
The formula for$sodium$ \[trioxalatoaluminate(iii)\] is:
A.$N{a_4}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
B.$N{a_5}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
C.$N{a_3}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
D.$N{a_2}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
Answer
590.4k+ views
Hint: The molecules which possess one or multiple metal centers that are bound to ligands are known as coordination compounds. Ligands are basically atoms, ions, or molecules which can donate electrons to metal. The coordination compounds can be neutral or charged.
Complete step by step answer:
The compounds which possess one or multiple metal centers which are surrounded by atoms, ions or molecules which can donate electrons are known as coordination compounds.
From this definition we can predict that $sodium$ \[trioxalatoaluminate(iii)\] is a coordination compound in which aluminium is the central metal atom, and the oxalate group is ligand.
So we can assume the complex as:
$N{a_x}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
We have to find the value of $x$ here.
The sum of oxidation states in a complex is always zero so we will use this concept to find out the value of $x$ .
$x + 3 - 6 = 0$
$x = 3$
Therefore the formula for $sodium$ \[trioxalatoaluminate(iii)\] is $N{a_3}\left[ {Al{{({C_2}{O_4})}_3}} \right]$.
We can also predict the value of $x$ by the given number at the end of the complex name. This number shows the charge or valency of the ligand so to counter it we have to use the positively charged ion
Hence option (C) is correct.
Note:
Oxalate ion (${C_2}{O_4}^{ - 2}$) is a weak field ligand. It causes only small splitting of energy levels. This ion is a bidentate ligand, which means that ion has two atoms which directly coordinate with the central atom and donate electrons.
Complete step by step answer:
The compounds which possess one or multiple metal centers which are surrounded by atoms, ions or molecules which can donate electrons are known as coordination compounds.
From this definition we can predict that $sodium$ \[trioxalatoaluminate(iii)\] is a coordination compound in which aluminium is the central metal atom, and the oxalate group is ligand.
So we can assume the complex as:
$N{a_x}\left[ {Al{{({C_2}{O_4})}_3}} \right]$
We have to find the value of $x$ here.
The sum of oxidation states in a complex is always zero so we will use this concept to find out the value of $x$ .
$x + 3 - 6 = 0$
$x = 3$
Therefore the formula for $sodium$ \[trioxalatoaluminate(iii)\] is $N{a_3}\left[ {Al{{({C_2}{O_4})}_3}} \right]$.
We can also predict the value of $x$ by the given number at the end of the complex name. This number shows the charge or valency of the ligand so to counter it we have to use the positively charged ion
Hence option (C) is correct.
Note:
Oxalate ion (${C_2}{O_4}^{ - 2}$) is a weak field ligand. It causes only small splitting of energy levels. This ion is a bidentate ligand, which means that ion has two atoms which directly coordinate with the central atom and donate electrons.
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