Can someone explain to me how to do each of these problems? I attempted them..but it seems that I don’t fully understand how to do them.
1. Determine the overall charge on each complex.
a) hexafluoroaluminate(III)
b) pentaamminefluoroplatinum(II)
c) diaquadichloroethylenediaminecobalt(III)
2. Give the oxidation state of the metal species in each complex.
a) [Co(NH3)5Cl]Cl2
b) [Ru(CN)3(CO)2]^3-
3. Give the oxidation state of each metal species.
Fe — (excess CO/heat) —> Fe(CO)5 — (heat/I2) —> Fe(CO)4I2
Fe = ?
Fe(CO)5 = ?
Fe(CO)4 I2 = ?
2 Answers
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1. Write the equations:
a) AlF6
b) Pt(NH3)5F
c) Co(H2O)2(Cl)2(en)
a) You know F has an oxidation state of -1 and the roman numeral tells you what the oxidation state of the central metal is, so you have (-6) + (3) = -3. Also the -ate ending is a big giveaway that this is an anion, so you should end up with a negative number
b) You know F = -1 and you know Pt = +2 and NH3 is neutral, so (-1) + (2) = +1
c) You know H2O and en is neutral, Cl has an oxidation state of -1 and you have two of them so 2(-1) + 3 = +1
2.
a) Work backwards. Cl2 has a charge of -2, so the other part in the brackets should have a charge of +2. NH3 is neutral, Cl is -1. Since it has to be +2, and you have a -1 so far, cobalt has to be +3.
b) The overall oxidation charge is -3 as they gave you, CN is -1 and CO is neutral. So so far you have a charge of -3. Ru would have to have an oxidation state of 0 (zero). Yes that’s possible…
3. Fe = 0 for the first, For Fe(CO)5, it’s also 0 and for Fe(CO)4I2, I has a oxidation state of -1, so Fe is 2+
I would double check on this, but that’s my best opinion on these answers 🙂
Source(s): Just took a UCD chemistry midterm on this… -
It depends on many factors
