Question

Order the following metals from shortest to longest expected bonds to an alkoxide ligand and explain your reasoning:
Ti(3+), Cr(3+), Mo(3+), W(3+), Mn(3+), Fe(3+), Co(3+), Ni(3+)

Answer 1

Ti, Cr, Mn, Fe, Co and Ni are transition metals that belong to the 3d series in the order from left to right across the periodic table. Mo and W belong to the 4d and 5d series respectively.

The valence electron configurations for the ions are:

Ti3+ = 3d¹

Cr3+ = 3d³ ; Mo3+ = 4d³ ; W3+ = 5d³

Mn3+ = 3d⁴

Fe3+ = 3d⁵

Co3+ = 3d⁶

Ni3+ = 3d⁷

The atomic size decreases across a period and increases going down the group.

The order from the shortest to the largest would be:

Ni3+ , Co3+, Fe3+, Mn3+, Cr3+, Ti3+, Mo3+, W3+