An atom $X$ has three electrons more than noble gas configuration. What type of ion will it form? Write the formula of its ‘Phosphate’

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Solution

An atom $X$ has three electrons more than noble gas configuration. Thus, it will lose $3$ electrons and form a cation: $X^{3 +}$
The charge on phosphate ion is $- 3$ i.e. $P O_{4}^{3 -}$ and hence the formula of its Phosphate is $X P O_{4}$

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An atom X has three electrons more than noble gas configuration. What type of ion will it form? Write the formula of its ‘Phosphate’

An atom X has three electrons more than noble gas configuration. Thus, it will lose 3 electrons and form a cation: X3+ The charge on phosphate ion is −3 i.e. PO3−4 and hence the formula of its Phosphate is XPO4

An atom $X$ has three electrons more than noble gas configuration. What type of ion will it form? Write the formula of its ‘Phosphate’

An atom $X$ has three electrons more than noble gas configuration. Thus, it will lose $3$ electrons and form a cation: $X^{3 +}$
The charge on phosphate ion is $- 3$ i.e. $P O_{4}^{3 -}$ and hence the formula of its Phosphate is $X P O_{4}$