{"id":10651,"date":"2019-12-09T14:41:22","date_gmt":"2019-12-09T09:11:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10651"},"modified":"2021-07-02T15:43:16","modified_gmt":"2021-07-02T10:13:16","slug":"2nd-puc-chemistry-question-bank-chapter-9","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-9\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds"},"content":{"rendered":"

You can Download Chapter 9 Coordination Compounds Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds<\/h2>\n
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2nd PUC Chemistry Coordination Compounds NCERT Textbook Questions and Answers<\/h3>\n

Question 1.
\nExplain the bonding in coordination compounds in terms of Werner\u2019s postulates.
\nAnswer:
\nWerner\u2019s postulates explain the bonding in coordination compounds as follows:
\n(i) A metal exhibits two types of valencies ‘ namely, primary and secondary valencies.
\nPrimary valencies are satisfied by negative ions while secondary valences are satisfied by both negative and neutral ions.
\n(In modem terminology the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion)<\/p>\n

(ii) A metal ion has a definite number of secondary valences around the central atom. Also, these valences project in a specific direction in the space assigned to the definite geometry of the coordination compound.<\/p>\n

(iii) Primary valences are usually ionisable, while secondary valences are non ionisable.<\/p>\n

Question 2.
\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:
\n\"2nd
\nBoth compounds i.e. FeSO4<\/sub> (NH4<\/sub>)2<\/sub>SO4<\/sub>.6H2<\/sub>O and [Cu(NH3<\/sub>)4<\/sub>]SO4<\/sub>.5H2<\/sub>O fall under the category of addition compounds with only one major difference i.e. the former is an example of a double salt, while the latter is a coordination compound. A double salt is an addition compound that is stable in the solid state bnK that which breaks up into its constituent ions in the dissolved state. These compounds exhibit individual properties of their constituents for e.g. FeSO4<\/sub> (NH4<\/sub>)2<\/sub>SO4<\/sub>.6H2<\/sub>O breaks into Fe2+<\/sup> NH4<\/sub>+<\/sup>, and SO4<\/sub>2-<\/sup>\u00a0ions. Hence, it gives a positive test for Fe2+<\/sup> ions. A coordination compound is an addition compound which retains its identity In the solid as well as in the dissolved state. However, the individual properties of the constituents are cost. This happens because [Cu(NH3<\/sub>)]4<\/sub>]SO4<\/sub>.5H2<\/sub>O does not show the test for Cu2+<\/sup> The ions present in the solution of [Cu(NH3<\/sub>)4<\/sub>]SO4<\/sub>.5H2<\/sub>O are [Cu(NH3<\/sub>)4<\/sub>]2+<\/sup> and SO4<\/sub>2-<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 3
\nExplain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic complexes;
\nAnswer:
\n(i) Coordination entity:
\nA coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion is surrounded by a suitable number of neutral molecules or negative ion (called, ligands).
\nFor examples:
\n[Ni(NH3<\/sub>)6<\/sub>]2+<\/sup>, [Fe(CN)6<\/sub>]4+<\/sup> = cationic complex
\n[PtCl4<\/sub>]2-<\/sup>, [Ag(CN)2<\/sub>]–<\/sup> – anionic complex
\n[Ni (CO)4<\/sub>], [CO (NH3<\/sub>)4<\/sub>Cl2<\/sub>] = neutral complex<\/p>\n

(ii) ligand:
\nThe neutral molecules or negatively charged ions that surround the metal atom in a coordination entity or a coordinal complex are known as ligands.
\nFor example
\n\"2nd
\nCl–<\/sup>, OH–<\/sup>. Ligands are usually polar in nature and possess at least one unshared pair of valence electrons.<\/p>\n

(iii) Coordination number:
\nThe total number of ligands (either neutral molecules or negative ions) that get attached to the central metal atom in the coordination sphere is called the coordination number of the central metal atom. It is also referred to as its ligaricy.
\nFor example:
\n(a) In the complex K2<\/sub>[PtCl6<\/sub>], there are six chloride ions attached to Pt in the coordinate sphere. Therefore, the coordination number of Pt is 6.
\n(b) similarly, in the complex [Ni (NH3<\/sub>)4<\/sub> Cl2<\/sub>], the cooordination number of the central atom (Ni) is 6.<\/p>\n

