Bihar Board - Class 12 Chemistry - Chapter 9: Coordination Compounds Long Answer Question
Long Question Answer
Q. 1. Write the formula for the following coordination compounds :
(i) Tetra-amine di-aqua cobalt (III) chloride.
(ii) Potassium tetracyanonickelate (II).
(iii) Amine bromochloride a nitrito-N-planinate (II)
(iv) Iron (III) hexacyanoferrate (II).
Answer ⇒ (i) Tetraamine di-aqua cobalt (III) chloride [Co(NH3)4(H2O)2]Cl3.
(ii) Potassium tetracyanonickelate (II) -K2[Ni(CN)4]
(iii) Amine bromido chlorido nitrito-N-platinate (II)- Pt(NH3)BrCl(NO2)]
(iv) Iron (III) hexacyanoferrate (II) Fe4[Fe(CN)6]3+
Q.2. Write the IUPAC names of the following coordination compounds :
(a) [CO(NH3)6]Cl3 ,
(b) [CO(NH3)5Cl]Cl2,
(c) K3[Fe(CN)6],
(d) K3[Fe(C2O4)],
(e) K2[PdCl4] ,
(f) Pt(NH3)2Cl(NH2CH3)]Cl.
Answer ⇒ (a) Hexa-amino cobalt (III) chloride.
(b) Penta amine chlorido cobalt (III) chloride.
(c) Potassium hexacyano ferrate (III).
(d) Potassium trioxalatoferrate (III).
(e) Potassium tetrachloride palladate (II).
(f) Diammine chlorido (methylamine) platinum (II) chloride.
Q. 3. Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers :
(i) K[Cr(H2O)2(C2O4)2],
(ii) [CO(en)3Cl3],
(iii) [CO(NH3)5(NO2)](NO3)2 ,
(iv) [Pt(NH3)H2O Cl2].
Answer ⇒ (i) Both geometrical (cis, trans) and optical isomers for Cis can exist.
(ii) Two optical isomers can exist.
(iii) There are 10 possible isomers. There are geometrical, ionization and linkage isomers possible.
(iv) Geometrical (Cis. trans) isomers can exist.
Q. 4. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Answer ⇒ In the presence of NH3, the 3d electrons pair up leaving two d'orbitals empty to be involved in d2sp3. hybridisation forming inner orbital complex in case of [Co(NH3)6]3+
In [Ni(NH3)6]2+ -, Ni is in +2 oxidation state and has d8 configuration, hybridisation involved in sp3d2 forming an outer orbital complex.
Q. 5. Write the formulas for the following coordination compounds:
(i) Tetra amino aqua chloro cobalt (HI) chloride.
(ii) Potassium tetra hydroxozincate (II).
(iii) Potassium tris oxalato aluminate (III).
(iv) Dichloro bis (ethane-1, 2-diamine cobalt(III).
(v) Tetracarbonyl nickel (O).
Answer ⇒ (i) [CO(NH3)4(H2O)Cl]Cl2
(ii) K2[Zn(OH)4]
(iii) K3[Al(C2O4)3]
(iv) [CoCl2(en)]+
(v) [Ni(CO)4]
Q. 6. Write the IUPAC names of the following Coordination compounds :
(i) [Pt(NH3)2Cl(NO2)]
(ii) K3[Cr(C2O4)3]
(iii) [CoCl2(en)2]Cl
(iv) [Co(NH3)6 (CO3)Cl
(v) Hg[Co(SCN)4].
Answer ⇒ (i) Diamine chloronitrito-N-platinum (II).
(ii) Potassium tris oxalato chromate (III).
(iii) Dichloro bis ethane-1 , 2-diamine cobalt (III) chloride.
(iv) Hexa amine carbonato cobalt (III) chloride.
(v) Mercury tetrathiocyanato cobaltate (III).
Q. 7. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer ⇒ Postulates are :
(i) In coordination compounds metals show two types of linkages (valencies)-primary and secondary.
(ii) The primary valencies are normally ionisable and are satisfied by negative ions.
(iii) The secondary valencies are non-ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a neutral.
(iv) The ions/ groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different numbers.
In the modern formulation, such spatial arrangements are called co-ordination entities or complexes and the ions outside the square bracket are called counter ions.
