# Give reasons: (i) Transition metals tend to be unreactive with increasing atomic numbers in a series. (ii) The largest number of oxidation states are exhibited by the elements in the middle of the first row transition elements. (iii) Transition elements show similarity in properties along a period and down the group.

Sunil Kumar FP
8 years ago

(1)fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in a d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s-orbital electrons since the 3d-orbital is located before the 4s-orbital in the periodic table.

For example: Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and therefore the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3.

The formula for determining oxidation states would be(with the exception of copper andchromium):

Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons)

(2)period trend become significant
group-similarity because of valency