Question

Arrange the following species according to the number of unpaired electrons they contain, starting with the one that has the greatest number: Fe, Sc3+, Ti2+, Mn4+, Cr, Cu2+. Rank species according to the number of unpaired electrons from greatest to smallest. To rank items as equivalent, overlap them.

Answer 1

Answer – We are given the element and ions for the transition metal and we need rank species according to the number of unpaired electrons from greatest to smallest, so first we need to write the electronic configuration of each –

Fe – [Ar] 4s² 3d⁶

So, Fe has 4 unpaired electrons in 3d⁶

Sc³⁺ - [Ar] 4s⁰ 3d⁰

So Sc³⁺ has no any unpaired electrons.

Ti²⁺ = [Ar] 4s² 3d⁰

So Ti²⁺ has no any unpaired electrons.

Mn⁴⁺ = [Ar] 4s² 3d¹

So Mn⁴⁺ has 1 unpaired electron.

Cr = [Ar] 4s¹ 3d⁵

So, Cr has 6 unpaired electrons

Cu²⁺ = [Ar] 4s¹ 3d⁸

So Cu²⁺ has 3 unpaired electrons

So, species according to the number of unpaired electrons from greatest to smallest as follow –

Cr > Fe > Cu²⁺ > Mn⁴⁺ > Sc³⁺ =Ti²⁺