He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus
[CO(NH3)6]3+ , [COCl(NH3)5]2+ and [CoCl(NH3)4]+ are octahedral while [Ni(CO)4] and [PtCl4]2- are tetrahedral and square planar respectively.
Q. 8. Using IUPAC norms write the formulae of the following:
(i) Hexa amine cobalt (III) sulfate.
(ii) Potassium tetrachloro palladate (II).
(iii) Potassium tri (oxalato) chromate (III).
(iv) Dimine dichloro platinum (II).
(v) Hexa amine platinum (IV).
(vi) Potassium tetracyano nickelate (II).
Answer ⇒ (i) [Co(NH3)2(SO4)3
(ii) K2[PdCl4]
(iii) K3[Cr(C2O4)3]
(iv) [Pt(NH3)2Cl2]
(v) [Pt(NH3)6]4+
(vi) K2[Ni(CN)4]
Q. 9. Using IUPAC norms write the systematic name of the following:
(i) [Co(NH3)6Cl3]
(ii) [Co(NH3)4Cl(NO2)]Cl
(iii) [Ni(NH3)6]Cl2
(iv) [Mn(H2O)6]2+
(v) [Co(en)3]3+
(vi) [Ti(H2O)6]3+
(vii) [NiCl4]2+
(viii) [Ni(Co)4]
Answer ⇒ (i) Hexa-amino cobalt (III) chloride.
(ii) Tetra-amino chloro nitro cobalt (III) chloride.
(iii) Hexammine nickel (II) chloride.
(iv) Hexa-aqua manganese (II) ion.
(v) Tris (1 , 2-ethane diamine) cobalt (III) ion.
(vi) Hexa-aqua titanium (III) ion.
(vii) Tetrachloronickelate (II) ion.
(viii) Tetracarbonyl nickel (O).
Q. 10. Draw the structures of optical isomers of
(i) [Cr(C2O4)3]3-
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Answer ⇒ (i) The two optical isomers of [Cr(C2O4)3]3- are C2O24 is in short ox :
Non-superimposable mirror image on the object
(ii) The two optical isomers of [PtCl2(en)2]2+ are:
(iii) [Cr(NH3)2Cl2(en)]+: The two optical isomers are
Q. 11. Draw all the isomers (geometrical and optical) of
(i) [CoCl2(en)2]+
(ii) [Co(NH3)Cl(en)2]2+
(iii) Co(NH3)2Cl2(en)]+
Answer ⇒ (i) [CoCl2(en)2]+
Geometrical isomers (cis and trans) of [CoCl2(en)2]
(ii) [Co(NH3)Cl(en)2]2+
Optical isomers d and l of cis [CoCl2(en)2]
Geometrical isomers cis and trans of [Co(NH3)Cl(en)2].
(iii) Co(NH3)2Cl2(en)]+
Q. 12. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory.
(i) [Fe(CN6]4-
(ii) [FeF6]3-
(iii) [Co(OX)3]3-
(iv) [CoF6]3-
Answer ⇒ (i) [Fe(CN6]4-
Outer electronic configuration of iron (Z=26) the ground state is 3d64s2 . Iron in this complex is in +2 oxidation state. Iron achieves +2 oxidation state by the loss of two 4s electrons. The resulting Fe2+ the ion has an outer electronic configuration of 3d6.
It has been experimentally observed that this complex is diamagnetic as such has no unpaired electron. To account for this two unpaired electron in the 3d-subshell pair up, this leaving two 3d-orbitals empty. These two vacant 3d-orbitals along with one 4s-orbital and three 4p-orbitals hybridize to give six equivalent d2sp3 hybridized orbitals. Six pairs of electrons, one from each cyanide ion, occupy the six vacant hybrid orbital so produced. The resulting complex ion has an octahedral geometry and is diamagnetic due to the absence of unpaired electrons.
(ii) Bonding in [FeF6]3-
The oxidation state of Fe is +3 and its coordination number is 6. The complex has octahedral geometry and experimental study shows that it is paramagnetic.
The bonding in this entity is explained on the basis of overlap of sp3d2 hybrid orbitals of Fe3+ ion and six lone pair orbitals of cyanide ligands. Five 3d-electrons are unpaired which make it paramagnetic. The outer electronic configuration of Fe3+ ion is 3d5 which is highly stable and no pairing of electrons takes place even in presence of strong field of F-ions. Here, 4d orbitals of Fe3+ (which are empty) are involved.
