Question

draw the lewis structure for the clo₃⁻ ion. predict its geometry and describe the hybridization state of the cl atom. part 1 of 3 draw the lewis structure for the clo₃⁻ ion. be sure to add in all non - zero formal charges. click and drag to start drawing a structure.

Explanation:

Step1: Count valence electrons

Chlorine (Cl) has 7 valence electrons, oxygen (O) has 6 valence electrons. For $ClO_3^ -$, the total number of valence electrons is $7 + 3\times6+1=26$ (the extra 1 is due to the - 1 charge).

Step2: Determine the central atom

Chlorine is the central atom as oxygen is more electronegative.

Step3: Connect atoms and distribute electrons

Connect Cl to 3 O atoms with single - bonds first, using 6 electrons. Then distribute the remaining 20 electrons as lone - pairs on the O atoms. But to minimize formal charges, we can form a double - bond between Cl and one of the O atoms.
The Lewis structure has Cl in the center, single - bonded to two O atoms (each with 3 lone - pairs) and double - bonded to one O atom (with 2 lone - pairs). The formal charge on Cl is $7-(2 + \frac{1}{2}\times6)=1$, and on the two singly - bonded O atoms is $6-(6+\frac{1}{2}\times2)= - 1$, and on the doubly - bonded O atom is $6-(4+\frac{1}{2}\times4)=0$.

Step4: Predict geometry

The number of bonding regions and lone - pairs around Cl is 4 (3 bonding and 1 lone - pair). According to VSEPR theory, the electron - pair geometry is tetrahedral and the molecular geometry is trigonal pyramidal.

Step5: Determine hybridization

Since there are 4 electron - domains around Cl, the hybridization of Cl is $sp^3$.

Answer:

The Lewis structure has Cl as the central atom, single - bonded to two O atoms (each with 3 lone - pairs) and double - bonded to one O atom (with 2 lone - pairs) and overall charge of - 1. The geometry is trigonal pyramidal and the hybridization of Cl is $sp^3$.