Oxidation states of transition elements

oxidation and reduction process

Oxidation:

losing electrons and the oxidation number increases Mn²⁺ oxidation Mn³⁺ + e-

Reduction:

Gaining electrons and the oxidation number decreases

Give reason:

It is easy to oxidize iron(II)ion to iron(III) ion where it is difficult to oxidize manganese (II)ion to manganese (II) ion?

-because the electronic configuration of an iron atom is :26Fe [Ar18]4s²,3d⁶

-Iron exhibits two oxidation numbers (a) +2 as iron(II) ion, Fe²⁺ (b) +3 as iron(III) ion, Fe³⁺

-An aqueous solution containing iron(II) ions, Fe²⁺ is pale green in color, whereas that containing iron(III) ions, Fe³⁺is yellow/yellowish-brown/ brown in color.

-Changing iron(II) ions to iron(III) ions is an oxidation and therefore requires an oxidizing agent.

Fe²⁺ → Fe³⁺ + e^– -On the other hand, changing iron(III) ions to iron(II) ions is a reduction and therefore requires a reducing agent. Fe²⁺ + e^– → Fe²⁺

-How to calculate the oxidation state of of $F e$ in $K_{2} F e O_{4}$ : $K_{2} F e O_{4}$ = $2 (+ 1) + x + 4 (- 2) = 0 2 + x - 8 = 0 x - 6 = 0 x = + 6$ Oxidation number of $F e$ is $+ 6 .$

Table of Contents

$Oxidation states of the first transition series :$

**oxidation states of first transition series1**

The common oxidation states of first transition series:

The oxidation states of transition elements in some of their compounds:

	group no. of transition elements
[ScO₂]^–	3=oxidation state=3
[TiO₃]^2-	4=oxidation state=4
[VO₃]^-5	5=oxidation state=5
[Cr₂O₇]^2-	6=oxidation state=6
[CrO₄]^2-	6=oxidation state=6
[MnO₄]^–	7=oxidation state=7

How to calculate the oxidation number ?

Write the electronic configurations of:

neutral iron,
iron(II) ion, and
iron(III) ion.

answer :
The atomic number of iron is 26 so there are 26 protons in the species.

1)Fe: [Ar] 4s² 3d⁶

2)Fe²⁺: [Ar] 3d⁶

3)Fe³⁺: [Ar] 3d⁵

Note that the s-orbital electrons are lost first, then the d-orbital electrons.

What makes zinc stable as Zn²⁺?

What makes scandium stable as Sc³⁺?

Zinc has the neutral configuration [Ar]4s²3d¹⁰. Losing 2 electrons does not alter the complete d orbital. Neutral scandium is written as [Ar]4s²3d¹. Losing 3 electrons brings the configuration to the noble state with valence 3p⁶.

Why is iron almost always Fe²⁺ or Fe³⁺?

Answer: Iron is written as [Ar]4s²3d⁶. Losing 2 electrons from the s-orbital (3d⁶) or 2 s- and 1 d-orbital (3d⁵) electron are fairly stable oxidation states.

Example :
Write manganese oxides in a few different oxidation states.

Which ones are possible and/or reasonable?

Answer:
Although Mn⁺² is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn_(s), Mn(II) as MnO, Mn(II,III) as Mn₃O₄, Mn(IV) as MnO₂, or manganese dioxide, Mn(VII) in the permanganate ion MnO₄^–, and so on.

Oxidation State of Transition Metals in Compounds
When given an ionic compound such as $AgCl$ , you can easily determine the oxidation state of the transition metal. In this case, you would be asked to determine the oxidation state of silver (Ag).

Since we know that chlorine (Cl) is in the halogen group of the periodic table, we then know that it has a charge of -1, or simply Cl^–.

In addition, by seeing that there is no overall charge for $AgCl$ , (which is determined by looking at the top right of the compound, i.e., AgCl^#, where # represents the overall charge of the compound) we can conclude that silver ( $Ag$ ) has an oxidation state of +1.

This gives us Ag⁺ and Cl^–, in which the positive and negative charge cancels each other out, resulting with an overall neutral charge; therefore +1 is verified as the oxidation state of silver (Ag).