The entity is strongly paramagnetic due to five unpaired electrons and is an outer orbital complex.
(iii) Bonding in [Co(OX)3]3-
As bidentate oxalate ions approach pairing of electrons in 3d-orbitals take place.
Because all electrons are paired. Hence diamagnetic.
(iv) [CoF6]3- oxidation state of Co is +3
As there are four unpaired electrons, the complex is paramagnetic in nature.
Q. 13. Discuss briefly the role of coordination compounds in
(a) biological systems and (b) analytical chemistry (c) medicinal chemistry and extraction/ metallurgy of metals.
Answer ⇒ (a) Biological system :
(i) Haemoglobin, the red blood cell which acts as an oxygen carrier to different parts of the body, is a complex of iron (II).
(ii) Vitamin B12 is a complex of cobalt metal. Trace metals also function through coordination processes.
(iii) A complex of zinc (II) called the enzyme CPA helps in digestion of food.
(iv) The chlorophyll, the green coloring plant pigment that plays an important role in photosynthesis is a complex of magnesium. In chlorophyll Mg is coordinated to four nitrogens of porphyrin units.
(b) Analytical chemistry :
(i) Multidentate ligand, EDTA (ethylene diamine tetra acetic acid) forms highly stable complexes with metal ions like Ca2+ and Mg2+. This fact is used to estimate the hardness of water by simple titration method using EDTA solution.
(ii) A confirmatory test for the detection of copper (II) involves the formation of a deep-blue coloured complex, [Cu(NH3)4]2+. in addition to an ammonia solution to a solution of copper (II) salt.
Cu2+ + 4NH3+ [Cu(NH3)4]2+]
Tetra amine copper (II) ion (Deep blue)
(iii) The separation of group 1 precipitate of AgCl2 , HgCl2 and PbCl2 involves addition of aqueous ammonia solution to the precipitate, when only silver chloride dissolves, due to the formation of the complex ion. [Ag(NH3)2]+
AgCl +2NH3 [Ag(NH3)2]++Cl-
HgCl2 and PbCl2 do not form complex ions with NH3 and hence, do not dissolve.
(iv) A confirmatory test for nickel consists in addition of a solution of dimethyl glyoxime, when a scarlet-red coloured precipitate is formed due to the formation of a chelate complex.
(c) Medicinal Chemistry : compounds are used in medicines. Some common examples are :
(i) Vitamin B12 used to prevent anemia is a complex compound of cobalt.
(ii) The complex of calcium and EDTA is used in the treatment of lead poisoning. Inside the body calcium in the complex is replaced by lead. The more soluble lead EDTA complex is eliminated in urine.
(iii) The platinum complex cis-[Pt(NH3)2Cl2] known as cis platin is used in the treatment of cancer.
(d) Extraction/metallurgy of Metals :
(i) Extraction metals like silver and gold is carried out by forming their soluble cyanide complex, e.g.,
4Au +16Cn- +6H2O+O24[Au(CN)2]- +12OH-
Dicyano gold (I) (soluble)
The solution containing cyanide complex is then treated with zinc, when gold is precipitated.
2[Au(CN)2]-Zn[Zn(CN)4]2- + 2 Au
(ii) Coordination compounds of silver and gold are used as the constituents of electroplating baths for the controlled delivery of Ag+ and Au ions, during electro refining of these metals.
Q. 14. Give the principle of Habber’s process for the manufacture of NH3. HOW NH3 is converted into HNO3 Write principle and equation only.
Answer⇒ Habber’s process: Ammonia is manufactured by direct combination of nitrogen and hydrogen at very high pressure (about 200 atm ) and low temp (450C) in the presence of a catalyst.
N2 +3H2 ⇌ 2NH3 +QKCal
The reaction is reversible, exothermic and there is contraction in volume. Therefore, maximum yield of ammonia will be at high pressure, low temperature and removal of ammonia from the system to make only forward reaction.
Ammonia is converted to nitric acid by catalytic oxidation of ammonia in presence of platinum asbestos to get nitric oxide.
