The correct order of following 3d metal oxides, according to their oxidation numbers is :
(a) $CrO_3$
(b) $Fe_2O_3$
(c) $MnO_2$
(d) $V_2O_5$
(e) $Cu_2O$
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Oxidation states in transition metal oxides usually correspond to the maximum valency for higher oxidation states (like Cr and V).
To determine the correct order of the metal oxides based on their oxidation numbers, we need to calculate the oxidation state of the metal in each given oxide.
$CrO_3$: In chromium trioxide, oxygen has an oxidation number of -2. The overall charge is zero, so:
x + 3(-2) = 0 \Rightarrow x - 6 = 0 \Rightarrow x = +6 Thus, for $CrO_3$, the oxidation number of Cr is +6.
$Fe_2O_3$: In iron(III) oxide, the oxidation number of oxygen is -2. The overall charge is zero, so:
2x + 3(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow 2x = +6 \Rightarrow x = +3 Thus, for $Fe_2O_3$, the oxidation number of Fe is +3.
$MnO_2$: In manganese dioxide, oxygen is -2. The overall charge is zero, so:
x + 2(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4 Thus, for $MnO_2$, the oxidation number of Mn is +4.
$V_2O_5$: In vanadium pentoxide, oxygen is -2. The overall charge is zero, so:
2x + 5(-2) = 0 \Rightarrow 2x - 10 = 0 \Rightarrow 2x = +10 \Rightarrow x = +5 Thus, for $V_2O_5$, the oxidation number of V is +5.
$Cu_2O$: In copper(I) oxide, oxygen has an oxidation number of -2. The total charge is zero, so:
2x + (-2) = 0 \Rightarrow 2x = +2 \Rightarrow x = +1 Thus, for $Cu_2O$, the oxidation number of Cu is +1.
Now, we can order the oxides by the oxidation numbers of their metals:
$CrO_3$: +6
$V_2O_5$: +5
$MnO_2$: +4
$Fe_2O_3$: +3
$Cu_2O$: +1
The correct order based on oxidation numbers is: (a) > (d) > (c) > (b) > (e).
Hence, the correct answer is: (a) > (d) > (c) > (b) > (e).
