Which ligand can bring highest oxidation state in metal?
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state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Almost all of the transition metals have multiple potential oxidation states.
Introduction
Oxidation results in an increase in the oxidation state. Reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and therefore a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants. On the other hand, lithium (Li) and sodium (Na) are incredibly strong reducing agents (likes to be oxidized), meaning that they easily lose electrons. In this module, we will precisely go over the oxidation states of transition metals.
Unpaired Electrons of d-orbitals
To fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in the d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s orbital electrons since the 3d orbital is located before the 4s orbital in the periodic table.
For example: Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and therefore the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3. The formula for determining oxidation states would be (with the exception of copper and chromium):
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 # representsthe 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).
REAL WORLD EXAMPLES
Magnets are used in electric motors and generators that allow us to have computers, light, telephones, televisions, and electric heat. Magnetism is a function of chemistry that relates to the oxidation state. The oxidation state determines if the element or compound is diamagnetic or paramagnetic. Diamagnetic substances have only paired electrons, and repel magnetic fields weakly. These substances are non-magnetic, such as wood, water, and some plastics. However, paramagnetic substances become magnetic in the presence of a magnetic field. Paramagnetic substances have at least one unpaired electron. Another stronger magnetic force is a permanent magnet called a ferromagnet. These are much stronger and do not require the presence of a magnetic field to display magnetic properties. These are the type of magnets found on your refrigerator.
Problems
Determine the oxidation states of the transition metals found in these neutral compounds. Note: The transition metal is underlined in the following compounds.
(A) Copper(I) Chloride: CuCl(B) Copper(II) Nitrate: Cu(NO3)2(C) Gold(V) Fluoride: AuF5(D) Iron(II) Oxide: FeO(E) Iron(III) Oxide: Fe2O3(F) Lead(II) Chloride: PbCl2(G) Lead(II) Nitrate: Pb(NO3)2 (H) Manganese(II) Chloride: MnCl2(I) Molybdenum trioxide: MoO3(J) Nickel(II) Hydroxide: Ni(OH)2(K) Platinum(IV) Chloride: PtCl4 (L) Silver Sulfide: Ag2S(M) Tungsten(VI) Fluoride: WF6(N) Vanadium(III) Nitride: VN(O) Zirconium Hydroxide: Zr(OH)4Determine the oxidation state of the transition metal for an overall non-neutral compound: Manganate (MnO42-)Why do transition metals have a greater number of oxidation states than main group metals (i.e. alkali metals and alkaline earth metals)?Which transition metal has the most number of oxidation states?Why does the number of oxidation states for transition metals increase in the middle of the group?What two transition metals have only one oxidation state?
Introduction
Oxidation results in an increase in the oxidation state. Reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and therefore a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants. On the other hand, lithium (Li) and sodium (Na) are incredibly strong reducing agents (likes to be oxidized), meaning that they easily lose electrons. In this module, we will precisely go over the oxidation states of transition metals.
Unpaired Electrons of d-orbitals
To fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in the d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s orbital electrons since the 3d orbital is located before the 4s orbital in the periodic table.
For example: Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and therefore the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3. The formula for determining oxidation states would be (with the exception of copper and chromium):
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 # representsthe 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).
REAL WORLD EXAMPLES
Magnets are used in electric motors and generators that allow us to have computers, light, telephones, televisions, and electric heat. Magnetism is a function of chemistry that relates to the oxidation state. The oxidation state determines if the element or compound is diamagnetic or paramagnetic. Diamagnetic substances have only paired electrons, and repel magnetic fields weakly. These substances are non-magnetic, such as wood, water, and some plastics. However, paramagnetic substances become magnetic in the presence of a magnetic field. Paramagnetic substances have at least one unpaired electron. Another stronger magnetic force is a permanent magnet called a ferromagnet. These are much stronger and do not require the presence of a magnetic field to display magnetic properties. These are the type of magnets found on your refrigerator.
Problems
Determine the oxidation states of the transition metals found in these neutral compounds. Note: The transition metal is underlined in the following compounds.
(A) Copper(I) Chloride: CuCl(B) Copper(II) Nitrate: Cu(NO3)2(C) Gold(V) Fluoride: AuF5(D) Iron(II) Oxide: FeO(E) Iron(III) Oxide: Fe2O3(F) Lead(II) Chloride: PbCl2(G) Lead(II) Nitrate: Pb(NO3)2 (H) Manganese(II) Chloride: MnCl2(I) Molybdenum trioxide: MoO3(J) Nickel(II) Hydroxide: Ni(OH)2(K) Platinum(IV) Chloride: PtCl4 (L) Silver Sulfide: Ag2S(M) Tungsten(VI) Fluoride: WF6(N) Vanadium(III) Nitride: VN(O) Zirconium Hydroxide: Zr(OH)4Determine the oxidation state of the transition metal for an overall non-neutral compound: Manganate (MnO42-)Why do transition metals have a greater number of oxidation states than main group metals (i.e. alkali metals and alkaline earth metals)?Which transition metal has the most number of oxidation states?Why does the number of oxidation states for transition metals increase in the middle of the group?What two transition metals have only one oxidation state?
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