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प्रश्न
Highest oxidation state of manganese in fluoride is \[\ce{+4 (MnF4)}\] but highest oxidation state in oxides is \[\ce{+7 (Mn2O7)}\] because ______.
विकल्प
fluorine is more electronegative than oxygen.
fluorine does not possess d-orbitals.
fluorine stabilises lower oxidation state.
in covalent compounds fluorine can form single bond only while oxygen forms double bond.
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उत्तर
Highest oxidation state of manganese in fluoride is \[\ce{+4 (MnF4)}\] but highest oxidation state in oxides is \[\ce{+7 (Mn2O7)}\] because in covalent compounds fluorine can form single bond only while oxygen forms double bond.
Explanation:
Oxygen has the capacity to form multiple bonds which enables it to form a variety of covalent compounds.
In \[\ce{(Mn2O7)}\] also, 6 oxygen are doubly bonded to two manganese atoms and one oxygen is forming bridge between two.
While in \[\ce{(MnF4)}\], four fluorine atoms are singly bonded to manganese atom giving it a +4 oxidation state.
Therefore, due to capability of oxygen to have multiple bonds in covalent compounds, manganese is having higher oxidation state of +7 in \[\ce{(Mn2O7)}\].
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संबंधित प्रश्न
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3d3
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(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
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3) Which ion is colourless and why?
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