Give the oxidation state of the metal for each of the...

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula $\mathrm{M}_{3} \mathrm{O}_{4}$ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format $\mathrm{MO} \cdot \mathrm{M}_{2} \mathrm{O}_{3},$ to permit estimation of the metal's two oxidation states.) (a) $\mathrm{Sc}_{2} \mathrm{O}_{3}$ (b) $\mathrm{Ti} \mathrm{O}_{2}$ (c) $\mathrm{V}_{2} \mathrm{O}_{5}$ (d) $\mathrm{CrO}_{3}$ (e) $\mathrm{MnO}_{2}$ (f) $\mathrm{Fe}_{3} \mathrm{O}_{4}$ (g) $\mathrm{Co}_{3} \mathrm{O}_{4}$ (h) $\mathrm{NiO}$ (i) $\mathrm{Cu}_{2} \mathrm{O}$

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula $\mathrm{M}_{3} \mathrm{O}_{4}$ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format $\mathrm{MO} \cdot \mathrm{M}_{2} \mathrm{O}_{3},$ to permit estimation of the metal's two oxidation states.)
(a) $\mathrm{Sc}_{2} \mathrm{O}_{3}$
(b) $\mathrm{Ti} \mathrm{O}_{2}$
(c) $\mathrm{V}_{2} \mathrm{O}_{5}$
(d) $\mathrm{CrO}_{3}$
(e) $\mathrm{MnO}_{2}$
(f) $\mathrm{Fe}_{3} \mathrm{O}_{4}$
(g) $\mathrm{Co}_{3} \mathrm{O}_{4}$
(h) $\mathrm{NiO}$
(i) $\mathrm{Cu}_{2} \mathrm{O}$

Key Concepts

Mixed Valence Compounds

Mixed valence compounds contain atoms of the same element in different oxidation states. This concept is particularly important when analyzing compounds like certain metal oxides, where the presence of more than one oxidation state can provide insights into electron delocalization and conductivity.

Transition Metal Chemistry

Transition metals are characterized by their ability to adopt multiple oxidation states due to the involvement of d-orbitals in bonding. This variable oxidation state is a key factor in the chemistry of their compounds, affecting color, magnetism, and catalytic properties.

Charge Balance Principle

The charge balance principle states that the sum of the oxidation states of all atoms in a neutral compound must equal zero. This principle is used to calculate unknown oxidation numbers when the oxidation states of other atoms, such as oxygen in oxides, are known.

Oxidation State Determination

This concept involves assigning hypothetical ionic charges to atoms in a compound, based on the idea that bonds can be treated as purely ionic. It is essential for understanding electron transfer in reactions and for predicting the reactivity and properties of compounds.

Transcript

00:03 All right, so in this video, we are looking at finding the oxidation state of the metal in some oxides.

00:14 All right, so one is oxidation states.

00:18 That is the number of electrons, an atom loses organ in order to form a bond.

00:27 And that is actually the charge of that atom and that given compound.

00:33 So the first one is c2 .03.

00:46 So in finding the oxidation state, oxygen, unless in peroxide, like hydrogen peroxide, oxygen should be taken as minus 2.

01:02 But in peroxide, it's minus 1.

01:04 Okay, so in this oxide, the oxygen state of oxygen is minus two.

01:11 So if we don't know that of scantium, we can represent it by a letter, let's say, x.

01:17 But you have two of the scandiums.

01:20 So you have two times x, then plus that of oxygen, which is minus two.

01:31 But we have three of the oxygen, so minus two times three.

01:38 So we have 2x minus 6.

01:42 But we have to equate this to zero because the total charge of the compound is zero.

01:48 The compound is neutral.

01:50 So 2x minus 6 equals 0.

01:54 So if you add 6 to both sides, so you have 2x equals 6.

02:06 So it divided both sides by 2x is going to 6 over 2, which is equal to plus 3.

02:13 So the oxidation state of scandinium and that oxide is plus 3.

02:20 Then the next one, the next one is titanium.

02:32 2.

02:33 Okay.

02:34 Yeah, so titanium, let's call that x.

02:37 So oxidation state of x, right, so that is x plus minus 2 times 2 equals 0.

02:56 So x minus 4 equals 0, so x equals plus 4.

03:02 So the oxidation state of titanium, and that oxide is plus 4.

03:10 Then the next one is 2 .05.

03:22 And oxide of the vanadium.

03:24 Alright, so vanadium we're represented by a letter to say x.

03:28 But you have 2 of the vanadium, so it's 2x, then plus minus 2 times 5 equals 0.

03:40 So 2x minus 10 equals 0.

03:47 So 2x equals 10.

03:50 So x equals 10 over 5, which is equal to 10 over 2, sorry, 10 over 2, which is equal to plus 5.

04:10 So stay instead of vanadium is plus 5...

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