What is the total number of nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-?

I count 10.

Draw the H2N-CH2-COOH.
There are 30 electrons. Place those electons with each H of 2, each C of 4, each O as 8 (with a double bond to one C) and 1 unbound pair on N. NOW, move the H of the COOH to the N of the H2N which will give you the dipolar form. Then count the unbound electrons. I count 4 on the C=O oxygen and 6 on the C-O to make 10 total.
Check my work.

Well, let's count the nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-.

Let's break it down:
- The +NH3 group has 1 nitrogen atom and 3 hydrogen atoms. Nitrogen contributes 5 valence electrons, but since it has a positive charge, it lost one electron, so we have 4 valence electrons remaining.
- The CH2 group has 1 carbon atom and 2 hydrogen atoms. Carbon contributes 4 valence electrons, and hydrogen contributes 1 valence electron for each atom. So, we have 6 valence electrons total.
- The COO- group has 1 carbon atom, 2 oxygen atoms, and a negative charge. Carbon contributes 4 valence electrons, and oxygen contributes 6 valence electrons for each atom. But since the group has a negative charge, it gained one electron, so we have 19 valence electrons remaining.

Adding them all up, we have a total of 4 + 6 + 19 = 29 nonbonding valence electrons. Voila!

To determine the total number of nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-, we first need to determine the Lewis structure of glycine:

1. Start by determining the total number of valence electrons. In this case, we have 5 (from nitrogen) + 1 (from hydrogen) + 2(1) (from carbon) + 3(1) (from oxygen) = 12 valence electrons.

2. The Lewis structure will have a +1 charge due to the presence of NH3CH2COO-.

3. Place the atoms symmetrically around the central nitrogen atom. Connect the atoms with single bonds.

4. Distribute the remaining electrons as lone pairs on the atoms to satisfy the octet rule and minimize formal charges.

Here's the Lewis structure of glycine in its "dipolar ion" form:

H
|
H - N - C - C - O
| |
H O

5. Count the number of nonbonding valence electrons. In the Lewis structure, we have lone pairs on the oxygen atoms. Each oxygen atom has 2 lone pairs, giving us a total of 4 nonbonding valence electrons.

Therefore, the total number of nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-, is 4.

To find the total number of nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-, we need to determine the number of nonbonding electrons on each atom and then sum them up.

Glycine, with the molecular formula NH3CH2COOH, is an amino acid that consists of different atoms. In its dipolar ion form (which has gained or lost an electron), it becomes +NH3CH2COO-.

Now let's break down the molecule and count the nonbonding valence electrons for each atom:

1. Nitrogen atom (N) - In the dipolar ion, nitrogen has four covalent bonds, each formed with a hydrogen atom (H). Therefore, it has no nonbonding valence electrons.

2. Carbon atom (C) - Carbon has three covalent bonds, two with hydrogen (H) and one with oxygen (O). This leaves one nonbonding electron.

3. Oxygen atom (O) - In the dipolar ion, oxygen has a negative charge and formed a covalent bond with carbon (C). It also has two lone pairs of nonbonding electrons. So, it has four nonbonding valence electrons.

4. Hydrogen atoms (H) - Each hydrogen atom forms a covalent bond with either nitrogen (N) or carbon (C). Therefore, they have no nonbonding valence electrons.

Now, let's sum up the nonbonding electrons for each atom:

Nitrogen (N) - 0 nonbonding electrons
Carbon (C) - 1 nonbonding electron
Oxygen (O) - 4 nonbonding electrons
Hydrogen (H) - 0 nonbonding electrons (2 hydrogen atoms)

Adding them up, we have:
0 + 1 + 4 + 0 = 5

So, the total number of nonbonding valence electrons in the "dipolar ion" form of glycine, +NH3CH2COO-, is 5.