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

Look up the electronegativities of each pair. The higher EN will have a negative charge; the smaller EN will have a positive charge. So you can write delta - on the negative end and delta + on the positive end. For example,

N-H. EN for N is about 3.0 and H is about 2.1. That makes N the negative end and H the positive end.

What about the N-P?

1. Si-Cl:

- The chlorine atom is more electronegative than the silicon atom, so it pulls the electrons towards itself.
- The arrow will point from the silicon atom (\delta^{+}) towards the chlorine atom (\delta^{-}).
- The dipole can be represented as: Si \delta^{+} -> Cl \delta^{-}.

2. Se-F:
- Fluorine is significantly more electronegative than selenium, so the electrons are pulled towards fluorine.
- The arrow will point from selenium (\delta^{+}) towards fluorine (\delta^{-}).
- The dipole can be represented as: Se \delta^{+} -> F \delta^{-}.

3. Br-F:
- Fluorine is more electronegative than bromine, so the electrons are pulled towards fluorine.
- The arrow will point from bromine (\delta^{+}) towards fluorine (\delta^{-}).
- The dipole can be represented as: Br \delta^{+} -> F \delta^{-}.

4. N-H:
- Nitrogen is more electronegative than hydrogen, so the electrons are pulled towards nitrogen.
- The arrow will point from hydrogen (\delta^{+}) towards nitrogen (\delta^{-}).
- The dipole can be represented as: H \delta^{+} -> N \delta^{-}.

5. N-P:
- Both nitrogen and phosphorus are non-metals and have similar electronegativities.
- In this case, the bond is nonpolar, meaning there is no significant dipole moment.
- The dipole for N-P will be represented as: N-P.

To determine the positive and negative ends of a bond and its corresponding dipole, we need to consider the electronegativity difference between the atoms involved in the bond.

Electronegativity is a measure of an atom's ability to attract electrons towards itself in a chemical bond. If the electronegativity difference between two atoms is significant, the bond will exhibit a polarity, meaning one end will be partially positive (\delta^{+}), and the other end will be partially negative (\delta^{-}).

To find the electronegativity difference, we can use the Pauling scale, where the values are given for each atom. The greater the difference in electronegativity, the more polar the bond will be.

Here are the electronegativity values for the elements involved in the given bonds:
1. Si-Cl: Si = 1.90, Cl = 3.16 (electronegativity difference = 1.26)
2. Se-F: Se = 2.55, F = 3.98 (electronegativity difference = 1.43)
3. Br-F: Br = 2.96, F = 3.98 (electronegativity difference = 1.02)
4. N-H: N = 3.04, H = 2.20 (electronegativity difference = 0.84)
5. N-P: N = 3.04, P = 2.19 (electronegativity difference = 0.85)

Now, let's analyze each bond one by one:

1. Si-Cl:
The electronegativity difference between Si and Cl is 1.26. Since Cl is more electronegative than Si, the bond will be polar, with the Cl atom being partially negative (\delta^{-}), and the Si atom being partially positive (\delta^{+}). To represent this dipole, draw an arrow pointing from Si to Cl.

\delta^{+} Si ——> Cl \delta^{-}

2. Se-F:
The electronegativity difference between Se and F is 1.43. Since F is more electronegative than Se, the bond will be polar, with the F atom being partially negative (\delta^{-}), and the Se atom being partially positive (\delta^{+}). To represent this dipole, draw an arrow pointing from Se to F.

\delta^{+} Se ——> F \delta^{-}

3. Br-F:
The electronegativity difference between Br and F is 1.02. Since F is more electronegative than Br, the bond will be polar, with the F atom being partially negative (\delta^{-}), and the Br atom being partially positive (\delta^{+}). To represent this dipole, draw an arrow pointing from Br to F.

\delta^{+} Br ——> F \delta^{-}

4. N-H:
The electronegativity difference between N and H is 0.84. Although there is a difference in electronegativity, it is not significant enough to create a fully polar bond. Instead, it exhibits a partial covalent character. However, when representing partial charges, we can consider the more electronegative atom, which is N, as partially negative (\delta^{-}), and H as partially positive (\delta^{+}). To represent this dipole, draw an arrow pointing from N to H.

\delta^{+} H <—— N \delta^{-}

5. N-P:
The electronegativity difference between N and P is 0.85. Similar to the N-H bond, it exhibits partial covalent character. When representing the dipole, we consider N as partially negative (\delta^{-}) and P as partially positive (\delta^{+}). Draw an arrow pointing from N to P.

\delta^{+} P <—— N \delta^{-}