Which of the following are likely to form an ionic bond?
To determine whether a bond is ionic, we need to consider the electronegativity difference between the atoms involved. Ionic bonds usually form between atoms with a large electronegativity difference.
The difference in electronegativity should be greater than 1.7 to consider the bond predominantly ionic.
Among the following examples, the atoms that are most likely to form an ionic bond are:
1. Sodium (Na) and Chlorine (Cl): The electronegativity difference between Na (0.93) and Cl (3.16) is 2.23, so they are likely to form an ionic bond.
2. Cesium (Cs) and Fluorine (F): The electronegativity difference between Cs (0.79) and F (3.98) is 3.19, so they are also likely to form an ionic bond.
On the other hand, all the other elements listed have electronegativity differences of less than 1.7, so they are less likely to form ionic bonds:
3. Carbon (C) and Oxygen (O): The electronegativity difference between C (2.55) and O (3.44) is only 0.89.
4. Hydrogen (H) and Sulfur (S): The electronegativity difference between H (2.20) and S (2.58) is only 0.38.
5. Nitrogen (N) and Oxygen (O): The electronegativity difference between N (3.04) and O (3.44) is only 0.40.
Therefore, only sodium and chlorine (NaCl) and cesium and fluorine (CsF) are likely to form ionic bonds from the options provided.