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Why can’t bromine give +7 oxidation state?
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+ Inorganic chemistry
+ Oxidation
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Marlene Dixon
Why can’t bromine give +7 oxidation state?
Actually, bromine can.
It is just very difficult to attain, as a matter of fact. Attempts to synthesise a bromine compound with the element in its +7 oxidation state can be traced back to the sixties, but only toward the end of the 20th century was a successful synthetic strategy devised and carried out. It involves the oxidation of bromate anion with a strong oxidiser in the presence of OH¯ ions:
BrO₃¯ + F₂ + 2OH¯ → BrO₄¯ + H₂O + 2F¯
Surprisingly enough, the perbromate anion BrO₄¯ is a very stable one, although it is a very strong oxidiser itself. The central Br atom displays a +7 oxidation state.
The reluctance of bromine to form compounds in its highest possible oxidation state follows the same trend displayed by other p-block elements[1] of the fourth Period, As and Se: this phenomenon arises from their position in the Periodic Table that follows d-block elements. As the d-orbitals have marked characteristics of directionality, a full shell of such a set of orbitals poorly screens the nuclear charge (the so-called lanthanide or scandide contraction)[2] and, as a result, makes some electrons less available to chemical bonding.
Here in Br, the 3d orbitals are full of electrons and cannot screen efficiently the nuclear charge: this, in turn, means less availability to form bonds of some of the outermost electrons.
It is just very difficult to attain, as a matter of fact. Attempts to synthesise a bromine compound with the element in its +7 oxidation state can be traced back to the sixties, but only toward the end of the 20th century was a successful synthetic strategy devised and carried out. It involves the oxidation of bromate anion with a strong oxidiser in the presence of OH¯ ions:
BrO₃¯ + F₂ + 2OH¯ → BrO₄¯ + H₂O + 2F¯
Surprisingly enough, the perbromate anion BrO₄¯ is a very stable one, although it is a very strong oxidiser itself. The central Br atom displays a +7 oxidation state.
The reluctance of bromine to form compounds in its highest possible oxidation state follows the same trend displayed by other p-block elements[1] of the fourth Period, As and Se: this phenomenon arises from their position in the Periodic Table that follows d-block elements. As the d-orbitals have marked characteristics of directionality, a full shell of such a set of orbitals poorly screens the nuclear charge (the so-called lanthanide or scandide contraction)[2] and, as a result, makes some electrons less available to chemical bonding.
Here in Br, the 3d orbitals are full of electrons and cannot screen efficiently the nuclear charge: this, in turn, means less availability to form bonds of some of the outermost electrons.
In case of bromine the penultimate shell ( 3S²P6d10 ) is weakly screening and has the energy required to promote an electron from 4s - orbital of the atom in the second exited state to 4d - orbital is markedly higher than that required in chlorine. Thus Br doesn't exhibit +7 oxidation state.
In case of bromine the penultimate shell ( 3S²P6d10 ) is weakly screening and has the energy required to promote an electron from 4s - orbital of the atom in the second exited state to 4d - orbital is markedly higher than that required in chlorine. Thus Br doesn't exhibit +7 oxidation state.
Actually, bromine can.
It is just very difficult to attain, as a matter of fact. Attempts to synthesise a bromine compound with the element in its +7 oxidation state can be traced back to the sixties, but only toward the end of the 20th century was a successful synthetic strategy devised and carried out. It involves the oxidation of bromate anion with a strong oxidiser in the presence of OH¯ ions:
BrO₃¯ + F₂ + 2OH¯ → BrO₄¯ + H₂O + 2F¯
Surprisingly enough, the perbromate anion BrO₄¯ is a very stable one, although it is a very strong oxidiser itself. The central Br atom displays a +7 oxidation state.
The reluctance of bromine to form compounds in its highest possible oxidation state follows the same trend displayed by other p-block elements[1] of the fourth Period, As and Se: this phenomenon arises from their position in the Periodic Table that follows d-block elements. As the d-orbitals have marked characteristics of directionality, a full shell of such a set of orbitals poorly screens the nuclear charge (the so-called lanthanide or scandide contraction)[2] and, as a result, makes some electrons less available to chemical bonding.
Here in Br, the 3d orbitals are full of electrons and cannot screen efficiently the nuclear charge: this, in turn, means less availability to form bonds of some of the outermost electrons.
Footnotes
[1] p-Block ElementsActually, bromine can.
It is just very difficult to attain, as a matter of fact. Attempts to synthesise a bromine compound with the element in its +7 oxidation state can be traced back to the sixties, but only toward the end of the 20th century was a successful synthetic strategy devised and carried out. It involves the oxidation of bromate anion with a strong oxidiser in the presence of OH¯ ions:
BrO₃¯ + F₂ + 2OH¯ → BrO₄¯ + H₂O + 2F¯
Surprisingly enough, the perbromate anion BrO₄¯ is a very stable one, although it is a very strong oxidiser itself. The central Br atom displays a +7 oxidation state.
The reluctance of bromine to form compounds in its highest possible oxidation state follows the same trend displayed by other p-block elements[1] of the fourth Period, As and Se: this phenomenon arises from their position in the Periodic Table that follows d-block elements. As the d-orbitals have marked characteristics of directionality, a full shell of such a set of orbitals poorly screens the nuclear charge (the so-called lanthanide or scandide contraction)[2] and, as a result, makes some electrons less available to chemical bonding.
Here in Br, the 3d orbitals are full of electrons and cannot screen efficiently the nuclear charge: this, in turn, means less availability to form bonds of some of the outermost electrons.
Footnotes
[1] p-Block ElementsMore
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In case of bromine the penultimate shell ( 3S²P6d10 ) is weakly screening and has the energy required to promote an electron from 4s - orbital of the atom in the second exited state to 4d - orbital is markedly higher than that required in chlorine. Thus Br doesn't exhibit +7 oxidation state.
In case of bromine the penultimate shell ( 3S²P6d10 ) is weakly screening and has the energy required to promote an electron from 4s - orbital of the atom in the second exited state to 4d - orbital is markedly higher than that required in chlorine. Thus Br doesn't exhibit +7 oxidation state.
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