The correct decreasing order of the following | vedclass

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(B) The oxidizing and fluorinating power of xenon fluorides depends on the ability of the $Xe$ atom to get reduced, which is directly related to the o

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$XeF_6 > XeF_4 > XeF_2$

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(B) The oxidizing and fluorinating power of xenon fluorides depends on the ability of the $Xe$ atom to get reduced, which is directly related to the oxidation state of $Xe$ in the compound.In $XeF_6$, the oxidation state of $Xe$ is $+6$.In $XeF_4$, the oxidation state of $Xe$ is $+4$.In $XeF_2$, the oxidation state of $Xe$ is $+2$.As the oxidation state increases, the tendency to get reduced increases, making the compound a stronger oxidizing and fluorinating agent.Therefore, the correct decreasing order is $XeF_6 > XeF_4 > XeF_2$.

List-$I$	List-$II$
$(A) \ XeF_4$	$(1) \ \text{Distorted octahedral}$
$(B) \ XeF_6$	$(2) \ \text{Tetrahedral}$
$(C) \ XeO_3$	$(3) \ \text{Square planar}$
$(D) \ XeO_4$	$(4) \ \text{Pyramidal}$

The correct decreasing order of the following $Xe$ compounds to act as both fluorinating and oxidizing agent is
$(I)$ $XeF_6$
$(II)$ $XeF_4$
$(III)$ $XeF_2$

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists:

List-$I$ List-$II$

$(A) \ XeF_4$ $(1) \ \text{Distorted octahedral}$

$(B) \ XeF_6$ $(2) \ \text{Tetrahedral}$

$(C) \ XeO_3$ $(3) \ \text{Square planar}$

$(D) \ XeO_4$ $(4) \ \text{Pyramidal}$

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The correct decreasing order of the following $Xe$ compounds to act as both fluorinating and oxidizing agent is$(I)$ $XeF_6$$(II)$ $XeF_4$$(III)$ $XeF_2$