB. FeC. MgD. Ca |
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Answer» Correct Answer - C The green pigment present in plants contains magnesium. |
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| 72. |
When ammonia is added to green aqueous solution of nickel (II) sulphate, the colour of the solution changes to the violet. This is caused byA. nickel ion undergoing a change in oxidation stateB. ammonia molecules replacing water molecules surrounding nickelC. change in co-ordination number of nickelD. change in pH value of the solution. |
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Answer» Correct Answer - B The colour change is due to replacement of water molecules surrounding Ni by the ammonia molecules. |
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| 73. |
The expected freezing point of `0.0100` m `[Co(NH_(3))_(5)Cl]Cl_(2)` isA. `+ 0.0279^(@)`CB. `-0.0279^(@)` CC. `0.0186^(@)` CD. `0.0558^(@)` C |
| Answer» See Comprehensive Review. | |
| 74. |
Berlin green isA. `[Pt(NH_(3))_(4)][PtCl_(4)]`B. `{:(III),(Fe[Fe(CN)_(6)]):}`C. `Cr_(2)O_(3)`D. `{:(III),(K_(3)[Co(NO_(2))_(6)]):}` |
| Answer» See Comprehensive Review. | |
| 75. |
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. `|V(gly)_(2)(OH)_(2)(NH_(3))_(2)|^(2+)`B. `|Fe(en)(bpy)(NH_(3))_(2)|^(3+)`C. `|Co(OX)_(2)(OH)_(2)|^(2-)`D. `|Ti(NH_(3))_(6)|^(3+)` |
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Answer» Correct Answer - C Oxidation state of V in (A) `x + 2 (-1) + 2(-1) +` = `+1` or x = `+5` Valence shell configuration `3d^(0)` Number of unpaired electron 1. Oxidatios state of Fe in (B) `x + 0 + 0 + 0 = +2` or = `+2` Valence shell configuration = `3d^(6)` (i.e. 4 unpaired electrons) Oxidation state of Co in (C) `x+ 2 (-2) + 2(-1) = -2 or x = +4` Valence shell configuration = `3d^(5)` (i.e. 5 unpaired electron) Oxidation state of Ti in (D) `x + 6(0) = + 3 or x = +3` Valence shell configuration = `3d^(1)` (i.e. 1 unpaired electron). Thus complex of cobalt has maximum unpaired electrons (viz 5 ) and hence has highest paramagnetism. |
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| 76. |
AgCl dissolves in `NH_(4)OH` due to the formation ofA. `[Ag(NH_(4))_(2)]Cl`B. `[Ag(NH_(4))_(3)]Cl`C. `[Ag(NH_(3))_(2)]Cl`D. `[Ag(NH_(3))_(2)]OH.` |
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Answer» Correct Answer - C `underset("White ppt.")(AgCl(s)) + 2NH_(4)OH (aq)rarrunderset("Soluble complex")([Ag(NH_(3))_(2)]Cl (aq)) + 2H_(2)O(l)` |
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| 77. |
The effective atomic number of Fe in `Fe(CO)_(5)` isA. 26B. 34C. 36D. 54 |
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Answer» Correct Answer - C `EAN = Z-O.N+2xxC.N.` =`26-O+2xx5=36` |
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| 78. |
Number of electrons gained by Pd (for coordination ) in `[PdCl_(4)]^(2-)` isA. 4B. 8C. 10D. zero |
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Answer» Correct Answer - B No. of electrons gained = `2xx` No. of monodentat ligands = `2 xx 4 = 8` |
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| 79. |
The effective atomic number of Cr(at. No. 24) in `[Cr(NH_(3))_(6)]Cl_(3)` isA. 35B. 27C. 33D. 36 |
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Answer» Correct Answer - C Effective atomic number = Elecrtrons on `Cr^(3+) +` Electrons from 6 `NH_(3)` ligands . `=21 +6 xx 2 = 21 + 12 = 33`. |
