Question

review each of the chemical compounds and classify each one as either polar or nonpolar
h₂te
sbr₄
ash₃
i₂

Explanation:

• For \(H_2Te\):
• The central atom is \(Te\). It has \(6\) valence electrons. It forms \(2\) bonds with \(H\) atoms and has \(2\) lone pairs (\(6 - 2= 4\) electrons in lone pairs, \(4\div2 = 2\) lone pairs).
• The molecular geometry is bent (similar to \(H_2O\) but with less electronegativity difference). The bond dipoles do not cancel out (because of the lone pairs and the difference in electronegativity between \(Te\) and \(H\)), so it is polar.
• For \(SiBr_4\):
• The central atom \(Si\) has \(4\) valence electrons. It forms \(4\) bonds with \(Br\) atoms (no lone pairs as \(4 - 4=0\)).
• The molecular geometry is tetrahedral. The bond dipoles (due to \(Si - Br\) bonds) cancel out because of the symmetric tetrahedral shape. So it is non - polar.
• For \(AsH_3\):
• The central atom \(As\) has \(5\) valence electrons. It forms \(3\) bonds with \(H\) atoms and has \(1\) lone pair (\(5 - 3 = 2\) electrons in lone pair).
• The molecular geometry is trigonal pyramidal. The bond dipoles (due to \(As - H\) bonds) do not cancel out (because of the lone pair), so it is polar.
• For \(I_2\):
• It is a diatomic molecule with two \(I\) atoms. The bond is non - polar (\(I - I\) bond, same atoms, so no electronegativity difference). So it is non - polar.

Answer:

• Polar: \(H_2Te\), \(AsH_3\)
• Nonpolar: \(SiBr_4\), \(I_2\)