The ratio of elimination to substitution is exactly the same (26% elimination) for 2−bromo−2−methylbutane and 2−iodo−2−methylbutane in 80% ethanol/20% water at 25°C. By what mechanism does elimination most likely occur in these compounds under these conditions?

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.

It is sure ( overimitation)