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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 ______.
पर्याय
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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`MnO_4^(-) + H_2O + I^(-) ->`
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(i) \[\ce{CrO3}\]
(ii) \[\ce{MoO3}\]
(iii) \[\ce{WO3}\]
(iv) \[\ce{CrO^{2-}4}\]
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| Column I (Aqueous solution of salt) |
Column II (Colour) |
| (i) \[\ce{FeSO2.7H2O}\] | (a) Green |
| (ii) \[\ce{NiCl2.4H2O}\] | (b) Light pink |
| (iii) \[\ce{MnCl2.4H2O}\] | (c) Blue |
| (iv) \[\ce{CoC12,6H2O}\] | (d) Pale green |
| (v) \[\ce{Cu2 Cl2}\] | (e) Pink |
| (f) Colourless |
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