Cu2+ + I- --> Cu2I
The compound created is Copper(I) Iodide
im not sure that is possible what you are saying is that 63- I have never seen this in my 8 years since first taking algebra. the negative sign outside the exponent can mean a negative charge on chemistry. lets say you have a copper ion... copper = Cu2- this means is has two extra electrons that it wants to share
Roman numerals are typically used in compound names to indicate the oxidation state of an element. This is necessary when the element can have multiple oxidation states and is written as a cation in the compound. For example, iron can exist in the +2 or +3 oxidation state, so the compound name "iron(II) chloride" specifies that it is the +2 oxidation state of iron.
Formula: Cu2+
For the cation the formula is Cu2+.
Cu(s) | Cu2+(aq) Au+(aq) | Au(s)
Cu2+ + BO3+ = Cu3BO2
Copper two
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V
The octet rule does not apply to transition metals.
The overall voltage for a redox reaction with the half reactions Mg s -- Mg2 plus plus 2e- and Cu2 plus -- Cu is 76 V.
Cu2+ + O2- + H+ + Cl- = Cu2+ + 2 Cl- + H2O
Copper does not form +3 oxidation number. So there cant be a formula like that.
E2xoy4=cu8
29 protons and 27 electrons.