Write the Lewis Structure for XeCl2.Indicate the number of bonding electron pairs and nonbonding electron pairs around the central atom.Bonding Electron Pairs Nonbonding Electron Pairs or Lone Electron

Step 1: Determine the total number of valence electrons. Xenon (Xe) has 8 valence electrons and each Chlorine (Cl) atom has 7 valence electrons. Since there are two Chlorine atoms, the total number of valence electrons is 8 + 7*2 = 22.

Step 2: Draw a skeleton structure of the molecule. Xenon is the central atom and is surrounded by the two Chlorine atoms.

Step 3: Connect the atoms with single bonds. Each single bond uses up 2 valence electrons, so we have used 4 out of 22 electrons.

Step 4: Complete the octets of the outer atoms (Chlorine atoms) by adding lone pairs. Each Chlorine atom needs 6 more electrons to complete its octet, so we use 12 more electrons. We have now used 4 + 12 = 16 electrons.

Step 5: Place the remaining electrons on the central atom (Xenon). We have 22 - 16 = 6 electrons left, which we place on Xenon.

Step 6: If the central atom does not have an octet, form double or triple bonds until it does. In this case, Xenon has more than an octet, which is allowed because it can have an expanded octet.

So, the Lewis structure for XeCl2 is:

   Cl - Xe - Cl
       :   :
       :   :

The central atom (Xe) has 3 bonding electron pairs (2 with each Cl and one lone pair) and 3 nonbonding electron pairs or lone pairs.