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| 80. |
Which of the following facts about the complex `[Cr(NH_(3))_(6)]Cl_(3)` is wrong ?A. The complex involves `d^(2)sp^(3)` hybrdisation and is octahedral in shapeB. The complex is paramagnetic in natureC. The complex gives a while precipitate with `` solution .D. The complex gives a white precipitate with `AgNO_(3)` solution. |
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Answer» Correct Answer - C It is an inner orbital complex since the state of hybridisation is `d^(2)sp^(3)` |
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| 81. |
The oxidation states of Cr in `[Cr(H_(2)O)_(6)]Cl_(3). [Cr(C_(6)H_(6))_(2)]` and `K_(2)[Cr(CN)_(2)(O)_(2)(O_(2))(NH_(3))]` respectively areA. `+3, +2` and `+4`B. `+3, 0` and `+6`C. `+3, 0 ` and `+4`D. `+3, +4` and `+6` |
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Answer» Correct Answer - B `[Cr(H_(2)O)_(6)]Cl_(3)` `rArr x+0xx6-1xx3=0` or `x=+3` `[Cr(C_(6)H_(6))_(2)]` `rArr x+2xx0=0` or `x=0` `[Cr(CN)_(2)(O)_(2)NH_(3)]^(2-)` `x+2(-1)+2(-2)+2(-1)+0=-2` `x-2-4-2=-2 or x=+6` |
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| 82. |
AgCl reacts with aqueous solution of sodium thiosulphate and potassium cyanide solutions to give :A. `Na[Ag(S_(2)O_(3))_(2)]` and `K[Ag(CN)_(2)]`B. `Na[Ag(S_(2)O_(3))_(2)]` and `K_(3)[Ag(CN)_(2)]`C. `Na_(3)[Ag(S_(2)O_(3))_(2)]` and `K[Ag(CN)_(2)]`D. `Na_(3)[Ag(S_(2)O_(3))_(2)]` and `K[Ag(CN)_(2)]` |
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Answer» Correct Answer - C c) it is the correct answer. |
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| 83. |
Which of the following complexes has lowest molar conductance ?A. `CoCl_(3).3NH_(3)`B. `CoCl_(3).4NH_(3)`C. `CoCl_(3).5NH_(3)`D. `CoCl_(3).6NH_(3)` |
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Answer» Correct Answer - A The complex [`CoCl_(3)NH_(3)`] has the lowest molar conductance since it furnishes no ion in the aqueous solution. |
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| 84. |
Which of the following will exhibit maximum ionic conductivity?A. `K_(4)[Fe(CN)_(6)]`B. `[Co(NH_(3))_(6)Cl_(3)`C. `[Cu(NH_(3))_(4)]Cl_(2)`D. `[Ni(CO)_(4)]` |
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Answer» Correct Answer - A `K_(4)[Fe(CH)_(6)](aq)hArr4K^(+)(aq)^(+)+[Fe(CN)]^(4-)(aq)` It gives five ions in solution . `[Co(NH_(3))_(6)]Cl_(3)(aq)hArr[Co(NH_(3))_(6)](aq)+3Cl^(-)(aq)` It gives four ions in solution . `[Cu(NH_(3))_(4)]Cl_(2)(aq)hArr[Cu(NH_(3))_(4)]^(2+)(aq)+2Cl^(-)(aq)` It gives three ions in solution. `[Ni(CO)_(4)]` is not an ionic compound . `K_(4)[Fe(CN)_(6)` gives maximum ions in solution and as such shows maximum ionic conductivity . |
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| 85. |
The oxidation state of `Cr` in `[Cr(NH_(3))_(4)Cl_(2)]^(+)` is:A. `+2`B. `+3`C. `+4`D. `+5` |
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Answer» Correct Answer - A `[Cr(NH_3)_4Cl_2]^(+)` x + 4(0) + 2(-1) + +1 x = +1 + 2 = +3 |
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| 86. |
The oxidation state of `Cr` in `[Cr(NH_(3))_(4)Cl_(2)]^(+)` is:A. `+3`B. `+2`C. `+1`D. 0 |