4NH3+5O2 
Nitric oxide with excess of oxygen is changed into NO2 and then into HNO3
2NO +O2- 2NO2
3NO2 + H2O 2HNO3 +NO
Q. 15. Arrange the following in the order of property indicated for each set :
(a) Fe2 , Cl2 , Br2 , I2 - Increasing order of bond energy.
(b) NH3 , PH3 , AsH3 , SbH3 , BiH3-Increasing base strength.
(c) HOCl , HOBr , HOI - Increasing acid strength
(d) H2O , H2S , NH3 , XeF2-Increasing bond angle
(e) HF , HCl , HBr , HI -Increasing thermal stability
Answer ⇒ (a) I2 < Br2 <F2 <Cl2
(b) BiH3 <SbH3<AsH3 <PH3 <NH3
(c) HOI < HOBr < HOCl
(d) H2S <H2O <NH3 <XeF2
(e) HI < HBr < HCl < HF
Q. 16. Account for the following:
(a) Sulphur in the vapor state exhibits paramagnetism.
(b) Noble gasses form compounds with fluorine and oxygen only.
(c) Bleaching of flowers by chlorine is permanent while that by sulfur dioxide is temporary. Why ?
Answer ⇒ (a) In vapor State (1000 k). Sulphur partly exists as S2 molecule like O2 has two unpaired electrons in the anti-bonding i molecular orbital and exhibits Paramagnetism.
(b) It is because F2 and O2 are the best oxidizing agents
(c) In presence of moisture, chlorine acts as an oxidizing agent and a bleaching agent
Cl2 +H2O HCl + HOCl
. Unstable
HOCl HCl +O Nascent oxygen.
Coloured mater +O Colorless matter.
The bleaching action is permanent due to nascent oxygen. Bleaching action of SO2 The presence of moisture is due to nascent hydrogen.
2SO2 +2H2OH2SO4 +2H
Coloured matter +H colorless matter.
Therefore bleaching action of SO2 is temporary.
Q. 17. Account for the following :
(a) HI is stronger acid than HF.
(b) The electron affinity of fluorine is less than that of chlorine.
(c) Formic acid is stronger than Acetic acid.
(d) Aniline is less basic than methylamine.
(e) PCl5 is known but NCl5 is not known.
(f) Fluorine shows only one oxidation state but the rest of the halogens show different oxidation states.
(g) H2O is a liquid but H2S is gas.
(h) Noble gasses are monatomic in nature.
(i) Bleaching of flowers by chlorine is permanent but that by sulfur dioxide is temporary.
(j) NH3 has higher S.P. than PH3.
(k) Write the structural formula of XeOF4.
Answer ⇒ (a) Due to the bigger atomic size of iodine with respect to fluorine. HI is stronger acid than H-F.
(b) Bond dissociation energy decreases as bond distance increases. The F-F bond dissociation energy is smaller than that of Cl-Cl. Due to this the electron affinity of fluorine to less than that of chlorine.
(c) Acidity of an acid mainly depends upon release of proton from the solution.
(d) One lone pair of electrons is present on the Nitrogen atom in Aniline which is overloaded with an -electron ring. Hence, Nitrogen atom is + only charged and not in the shape that can give methyl amine.Hence aniline is less basic than methyl amine.
(e) Due to last of d-sub-shell Nitrogen cannot from NCl5.
(f) Due to lack of d-subshell fluorine shows only one oxidation state. While rest shows different oxidation states.
(g) Due to Hydrogen bonding water is liquid, but H2S is gas.
(h) Noble gasses are non-atomic, bond order is found in this case zero.
(i) Chlorine shows bleaching action, through oxygen which is permanent.
But SO2 shows bleaching action through Hydrogen which is temporary.
(j) NH3 is associated molecule due to H-bonding and so molecular mass is high and as such it is liquid which PH3 is monomeric law mol. mass and hence a gas.
(k) Of the noble only Zeron from chemical compounds. Xe has comparable I.P. than oxidized cation so Xenon forms compounds with highly electron element fluorine.
Q. 18. I. Arrange following in the increaing properties :
(a) HF , HCl , HBr , HI (acidic strength)
(b) H2O , H2S , H2Sc , H2Tc (B.P.)
