Transition metals, oxidations states and numbers

A Transition Metal is defined as a metal with an incomplete d-subshell in at least one of its ions.

Chromium and Copper These two metals have an electronic configuration which does not follow the Aufbau Principle. Cr = [Ar], 3d5, 4s1 rather than [Ar], 3d4, 4s2 Cu = [Ar], 3d10, 4s1 rather than [Ar], 3d9, 4s2 This is due to the stability of the ions due to having a half-filled subshell in the case of Cr and a full sub-shell in the case of Cu.

Transition metal ions When transition metal atoms lose electrons to form ions, the Aufbau Principle is followed in all cases i.e. the 4s electrons are lost before the 3d due to the 4s subshell having lower energy than the 3d subshell. e.g. Cu2+ = [Ar], 3d9 Mn2+ = [Ar], 3d5

Exercise Write the electronic configuration for (a) Fe+ (b) Fe3+ (c) Cr2+ (d) Co2+ (e) Mn5+ (f) Ni3+ Why can zinc be said to not be a Transition metal?

Oxidation states and numbers The oxidation state of a Transition metal is the number in front of the charge of the ion written as a Roman Numeral. e.g. the oxidation state of Cu+ is I the oxidation state of Mn7+ is VII The oxidation number is the charge on the ion but written the opposite way around with the charge first followed by the number. e.g. the oxidation number of Cu2+ is +2

The oxidation states of some Transition metals Element Most common oxidation numbers Ti +2, +3, +4 V +1, +2, +3, +4, +5 Cr +1, +2, +3, +4, +5, +6 Mn +1, +2, +3, +4, +5, +6, +7 Fe Co Ni +1, +2, +3, +4 Cu +1, +2, +3

Rules for calculating oxidation numbers The oxidation number of an element is 0 regardless of whether it is monatomic or diatomic. In compounds, oxygen tends to have an oxidation number = -2 and hydrogen an oxidation number of +1. In metal hydrides, H has an oxidation number of -1. The sum of all the oxidation numbers of the atoms or ions in a compound must equal 0 as all compounds are neutrally charged. The sum of all the oxidation numbers of the atoms in a group ion must add up to give the charge on the group ion.

Calculating oxidation numbers Example 1 Calculate the oxidation number and oxidation state of Mn in MnO4-. (1 x Mn) + (4 x -2) = -1 Mn = -1 + 8 = +7 The oxidation number of Mn is +7 The oxidation state of Mn is VII

Example 2 Calculate the oxidation number and oxidation state of Ti in TiO2. (1 x Ti) + (2 x -2) = 0 Ti = 0 + 4 = +4 The oxidation number of Mn is +4 The oxidation state of Mn is IV

Exercise Calculation the oxidation number and state of the following: (a) V in V2O5 (b) Cr in Cr2O72- (c) Mn in MnO (d) Fe in Fe2O3 (e) Ni in NiCl2 (f) Cu in Cu2O