The correct shape and hybridization of \(XeF_2\) are:

linear; \(sp^3d\)

The correct answer is option 1. linear; \(sp^3d\). 

Xenon (Xe) is a noble gas with 8 valence electrons. Fluorine (F) has 7 valence electrons, and it needs 1 electron to complete its octet. In \( \text{XeF}_2 \), xenon forms bonds with two fluorine atoms, using 2 of its 8 valence electrons. This leaves xenon with 6 remaining electrons, which will be arranged as 3 lone pairs.

The total number of electron pairs (bond pairs + lone pairs) around xenon is 5 (2 bond pairs + 3 lone pairs). To accommodate 5 electron pairs, xenon undergoes \( sp^3d \) hybridization. In \( sp^3d \) hybridization, one \( s \), three \( p \), and one \( d \) orbital mix to form five hybrid orbitals.

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the arrangement of electron pairs around the central atom will minimize repulsion. 

With 5 regions of electron density, the electron pair geometry would be trigonal bipyramidal. However, the lone pairs prefer to occupy the equatorial positions in this geometry because these positions experience less repulsion compared to the axial positions. This results in the two fluorine atoms occupying the axial positions, and the lone pairs occupying the equatorial positions. The axial positions are 180° apart, leading to a linear molecular shape.

The 3 lone pairs occupy the equatorial positions in a trigonal bipyramidal arrangement to minimize repulsion. The 2 fluorine atoms occupy the axial positions, 180° apart. This gives \( \text{XeF}_2 \) a linear shape. The hybridization of xenon is \( sp^3d \) because 5 regions of electron density correspond to this hybridization.

Thus, the correct description of \( \text{XeF}_2 \) is linear with \( sp^3d \) hybridization. The correct option is: 1. linear; \( sp^3d \).