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If an element $\ce{X}$ forms the highest oxide of the formula $\ce{XO3}$, then it belongs to group:
A) 14
B) 15
C) 16
D) 17

How is an oxidation number of an element related to its group?

I tried comparing $\ce{X}$ with $\ce{N}$-, $\ce{S}$-, $\ce{As}$-compounds, but all 3 come in the form $\ce{XO3}$ ($\ce{NO3}$, $\ce{SO3}$, $\ce{AsO3}$), so I can't guess the group number.

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    $\begingroup$ Read the text more carefully. It says that the "highest oxide" has the formula $\ce{XO_3}$, while the 3 possibilities you mention contain only one oxide and 2 oxyanions. $\endgroup$
    – sabinagio
    Commented Jul 13, 2014 at 11:45
  • $\begingroup$ Isn't $\ce{SO3}$ the highest oxide of $\ce{S}$? $\endgroup$ Commented Jul 25, 2014 at 20:20
  • $\begingroup$ actually SO4- is the highest oxide. Oxide means that the oxygen is more electronegative, not necessarily that the compound is also neutral. $\endgroup$
    – Caters
    Commented Aug 16, 2014 at 21:17

1 Answer 1

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For main group elements (groups 1, 2, 13-18), the highest possible oxidation state $O_\text{max}$ is given by the group number $G$ (for groups 1 and 2) or by the group number - 10 (for the rest): $$ O_\text{max} = \begin{cases} G &\text{if } G \in \{1,2\} \\ G - 10 &\text{if } 13 \leq G \leq 18 \end{cases} $$

This means that for your compound $\ce{\stackrel{VI}{X}O3}$, so with the oxidation state VI, the corresponding group would be group 16.

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