(iv) Coordination polyhedron:
\nCoordination polyhedrons, about the central atom can be defined as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere.
\nFor example:
\n(a) Square planar
\n\"2nd
\n(b) Square planar
\n\"2nd<\/p>\n

(v) Homoleptic complexes:
\nThese are those complexes in which the metal ion is bound to only one kind of a donor group For e.g.:
\n[CO (NH3<\/sub>)6<\/sub>]3+<\/sup>
\n[pt Cl4<\/sub>]2<\/sup><\/p>\n

(vi) Heteroleptic complexes:
\nHeteroleptic complexes are those complexes where the central metal is bound to more than one type of a donor group.
\nFor e.g.:
\n[CO(NH3<\/sub>)4<\/sub>C2<\/sub>]+<\/sup>
\n[CO(NH3<\/sub>)5<\/sub> Cl]2+<\/sup><\/p>\n

Question 4.
\nWhat is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
\nAnswer:
\nA ligand may contain one or more unshared pairs of electrons which are called the donor sites of ligands. Now depending on the number of these donor sites, ligands can be classified as follows.
\n(a) Unidentate ligands: Ligands with only one donor sites are called unidentate ligands.
\n\"2nd
\n(b) Didentate ligands: Ligands that have two donor sites are called didentate ligands.
\nFor e.g.:
\n(a) Ethane – 1, 2 – diamine.
\n\"2nd
\n(b) enolic acid
\n\"2nd
\nc) Ambidentate ligand:
\nLigands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands.
\nFor Example :
\n(a)
\n\"2nd
\n(b)<\/p>\n

\"2nd<\/p>\n

Question 5.
\nSpecify the oxidation numbers of the metals in the following coordination entities:
\n(i) [CO(H2<\/sub>O)(CN)(en)2<\/sub>]2+<\/sup>
\n(ii) [COBr2<\/sub>(en)2<\/sub>]+<\/sup>
\n(iii) [PtCl4<\/sub>]2-<\/sup>
\n(iv) K3<\/sub>[Fe(CN)6<\/sub>]
\n(v) [Cr(NH3<\/sub>)3<\/sub>Cl3<\/sub>]
\nAnswer:
\n(i) [CO(H2<\/sub>O) (CN) (en)2<\/sub>]2+<\/sup>
\nLet the oxidation number of CO be x.
\nThe charge on the complex is + 2<\/p>\n

\"2nd<\/p>\n

(ii) [PtCl4<\/sub>]2-<\/sup>
\nx + 4 (-1) = -2 x = +2<\/p>\n

(iii) [COBr2<\/sub> (en)2<\/sub>]+<\/sup>
\nx + 2 (-1) + 2.(0) = + 1
\nx = + 3<\/p>\n

(iv) K3<\/sub>[Fe(CN)6<\/sub>]
\nx + 6 (-1) = -3
\nx = + 3<\/p>\n

(v) [Cr (NH3<\/sub>)3<\/sub> Cl3<\/sub>]
\nx + 3 (0) +3 (-1) = 0
\nx – 3 = 0
\nx = 3<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nUsing IUPAC norms write the formula for the following:
\n(i) Tetrahydroxozincate (II)
\n(ii) Potassium tetrachloridopalladate(II)
\n(iii) piamminedichloridoplatinum(II)
\n(iv) Potassium tetracyanonickelate(II)
\n(v) Pentaamminenitrito-O-cobalt(III)
\n(vi) Hexaamminecobalt(III) sulphate
\n(vii) Potassium tri(oxalato)chromate(III) (viii) Hexaammineplatinum(IV)
\n(ix) Tetrabromidocuprate(II)
\n(x) Pentaamminenitrito-N-cobalt(III)
\nAnswer:
\n(i) [Zn (O4<\/sub>)]2-<\/sup>
\n(ii) K2 <\/sub>[Pd Cl4<\/sub>]
\n(iii) [Pt (NH3<\/sub>)2<\/sub> Cl2<\/sub>]
\n(iv) K2<\/sub>[Ni(CN)4<\/sub>]
\n(v) [CO (NH3<\/sub>)5<\/sub> (ONO)]2+<\/sup>
\n(vi) [CO (NH3<\/sub>)6<\/sub>]2<\/sub>(SO4<\/sub>)3<\/sub>
\n(vii) K3<\/sub> [Cr (C2<\/sub>O4<\/sub>)3<\/sub>]
\n(viii) [Pt (NH3<\/sub>)6<\/sub>]4+<\/sup>
\n(ix) [Cu (Br4<\/sub>)2-<\/sup>
\n(x) [CO (NO2<\/sub>) (NH3<\/sub>)5<\/sub>]2+<\/sup><\/p>\n

