2.2.1 (d) Deducing Electronic configurations of atoms and ions
Syllabus
(d) deduction of the electron configurations of:
(i) atoms, given the atomic number, up to Z = 36
(ii) ions, given the atomic number and ionic charge, limited to s- and p-blocks up to Z = 36.
{Learners should use sub-shell notation, i.e. for Oxygen: 1s22s22p4.}
What does this mean?
Atoms
Given the Atomic Number of an atom you know how many electrons it has.
You could simply start from 1s2
Then add 2s2
Then 2P6 etc until your electronic configuration adds up to the Atomic Number (Z)
This will work up to Z = 18 =Neon.
eg 12Mg = 1s22s22p63s2
Or, you can look at the Periodic Table
Providing you recall that Groups 1 and 2 are the s-block and the rest are p-block, you can read the electronic configuration directly.
eg F is 5 elements into Period 2's p-block
So, its configuration ends with 2p5
So, it must be 1s22s22p5
eg Si is 2 elements into Period 3's p-block
So, its configuration ends with 3p2
So, it must be 1s2 2s2 2p6 3s2 3p2
eg Na is 1 element into Period 3's s-block
So, its configuration ends with 3s1
So it must be 1s2 2s2 2p6 3s1
Beyond element 18 it is less obvious, although you can do the same for most elements.
eg Ga is 1 element into Period 4's p block making it 1s2 2s2 2p6 3s23p6 4s1 3d10 4p1
eg I is 5 elements into Period 5's p-block making it 1s2 2s2 2p6 3s23p6 4s1 3d10 4p6 5s2 4d10 5p5
The exceptions that you need to recall are Cr and Cu - see 2.2.1 (a,b,c)
PowerPoint
Ions
Simple ions
For the first 20 elements the easiest way to find the electronic configuration of an ion is to first find the configuration of the atom and then adding electrons if it is a (negative) anion, or subtracting electrons if it is a (positive) cation.
eg What is the electronic configuration of Iodide (I-) ions?
Electronic configuration of Iodine = 1s2 2s2 2p6 3s23p6 4s1 3d10 4p6 5s2 4d10 5p5
Electronic configuration of Iodide (I-) = 1s2 2s2 2p6 3s23p6 4s1 3d10 4p6 5s2 4d10 5p6 - (one extra electron)
eg What is the electronic configuration of Calcium (Ca2+) ions?
Electronic configuration of Calcium atoms= 1s2 2s2 2p5 3s23p6 4s2
Electronic configuration of Calcium ions (Ca2+) = 1s2 2s2 2p5 3s23p6 - (two fewer electrons)
Transition metal ions
If you can already write a full electronic configuration for a transition metal atom - see 2.2.1 (a,b,c) then all you have to remember is that Transition Elements always lose their 4s electrons first - despite them being at lower energy!
Annoying, eh!
Copper atoms are 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Cu+ ions are 1s2 2s2 2p6 3s2 3p6 4s0 3d10
Cu2+ ions are 1s2 2s2 2p6 3s2 3p6 4s0 3d9
Exam-style Questions
1. (a) Explain why certain elements in the Periodic Table are classified as p-block elements. Illustrate your answer with an example of a p-block element and give its electronic configuration.
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Answers
1. (a) Elements in the p block have their outer electron(s) in p orbital(s) or levels or sub-shells (1)
example of element (1)
correct electronic configuration (1)
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