(c) F2 , Cl2 , Br2 , I2 (Bond dissociation energy)
II. Predict the probable structure of BrF3 on the basis of VSEPR theory.
Answer⇒ I. (a) HF < HCl < HBr < HI
(b) H2S <H2Sc<H2Tc <H2O
(c) I2 < F2 < Br2 < Cl2
II. The central atom Br has 7 Valence Electron. Three of these will form covalent bond with three F-atoms leaving behind electrons. Thus there are three bond pairs and two lone pairs. According to VSEPR — Theory, these bonds will occupy the corners of a trigonal bipyramid; two lone pairs will occupy equatorial position to minimize lone pair – lone pair and B.P.-L.P. repulsion, which are greater than B.P. – B.P. repulsion. In addition, the axial F-atom will be bent towards the equatorial F in order to minimize the L.P. – L.P. repulsion Therefore shape would be slightly bent T.
Q. 19. (i) Write the electronic configuration of Cu2+ , Z=29 for Cu.
(ii) Discuss the oxidation states of lanthanides.
(iii) What is the trend in the ionic radii of transition elements ?
(iv) Account for the fact that the second ionization energies of both chromium and copper are higher than those of the next element.
(V) Why is manganese more stable in the +2 state than the +3 state while the reverse is true for iron ?
Answer ⇒ (i) Cu atom contains 29 electrons but Cu2+ ion contains 27 electrons. These electrons are arranged as
Cu2+ = 1s2 2s2 2p63s23p63d9
(ii) Typical oxidation state of all lanthanides is +3. Some lanthanides also exhibit +2 and +4 oxidation states in addition to +3. For example cerium forms salt in +4 oxidation state Europium (II) salts are also known. These additional oxidation states are possible only because of the extra stability of an empty half-filled or completely filled orbitals in some lanthanides. For example, in Ce4+ f-orbital is empty in Eu2+ and Yb2+ f-orbital is completely filled ( 4f14 5d06s0).
(iii) The ionic radii decrease with the increase in the atomic numbers of transition elements in horizontal row. The decreasing trend is attributed to the increasing pull on the d-electrons by the nucleus.
(iv) In chromium and copper the second electron is lost from (n-1) d-orbitals, whereas in the next higher element it is from ns.
(v) Mn2+ and Fe3+ have half-filled 3d-orbitals that make them more stable than Mn3+ and Fe2+ respectively.
Q. 20. Illustrate with an example each of the following :
(a) Ionization isomerism
(b) Linkage isomerism
(c) Coordination isomerism.
Answer ⇒ (a) Ionization isomerism : The compounds which have the same molecular formula but give different ions in solution are called ionization isomers. In this type of isomerism, the difference arises from the interchange of groups within or outside the coordination spheres. Therefore, these isomers give different ions in solution. For example :
[Co(NH3)5 Br]SO4 [Co(NH3)5 SO4]Br
Bromo penta-amine cobalt (III) Sulfate penta-amine cobalt
sulfate (violet in color) (III) bromide (red in color)
(b) Linkage isomerism : The compounds which have the same molecular formula but differ in the mode of attachment of a ligand to the metal atom or ion are called linkage isomers. For example, in NO2- ion, the nitrogen atom as well as the oxygen atom can donate their lone pairs. This gives rise to isomerism. If nitrogen donates its lone pair, one particular compound will be formed. On the other hand, if oxygen donates its lone pair, a different compound is obtained. If the bonding is through N , the ligand is named as nitro and if it is through O , it is named as nitrato.
NO2- nitro ONO- nitrato
For example, two different penta-amine cobalt (III) chlorides each containing the NO2 , group in the complex ion have been prepared. These are :
[Co(NH3)5(NO)2]Cl2 Penta-amine nitro cobalt
Yellow brown (III) chloride
[Co(NH3)5(ONO)]Cl2 Penta-amine nitrato cobalt (III) chloride.
(c) Coordination isomerism : This type of isomerism occurs in compounds containing both cationic and anionic complexes and isomers differ in the distribution of ligands in the coordination sphere of cationic and anionic parts. The example are :
(i) [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6[Co(CN)6]
(ii) [Cu(NH3)4][PtCl4] and [Pt(NH3)4[CuCl4]
This type of isomerism is also shown by compounds in which the metal ion is the same in both cationic and anionic complexes. For example :
(i) [Co(NH3)6][Cr(CN)6] and
[Cr(NH3)4(CN)4[Cr(NH3)2(CN)4]
(ii) [Pt(NH3)4][PtCl4] and [PtCl(NH3)3[PtCl3(NH3)]
Q. 21. Explain the bonding in coordination compounds in terms of Werner’s postulates
Answer ⇒ Werner’s Postulates for bonding in coordination compounds :
1. Every metal atom possesses two types of valencies (or linkages).
(a) Primary linkage and
(b) Secondary linkage.