Question 7.
\nUsing IUPAC norms write the systematic names of the following:
\n(i) [CO(NH3<\/sub>)6<\/sub>]Cl3<\/sub>
\n(ii) [Pt(NH3<\/sub>)2<\/sub>Cl(NH2<\/sub>CH3<\/sub>)]Cl
\n(iii) [Ti(H2<\/sub>O)6<\/sub>]3+<\/sup>
\n(iv) [CO(NH3<\/sub>)4<\/sub>Cl(NO2<\/sub>)]Cl
\n(v) [Mn(H2<\/sub>O)6<\/sub>]2+<\/sup>
\n(vi) [NiCl4<\/sub>]2-<\/sup>
\n(vii) [Ni(NH3<\/sub>)6<\/sub>]Cl2<\/sub>
\n(viii) [CO(en)3<\/sub>]3+<\/sup>
\n(ix) [Ni(CO)4<\/sub>]
\nAnswer:
\n(i) Hexaamminecobalt (III) chloride.
\n(ii) Diamine chlorido (methylamine) platinum (II) chloride.
\n(iii) Hexaaquatitanium (III) ion.
\n(iv) Tetraamminichloridonitrite – N – cobalt (HI) chloride.
\n(v) Flexaquamanganese (II) ion.
\n(vi) Tetrachloridononickeiate (II) ion.
\n(vii) Hexaammmenickei (11) chloride,.
\n(viii)Tris (ethane -1,2 – diammine) cobalt (III) ion.
\n(ix) Tetracarbonylnickel (O)<\/p>\n

Question 8.
\nList various types of isomerism possible for coordination compounds, giving an example of each.
\nAnswer:
\n\"2nd
\n(a) Geometric isomerism:
\nThis type of isomerism is common in heteroleptic complexes It arises dire to the different possible geometric arrangements of the li sands.
\nFor Example
\n\"2nd<\/p>\n

(b) Optical isomerism:
\nThis type of isomerism arises in chiral molecules Isomers are mirror images of each other and are non-superimposable.
\n\"2nd<\/p>\n

(c) Linkage isomerism:
\nThis type of isomerism is found in complexes that contain ambidentate ligands.
\nFor example:
\n[CO (NH3<\/sub>)5<\/sub> (NO2<\/sub>)]Cl2<\/sub> and
\n[CO (NH3<\/sub>)5<\/sub> (ONO)] Cl2<\/sub><\/p>\n

(d) Coordination isomerism:
\nThis type of isomerism arises when the ligands are inter changed between cationic and anionic entities of different metal ions present in the complex.
\n[CO (NH3<\/sub>)6<\/sub>] [Cr (CN)6<\/sub>] and
\n[CO (CN)6<\/sub>][Cr (CN)6<\/sub>]<\/p>\n

(e) Ionisation isomerism:
\nThis type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. Thus, complexes that have the same composition, but furnish different ions when dissolved in water are called ionisation isomers.
\nFor example
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nHow many geometrical isomers are possible in the following coordination entities?
\n(i) [Cr(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n(ii) [CO(NH3<\/sub>)3<\/sub>Cl3<\/sub>]
\nAnswer:
\n(i) For [Cr(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>\u00a0No geometric isomer is possible as it is a bidentate ligand.
\n\"2nd
\n(ii) [CO(NH3<\/sub>)3<\/sub>Cl3<\/sub>]
\nTwo geometrical isomers are possible.
\n\"2nd<\/p>\n

Question 10.
\nDraw the structures of optical isomers of:
\n(1) [Cr(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n(ii) [PtCl2<\/sub>(en)2<\/sub>]2+<\/sup>
\n(iii) [Cr(NH3<\/sub>)2<\/sub>Cl2<\/sub>(en)]+<\/sup>
\nAnswer:
\n(i) [Cr(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n\"2nd
\n(ii) [PtCl2<\/sub>(en)2<\/sub>]2+<\/sup>
\n\"2nd<\/p>\n

