Please determine the number of SO4^2- ions present in the complex of an ethylene
ID: 797239 • Letter: P
Question
Please determine the number of SO4^2- ions present in the complex of an ethylenediamine bidentate ligand (nickle),
the overall charge on the complex, [Niv(en)w(H2O)x]^2+ ySO4^2- zH2O is zero; the charge on sulfate
ion is 2-; en and water ligands both have a charge of zero; and nickel has a charge of 2+. Using
this information and the fact that there is 56.27grams en and 17.89 grams Ni present in 100g of the compound and there are .0800 M of en and .02700 M Ni
calculate the number of moles of SO42- in 100 grams of the complex
Then determine the emperical formula (Niv enw H2Ox ySO4 zH2O).
Explanation / Answer
Moles of Ni in 100g complex = 17.89/58.7 = 0.3048moles
Number of moles of en ligand in 100g complex = 56.27/60.10 = 0.9363 moles
Mole ratio of en to Ni = 0.9363/0.3048 = 3 (approximately)
This implies w = 3v
Also, number of Ni atoms equals to number of sulphate ions, as there must be charge balance to attain a electrically neutral complex.
=> v=y
Therefore, moles of sulphate ion in the complex = v = 0.3048 moles
en is a bidendate ligand.
Considering that 3 en form a complex with an Ni atom, there is no scope for the existence of H2O as a ligand in the complex.
This is because, the octahedral geometry is achieved with 3 en ligands around Ni.
So, y=z=0
The empirical formula is [Ni(en)3]2+ SO42-