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Question

Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
  1. True
  2. False

A
False
B
True
Solution
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1)In case of oxygen, Mn shows the highest oxidation state of +7. This is because Mn forms pπdπ multiple bonds using 2p orbitals of oxygen and 3d orbitals of Mn.

2) With F,Mn displays an oxidation state of +4 because of the single bond formation caused by the unavailability of 2p orbitals in F for multiple bonding.

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Similar Questions
Q1
Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
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Q2
Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4 .
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.
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Q3
(a) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Zirconium and Hafnium exhibit similar properties.
(iii) Transition metals act as catalysts.
(b) Complete the following equations:

(i)2MnO2+4KOH+O2 (ii)Cr2O27+14H++6I
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Q4
(a) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Zirconium and Hafnium exhibit similar properties.
(iii) Transition metals act as catalysts.
(b) Complete the following equations:

(i)2MnO2+4KOH+O2 (ii)Cr2O27+14H++6I

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Q5
(A) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows oxidation state of +4.
(ii) Cr2+ is a strong reducing agent.
(iii) Cu2+ salts are coloured, while Zn2+ salts are white.
(B) Complete the following equations:
(i) 2MnO2+4KOH+O2Δ
(ii)
Cr2O27+14H++6e
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