(iii) [Cr(NH3<\/sub>(en)2<\/sub>]2+<\/sup>
\n\"2nd
\n\"2nd<\/p>\n

Question 11.
\nDraw all the isomers (geometrical and optical) of:
\n(i) [COCl2<\/sub>(en)2<\/sub>]+<\/sup>
\n(ii)[CO(NH3<\/sub>)Cl(en)2<\/sub>]2+<\/sup>
\n(iii) [CO(NH3<\/sub>)2<\/sub>Cl2<\/sub>(en)]+<\/sup>
\nAnswer:
\n(i) [COCl2<\/sub>(en)2<\/sub>]+<\/sup>
\n\"2nd
\n\"2nd
\n(ii)[CO(NH3<\/sub>)Cl(en)2<\/sub>]2+<\/sup>
\n\"2nd
\n(iii) [CO(NH3<\/sub>)2<\/sub>Cl2<\/sub>(en)]+<\/sup>
\n\"2nd<\/p>\n

Question 12.
\nWrite all the geometrical isomers of [Pt(NH3<\/sub>)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
\nAnswer:
\n[Pt (NH3<\/sub>) (Br) (eC) (Pg)]
\n\"2nd<\/p>\n

From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerisation. They do so only in presence of unsymmetric chelating agents.<\/p>\n

Question 13.
\nAqueous copper sulphate solution (blue in colour) gives:
\n(i) a green precipitate with aqueous potassium fluoride and
\n(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
\nAnswer:
\nAqueous CuSO4<\/sub> exists as [Cu(H2<\/sub>O)4<\/sub>]SO4<\/sub>. It is blue in colour due to presence of [CU(H2<\/sub>O)4<\/sub>]2+<\/sup> ions.
\n(i) When KF is added
\n\"2nd
\n(ii) When KCl is added.<\/p>\n

[Cu (H2<\/sub>O)4<\/sub>]2+<\/sup> 4Cl–<\/sup>\u00a0<\/sub>\u2192 [CuCl4<\/sub>]2-<\/sup> + 4H2<\/sub>O
\nIn both the cases, the weak field ligand water is replaced by the F–<\/sup> and Cl–<\/sup> ions.<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\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:
\nCuSO4(aq)<\/sub> + 4KCN (aq)<\/sub> \u2192 K2<\/sub>Cu (CN)4<\/sub>](aq)<\/sub> + K2<\/sub>SO4<\/sub>
\ni.e. [Cu (H2<\/sub>O)4<\/sub>]2+<\/sup> + 4CN–<\/sup> \u2192 [Cu (CN)4<\/sub>]2-<\/sup> + 4H2<\/sub>O
\nThus, the coordination entity formed in the process is K2<\/sub> [Cu (CN)4<\/sub>]. K2<\/sub> [Cu (CN)4<\/sub>] is a very stable complex which does not ionize to give Cu2+<\/sup> ions when added to water. Hence Cu2+<\/sup> ions are not precipitated when H2<\/sub>S(g)<\/sub> is passed through the solution.<\/p>\n

Question 15.
\nDiscuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
\n(i) [Fe(CN)6<\/sub>]4-<\/sup>
\n(ii) [FeF6<\/sub>]3-<\/sup>
\n(Hi) [CO(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n(iv) [COF6<\/sub>]3-<\/sup>
\nAnswer:
\n(i) [Fe(CN)6<\/sub>]4-<\/sup>
\nIn the above coordination complex. Iron exists in +2 oxidation state.
\nFe2+<\/sup>: electronic configuration is 3d6<\/sup>
\n\"2nd
\nAs CN–<\/sup> is a strong field ligand, it causes pairing of the unpaired 3d electrons.
\n\"2nd
\nSince there are six ligands aound the central metal ion, the most feasible hybridisation is d2<\/sup>sp3<\/sup> 6 electrons pairs from CN–<\/sup> ions occupy the six hybrid d2<\/sup>sp3<\/sup> orbitais .
\n\"2nd
\nHence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).<\/p>\n

