Respuesta :
Working out the electronic structures of ions
Ions are atoms (or groups of atoms) which carry an electric charge because they have either gained or lost one or more electrons. If an atom gains electrons it acquires a negative charge. If it loses electrons, it becomes positively charged.
The electronic structure of s- and p-block ions
Write the electronic structure for the neutral atom, and then add (for a negative ion) or subtract electrons (for a positive ion).
To write the electronic structure for Cl -:
Cl 1s22s22p63s23px23py23pz1 but Cl- has one more electron
Cl- 1s22s22p63s23px23py23pz2
To write the electronic structure for O2-:
O 1s22s22px22py12pz1 but O2- has two more electrons
O2- 1s22s22px22py22pz2
To write the electronic structure for Na+:
Na 1s22s22p63s1 but Na+ has one less electron
Na+ 1s22s22p6
To write the electronic structure for Ca2+:
Ca 1s22s22p63s23p64s2 but Ca2+ has two less electrons
Ca2+ 1s22s22p63s23p6
The electronic structure of d-block ions
Here you are faced with one of the most irritating facts in chemistry at this level! When you work out the electronic structures of the first transition series (from scandium to zinc) using the Aufbau Principle, you do it on the basis that the 3d orbitals have a higher energy than the 4s orbital.
That means that you work on the assumption that the 3d electrons are added after the 4s ones.
However, in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first.
You must remember this:
When d-block elements form ions, the 4s electrons are lost first.
Provided you remember that, working out the structure of a d-block ion is no different from working out the structure of, say, a sodium ion.
Note: The problem here is that the Aufbau Principle can only really be used as a way of working out the electronic structures of most atoms. It is a simple way of doing that, although it fails with some, like chromium or copper, of course, and you have to learn these.
There is, however, a flaw in the theory behind it which produces problems like this. Why are the apparently higher energy 3d electrons not the ones to get lost when the metal ionises?
I have written a detailed explanation of this on another page called the order of filling 3d and 4s orbitals. If you are a teacher or a very confident student then you might like to follow this link.
If you aren't so confident, or are coming at this for the first time, I suggest that you ignore it. Learn how to work out the structures of these atoms using the Aufbau Principle on the assumption that the 3d orbitals fill after the 4s, and learn that when the atoms ionise, the 4s electrons are always lost first. Just ignore the contradictions between these two ideas!
To write the electronic structure for Cr3+:
Cr 1s22s22p63s23p63d54s1
Cr3+ 1s22s22p63s23p63d3
The 4s electron is lost first followed by two of the 3d electrons.
To write the electronic structure for Zn2+:
Zn 1s22s22p63s23p63d104s2
Zn2+ 1s22s22p63s23p63d10
This time there is no need to use any of the 3d electrons.
To write the electronic structure for Fe3+:
Fe 1s22s22p63s23p63d64s2
Fe3+ 1s22s22p63s23p63d5
The 4s electrons are lost first followed by one of the 3d electrons.
The rule is quite simple. Take the 4s electrons off first, and then as many 3d electrons as necessary to produce the correct positive charge.
