Question

How is the electron configuration of Cr³⁺. 

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ 

= [Ar]3d⁵ 4s¹ 

= [Ar]3d³

Answer 1

The atomic number of Cr is 24. There are 6 electrons in the outer two shells of Cr that is, 3d and 4s.

The assumed electronic configuration of Cr is [Ar]3d⁴4s² 

Filling of electron take place in increasing order of n + l value of orbital.

For 3d orbital,

Substitute, 3 for n and 2 for l thus,

n + l = 3 + 2 = 5

Similarly, for 4s orbital substitute, 4 for n and 0 for l thus,

n + l = 4 + 0 = 4

Hence, 4s should be filled first as 4s² and remaining 4 electrons goes 3d as 3d⁴. But while writing 4s² is written after 3d⁴.

The actual electronic configuration of Cr is [Ar]3d⁵4s¹. This is because half or full-filled orbitals are more stable than partially filled.

The d subshell has five orbitals. This subshell is extremely stable when it is empty or half-filled. So, Cr exists as [Ar]3d⁵4s¹  and not as [Ar]3d⁴4s²  to acquire half-filled configuration.

Removing 3 electrons from Cr, Electronic configuration of Cr⁺³ is [Ar]3d³4s⁰..

Electronic configuration of Cr⁺³ is [Ar]3d³4s⁰.

To form a trivalent cation of Cr atom, 3 electrons are removed from its outer shells. One electron of the 4s shell and two electrons from the 3d shells.