(ii) [Fe<\/sub>F6<\/sub>]3-<\/sup>
\nOxidation state of Fe is + 3
\nOrbitais of Fe+3<\/sup> ion.
\n\"2nd<\/p>\n

There are 6F–<\/sup> ions. Thus, it will undergo d2<\/sup>Sp3<\/sup>\u00a0or sp3<\/sup>d2<\/sup> hybridisation. As F–<\/sup> is a weak ligand, it does not cause pairing of electrons in the 3d orbitals. Hence the most feasible hybridisation is Sp3<\/sup>d2<\/sup>.
\n\"2nd
\nHence the geometry of the complex is found to be octahedral.<\/p>\n

(iii) [CO(C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\nCobalt exists + 3 oxidation state in the complex.
\n\"2nd
\nOxalate is a weak field ligand. Therefore, it cannot cause pairing of the 3d orbital electrons.
\nAs there are 6 ligands, hybridisation has to be either Sp3<\/sup>d2<\/sup> or d2<\/sup>Sp3<\/sup> hybridization.
\nAs there are no pairing of electrons Sp3<\/sup>d2<\/sup> is the most stable hybridisation.<\/p>\n

\"2nd
\nOxalate being a bidentate ligand donates 2 pairs of electrons each.
\nHence, the geometry of the complex is found to be octahedral.<\/p>\n

(iv) [COF6<\/sub>]3-<\/sup>
\ncobalt exists in the +3 oxidation state.
\n\"2nd
\nAgain, fluoride ion is a weak ligand. It cannot cause the pairing of the 3d electrons. Sp3<\/sup>d2<\/sup> hybridised orbitals of CO3+<\/sup> ion are
\n\"2nd
\nHence, the geometry of the complex is octahedral and paramagnetic.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nDraw figure to show the splitting of d orbitals in an octahedral crystal field.
\nAnswer:
\n\"2nd<\/p>\n

The splitting of d orbitals in an octahedral field takes place in such a way that dx2<\/sup>-y2<\/sup>, dz2<\/sup> experience a rise in energy and form the e.g. level, dry, dyz, dzx experience a fall in energy and form t2g level.<\/p>\n

Question 17.
\nWhat is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
\nAnswer:
\nA spectrochemical series is the arrangement of common ligands in the increasing order of their crystal field splitting energy (CSFE) values the ligands present on the RHS of the series are strong fields ligands while that on the LHS are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.
\nI–<\/sup> < Br–<\/sup> < S2-<\/sup> < SCN–<\/sup>< Cl–<\/sup> < N3<\/sub> < F–<\/sup> < OH– <\/sup>< C2<\/sub>O4<\/sub>2-<\/sup> ~ H2<\/sub>O
\n< NCS–<\/sup> ~ H–<\/sup> < CN–<\/sup> < NH3<\/sub>, < en ~ SO3<\/sub>2-<\/sup>2- < NO2<\/sub>–<\/sup> < phen < CO<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nWhat is crystal field splitting energy? How does the magnitude of \u2206o<\/sub> decide the actual configuration of d orbitals in a coordination entity?
\nAnswer:
\nThe degenerate d-orbitals (in a spherical field environment) split two level i.e. eg<\/sub> and t2<\/sub>g in the presence of ligands. The splitting of the degenerate orbitals in the presence of ligands is called crystal field splitting and the energy difference between the two levels (e and t2<\/sub>g) is called the crystal field splitting energy. It is denoted by \u2206o<\/sub>. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has filled in the three t2<\/sub>g orbitals, the filling of the electrons takes place in 2 ways.<\/p>\n

It can enter the orbital (giving) rise to t3<\/sup>g eg<\/sub>, like electronic configuration on the pairing of the electrons can take place in the t2<\/sub>g orbitals
\n(giving rise to t4<\/sup>2<\/sub>g eg0<\/sup> like electronic configuration). If the \u2206o<\/sub> value of a ligand is less than the pairing energy, then the electrons enter the eg orbital. On the other hand, if the \u2206o<\/sub> value of a ligand is more than the pairing energy, then the electrons enter the t2<\/sub>g orbitals.<\/p>\n

