Questions: Coordination Compounds
Coordination Compounds25 min readLesson 5 of 6
Q1 State Werner's postulates and explain how they account for the number of chloride ions precipitated from cobalt ammine chlorides.
Answer (English): Werner proposed that a metal in a coordination compound shows two kinds of valence. The primary valence is the ionic valence, satisfied by negative ions, and the secondary valence is the coordination number, satisfied by neutral molecules or ions that donate lone pairs. In CoCl3·6NH3 all three chloride ions are ionised and precipitated by silver nitrate because ammonia molecules occupy all six secondary sites. As the number of ammonia molecules falls in the series, some chloride ions move into the coordination sphere to fill the secondary sites, and fewer chloride ions are precipitated.
Hinglish: Werner ke do valence hain: primary valence ionisable hoti hai aur negative ions se poori hoti hai; secondary valence coordination number hoti hai aur ligands se poori hoti hai. CoCl3·6NH3 mein saare teen Cl- ionise ho jaate hain kyunki ammonia ne saari coordination sites le li hain. Ammonia kam hone par chloride ions coordination sphere mein aa jaate hain, isliye precipitate kam hota hai.
Q2 Find the oxidation state of cobalt and the coordination number in [Co(H2O)(CN)(en)2]2+. Then calculate the EAN of the metal.
Answer (English): Water is neutral and contributes zero, cyanide contributes minus one, and each ethane-1,2-diamine is neutral, so the ligands total minus one. The complex has a charge of plus two, so the cobalt oxidation state must be plus three. The coordination number counts the donor atoms: one from water, one from cyanide and two each from the two bidentate en ligands, giving six. The effective atomic number is (27 minus 3) plus twice six, which is 24 plus 12, equal to 36, the krypton count.
Hinglish: H2O neutral, CN- minus one aur en neutral, isliye ligands ka total minus one aur cobalt +3 hai. Donor atoms se coordination number 6 milta hai aur EAN = (27 - 3) + 2(6) = 36.
Q3 Write the IUPAC names of [Co(NH3)6]Cl3 and K[Cr(H2O)2(C2O4)2]·H2O.
Answer (English): For [Co(NH3)6]Cl3 the complex is a cation named first: six ammonia ligands give hexaammine, the cobalt is +3, so the name is hexaamminecobalt(III) chloride. For the second compound the complex is an anion, so the metal takes the suffix ate: chromium becomes chromate. The ligands are two water molecules and two bidentate oxalate ions, giving diaqua and dioxalato, so the complex ion is diaquadioxalatochromate(III). The potassium ion is the counter cation, and one water of crystallisation is shown as a monohydrate, so the full name is potassium diaquadioxalatochromate(III) monohydrate.
Hinglish: [Co(NH3)6]Cl3 ka naam hexaamminecobalt(III) chloride hai. Second compound mein complex anion hai, isliye chromium ka naam chromate ho jaata hai aur water of crystallisation monohydrate ke roop mein likha jaata hai: potassium diaquadioxalatochromate(III) monohydrate.
Q4 Write the formulas of tetrahydroxidozincate(II) and pentaamminecarbonatocobalt(III) chloride.
Answer (English): In tetrahydroxidozincate(II) the complex ion carries a charge of minus two because the zinc is +2 and four hydroxide ligands contribute minus four, so the formula is [Zn(OH)4]2-. In pentaamminecarbonatocobalt(III) chloride the complex ion is a cation: five neutral ammonia ligands plus one carbonate ligand of charge minus two give an overall plus one, which balances the chloride, so the formula is [Co(NH3)5(CO3)]Cl.
Hinglish: Tetrahydroxidozincate(II) mein Zn +2 aur chaar OH- mila kar charge minus two banta hai, isliye [Zn(OH)4]2-. Pentaamminecarbonatocobalt(III) chloride mein complex cation plus one hota hai, isliye formula [Co(NH3)5(CO3)]Cl hai.
Q5 Explain the types of structural isomerism in coordination compounds with one example of each.
Answer (English): Ionisation isomerism arises when a counter ion and a ligand exchange places, as in [Co(NH3)5SO4]Cl and [Co(NH3)5Cl]SO4, which give different ions in solution. Hydration isomerism is the same effect with water, as in the different hydrates of chromium chloride. Linkage isomerism occurs with ambidentate ligands that can bind through different atoms, such as the nitrite ion binding through N or O in [Co(NH3)5(NO2)]2+ and [Co(NH3)5(ONO)]2+. Coordination isomerism occurs when both the cation and the anion are complexes and the ligands exchange between the metals, as in [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
Hinglish: Ionisation isomerism mein counter ion aur ligand jagah badalte hain, jaise [Co(NH3)5SO4]Cl aur [Co(NH3)5Cl]SO4. Hydration isomerism mein paani jagah badalta hai. Linkage isomerism ambidentate ligand ke different donor atoms se aata hai, jaise NO2 mein N ya O se binding. Coordination isomerism mein dono complex ions ke beech ligands exchange hote hain.
