Draw the Lewis Structure of H2S2 and Identify Its Geometry

To draw the Lewis structure of H₂S₂ (dihydrogensulfide), we first need to account for the valence electrons in the molecule. Hydrogen has 1 valence electron and sulfur has 6 valence electrons. Thus, for H₂S₂, the total number of valence electrons is:

2 (from H) + 2 × 6 (from S) = 14 valence electrons.

Next, we arrange the atoms in the structure. The two sulfur atoms will be connected to each other, and each sulfur will be bonded to one hydrogen atom. The basic structure looks like this:

    H   H
     |   |
   S - S

Now, we will distribute the remaining valence electrons. Each H atom uses 1 electron to form a bond with S, consuming 2 electrons total. The remaining 12 electrons (14 total – 2 used in bonding) will be used to complete the octets of the sulfur atoms.

We can put 6 electrons (3 lone pairs) around each sulfur atom:

    H   H
     |   |
   S:  :S
     :   :

This gives us a Lewis structure with the following arrangement:

   H   H
    |   |
   S:S
    :   :

Now we can identify the geometry of the molecule. The central S-S bond is formed, and the angles around each sulfur atom can be determined. Since there are two hydrogen atoms bonded to one sulfur and two lone pairs of electrons around each sulfur, we start by noting that the geometry of each sulfur is determined by the 2 bonded pairs and 2 lone pairs, which results in a bent or V-shaped geometry. The bond angle is approximately 104.5°.

In summary, the Lewis structure for H₂S₂ shows two sulfur atoms bonded to each other with one hydrogen atom each, and the molecular geometry is bent due to lone pair repulsions.