Kerala Plus Two<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nKerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 9 Coordination Compounds<\/h2>\nPlus Two Chemistry Coordination Compounds One Mark Questions and Answers<\/h3>\n
Question 1.
\nThe ions or molecules bound to the central atom\/ion in the coordination entity are called ___________.
\nAnswer:
\nLigands<\/p>\n
Question 2.
\nIn [Co(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>, the coordination number of cobalt is _________.
\nAnswer:
\nsix<\/p>\nQuestion 3.
\nWhich complex has a square planar structure?
\n(a) [Ni(CO)4<\/sub>]
\n(b) [NiCI4<\/sub>]2-<\/sup>
\n(c) [Ni(H2<\/sub>O)6<\/sub>]2+<\/sup>
\n(d) [CU(NH3<\/sub>)4<\/sub>]2+<\/sup>
\nAnswer:
\n(d) [CU(NH3<\/sub>)4<\/sub>]2+<\/sup><\/p>\nQuestion 4.
\nSay TRUE or FALSE.
\n[Co(NO3<\/sub>)(NH3<\/sub>)5<\/sub>]SO4<\/sub> and [Co(NO3<\/sub>)(NH3<\/sub>)4<\/sub>(SO4<\/sub>)](NH3<\/sub>) are ionisation isomers.
\nAnswer:
\nFALSE<\/p>\nQuestion 5.
\nThe formula of Wilkinson’s catalyst is _________.
\nAnswer:
\n[RhCl(PPh3<\/sub>)3<\/sub>]<\/p>\nQuestion 6.
\nThe charge of Ni in [Ni(CO)4<\/sub>] is
\n(a) +1
\n(b) +2
\n(c) 0
\n(d) +4
\nAnswer:
\n(c) 0<\/p>\nQuestion 7.
\nThe central metal ion present in chlorophyll?
\n(a) Fe2+<\/sup>
\n(b) Cu2+<\/sup>
\n(c) Mg2+<\/sup>
\n(d) CO2+<\/sup>
\nAnswer:
\n(c) Mg2+<\/sup><\/p>\nQuestion 8.
\nEDTA is a dentate ligand
\n(a) uni dentate
\n(b) bidentate
\n(c) Tridentate
\n(d) hexadentate
\nAnswer:
\n(d) hexadentate<\/p>\n
Question 9.
\nWhich is an example for homoleptic complexes
\n(a) [Co(NH3<\/sub>Cl2<\/sub>]+<\/sup>
\n(b) [CoNH3<\/sub>)6<\/sub>]3+<\/sup>
\n(c) [Cr(NH3<\/sub>)(H2<\/sub>O)3<\/sub>]Cl3<\/sub>
\n(d) [CoCl2<\/sub>(en)2<\/sub>]
\nAnswer:
\n(b) [CoNH3<\/sub>)6<\/sub>]3+<\/sup><\/p>\nQuestion 10.
\nAmmonia will not form complex with
\n(a) Ag2+<\/sup>
\n(b) Pb2+<\/sup>
\n(c) Cu2+<\/sup>
\n(d) Cd2+<\/sup>
\n(e) Fe2+<\/sup>
\nAnswer:
\n(b) Pb2+<\/sup><\/p>\nPlus Two Chemistry Coordination Compounds Two Mark Questions and Answers<\/h3>\n
Question 1.
\nIn a seminar, Jishnu argued that the \u201chexaflourocobaltate(III) ion is highly paramagnetic than hexacyanoferrate(III) ion.<\/p>\n
\n- Do you agree with these words?<\/li>\n
- Explain it.<\/li>\n
- Write the formulae of the given coordination compounds.<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Yes<\/li>\n
- CN–<\/sup> is a strong field ligand so paring occurs. Number of unpaired electron decreases, paramagnetism decreases. F–<\/sup> is a weak field ligand so no paring occurs, paramagnetism increases.<\/li>\n
- [CoF6<\/sub>]3-<\/sup>and [Co(CN)6<\/sub>]3-<\/sup><\/li>\n<\/ol>\n
Question 2.
\nRaju: Coordination compounds are coloured.
