Explanation:
The potassium fluoride will dissociate into potassium ions and fluoride ions in their aqueous solution.
So, when 1 mol of hexaaqua aluminium (III) reacts with 6 moles of fluoride ion it gives 1 mole of hexafluoroaluminate(III).
The reaction is given as:
The amine here is the easiest to spot since there’s only one structure that has a nitrogen atom, which would be the first (the first structure is a primary amine).
The distinguishing functional group of an alcohol is the hydroxy group (—OH). Both the second and third structures have an —OH group, but the —OH in the third structure is part of a carboxyl group (—COOH or —C(=O)OH). A carboxyl group takes priority over hydroxy group. Thus, the second structure would be an alcohol and the third structure would be a carboxylic acid.
That leaves us with the fourth structure, a hydrocarbon with a halogen substitutent, or, aptly named, a halocarbon.
Answer : The formula of hexaaquamanganese(II)sulfate is ![[Mn(H_2O)_6]SO_4](https://tex.z-dn.net/?f=%5BMn%28H_2O%29_6%5DSO_4)
.
Explanation :
Rules for writing formulas of coordination complexes :
- Central metal is written first and then the ligands are written along with their prefixes.
- The anion ligands are written before the neutral ligands.
- If there are more than one anion ligands or neutral ligands then they are written in alphabetical order.
The given name of complex is, hexaaquamanganese(ii) sulfate.
In this, the central metal is manganese which is written as Mn and the ligand is water (which is a neutral ligand) is written as 
and an anion is sulfate which is written as 
.
Water is a neutral ligand and sulphate has a charge of -2. As the overall complex is neutral the metal bears a +2 charge which is represented in Roman numeral after the metal.
Therefore, the formula of hexaaquamanganese(II)sulfate is written as .
