SME
Grade 10 Science · Problem type
Configurations of ions
Use this for an ion’s configuration, such as Fe³⁺: electrons taken from the highest shell first.
Diagram
23 electrons (Iron ion, charge +3): [Ar] 3d⁵.
Boxes fill from the lowest energy up; each box of a subshell takes one arrow before any takes two, and two arrows in a box point opposite ways.
Unpaired electrons (single arrows, lit): 5.
This is an example. Type your own values to start a new problem.
q
=
Z
−
e
u
Unpaired electrons
worked out (not typed)
Formulas
q
=
Z
−
e
+3 = 26 − 23
u
=
unpaired
electrons
of
element
Z
with
e
electrons
5 = unpaired electrons of element 26 with 23 electrons
Assumptions
•
A positive ion loses from the highest shell first, so iron loses 4s before 3d.
•
A negative ion gains electrons in the next empty places.
Step-by-step
WE KNOW
Z = 26, q = +3
FIND
electrons (e), unpaired electrons (u)
Step 1 · Find electrons (e)
Use q = Z − e
A positive ion has lost electrons; a negative ion has gained them.
e
=
Z
−
q
e
=
26
−
3
e
=
23
Step 2 · Find unpaired electrons (u)
Use u = unpaired electrons of element Z with e electrons
Fill the boxes in order, one arrow in each box of a subshell before pairing, then count the single arrows.
u
=
unpaired
electrons
of
element
26
with
23
electrons
Z
=
26
(iron)
with
23
electrons:
1s²
2s²
2p⁶
3s²
3p⁶
3d⁵
u
=
5
Check
Put the numbers back in. Both sides should match.
+3 = 26 − 23 ✓
5 = unpaired electrons of element 26 with 23 electrons ✓
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