Q6 Using valence bond theory, explain why [Co(NH3)6]3+ is an inner orbital complex while [Ni(NH3)6]2+ is an outer orbital complex.
Answer (English): Co3+ is a d6 ion. Ammonia is a strong enough ligand to force pairing, so the six d electrons occupy the three t2g orbitals and one inner d orbital becomes available for d2sp3 hybridisation. The complex is therefore inner orbital, octahedral and diamagnetic because all the electrons are paired. Ni2+ is a d8 ion, and with ammonia as the ligand the pairing does not free an inner d orbital, so the hybridisation uses the outer d orbital in sp3d2. The complex is outer orbital, octahedral and paramagnetic with two unpaired electrons.
Hinglish: Co3+ d6 hai; NH3 strong ligand hai isliye electrons pair ho jaate hain aur d2sp3 inner orbital complex banta hai, jo diamagnetic hai. Ni2+ d8 hai; NH3 pairing nahi karata, isliye sp3d2 outer orbital complex banta hai jo do unpaired electrons ke saath paramagnetic hai.
Q7 Describe how an octahedral ligand field separates the d orbitals, and use that picture to decide between high spin and low spin for a d4 ion.
Answer (English): The six ligands of an octahedral field approach along the axes and repel the d orbitals that point along the axes, namely dz2 and dx2-y2, raising them in energy into the eg set. The dxy, dyz and dxz orbitals point between the axes and form the lower t2g set. The gap between the sets is Δo. For a d4 ion, three electrons fill t2g singly. The fourth electron has a choice: enter eg and pay Δo, or pair in a t2g orbital and pay the pairing energy P. If Δo is less than P, as with weak field ligands such as chloride, the electron enters eg and the complex is high spin with four unpaired electrons. If Δo is greater than P, as with cyanide, it pairs and the complex is low spin with two unpaired electrons.
Hinglish: Octahedral field mein axis ke along waale dz2 aur dx2-y2 upar eg set banate hain, aur beech waale dxy, dyz, dxz neeche t2g set; gap Δo hai. d4 ion mein teesre electron ke baad chautha electron ya eg mein jaata hai ya pair hota hai. Δo < P ho to high spin (chaar unpaired), Δo > P ho to low spin (do unpaired).
Q8 Calculate the crystal field stabilisation energy of a d3 octahedral complex in terms of Δo, and of a high spin d5 octahedral complex.
Answer (English): For a d3 octahedral complex the three electrons occupy the t2g set singly, so the CFSE is minus 0.4Δo times three, which is minus 1.2Δo. For a high spin d5 complex three electrons occupy t2g and two occupy eg, so the CFSE is minus 0.4Δo times three plus 0.6Δo times two, which is minus 1.2Δo plus 1.2Δo, equal to zero.
Hinglish: d3 mein teeno electrons t2g mein hain, isliye CFSE = -0.4Δo × 3 = -1.2Δo. High spin d5 mein teen t2g aur do eg mein hote hain, isliye CFSE = -1.2Δo + 1.2Δo = 0. Isi wajah se Mn2+ aur Fe3+ jaise high spin d5 ions outer orbital complexes banate hain.
Q9 Why is [Ti(H2O)6]3+ violet while [Ni(CN)4]2- is colourless?
Answer (English): Ti3+ is a d1 ion. Its single electron absorbs light in the visible region to move from the t2g set to the eg set, and the colour observed is the complement of the absorbed wavelength, which is violet. Ni2+ in [Ni(CN)4]2- is d8, and cyanide is a strong field ligand that pairs all the d electrons, so the ion has a completely filled d subshell and no d-d transition is possible in the visible region. The complex is therefore colourless.
Hinglish: Ti3+ d1 hai; ek electron visible light absorb karke t2g se eg mein jaata hai, isliye violet colour dikhta hai. [Ni(CN)4]2- mein Ni2+ d8 hai aur CN- strong field ki wajah se saare electrons pair ho jaate hain, isliye d-d transition nahi hota aur complex colourless hai.
Q10 What is the chelate effect? Give two applications of coordination compounds in medicine or everyday life.
Answer (English): The chelate effect is the extra stability that a complex gains when a polydentate ligand binds the metal at two or more points and forms rings. Binding is favoured because one chelating ligand replaces several solvent or monodentate ligands. EDTA is hexadentate and binds calcium and iron strongly, which is why it removes scale and keeps iron available in food fortification. In medicine, cisplatin is a square planar platinum complex used in cancer chemotherapy, and haemoglobin carries oxygen using an iron complex.
Hinglish: Chelate effect tab hota hai jab polydentate ligand metal ko do ya zyada points se pakad kar rings banata hai; entropy change ki wajah se complex zyada stable hota hai. EDTA calcium aur iron ko strongly bind karta hai. Cisplatin cancer chemotherapy mein use hota hai, haemoglobin mein iron aur chlorophyll mein magnesium complex ka role hai.
The quiz on this page covers coordination number, isomerism, magnetic moment, colour and stability. Attempt it after the ten answers above.