For each of the following bonds, indicate the positive end with \delta^{+} and the negative end with \delta^{-}. Write an arrow to show the dipole for each.
1. Si-Cl
2. Se-F
3. Br-F
4. N-H
5. N-P
I thought we went over this a couple of days ago. What's the problem?
To determine the positive (+) and negative (-) ends of a bond, we need to consider the electronegativity difference between the atoms involved in the bond. The more electronegative atom will have a partial negative charge (\(\delta^{-}\)), while the less electronegative atom will have a partial positive charge (\(\delta^{+}\)).
1. Si-Cl:
Silicon (Si) is less electronegative than chlorine (Cl) because chlorine is higher in the periodic table and has a higher electronegativity value. Thus, Si-Cl bond is polar, and the arrow indicating the bond dipole points towards the chlorine (Cl) end (\(\delta^{+}\) at Silicon and \(\delta^{-}\) at Chlorine).
2. Se-F:
Selenium (Se) has a lower electronegativity than fluorine (F). Consequently, the Se-F bond is polar, and the arrow indicating the bond dipole is directed towards fluorine (F) (\(\delta^{+}\) at Selenium and \(\delta^{-}\) at Fluorine).
3. Br-F:
Bromine (Br) is less electronegative than fluorine (F). Therefore, the Br-F bond is polar, and the arrow indicating the bond dipole is directed towards fluorine (F) (\(\delta^{+}\) at Bromine and \(\delta^{-}\) at Fluorine).
4. N-H:
Nitrogen (N) is more electronegative than hydrogen (H). Hence, the N-H bond is polar, and the arrow indicating the bond dipole points towards nitrogen (N) (\(\delta^{+}\) at Hydrogen and \(\delta^{-}\) at Nitrogen).
5. N-P:
Nitrogen (N) is more electronegative than phosphorus (P). Thus, the N-P bond is polar, and the arrow indicating the bond dipole points towards nitrogen (N) (\(\delta^{+}\) at Phosphorus and \(\delta^{-}\) at Nitrogen).
Remember, the positive and negative charges are denoted by \(\delta^{+}\) and \(\delta^{-}\) because the charges are partial, meaning they are smaller than a full +1 or -1 charge.