2. (a) Primary linkage corresponds to the oxidation state of the metal atom.
(b) Primary linkage is ionisable.
(C) Primary linkages are satisfied with negative ions.
3. (a) Secondary linkage corresponds to the coordination number of the metal atom.
(b) Secondary linkages are non-ionisable. The coordination number of a metal is a fixed number
(c) Secondary linkages are satisfied either by negative ions or neutral ligands.
(d) Secondary linkages are directed towards fixed positions in space about the central metal atom or ion.
As a result, the coordination compound acquires a definite shape or geometry.
In the light of the postulates above, Werner, who had prepared the following compounds by reacting Cobalt (III) chloride with NH3, successfully explained the structure of these coordination compounds.
CoCl3.6NH3 Orange yellow
CoCl3.5NH3.H2O Pink
CoCl3.5NH3 Violet
CoCl3.4NH3 Violet }
CoCl3.4NH3 Green}
Different colors of CoCl3.4NH3 compound is due to the fact that they are cis and trans isomers.
Q. 22. What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes.
Answer ⇒ The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association; quantitatively expresses the stability. thus, if we have a reaction of the type.
M + 4L ML4
K= [ML4][M][L]4
Larger the stability constant, the higher the proportion of ML4 that exists in solution.
Factors on which stability of the complex depends:
(i) Charge on the central metal ion : Greater the charge on the central metal ion, greater is the stability of the complex.
(ii) Nature of the metal ion : Groups 3 to 6 and inner transition elements form stable complexes when donor atoms of the ligands are N , O and F. The elements after group 6 of the transition metals which have relatively d-orbitals (e.g., Rh , Pd , Ag , Au , Hg etc.) form stable complexes when the donor atoms of the ligands are heavier members of N ,O and F family.
(iii) Basic nature of the ligand : Greater the basic strength of the ligand, greater is the stability of the complex.
(iv) Presence of chelate rings in the complex increases its stability. It is called the chelate effect. It is maximum for the 5- and 6-membered rings.
(v) Effect of multidentate cyclic ligands : If the ligands happen to be multidentate and cyclic without any steric effect, the stability of the complex is further increased.
Q. 23. Specify the oxidation number of the metals in the following coordination entities :
(i) [Co(H2O)(CN)(en)2]2+
(ii) [PtCl4]2-(iii) [CoBr2(en)2]+
(iv) K3[Fe(CN)6]
Answer ⇒ (i) x+0+(-1)+0=+2
∴ x=3
oxidation no. of Co. =3
(ii) x +4 (-1)=-2
or, x=2
oxidation no. of Pt=2
(iii) x+2(-1)+0=+1
or, x=+3
oxidation no. of Co=3
(iv) 3 (+1)+x+6(-1)=0
or, x=3
oxidation no. of Fe=3
Q. 24. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :
(i) K3[Co(C2O4)3]
(ii) Cis-[Cr(en)2Cl2]Cl
(iii) (NH4)2[CoF4]
(iv) [Mn(H2O)6]SO4
Answer ⇒ (i) Oxidation state =+3 Coordination number =6 d-orbital occupation is t2g6 eg0
(ii) Oxidation state =+3 Coordination number =6 d-orbital occupation is d3(t2g3)d3(t2g3)
(iii) Oxidation state =+2 Coordination number=4 d-orbital occupation is sd7 (t2g5 eg2)
(iv) Oxidation state =+2 Coordination number =6 d-orbital occupation is d5 (t2g3 eg2)
Q.25. Discuss the nature of bonding in metal carbonyls.
Answer ⇒ In metal carbonyls, the metal-carbon bond contains both the and bond characters. A bond forms when a lone pair of electrons are donated to the empty orbital of the metal by the carbonyl carbon. A π bond forms when a pair of electrons are donated to the empty antibonding orbital by the filled d orbital of the metal. This, in its entirety, stabilizes and strengthens the metal-ligand bonding.
The above two types of bonding are represented as :
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