# What is the oxidation state of copper in Gilman's Reagent? [closed]

What is the oxidation state of copper in Gilman's reagent? In $$\ce{R-Cu}$$ bond the oxidation state is +1. Thus in $$\ce{R-Cu-R}$$ bond the oxidation state has to be +2. But if that so why positive $$\ce{Li}$$ act as counter ion in Gilman's reagent? And why is it named as dialkyl cuprate (-ate for anionic complexes)?

• All of your questions can be answered by the fact that R-Cu-R in a Gilman reagent has a negative charge on it. – orthocresol Dec 5 '19 at 13:05
• Why it has a negative charge – Grace Dec 5 '19 at 13:07
• Because copper is not in the +2 oxidation state; it's +1. – orthocresol Dec 5 '19 at 13:15
• Since it make bond with two alkyl group ,should it be +2 – Grace Dec 5 '19 at 13:19
• Oxidation state isn't determined by how many bonds to alkyl groups you have. By that logic the Mn in permanganate ion $\ce{MnO4-}$ would have an oxidation state of +8, since it's bonded to four oxygens. – orthocresol Dec 5 '19 at 13:23

Copper has $$11$$ valence electrons as a neutral atom, as the $$d$$ orbitals are able to participate in reactions. In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets. But then the lithium, bonded to copper (bonding lithium to the alkyl groups would give a separate organolithium molecule instead), adds an electron. So the copper goes from $$11$$ valence electrons down to $$10$$ in the oxidation state formalism, losing one overall; the oxidation state of copper is thereby $$+1$$.