Iron(II) is easily oxidized while manganese(II) is not

The half-filled 3d⁵ decides

1. Electronic configurations and oxidation states

Iron(II) is easily oxidized while manganese(II) is not

2. Who gives up one more electron?

Fe²⁺ Fe X^{2 +}

and

Mn²⁺ Mn X^{2 +}

can each lose one more electron. One does it easily; the other does not.

Pick the right answer

Predict: which one is easily oxidized to the 3+ ion?

A. $Fe²⁺ Fe X^{2 +}$
B. $Mn²⁺ Mn X^{2 +}$
C. Both equally: they have the same charge

الشرح

Fe²⁺ Fe X^{2 +}

is easily oxidized to

Fe³⁺ Fe X^{3 +}

: losing one 3d electron turns its

3d⁶ 3 d^{6}

into the stable half-filled

3d⁵ 3 d^{5}

Mn²⁺ Mn X^{2 +}

is already

3d⁵ 3 d^{5}

, so it is hard to oxidize to

Mn³⁺ Mn X^{3 +}

3. Make the 2+ ions

Remove electrons from Fe and Mn. Remember: the 4s electrons leave first.

electrons lost 0

Iron (Z = 26)

↑↓

↑↓↑↑↑↑

Fe:

[Ar]\ 4s²\ 3d⁶

Manganese (Z = 25)

↑↓

↑↑↑↑↑

Mn:

[Ar]\ 4s²\ 3d⁵

4. Iron (II) → iron (III)

Fe²⁺ \to[oxidation] Fe³⁺ + e- Fe X^{2 +} oxidation Fe X^{3 +} + e X^{-}

Fe²⁺ Fe X^{2 +}

↑↓↑↑↑↑

[Ar]\ 3d⁶

Fe³⁺ Fe X^{3 +}

↑↑↑↑↑

[Ar]\ 3d⁵

: half filled, more stable

Iron (III) ion is more stable, as its 3d sublevel is half filled (

3d⁵ 3 d^{5}

). So the reaction goes toward the more stable ion.

The electron that leaves is the paired one. Now every 3d orbital holds one electron.

5. Your turn: manganese (II) → manganese (III)

Same steps for Mn.

Mn²⁺ Mn X^{2 +}

[Ar]\ 3d⁵

. Oxidation removes one more electron.

Fill in the blanks

Mn³⁺ Mn X^{3 +}

= [Ar] 3d_____ . After oxidation, the 3d sublevel is _____ half filled (type: still / no longer).

الشرح

Mn³⁺ Mn X^{3 +}

[Ar]\ 3d⁴

: one 3d orbital is now empty.

Show the boxes and the book's answer

Mn²⁺ Mn X^{2 +}

↑↑↑↑↑

[Ar]\ 3d⁵

: half filled, more stable

Mn³⁺ Mn X^{3 +}

↑↑↑↑

[Ar]\ 3d⁴

The 3d sublevel is half filled (

3d⁵ 3 d^{5}

) in

Mn²⁺ Mn X^{2 +}

, so it is more stable than

Mn³⁺ Mn X^{3 +}

. So

Mn²⁺ Mn X^{2 +}

is not readily oxidized to

Mn³⁺ Mn X^{3 +}

6. Iron's important oxidation states

Fe: [Ar] 4s² 3d⁶ → Fe²⁺ (+2): the two 4s electrons lost (lose 4s²)
Fe²⁺ (+2): the two 4s electrons lost → Fe³⁺ (+3): half-filled 3d⁵, stable (lose one 3d electron)

Book pp.20-21: iron's two important oxidation states

Iron has 8 electrons in 4s and 3d, but it does not reach +8. The book says all its states above +3 are not important.

7. Same 3d, different ions

Pick the right answer

Which two ions both have the stable half-filled

3d⁵ 3 d^{5}

sublevel?

A. $Fe³⁺ Fe X^{3 +}$ and $Mn²⁺ Mn X^{2 +}$
B. $Fe²⁺ Fe X^{2 +}$ and $Mn³⁺ Mn X^{3 +}$
C. $Fe³⁺ Fe X^{3 +}$ and $Mn³⁺ Mn X^{3 +}$
D. $Fe²⁺ Fe X^{2 +}$ and $Mn²⁺ Mn X^{2 +}$

الشرح

Electrons:

Fe³⁺ Fe X^{3 +}

26 − 3 = 23,

Mn²⁺ Mn X^{2 +}

25 − 2 = 23. Both are

[Ar]\ 3d⁵

True or false?

Fe³⁺ Fe X^{3 +}

is more stable than

Fe²⁺ Fe X^{2 +}

because its charge is bigger.

الشرح

The book's reason is the half-filled 3d sublevel (

3d⁵ 3 d^{5}

), not the size of the charge.

8. Build the exam answer

Put in order

Put the answer to 'Iron (II) is easily oxidized to iron (III)' in order.

$₂₆Fe X_{26} X_{226}^{2} Fe$ : $[Ar]\ 4s²\ 3d⁶$
$Fe²⁺ Fe X^{2 +}$ : $[Ar]\ 3d⁶$ (the two 4s electrons are lost first)
$Fe³⁺ Fe X^{3 +}$ : $[Ar]\ 3d⁵$
The 3d sublevel of $Fe³⁺ Fe X^{3 +}$ is half filled, so it is more stable
So the reaction goes toward forming $Fe³⁺ Fe X^{3 +}$

Pick the right answer

Which oxidation states of iron does the book call important?

A. +2 and +3
B. +2, +3 and +8
C. +3 only
D. +2, +3 and +6

الشرح

The book calls +2 and +3 the important oxidation states of iron: +2 from losing the two 4s electrons, and +3 with the stable half-filled

3d⁵ 3 d^{5}

. The states above +3 are not important.

9. Summary

$Fe²⁺ Fe X^{2 +}$ $3d⁶ 3 d^{6}$ → $Fe³⁺ Fe X^{3 +}$ $3d⁵ 3 d^{5}$ : oxidation gives a half-filled 3d, so it is easy.
$Mn²⁺ Mn X^{2 +}$ is already $3d⁵ 3 d^{5}$ . Losing one more electron would break it.
Iron's important states: +2 and +3. States above +3 are not important.

Back to the hook: the same

3d⁵ 3 d^{5}

makes one ion change easily and the other not. Can you explain why, with 3d boxes?

Reveal

Fe²⁺ Fe X^{2 +}

(

3d⁶ 3 d^{6}

) loses one 3d electron and becomes

Fe³⁺ Fe X^{3 +}

(

3d⁵ 3 d^{5}

): half filled, so the oxidation is easy.

Mn²⁺ Mn X^{2 +}

is already

3d⁵ 3 d^{5}

: losing one more electron would break the half-filled 3d and give

Mn³⁺ Mn X^{3 +}

(

3d⁴ 3 d^{4}

↑↑↑↑↑

Fe³⁺ Fe X^{3 +}

[Ar]\ 3d⁵

↑↑↑↑

Mn³⁺ Mn X^{3 +}

[Ar]\ 3d⁴

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