The correct decreasing order of oxidation sta | vedclass

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(B) To find the oxidation state of the underlined atoms:$1$. In $P_4O_{10}$, $4x + 10(-2) = 0 \Rightarrow 4x = 20 \Rightarrow x = +5$.$2$. In $SO_3$,

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$N_2O_5 > Al_2O_3 > H_2S$

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(B) To find the oxidation state of the underlined atoms:$1$. In $P_4O_{10}$, $4x + 10(-2) = 0 \Rightarrow 4x = 20 \Rightarrow x = +5$.$2$. In $SO_3$, $x + 3(-2) = 0 \Rightarrow x = +6$.$3$. In $H_2O$, $2(+1) + x = 0 \Rightarrow x = -2$.Order: $+5, +6, -2$ (Not decreasing).$1$. In $N_2O_5$, $2x + 5(-2) = 0 \Rightarrow 2x = 10 \Rightarrow x = +5$.$2$. In $Al_2O_3$, $2x + 3(-2) = 0 \Rightarrow 2x = 6 \Rightarrow x = +3$.$3$. In $H_2S$, $2(+1) + x = 0 \Rightarrow x = -2$.Order: $+5, +3, -2$ (Decreasing order).$1$. In $PbO_2$, $x + 2(-2) = 0 \Rightarrow x = +4$.$2$. In $N_2O_3$, $2x + 3(-2) = 0 \Rightarrow 2x = 6 \Rightarrow x = +3$.$3$. In $SO_3$, $x + 3(-2) = 0 \Rightarrow x = +6$.Order: $+4, +3, +6$ (Not decreasing).$1$. In $P_4O_6$, $4x + 6(-2) = 0 \Rightarrow 4x = 12 \Rightarrow x = +3$.$2$. In $Cl_2O_7$, $2x + 7(-2) = 0 \Rightarrow 2x = 14 \Rightarrow x = +7$.$3$. In $AlH_3$, $x + 3(-1) = 0 \Rightarrow x = +3$.Order: $+3, +7, +3$ (Not decreasing).Thus, option $B$ is correct.

The correct decreasing order of oxidation state of the underlined atom in each molecule is:

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