Question

What is the hybridisation in the central atom of ${\mathrm{H}}_3{\mathrm{O}}^{+}$?

Answer 1

Step 1: Formula used for calculation:

• Hybridization$=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where $\mathrm{V}$ is the number of valence electrons of the central atom, $\mathrm{M}$ is the monovalent atoms attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$represents the negative charge.

Step 2: Calculation of hybridization of Hydronium ion ${\mathrm{H}}_3{\mathrm{O}}^{+}$:

• The central atom in ${\mathrm{H}}_3{\mathrm{O}}^{+}$ is $\mathrm{O}$. The atomic number of $\mathrm{O}$ is $8$ so its electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4$.
• So $\mathrm{O}$ has $6$ valence electrons and there are $3$ monovalent $\mathrm{H}$ atoms attached to it.
• So, the hybridization of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ $=\frac{1}{2}\left[6+3-1+0\right]=4$
• The hybridization number is $4$ thus the hybridization of ${\mathrm{H}}_3{\mathrm{O}}^{+}$is ${\mathrm{sp}}^3$.
• The geometry of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ is trigonal pyramidal due to the presence of one lone pair of electrons.

Hence, the hybridisation in the central atom of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ is ${\mathrm{sp}}^3$.