Question 19.
\n[Cr(NH3<\/sub>)6<\/sub>]3+<\/sup> is paramagnetic while [Ni(CN)4<\/sub>]2-<\/sup> is diamagnetic. Explain why?
\nAnswer:
\nCr is in the +3 oxidation state i.e. d3 configuration. Also, NH3<\/sub> is a weak field ligand that does not cause the pairing of the electrons in the 3d orbitals.
\n\"2nd
\nTherefore, it undergoes d2<\/sup>sp3<\/sup> hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, ibis paramagnetic in nature.
\nIn [Ni CCN)4<\/sub>]2-<\/sup>, Ni exists in the +2 oxidation state.
\ni.e, d8<\/sup> Configuration.
\n\"2nd<\/p>\n

CN–<\/sup> is a strong field ligand. It causes the pairing of 3d orbital electrons. Then Ni2+<\/sup> undergoes dsp2<\/sup> hybridisation.
\n\"2nd
\nAs there are no unpaired ejecirons, it is diamagnetic.<\/p>\n

Question 20.
\nA solution of [Ni(H2<\/sub>O)6<\/sub>]2+<\/sup> is green but a solution of [Ni(CN)4<\/sub>]2-<\/sup> is colourless. Explain.
\nAnswer:
\nIn [Ni(H2<\/sub>O)6<\/sub>]2+<\/sup> is a weak field ligand. Therefore there are unpaired electrons in Ni2+<\/sup>. In this complex the cl electrons from the lower energy level can be excited to the higher energy level is the possib.lity of d- d transition is present. Hence [Ni(H2<\/sub>O)6<\/sub>]2+<\/sup>\u00a0is coloured.<\/p>\n

In [Ni(CN)4<\/sub>]2-<\/sup> CN is a strong field ligand. Therefore, d- d transition is not possible due to absence of unpaired electron (CN initiates pairing of d elcctrons ). Hence it is colourless.<\/p>\n

Question 21.
\n[Fe(CN)6<\/sub>]4-<\/sup> and [Fe(H2<\/sub>O)6<\/sub>]2-<\/sup> are of different colours in dilute solutions. Why?
\nAnswer:
\nThe colour of a particular coordination compound depends on the magnitude of crystal field splitting energy \u2206. This CFSE in turn depends on the nature of the ligand. In case of [Fe (CN)6<\/sub>)4-<\/sup> and [Fe (H2<\/sub>O)6<\/sub>]2-<\/sup> the colour differs because there is a difference in the CFSE. Now, CN–<\/sup> is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that absorption of energy for the extra d- d transition also differed. Hence, the transmitted colour also different.<\/p>\n

Question 22.
\nDiscuss the nature of bonding in, metal carbonyls.
\nAnswer:
\nThe metal carbon bonds in’ metal carbonyls have both \u03c3 and \u03c0 characteristics. A \u03c3 bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of metal. A \u03c0 bond is formed by the donation of a pair of vacant anti-bonding \u03c0 orbital (also known as back bonding of the carbonyl group). The \u03c3 bond strengthens the \u03c0 bond and vice-versa. This synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and metal.
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nGive the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:
\n(i) K3<\/sub>[CO(C2<\/sub>O4<\/sub>)3<\/sub>]
\n(ii) cis-[Cr(en)2<\/sub>Cl2<\/sub>]Cl
\n(iii) (NH4<\/sub>)2<\/sub>[COF4<\/sub>]
\n(iv) [Mn(H2<\/sub>O)6<\/sub>]SO4<\/sub>
\nAnswer:
\n(i) K3<\/sub>[CO(C2<\/sub>O4<\/sub>)3<\/sub>]
\nThe central metal ion is CO
\ncoordination number: 6
\noxidation state: +3
\nd orbital occupation : t6<\/sup>2g<\/sub> eg0<\/sup><\/p>\n

(ii) cis-[Cr(en)2<\/sub>Cl2<\/sub>]Cl
\ncentral metal: Cr
\ncoordination number: 6
\noxidation state:+3 .
\nd orbital occupation : t3<\/sup>2g<\/sub> eg0<\/sup><\/p>\n

(iii) (NH4<\/sub>)2<\/sub>[COF4<\/sub>]
\ncentral metal: Co oxidation state: +2 coordination number: 4
\nd orbital occupation : eg4<\/sup> t3<\/sup>2g<\/sub><\/p>\n