\nRamu: No, co-ordination compounds are colourless.<\/p>\n
\n- Whose statement is correct?<\/li>\n
- Explain the reason for your answer.<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Raju\u2019s statement is correct. Coordination compounds are usually coloured.<\/li>\n
- The colour of coordination compounds is due to d-d transition.<\/li>\n<\/ol>\n
Question 3.
\nWhy is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
\nAnswer:
\nTetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.<\/p>\n
Question 4.
\nFeSO4<\/sub> solution mixed with (NH4<\/sub>)2<\/sub>SO4<\/sub> solution in 1:1 molar ratio gives the test of Fe2+<\/sup> ion but CuSO4<\/sub> solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+<\/sup> ion. Explain why?
\nAnswer:
\nFeSO4<\/sub> does not form any complex with (NH4<\/sub>)2<\/sub>SO4<\/sub>. Instead, it forms a double salt FeSO4<\/sub>. (NH4<\/sub>)2<\/sub>SO4<\/sub>.6H2<\/sub>O which dissociates completely into ions. CuSO4<\/sub> when mixed with NH3<\/sub> forms a complex [CU(NH3<\/sub>)4<\/sub>]SO4<\/sub> in which the complex ion [CU(NH3<\/sub>)4<\/sub>]2+<\/sup> does not dissociate to give Cu2+<\/sup> ion.<\/p>\nQuestion 5.
\nWrite the geometrical isomers of [Pt(NH3<\/sub>)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
\nAnswer:
\nThe complex [Pt(NH3)(Br)(Cl)(py)] will form three geometrical isomers:
\n
\nSquare planar complexes of this type will not show geometrical isomerism.<\/p>\nQuestion 6.
\nAqueous copper sulphate solution (blue in colour) gives:<\/p>\n
\n- a green precipitate with aqueous potassium fluoride and<\/li>\n
- a bright green solution with aqueous potassium chloride. Explain these experimental results.<\/li>\n<\/ol>\n
Answer:
\nAqueous copper sulphate contains a coordination entity, [Cu(H2<\/sub>O)4<\/sub>]2+<\/sup> which is blue in colour. Water molecule is a weaker ligand than Cl–<\/sup> and F–<\/sup>.
\n1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.
\n<\/p>\n2. On addition of aqueous solution of KCl, an another bright green complex entity is formed which is soluble in water.
\n<\/p>\n
Question 7.
\nWhat is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2<\/sub>S(g) is passed through this solution?
\nAnswer:
\nAqueous solution of copper sulphate contains Cu2+<\/sup> ions in the form of complex entity, [Cu(H2<\/sub>O)4<\/sub>]2+<\/sup> and H2<\/sub>O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)4<\/sub>]2-<\/sup> is formed due to the following reaction:
\n<\/p>\nCyanide ligand, CN–<\/sup> is a strong field lignad and stability constant of [Cu(CN)4<\/sub>]2-<\/sup> is quite large and thus practically no Cu2+<\/sup> ions are left in solution. On passing H2<\/sub>S gas, no CuS is formed due to non-availability of Cu2+<\/sup> ions in solution.<\/p>\nQuestion 8.
\nOptical isomerism is usually exhibited by complexes containing polydentate ligand. What do you mean by ligand?
\nAnswer:
\nLigand is a neutral molecule or charged ion which can donate a lone pair of electron to the metal.<\/p>\n
Question 9.
\nCoordination complexes are of different types. Name the compounds.<\/p>\n
\n- [Cr(H2<\/sub>O)5<\/sub>Cl2<\/sub>]<\/li>\n
- K3<\/sub>[Cr(C2<\/sub>O4<\/sub>)3<\/sub>]<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Pentaaquadichloridochromium(II)<\/li>\n
- Potassiumtrioxalatochromate(III)<\/li>\n<\/ol>\n
Question 10.
\nWrite the IUPAC names of the following compounds.<\/p>\n
\n- K3M<\/sub>[Fe(CN)6<\/sub>]<\/li>\n
- [C0(NH3<\/sub>)5<\/sub>(CO3<\/sub>)]Cl<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Potassium hexacyanoferrate(III)<\/li>\n
- Pentaamminecarbanatocobalt(III) chloride<\/li>\n<\/ol>\n
Question 11.