Example :

Determine the oxidation state of cobalt in ${CoBr}_{2}^{}$ .

Answer:
Similar to chlorine, bromine ( $Br$ ) is also a halogen with an oxidation charge of -1 ( ${Br}^{-}$ ).
Since there are two bromines each with a charge of -1.
In addition, we know that ${CoBr}_{2}^{}$ has an overall neutral charge, therefore we can conclude that the cation (cobalt), $Co$ must have an oxidation state of +2 to neutralize the -2 charge from the two bromine anions.

Example:

What is the oxidation state of zinc in ${ZnCO}_{3}^{}$ . (Note: the ${CO}_{3}^{}$ anion has a charge state of -2)

Answer
Knowing that ${CO}_{3}^{}$ has a charge of -2 and knowing that the overall charge of this compound is neutral, we can conclude that zinc has an oxidation state of +2.
This gives us ${Zn}^{2 +}$ and ${CO}_{3}^{- 2}$ , in which the positive and negative charges from zinc and carbonate will cancel with each other, resulting in an overall neutral charge expected of a compound.

Electron configurations of unpaired electrons are said to be paramagnetic and respond to the proximity of magnets.

Fully paired electrons are diamagnetic and do not feel this influence. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is.

${Mn}_{2}^{} O_{3}^{}$ is manganese(III) oxide with manganese in the +3 state. 4 unpaired electrons means this complex is paramagnetic.

$[Ar] 4 s^{0} 3 d^{4}$

${MnO}_{2}^{}$ is manganese(IV) oxide, where manganese is in the +4 state. 3 unpaired electrons means this complex is less paramagnetic than Mn³⁺.

$[Ar] 4 s^{0} 3 d^{3}$

${KMnO}_{4}^{}$ is potassium permanganate, where manganese is in the +7 state with no electrons in the 4s and 3d orbitals.

$[Ar] 4 s^{0} 3 d^{0}$

Since the 3p orbitals are all paired, this complex is diamagnetic.

Oxidation state of iron in [Fe(H₂O)₅NO]SO₄ is :

Example: Find oxidation number of $F e$ in $K_{3} [F e (C N)_{6}]$ .

Answer::

In the complex $K_{3} [F e (C N)_{6}]$ K(Potassium) is Alkaline metal with +1 oxidation state, then for the $K_{3}$ it’s +3. CN is a Negative ligand ,for $C N_{6}$ it’s -6 Let, oxidation state of $F e$ will be $x$ .

The total oxidation state will be zero. $∴ 3 + x + - 6 = 0$ $∴ x = + 3$ thus the oxidation state of $F e$ is +3.

Find the oxidation number of Fe in $[F e (C N)_{6}]^{4-}$

Find the Oxidation number of $F e$ in $K_{2} F e O_{4}$ ?

$2 (+ 1) + x + 4 (- 2) = 0 2 + x - 8 = 0 x - 6 = 0 x = + 6$ Oxidation number of $F e$ is $+ 6$

Oxidation number of $F e$ in $F e_{3} O_{4}$ are:

$F e_{3} O_{4}$ contains $F e$ atoms of both $+ 2$ and $+ 3$ oxidation number.

It is a mixture of Ferrous $(F e O)$ and Ferric $(F e_{2} O_{3})$ oxides combined as $F e O . F e_{2} O_{3}$ .

-Oxidation state of oxygen in $F e O$ and $F e_{2} O_{3}$ is -2.

-Let the oxidation state of Fe in $F e O$ and $F e_{2} O_{3}$ be x and y respectively.

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Therefore, Oxidation state of $F e$ in $F e O$ – $x + (- 2) = 0$ $\Rightarrow x = 2$ Oxidation state of $F e$ in $F e_{2} O_{3}$ – $2 (y) + 3 (- 2) = 0$ $\Rightarrow 2 y = 6$ $\Rightarrow y = 3$ Hence the oxidation number of $F e$ in $F e_{3} O_{4}$ are +2 and +3 respectively.

Chemistry For Third Secondary

Curriculum

Oxidation states of transition elements

Oxidation states of the first transition series :

The common oxidation states of first transition series:

The oxidation states of transition elements in some of their compounds:

How to calculate the oxidation number ?

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$Oxidation states of the first transition series :$