Question

An atom X has three electrons more than the noble gas configuration. (a) What type of ion will it form? (b) Write the formula of its (i) sulphate (ii) nitrate (iii) phosphate (iv) carbonate (v) hydroxide

Answer 1

X has three electrons more than the noble gas configuration. (Given)

It indicates that the number of valence electrons= 3 for X.

As it has three electrons, it has more tendency to lose electrons and attain a stable electronic configuration.

• It will form an ion by the loss of 3 electrons.
• Hence, X will be a tripositive cation.
• Further, number of valence electrons is 3 i.e.; less than 4, therefore Valency= Valence electrons. So, X is trivalent.

Formula of Sulphate of X:

Formula of Sulphate= ${\mathrm{SO}}_4^{-2}$
Valency of Sulphate= 2
Using criss-cross method, where the valency of cation is transposed with the valency of anion and written as the subscript on the symbol.

$$\begin{gathered} \mathrm{Symbol}\mathrm{X}{\mathrm{SO}}_4 \\ \searrow \swarrow \\ \mathrm{Valency}32 \end{gathered}$$
The formula comes out to be ${\mathrm{X}}_2{\left({\mathrm{SO}}_4\right)}_3$

Formula of Nitrate of X:

Formula of Nitrate= ${\mathrm{NO}}_3^{-}$
Valency of Sulphate= 1
Using criss-cross method:

$$\begin{gathered} \mathrm{Symbol}\mathrm{X}{\mathrm{NO}}_3 \\ \searrow \swarrow \\ \mathrm{Valency}31 \end{gathered}$$
The formula comes out to be $\mathrm{X}{\left({\mathrm{NO}}_3\right)}_3$

Formula of Phosphate of X:

Formula of Phosphate= ${\mathrm{PO}}_4^{3-}$
Valency of Phosphate= 3
Using criss-cross method:

$$\begin{gathered} \mathrm{Symbol}\mathrm{X}{\mathrm{PO}}_4 \\ \searrow \swarrow \\ \mathrm{Valency}33 \end{gathered}$$
The formula comes out to be ${\mathrm{X}}_3{\left({\mathrm{PO}}_4\right)}_3$ or ${\mathrm{XPO}}_4$

Formula of Carbonate of X:

Formula of Carbonate= ${\mathrm{CO}}_3^{2-}$
Valency of Carbonate= 3
Using criss-cross method:

$$\begin{gathered} \mathrm{Symbol}\mathrm{X}{\mathrm{CO}}_3 \\ \searrow \swarrow \\ \mathrm{Valency}32 \end{gathered}$$
The formula comes out to be ${\mathrm{X}}_2{\left({\mathrm{CO}}_3\right)}_3$
Formula of Hydroxide of X:
Formula of Hydroxide= ${\mathrm{OH}}^{-}$
Valency of Hydroxide= 1
Using criss-cross method:

$$\begin{gathered} \mathrm{Symbol}\mathrm{X}\mathrm{OH} \\ \searrow \swarrow \\ \mathrm{Valency}31 \end{gathered}$$
The formula comes out to be $\mathrm{X}{\left(\mathrm{OH}\right)}_3$