Which molecules represented by the underlined atoms are in their highest oxidation state?

H_{2} {S - -}_{2} O_{8}

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{P - -}_{4} O_{10}

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F_{2} {O - -}_{2}

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{M n - -- -}_{2} O_{7}

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Solution

The correct options are
A ${P - -}_{4} O_{10}$
C ${M n - -- -}_{2} O_{7}$
D $H_{2} {S - -}_{2} O_{8}$
${M n}_{2} O_{7}$
+7
$F_{2} {O - -}_{2}$
+1
$P_{4} O_{10}$
+5
$H_{2} S_{2} O_{8}$
+6

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Similar questions

Q. Draw the structure of the following:
a.

{S F}_{4}

H_{2} S_{2} O_{7}

H_{2} S_{2} O_{8}

Q. Which molecules represents the unerlined atoms in their lowest oxidation state?

Q. What is the oxidation number of sulphur in

H_{2} S_{2} O_{8}

H_{2} S_{2} O_{8}

and

H_{2} S O_{5}

, both have

+ 6

oxidation state of sulphur. It is due to the :

Column-I-Compound, Column-II-Oxidation state, Column-III-Type of linkage

$\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & N a_{2} S_{2} O_{3} & (i) & + 5 & (P) & S - O - S (I I) & N a_{2} S_{4} O_{6} & (i i) & - 2 & (Q) & S - S (I I I) & H_{2} S_{2} O_{7} & (i i i) & + 6 & (R) & - (O - O) - (I V) & H_{2} S_{2} O_{8} & (i v) & + 4 & (S) & S = S \end{matrix}$

The only correct combination for Marshall’s acid:

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Which molecules represented by the underlined atoms are in their highest oxidation state?

The correct options are A P––4O10 C Mn––––2O7 D H2S––2O8Mn2O7+7F2O––2 +1P4O10 +5H2S2O8 +6

The correct options are
A ${P - -}_{4} O_{10}$
C ${M n - -- -}_{2} O_{7}$
D $H_{2} {S - -}_{2} O_{8}$
${M n}_{2} O_{7}$
+7
$F_{2} {O - -}_{2}$
+1
$P_{4} O_{10}$
+5
$H_{2} S_{2} O_{8}$
+6