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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: atadayyon on December 30, 2012, 08:21:53 PM

Title: Atom Configurations for V and Cr
Post by: atadayyon on December 30, 2012, 08:21:53 PM

Okay, so I understand that the atom configuration for chromium (Cr) is [Ar] 4s¹3d⁵ because the d orbitals like to stay "half filled."

Well according to that theory, then shouldn't the atom configuration for Vanadium (V) be [Ar] 4s⁰3d⁵? This way the s orbital stays empty and the d is half filled?

Why is this incorrect?

Title: Re: Atom Configurations for V and Cr
Post by: Radd on December 31, 2012, 03:04:43 AM

As far as I understand it Cr (I think Mo, Cu, Ag, too) is one of the exceptions to the aufbau principle in the d-block transition metals and Vanadium isn't.

So, for Vanadium, it's business as usual when using the block method to decipher its electron configuration.

Title: Re: Atom Configurations for V and Cr
Post by: fledarmus on January 01, 2013, 10:10:22 AM

You are looking at a balance of two energies, and your atom will, of course, pick the lowest.

First, the s orbital is lower energy than the d orbital.

Second, having a half-filled orbital shell gives you a small lowering of energy.

In general, the s level is close enough to the d level that the small lowering of energy for a half filled s level and a half filled d level is enough that the transition from ns₂(n-1)d₄ to ns₁(n-1)d₅ is favorable. You give up a little energy from the difference in s and d, but pick up a little more from having two half filled shells. If you are starting with only five electrons, however, you are losing the energy difference of two electrons going into s instead of d, not just one, and only picking up the half filled bonus for the d shell, not the s shell. The process you are looking at is ns₂(n-1)d₃ to ns₁(n-1)d₄ to ns₀(n-1)d₅, and neither of those steps is favorable energetically.

I hope that description made sense to somebody besides me  :)