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Answer» Correct Answer - A `[Cr(NH_(3))_(4)Cl_(2)]^(+)` Let O.S. of Cr = x `x + 4 xx 0 + 2 xx (-1) = +1` `x -2 = 1` x = `+3` |
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| 87. |
EAN of iron in `K_(4)[Fe(CN)_(6)]` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B Primary valency of iron in `K_(4)[Fe(CN)_(6)]` is 2 . This complex contains `Fe^(2+)` in the coordination sphere . |
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| 88. |
The geometry of the compound `[Pt(NH_(3))_(2)Cl_(2)]` isA. Square planarB. PyramidalC. TetrahedralD. octahedral |
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Answer» Correct Answer - A O.S.of Pt (Z = 78) in the complex is `+2(Pt^(2+))Pt^(2+) : [Xe]^(54)4f^(14)5d^(8)` The eight electrons in 5d-subshell pair up in four d-orbitals followed by `dsp^(2)`-hybridisation. The shape of the complex is square planar. |
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| 89. |
Which of the following is paramagnetic?A. `[Co(NH_(3))_(6)]^(3+)`B. `[Ni(CO)_(4)]`C. `[Ni(NH_(3))_(4)]^(2+)`D. `[Ni(CN)_(4)]^(2-)` |
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Answer» Correct Answer - C `Ni(Z = 28) : [Ar]^(18)4s^(2)3d^(8)` `Ni^(2+) : [Ar]^(18)3d^(8)` `[Ni(NH_(3))_(4)]^(2+)` : `Ni^(2+)` undergoes `sp^(3)`-hybridisation the resulting ion is paramagnetic due to the presence two unpaired electrons . `Ni(CO)_(4)` : Two 4s-electrons in Ni pair up with 3d-electrons followed by `sp^(3)`-hybridisation . The rasulting complex has no unpaired electron and is diamagnetic. `[Ni(CN)_(4)]^(2-)` : `CN^(-)` being a strong ligand , pairs up the two unpairad electrons in d-subshell of `Ni^(2+)` . This is followed by `dsp^(2)`-hybridisation . The complex ion has on unpaired electron and is diamagnetic. `[Co(NH_(3))_(6)]^(3+)` : `Co^(3+)` with `3d^(6)`-configuration undergoes `d^(2)sp^(3)`-hybridisation , after pairing up of all the six d-electrons in three d-orbitals . As the ion has no unpaired electron , it is diamagnetic. |
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| 90. |
The correct statement with respect to the complexes `Ni(CO)_(4) and |Ni(CN)_(4)|^(2-)` isA. nickel is in the same oxidation state in bothB. both have tetrahedral geometryC. both have square planar geometryD. have tetrahedral and square planar geometry respectively. |
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Answer» `Ni(CO)_(4)rarrsp^(3)"hybridisation"rarr"tatrahedral"` `|Ni(CN)_(4)|^(2-)rarrdsp^(2)"hybridisation"rarr "square planar"` |
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| 91. |
IUPAC name of `Na[CoCl(NO_(2))_(5)]` isA. chloronitrodis (ethylendiamine) cobaltic (III) chlorideB. chloronitrobis(ethylendiamine) cobalt (II) chlorideC. chlorodis (ethylediamine)nitrocobalt(III)chorideD. bis(ethylenediamine)chloronitrocobalt(III) chloride. |
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Answer» Correct Answer - C See Comprehensive Review. (IUPAC system of nomenclature). |
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| 92. |
Which of the following cations does not form complex with ammonia?A. `Al^(3+)`B. `Ag^(+)`C. `Cu^(2+)`D. `Cd^(2+)` |
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Answer» Correct Answer - A `Al^(3+)` does not form complexes with `NH_(3)` . |
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| 93. |