(iv)[Mn(H2<\/sub>O)6<\/sub>]SO4<\/sub>
\nCentral metal: Mn
\ncoordination number: 6
\noxidation state : + 2
\nd orbital occupancy : t3<\/sup>2g<\/sub> eg2<\/sup><\/p>\n

Question 24.
\nWrite down the IUPAC name for-each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
\n(i) K[Cr(H2<\/sub>O)2<\/sub>(C2<\/sub>O4<\/sub>)2<\/sub>].3H2<\/sub>O
\n(ii) [CO(NH3<\/sub>)5<\/sub>Cl.]Cl2<\/sub>
\n(iii) CrCl3<\/sub>(py)3<\/sub>
\n(iv) Cs[FeCl4<\/sub>]
\n(v) K4[Mn(CN)6<\/sub>]
\nAnswer:
\n(i) Potassium diaquadioxatatochromate (III) trihydrate oxidation state of chromium = +3
\nElectronic configuration: 3d3<\/sup>: t3<\/sup>2g<\/sub>
\ncoordination number: 6
\nShape: octahedral
\n\"2nd
\n\"2nd<\/p>\n

(ii) [Co (NH3<\/sub>)5<\/sub>Cl]Cl2<\/sub>
\nPentaamminechloridocobalt (III) chloride
\noxidation state =+3
\ncoordination number = 6
\nshape: octahedral
\nelectronic configuration : d6<\/sup>: t6<\/sup>2g<\/sub>
\n\"2nd
\nMagnetic moment = 0 (\u2235 n = 0)<\/p>\n

(iii) CrCl3<\/sub>(py)3<\/sub>
\nTrichloridopyridinechromium (III)
\noxidation state = +3
\nelectronic configuration : d3<\/sup>: t3<\/sup>2g<\/sub>
\ncoordihation number = 6
\nshape = octahedral
\nStereochemistry:
\n\"2nd<\/p>\n

Both isomers are optically active. Therefore, a total of 4 isomers exist.
\n\"2nd<\/p>\n

(iv) Cs [Fe Cl4<\/sub>]
\nIVPAC name: Caesium tetrachloroferrate (III)
\noxidation state = +3
\ncoordination number = 4
\nshape: tetrahedral
\nstereochemistry: optically inactive
\nMagnetic moment
\n\"2nd<\/p>\n

(v) K4<\/sub>[Mn(CN)6<\/sub>]
\nPotassium hexacyanomanganate (II)
\noxidation state =+2
\nelectronic configuration = d5<\/sup> : t5<\/sup>2g
\ncoordination number = 6
\nshape : octahedral
\nstereochemistry: optically inactive
\nMagnetic moment
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 25.
\nWhat is meant by stability of a coordination compound in solution? State the factor^ which govern stability of complexes.
\nAnswer:
\nThe stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium stability can be expressed quantitatively in terms of stability constant pr formation constant
\n\"2nd
\nFor this reaction, the greater the value of the stability constant, the greater is the proportion of ML3<\/sub> in the solution.
\nstability can be two types<\/p>\n

(a) Thermodynamic stability:
\nThe extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.<\/p>\n

(b) Kinetic stability:
\nThis helps in determining the speed with which the transformation will occur to attain the state of equiftbrium.
\nFactors that affect the stability of a complex are: ~ ;
\n(i) charge on the central metal ion: The greater the charge on the central metal ion, the greater in the stability of the complex.
\n(ii) Basic nature of ligand: A more basic ligand will form a more stable complex.<\/p>\n

(c) Presence of chelate rings: Chelation increases the stability of complexes<\/p>\n

Question 26.
\nWhat is meant by the chelate effect? Give an example.
\nAnswer:
\nWhen a ligand attaches to the metal ion in a manner that form’s a ring, then the metal ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings more stable than complexes without rings. This is known as chelate effect.
\nFor example:
\n\"2nd
\n\"2nd<\/p>\n

Question 27.
\nDiscuss briefly giving an example in each case the role of coordination compounds in:
\n(i) biological systems
\n(ii) medicinal chemistry and
\n(iii) analytical chemistry
\n(iv) extraction\/metaliurgy of metals.
\nAnswer:
\n(i) Role of coordination compounds in biological systems:
\nWe know that photosynthesis is possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example the oxygen – carrier of blood, i.e haemoglobin is a coordination compound of iron.<\/p>\n

