Question

How to find the Hybridization of Chlorine in Chlorate ion (${\mathrm{ClO}}_3^{-}$)?

Answer 1

VSEPR Theory (Valence shell electron pair repulsion theory)

• The valence shell electron pair repulsion theory is based on premise that there is repulsion between the pairs of valence electrons in all atoms.
• The atoms will always tend to arrange themselves in a manner in which this electron pair repulsion is minimized.
• This arrangement of atom determines the geometry of the molecule.
• The geometry of molecules can be determined by determining its number of hybrid orbital which is given as:$\frac{1}{2}\left\{\right(\mathrm{number}\mathrm{of}\mathrm{valence}\mathrm{electrons}\mathrm{on}\mathrm{central}\mathrm{atom})+(\mathrm{number}\mathrm{of}\mathrm{monovalent}\mathrm{atoms})-(\mathrm{charge}\mathrm{on}\mathrm{cation})+(\mathrm{charge}\mathrm{on}\mathrm{anion}\left)\right\}$
• After finding the number of hybrid orbital we can easily find the hybridization For example if the number of hybrid orbital is two the hybridization should be $\mathrm{sp}$.

Step 1: Concept of hybridization in Chlorate Ion

• Hybridization is characterized as the concept of mixing two atomic orbitals to give a degenerated new form of orbitals with the same energy levels.
• In Chlorate ion (${\mathrm{ClO}}_3^{-}$) the central atom is chlorine and it is surrounded by three oxygen atoms and has one lone pair of electrons.

Step 2: Calculation of number of Hybrid orbital

• The number of Hybrid orbital can be calculated as: $\frac{1}{2}\left\{\right(\mathrm{number}\mathrm{of}\mathrm{valence}\mathrm{electrons}\mathrm{on}\mathrm{central}\mathrm{atom})+(\mathrm{number}\mathrm{of}\mathrm{monovalent}\mathrm{atoms})-(\mathrm{charge}\mathrm{on}\mathrm{cation})+(\mathrm{charge}\mathrm{on}\mathrm{anion}\left)\right\}$
• The central atom Chlorine has $7$ valence electrons. oxygen is divalent atom so no monovalent atoms are present.
• In ${\mathrm{ClO}}_3^{-}$ has no cationic charge. The cationic charge is $-1$.
• On substituting all these values in the above equation we get:
• $$\begin{gathered} \frac{1}{2}(7+0-0+1) \\ =\frac{8}{2} \\ =4 \end{gathered}$$

Step 3: Determination of hybridization in Chlorate Ion

• As the number of hybrid orbital in Chlorate ion is Four, so there will be one $\mathrm{s}$ and three $\mathrm{p}$ orbital.
• The hybridization of chlorate ion is ${\mathrm{sp}}^3$.

Therefore, The hybridization of Chlorate ion (${\mathrm{ClO}}_3^{-}$) is ${\mathrm{sp}}^3$.