Which of the following pairs represents linkage isomers ?A. `[Cu(NH_(3))_(4)][PtCl_(4)]` and `[Co(en)_(3)]^(2+)`B. `[Pd(PPh_(3))_(2)(NCS)_(2)]` and `[Pd(PPh_(3))_(2)(SCN)_(2)]`C. `[Co(NH_(3))_(5)(NO_(3))]SO_(4)` and `[Co(NH_(3))_(5)(SO_(4))]NO_(3)`D. `[PtCl_(2)(NH_(3))_(4)]Br_(2)` and `[PtBr_(2)(NH_(3))_(4)]Cl_(2)` |
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Answer» Correct Answer - B The pair exhibits linkage isomerism. |
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| 94. |
The oxidation number of metal atom is zero inA. `[Ni(CO)_(4)]`B. `[Fe_(2)(CO)_(5)]`C. `Na[Co(CO)_(4)]`D. Cu |
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Answer» Correct Answer - a,b,c,d (a,b,c) are correct. |
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| 95. |
The geometry and magnetic behaviour of the complex `[Ni(CO)_4]` areA. square planar and geometry and diamagneticB. tetrahedral geometry and diamagneticC. tetrahedral geometry and paramagneticD. tetrahedral geometry and paramagnetic. |
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Answer» Correct Answer - B `[Ni(CO)_4]` is tetrahedral and diamagnetic in nature. For details , consult section 10 . |
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| 96. |
`K_(4)[Fe(CN)_(6)]` is used for detection ofA. `Fe^(+)` ionsB. `Fe^(3+)` ionsC. `Zn^(2+)` ionsD. `Cu^(2+)` ions. |
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Answer» Correct Answer - B::C::D `K_(4)[Fe(CN)_(6)]` or potassium ferrocyanide gives dark bluk colouration with `Fe^(3+)` ions , with `Cu^(2+)` ions `Zn^(2+)` ions and reddish brown ppt. with `Cu^(2+)` ions. |
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| 97. |
The oxidation state of Fe in brown complex `[Fe(H_(2)O)_(5)NO]SO_(4)` isA. `+1`B. `+2`C. `+3`D. `+4` |
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Answer» Correct Answer - A Oxidation state of Fe in `[Fe(H_(2)O)_(5)NO]SO_(4) is +1` as water `(H_(2)O)` is a neutral ligands with `NO^(+) -a` postivie ligand NO is an odd electron molecule . It gives its one unpaired electron to `Fe^(2+)` |
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| 98. |
Which of the following statements are correct regarding `[Fe(H_(2)O)_(5)NO]SO_(4)` ?A. it gives a brown ring test for nitratesB. oxidation state of Fe is +2C. charge on NO is +1D. C.N . Fe is 6. |
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Answer» Correct Answer - a,b,d (a,b,d) are correct . |
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| 99. |
Which of the following statements about `[Fe(H_(2)O)_(5)NO]^(2-)` are correct ?A. O.S. of Fe is `+1`B. C.N. of Fe is 6C. Charge on NO is `+1`D. O.S. of Fe is `+2` |
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Answer» Correct Answer - A::B::C In `[Fe(H_(2)O)_(5)NO]^(2+)`, O.S. of Fe is `+1` . There is one `NO^(+)` and five `H_(2)O` ligand . Coordination number of Fe is six in this complex as `H_(2)O` and `NO^(+)` are both monodentate. |
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| 100. |
violet coloured complex obtained in the detectiomn of sulphur is :A. `Na_(2)[Fe(NO)(CN)_(5)]`B. `Na_(3)[Fe(ONSNa)(CN)_(5)]`C. `Na_(4)[Fe(CN)_(5)NOS]`D. Both (b) and (c) |
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Answer» Correct Answer - D Both (b) and (c ) are the same . |
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