
Using IUPAC norms, write the formula for the following complexes:
(A)- Potassium tri(oxalato) chromate (III)
(B)- Hexaaquamanganese (II) sulphate
Answer
564.9k+ views
Hnt: The formula is written in order of the metal followed by anionic ligand and the neutral ligand attached to the central metal atom in the coordination sphere. The ion outside is the ionisation sphere which balances the charge over the coordination sphere.
Complete step by step answer:
In order to obtain the formula for the given coordination compounds, the IUPAC nomenclature will be used.
- According to which, in the complex the cation is named first, followed by the anionic part. If the central metal atom within the coordination sphere forms the anion, then the name of the metal ends with the suffix -ate or the Latin name if available.
- Also, the ligands attached to the central atom, are named first before the metal in the IUPAC nomenclature. But in the formula the ligands are written after the metal, where the anionic ligands are given preference over the neutral ligand.
-The ligands are named in alphabetical order, and the neutral ligands have a common name, whereas in the anionic ligands, the end is changed as follows: -ide to -o, -ate to -ato and -ite to -ito.
- The numerical prefix in front of the ligand, designated the number of each ligand present in the complex.
- The oxidation state in Roman numerical of the metal is also specified in the parentheses.
-So, in compound (A) it is seen that the potassium forms the cation, and the coordination sphere is anion, with the Cr as the central metal atom in (+3) oxidation state. The oxalato is the anionic ligand, with the prefix tri- indicating that three oxalate ligands are present in the complex. Then, the charge over the coordination sphere is (-3) which is balanced by three ${{K}^{+}}$ cations. Thus, the formula will be ${{K}_{3}}\left[ Cr{{\left( {{C}_{2}}{{O}_{2}} \right)}_{3}} \right]$ .
-Similarly, in compound (B) the central atom, Mn is part of the cation, with (+2) oxidation state. So, there is no change in its name. The aqua is the neutral ligand, with a prefix hexa- indicating that six aqua ligands are present. Then, the charge over the coordination sphere is (+2) which is balanced by the sulphate anion, $SO_{4}^{2-}$ . Thus, the formula will be $\left[ Mn{{({{H}_{2}}O)}_{6}} \right]S{{O}_{4}}$ .
Note: The coordination compound has a coordination sphere and the ionisation sphere, where the ionisable ion is named according to the common name as in its cation or anion form.
Complete step by step answer:
In order to obtain the formula for the given coordination compounds, the IUPAC nomenclature will be used.
- According to which, in the complex the cation is named first, followed by the anionic part. If the central metal atom within the coordination sphere forms the anion, then the name of the metal ends with the suffix -ate or the Latin name if available.
- Also, the ligands attached to the central atom, are named first before the metal in the IUPAC nomenclature. But in the formula the ligands are written after the metal, where the anionic ligands are given preference over the neutral ligand.
-The ligands are named in alphabetical order, and the neutral ligands have a common name, whereas in the anionic ligands, the end is changed as follows: -ide to -o, -ate to -ato and -ite to -ito.
- The numerical prefix in front of the ligand, designated the number of each ligand present in the complex.
- The oxidation state in Roman numerical of the metal is also specified in the parentheses.
-So, in compound (A) it is seen that the potassium forms the cation, and the coordination sphere is anion, with the Cr as the central metal atom in (+3) oxidation state. The oxalato is the anionic ligand, with the prefix tri- indicating that three oxalate ligands are present in the complex. Then, the charge over the coordination sphere is (-3) which is balanced by three ${{K}^{+}}$ cations. Thus, the formula will be ${{K}_{3}}\left[ Cr{{\left( {{C}_{2}}{{O}_{2}} \right)}_{3}} \right]$ .
-Similarly, in compound (B) the central atom, Mn is part of the cation, with (+2) oxidation state. So, there is no change in its name. The aqua is the neutral ligand, with a prefix hexa- indicating that six aqua ligands are present. Then, the charge over the coordination sphere is (+2) which is balanced by the sulphate anion, $SO_{4}^{2-}$ . Thus, the formula will be $\left[ Mn{{({{H}_{2}}O)}_{6}} \right]S{{O}_{4}}$ .
Note: The coordination compound has a coordination sphere and the ionisation sphere, where the ionisable ion is named according to the common name as in its cation or anion form.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

