Experiment 4 Electron Configuration Of Elements: Fill & Download for Free

Ionization energy is the amount of energy required to take one electron out of the elemental shells.So, whenever there is a stable electronic configuration from which the electron has to be removed the energy needed will be higher than an unstable electronic configuration. stable electronic configuration means, the full occupancy of the orbitals; i.e 2 for s, 6 for p, 10 for d and 14 for f-orbital. It is also considered to be a stable configuration if the orbitals are half filled. i.e s1, p3, d5, f7. It is to be noted that the the orbitals are more stable when completely filled than when it is half filled. So, whenever there is a comparison, you have to look at the electronic configuration.The 1st IE for Zinc is 906.4 KJ/mol and for Copper is 745.5 KJ/mol. This is because the outer shell orbital configuration of Zinc is 4s2 3d10,which is fully filled, whereas for copper it is 4s1 3d10. Hence, removal of the 1st electron from 4s orbital for copper is easy compared to 4s orbital of Zinc. Now when the 2nd electron has to be removed, the electronic configuration of Zn+ is 4s1 3d10 and that of Cu+ is 4s0 3d10. So, the second electron from Zinc will be removed easily from a half filled 4s1 orbital compared to 3d10 of Cu. 3d10 is fully filled and is also the inner orbital of Cu. Though 4s1 is half filled and ought to be thought as a stable configuration; 3d10 is more stable due to two facts. First d orbital is fully filled, secondly the principal quantum number fro this 3d orbital is 3, whereas for 4s, it is 4. So, lower the principal quantum number, the orbital is closer to the nucleus, which means it will experience more effective charge from the nucleus.Hence, 2nd IE of Zinc is 1733.3KJ/mol and Cu is 1957.9 KJ/mol.So, for any kind of comparison look at following aspects.Whether orbital is fully filled or half filled.What is the principal quantum number of the orbital concerned.(Lower pricipal quantum number means more stable and hence more energy required to remove the electron)

This is actually pretty cool!Look at the electron configuration of beryllium vs. boron. Be is 1s^2 2s^2 B is 1s^2 2s^2 2p^1.Wups! Wait a minute--I need to digress: Ionization energy is the energy required to remove an electron from a gaseous atom. It is purely a measure of the electronic charge attraction a given electron experiences toward the positively charged nucleus.There are actually two ways of looking at this, pick whichever one makes the most sense to you:Now the charge an electron feels is the nuclear charge minus the charge of all the electrons in the next lower shell. So in the case of Li, electron configuration 1s^2 2s^1, the electron in the 2s orbital feels the +3 charge of the nucleus minus the charges of the two electrons in the 1s orbital. You can think of it as that the 1s electrons are closer in to the nucleus and get in the way of the 2s electron's "view" of the nucleus. Net charge (called the effective nuclear charge) = +1. Similarly the 2s electrons in the Be atom feel a net charge of +2. But for boron while the 2 electrons in the 2s orbital feel the full +3 charge, the electron in the 2p orbital feels the +3 charge minus some fraction because the 2s electrons get in the way of some of the charge. So the 2p electron actually only feels slightly more than a +1 charge. Hence, it is easier to pull the electron away from the atom.The other way of looking at this is that the 2p orbitals are higher in energy than the 2s orbital. Hence, the electron in a 2p orbital has a higher potential energy and is thus already closer to the energy needed to strip it from the atom. As the effective nuclear charge increases, the ionization potential increases and as all 3 p orbitals are equivalent energy, there are no more bumps--except for the one caused by starting to fill half-filled orbitals (there is an energy penalty to pairing up electrons and this is seen in the slight decrease in ionization energy for the p^4 elements.The really awesome thing is that these orbitals were predicted on the basis of the hydrogen atom wave function, but predictions from this very simple model of the atom hold even as the number of protons and electrons in the element increase beyond any hope of calculation!Cheers!

Manganese has a low melting and boiling point because of its electronic configurationManganese has a half-filled 3d shell ( 5 electrons in 5 degenerate orbitals). This imparts an added stabilization to those 5 electrons since the atom as a whole has no angular momentum, because it is spherically symmetric. Similar phenomena take place for technetium and rhenium. Zinc, cadmium and mercury experience a similar but more pronounced stabilization due to a full d shell which experiences not just a stabilization due to spherical symmetry (To put it simply) but also because all of their electrons are paired up. All of the elements just listed also experience an increase in ionization energy since going from a 5 to a 4-electron system would result in a more significant destabilization as compared to going from, say, a 3 to a 2-electron system.