(ii) Role of coordination compounds in Medicinal chemistry: Certain coordination compounds of platinum (for example cis-platin) are used for inhibiting the growth of tumors.<\/p>\n

(iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.<\/p>\n

(iv) Role of coordination compounds in interaction or metallurgy of metals: The process of extraction of some of the metals from their ores involves the formation of complexes. For example in aqueous solution, gold combines with cyanide ions to form [Au (CN)2<\/sub>]. From this solution, gold is later extracted by the addition of Zn metal.<\/p>\n

\"KSEEB<\/p>\n

Question 28.
\nHow many ions are produced from the complex CO(NH3<\/sub>)6<\/sub>Cl2<\/sub> in solution?
\n(i) 6
\n(ii) 4
\n(iii) 3
\n(iv) 2
\nAnswer:
\n(iii) The given complex can be written as<\/p>\n

[CO(NH3<\/sub>)6<\/sub>]Cl2<\/sub> Thus, [CO (NH3<\/sub>)6<\/sub>]+<\/sup> along with two Cl–<\/sup> ions are produced.<\/p>\n

Question 29.
\nAmongst the following ions which one has the highest magnetic moment value?
\n(i) [Cr(H2<\/sub>O)6<\/sub>]3+<\/sup>
\n(ii) [Fe(H2<\/sub>O)6<\/sub>]2+<\/sup>
\n(iii) [Zn(H2<\/sub>O)6<\/sub>]2+<\/sup>
\nAnswer:
\n( i) n = 3
\n\"2nd<\/p>\n

(ii) n = 4
\n\"2nd<\/p>\n

(iii) n = 0
\n\u03bc = 0
\nHence [FeCH2<\/sub>O)6<\/sub>]2+<\/sup>\u00a0has the highest magnetic moment value.<\/p>\n

Question 30.
\nThe oxidation number of cobalt in K[CO(CO)4<\/sub>] is
\n(i) +1 (ii) +3 (iii) -1 (iv) -3
\nAnswer:
\nWe know that CG is a neutral ligand and K carries a charge of + 1. Therefore, the complex is written as K+<\/sup> [Co (CO)4<\/sub>]–<\/sup>. Therefore, the oxidation number of Co in the given complex is -1.
\nHence (III) is correct.<\/p>\n

Question 31.
\nAmongst the following, the most stable complex is
\n(i) [Fe(H2<\/sub>O)6<\/sub>]3+<\/sup>
\n(ii) [Fe(NH3<\/sub>)6<\/sub>]3+<\/sup>
\n(iii) [FeiC2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n(iv) [FeCl6<\/sub>]3-<\/sup>
\nAnswer:
\nWe know that the stability of a complex substance increases by chelation. Therefore, the most stable complex is [Fe (C2<\/sub>O4<\/sub>)3<\/sub>]3-<\/sup>
\n\"2nd<\/p>\n

Question 32.
\nWhat will be the correct order for the wavelengths of absorption in the visible region for the following:
\n[Ni(NO2<\/sub>)6<\/sub>]4-<\/sup>, [Ni(NH3<\/sub>)6<\/sub>]2+<\/sup>, [Ni(H2<\/sub>O)6<\/sub>]2+<\/sup>?
\nAnswer:
\nThe central metal ion in all the three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series as follows.
\nH2<\/sub> O < NH3<\/sub> < NO2<\/sub>–<\/sup>
\nThus, the amount of crystal field splitting observed will be in the following order.
\n\"2nd
\nHence, the wavelengths of absorption in the visible region will be in the order
\n[Ni(H2<\/sub>O)6<\/sub>]2+<\/sup> > [Ni(NH3<\/sub>)6<\/sub>]2+<\/sup> > [Ni (NO2<\/sub>)6<\/sub>]4-<\/sup><\/p>\n

\"KSEEB<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

You can Download Chapter 9 Coordination Compounds Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds 2nd PUC Chemistry Coordination Compounds NCERT Textbook Questions and Answers Question …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10651"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10651"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10651\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10651"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10651"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10651"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}