Final answer:
The Lewis structure for XeO₂ has xenon as the central atom with two double bonds to oxygen atoms and two lone pairs, resulting in the electron-pair arrangement AX₂E₂ with a tetrahedral electron-pair geometry. Option c is correct.
Step-by-step explanation:
When determining the Lewis structure for XeO₂, it is clear that xenon (Xe) is the central atom with oxygen (O) atoms bonded to it. Xenon, being a noble gas, can exceed the octet rule and form multiple bonds. From the references provided, we understand that in analog molecules such as XeF₂, the central atom Xe can have two bonding pairs and multiple lone pairs.
In the case of XeO₂, we would draw a Lewis structure with xenon as the central atom, forming double bonds with each of the two oxygen atoms. This results in the usage of four electrons for bonds with the oxygens (two pairs). Additionally, xenon will have lone pairs of electrons. As xenon has eight valence electrons and four are used in bonding with oxygen, it's left with two lone pairs to be accommodated on the xenon atom.
The electron-pair arrangement is thus AX₂E₂, with a steric number (SN) of 4, indicating that there are four regions of electron density around the xenon atom (two bonding pairs and two lone pairs). This results in a tetrahedral electron-pair geometry, even though the actual molecular shape will be bent due to the lone pairs causing the bonding pairs to be pushed together.
Therefore, the correct information about the electron-pair arrangement for XeO₂ is: Electron-pair arrangement: AX₂E₂, SN = 4, and tetrahedral. Option c is the correct choice.