The starting material in this synthesis experiment Fe(NH4)2(SO4)2*6H2O. What is the ionic charge of the atom? Show your calculations
Which atom are you talking about?
Plz answer this I need this
To determine the ionic charge of the atom in Fe(NH4)2(SO4)2*6H2O, you need to consider the charges of the individual ions present in the compound.
First, let's break down the compound into its constituent ions:
Fe(NH4)2(SO4)2*6H2O
Fe²⁺ (from Fe)
2(NH₄⁺) (from NH₄)
2(SO₄²⁻) (from SO₄)
6(H₂O) (from H₂O)
Now, let's determine the charges of each ion:
- The Fe²⁺ ion has a charge of +2. This is because Fe is a transition metal found in Group 8 of the periodic table, so it loses two electrons to achieve a stable electron configuration.
- The NH₄⁺ ion has a charge of +1. This is because NH₄ is a polyatomic ion called ammonium, which has a single positive charge. The positive charge comes from the lone pair of electrons on the nitrogen atom.
- The SO₄²⁻ ion has a charge of -2. This is because SO₄ is a polyatomic ion called sulfate, which has a double negative charge. The negative charge comes from the overall composition and distribution of the electrons in the ion.
- The H₂O molecule has no net charge. This is because water is a polar molecule, meaning it has a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom. Overall, the charges of the hydrogen atoms cancel out the charge of the oxygen atom, resulting in a neutral molecule.
Based on the above charges, we can summarize the ionic charge calculation:
Fe²⁺ + 2(NH₄⁺) + 2(SO₄²⁻) + 6(H₂O) = ?
Since Fe(NH4)2(SO4)2*6H2O is a neutral compound, the sum of all the charges must be zero. Therefore:
+2 + 2(+1) + 2(-2) + 0 = 0
Simplifying the equation:
+2 + 2 + (-4) + 0 = 0
Finally:
0 = 0
Hence, the ionic charge of the atom in Fe(NH4)2(SO4)2*6H2O is zero.
Note: This calculation assumes that the Fe(NH4)2(SO4)2*6H2O compound is overall neutral and does not dimerize or form complex ions.