\n[Fe(CN)6<\/sub>]3-<\/sup> is paramagnetic, while [Fe(CN)6<\/sub>]4-<\/sup> is diamagnetic. Explain with the help of VB theory.
\nAnswer:
\nIn [Fe(CN)6<\/sub>]3-<\/sup> iron is in +3 state and in [Fe(CN)6<\/sub>]4<\/sup>, iron is in +2 state. [Fe(CN)6<\/sub>]3-<\/sup> contains five electrons in d-level (3d5<\/sup>). In this complex iron undergoes d2<\/sup>sp3<\/sup> hybridisation.<\/p>\nDue to the presence of one unpaired electron, [Fe(CN)6<\/sub>]3-<\/sup>, is paramagnetic. In [Fe(CN)6<\/sub>]4-<\/sup> iron contains six electrons in d-level (3d6<\/sup>). It undergoes d2<\/sup>sp3<\/sup> hybridisation and has no unpaired electrons. Hence, [Fe(CN)6<\/sub>]4-<\/sup> is diamagnetic.<\/p>\nPlus Two Chemistry Coordination Compounds Three Mark Questions and Answers<\/h3>\n
Question 1.
\nLook at the following two diagrams.
\n<\/p>\n
\n- Are diagrams I and II correct? Justify. If the figure is not correct, redraw it.<\/li>\n
- Which theory is related to this?<\/li>\n
- Explain briefly, how this theory is applicable to octahedral complexes.<\/li>\n<\/ol>\n
Answer:
\n1. No, Figure (II) is wrong.
\n<\/p>\n
2. Crystal field theory.<\/p>\n
3. In the case of octahedral complexes, the ligands are approaching the \u2018d\u2019 orbitals through the axis. As a result of this the energy of dx\u00b2-y\u00b2<\/sub> and dz\u00b2<\/sub> orbitals increases and the energy of the remaining three orbitals decreases. The orbitals which possess high energy are represented as \u2018eg<\/sub>\u2019 levels and the orbitals which possess less energy are represented as \u201ct2g<\/sub>\u201d levels.<\/p>\nQuestion 2.
\nA list of coordination compounds are given below:
\n[Cr(H2<\/sub>O)6<\/sub>] Cl3<\/sub>, [Co(NH3<\/sub>)5<\/sub>Br] SO4<\/sub>, [Co(NH3<\/sub>)5<\/sub>NO2<\/sub>]2+<\/sup> and [Pt(NH3<\/sub>)2<\/sub>Cl2<\/sub>]. Which type of isomerism do these compounds exhibit?
\nAnswer:
\nHydrate Isomerism, Ionisation Isomerism, Linkage Isomerism, Geometrical Isomerism.<\/p>\nQuestion 3.
\nThe following are examples of coordination compounds. Identify the type of isomerism exhibited by each of them and write their possible isomers,<\/p>\n
\n- [Cr(NH3<\/sub>)5<\/sub>Br]SO4<\/sub><\/li>\n
- CrCl3<\/sub>.6H2<\/sub>O<\/li>\n
- [PtCl2<\/sub>(NH3<\/sub>)2<\/sub>]<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Cr(NH3<\/sub>)5<\/sub>Br]SO4<\/sub> – Ionisation isomerism – [Cr(NH3<\/sub>)5<\/sub>SO4<\/sub>]Br<\/li>\n
- CrCl3<\/sub>.6H2<\/sub>O\u00a0– Hydrate isomerism – [Cr(H2<\/sub>O)5<\/sub>Cl]Cl2<\/sub>.H2)<\/sub><\/li>\n
- PtCl2<\/sub>(NH3<\/sub>)2<\/sub>] – Geometrical isomerism<\/li>\n<\/ol>\n
<\/p>\n
Question 4.
\nWhat will be the correct order for the wavelengths of absorption in the visible region for the following:
\n[Ni(NO2<\/sub>)6<\/sub>]