Consider the following four elements,which ar | vedclass

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(D) Given that $X$ is in the $16^{th}$ group and $3^{rd}$ period,$X$ is Sulphur $(S)$.Based on the relative positions:$Y$ is above $X$ in the same gro

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$Y$ exhibits variable covalency

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(D) Given that $X$ is in the $16^{th}$ group and $3^{rd}$ period,$X$ is Sulphur $(S)$.Based on the relative positions:$Y$ is above $X$ in the same group ($16^{th}$ group,$2^{nd}$ period),so $Y$ is Oxygen $(O)$.$W$ is to the left of $X$ in the same period ($15^{th}$ group,$3^{rd}$ period),so $W$ is Phosphorus $(P)$.$Z$ is to the right of $X$ in the same period ($17^{th}$ group,$3^{rd}$ period),so $Z$ is Chlorine $(Cl)$.Evaluating the statements:$(a)$ Electronegativity order: $O > Cl > S > P$. Thus,$Y$ $(O)$ has the maximum electronegativity. (Correct)$(b)$ Catenation property order: $S > P > O > Cl$. Thus,$X$ $(S)$ has the maximum catenation property. (Correct)$(c)$ Electron affinity order: $Cl > S > O > P$. Note: $Cl$ has higher electron affinity than $O$. Thus,$Z$ $(Cl)$ has the maximum electron affinity. (Correct)$(d)$ Oxygen $(Y)$ has no $d$-orbitals and can only exhibit a covalency of $2$. It does not exhibit variable covalency. (Incorrect)Therefore,the incorrect statement is $(d)$.

Column $A$	Column $B$
$(i) \; Na_{2}O$	$(a) \; Neutral$
$(ii) \; Al_{2}O_{3}$	$(b) \; Basic$
$(iii) \; N_{2}O$	$(c) \; Acidic$
$(iv) \; Cl_{2}O_{7}$	$(d) \; Amphoteric$

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