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Solved Examples of Coordination Compounds

Question 1:

Write down the molecular formulae of the following co-ordination compounds.

(i) Hexaammine iron (III) nitrate

(ii) Ammonium tetrachlorocuprate (II)

(iii) Sodium monochloropentacyanoferrate (III)

(iv) Potassium hexafluorocabaltate (III)

Solution:

(i) [Fe(NH3)6](NO3)3

(ii) (NH4)2[CuCl4]

(iii) Na3[FeCl.(CN)5]

(iv) K3[CoF6]

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Question 2:

Write the IUPAC names of following compounds?

(i) [CoBr(NH3)5]SO4

(ii) [Fe(NH3)6][Cr(CN)6]

(iii) [Co(SO4)(NH3)5]+

(iv) [Fe(OH)(H2O)5]2+

Solution:

(i) Pentaamminebromocobalt (III) sulphate

(ii) Hexaammineiron (III) hexacyanochromate (III)

(iii) Pentaammine sulphate cobalt (III) ion

(iv) Pentaaquahydroxoiron (III) ion

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Question 3:         

Write down the IUPAC name of the complex K4[Fe(CN)6].

Solution:

Firstly the +ve part should be named followed by the negative part in which the name of the ligand should be given in alphabetical order with the metal in the negative part ending in - ate (oxidation state in parenthesis) Thus the name is Potassiumhexacyano - C-ferrate (II). Hyphen C is shown to indicate CN– bonded via carbon

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Question 4:

Write down the IUPAC name  of K2[Fe(NC)3Cl2(NH3)2].

Solution:

The positive part is named first followed by the negative part. In the negative part the names are written in alphabetical order followed by metal. So the name is Potassiumdiamminedichlorotricyano-N-  ferrate (III).

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Question 5:

Write the IUPAC name of [Co(NH3)4(NO2)2]Cl

Solution:

In the earlier two examples the negative part was the complex part while in this case the positive part is the complex. So it is named first with ligands in alphabetical order followed by metal (but not ending in –ate as the metal belong to the positive part of the complex). This is followed by the negative part.So the name is Tetraamminedinitrocobalt (III) chloride.

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Question 6:

Write down the IUPAC name of [Co(NH3)2Cl(en)2]Cl2

Solution:

The approach is same as the earlier one with the exception that in case of  -en which is actually ethylene diammine the term  bis –  comes to indicate two – en groups instead of – bi.

The name is Diamminechlorobis (ethylene diamine) cobalt(III) chloride.

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Question 7:     

Write IUPAC name of [Pt(Py)4][PtCl4].

Solution:

Here both the positive and negative part has the same metal. Procedure is same as earlier for the IUPAC name. Tetrapyridineplatinum(II) tetrachloroplatinate(II).

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Question 8: 

Write the IUPAC name of [Fe(NH3)4O2C2O4]Cl

Solution:

In this charge on the complex part is +1. The ligand oxalato has a charge of –2, so iron should be in  +3 state meaning O2 to be neutral. Now had O2 been superoxo (O2–) or peroxo (O2– – ) the negative charge of the ligands should have been –3 and –4 respectively. In that case Iron has to be +4 and +5 which is not possible. So O2 will behave as a neutral ligand and IUPAC name is Tetraammineoxalatodioxygeniron (III) chloride.

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Question 9:

Amongst Ni(CO)4, [Ni(CN)4]2– and NiCl42-

(a) Ni(CO)4 and NiCl42 are diamagnetic and [Ni(CN)4]2– is paramagnetic

(b) [NiCl4]2– and [Ni(CN)4]2– are diamagnetic and Ni(CO)4 is paramagnetic

(c) Ni(CO)4 and [Ni(CN)4]2– are diamagnetic and [NiCl4]2– is paramagnetic

(d) Ni(CO)4 is diamagnetic and [NiCl4]2– and [Ni(CN)4]2– are paramagnetic

Solution:

In Ni(CO)4, Ni has 3d10 configuration, diamagnetic.

In [Ni (CN)4]2–, Ni has 3d8 configuration but due to strong ligand field, all the d-electrons are spin paired giving dsp² hybridization, diamagnetic.

In [NiCl4]2–, Ni has 3d8 configuration and there is two unpaired electrons (weak chloride ligand do not pair up d-electrons) hence, paramagnetic.

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Question 10:

Among the following ions, which one has the highest paramagnetism ?       

(a) [Cr(H2O)6]3+

(b) [Fe(H2O)6]2+

(c) [Cu(H2O)6]2+ 

(d) [Zn(H2O)6]2+

Solution:

Fe in [Fe(H2O)6]2+ has maximum (four) unpaired electrons, has highest paramagnetism.  

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