Question

Explain with reasons:
(i) Transition metals show variable valency (variable oxidation states).
(ii) Copper is considered a transition metal even though it has a completely filled d-orbital \((3d^{10})\) in its \(+1\) state.

Hint:

Variable valency comes from the close energy of \((n-1)d\) and \(ns\) electrons; copper qualifies because \(\text{Cu}^{2+}\) is \(3d^{9}\), a partly filled d-orbital.

Answer 1

Part (i): Variable valency of transition metals.
Step 1: Transition metals have valence electrons in both the outer \(ns\) and the inner \((n-1)d\) sub-shells. For example, for the first series the configuration is \(3d^{1-10}\,4s^{1-2}\).
Step 2: The energies of the \((n-1)d\) and \(ns\) orbitals are very close, so the \(4s\) and \(3d\) electrons can both take part in bonding. Electrons are lost in steps, giving several stable oxidation states (for example Fe shows \(+2\) and \(+3\); Mn shows \(+2\) up to \(+7\)).
Step 3: Because successive oxidation states differ by removing one electron at a time and the energy required is small, the metals display variable valency.

Part (ii): Why copper is a transition metal.
Step 1: Definition. A transition element is one that has an incompletely (partially) filled d-orbital in its ground state or in at least one of its common oxidation states.
Step 2: The ground state configuration of copper is \([\text{Ar}]\,3d^{10}\,4s^{1}\). In its common and stable \(+2\) oxidation state, \(\text{Cu}^{2+}\) has the configuration \([\text{Ar}]\,3d^{9}\), which is a partially filled d-orbital.
Step 3: Since copper shows an incompletely filled d-orbital \((3d^{9})\) in its \(+2\) state, it satisfies the definition and is classed as a transition metal, even though \(\text{Cu}^{+}\) happens to be \(3d^{10}\).
\[\boxed{\text{Cu}^{2+}: [\text{Ar}]\,3d^{9} \Rightarrow \text{transition metal}}\]