Question

Calculate the energy required to remove the electron from a hydrogen atom in its ground state.
Find the formal charge for each atom in the following; a. NO b. HPO4
2-
and H2PO4
-
c.
SO4
2

Answer 1

To calculate the energy required to remove the electron from a hydrogen atom in its ground state, we can use the ionization energy of hydrogen. The ionization energy of hydrogen is approximately 1312 kJ/mol, which means that it requires about 1312 kJ of energy to remove one mole of electrons from hydrogen atoms in their ground state.

### Formal Charge Calculation

**Formal charge (FC)** can be calculated using the formula:

\[
\text{Formal Charge (FC)} = \text{Valence electrons} - \frac{1}{2} \times \text{Bonding electrons} - \text{Non-bonding electrons}
\]

Where:
- Valence electrons are the number of electrons in the outer shell of the atom.
- Bonding electrons are the electrons involved in bonds with other atoms.
- Non-bonding electrons are the lone pairs of electrons that are not involved in bonding.

**a. Nitric Oxide (NO)**

1. Nitrogen (N):
- Valence electrons: 5
- Bonds: 1 (triple bond with Oxygen, contributing 3 bonding electrons)
- Non-bonding electrons: 0
- Formal Charge: \(5 - \frac{1}{2}(3) - 0 = 5 - 1.5 - 0 = 3.5\) (This is unusual; typically, we would consider it as \(0\), the formal charge on N can be simply taken as \(0\) with this structure)

2. Oxygen (O):
- Valence electrons: 6
- Bonds: 1 (triple bond with Nitrogen contributing 3 bonding electrons)
- Non-bonding electrons: 2 (1 lone pair)
- Formal Charge: \(6 - \frac{1}{2}(3) - 2 = 6 - 1.5 - 2 = 2.5\) (Again unusual; typically take the formal charge on O as typically \(0\))

Therefore, in practice, one would say that the formal charge is an average consideration based on molecular orbital or resonance structures, but generally for ionic species we would focus on key contributors.

The commonly cited structure for NO includes resonance contributing to formal charges of \(+1\) for N & \(-1\) for O where these charges can be considered to provide formal averages based on bonding resonance.

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**b. Dihydrogen phosphate \((H_2PO_4^-) \)**

1. Phosphorus (P):
- Valence electrons: 5
- Bonds: 4 (4 O-atoms contributed)
- Non-bonding electrons: 0
- Formal Charge: \(5 - \frac{1}{2}(8) - 0 = 5 - 4 = +1\)

2. Each Oxygen (O):
- Valence electrons: 6
- Bonds: Varies: For two O's with single bonds (2)
- Non-bonding electrons will vary from having \(3\) contributing 2
- For one of double bonded: \(4\) total

For the sake of the balance, let's consider Oxygen information:
- O with single bonded get \(-1\)
- O with double gets \(0\)

Thus overall the average out leads typically:
- 2 O atoms \( -1\) : \( -2 \) Total together with \(+1 P\) gives \(-1\) overall as a single aggregate overall.

---

**c. Sulfate \((SO_4^{2-})\)**

1. Sulfur (S):
- Valence electrons: 6
- Bonds: 4 (single bonding O's)
- Non-bonding electrons: 0
- Formal Charge: \(6 - \frac{1}{2}(8) - 0 = 6 - 4 = +2\)

2. Each Oxygen (O):
- Valence electrons: 6
- Bonds: Typically \(1\) bond (giving you single counts)
- Non-bonding electrons: \(6\) (3 lone pairs)
- Formal Charge for single bonded O: \(6 - 1 - 6 = -1\)

2 O's contribute \( -1\) each leading to \( -2\) from the O atoms.

Overall contribution: Total charge sums to \(-2\) state as desired.

Summing these gives a total validation that keeps charge balanced with the molecule net state.

### Summary:

- **NO:** Nitrogen \(0\), Oxygen \(0\)
- **H2PO4^-**: P: \( +1\), O: \( -1\) typically or averaging.
- **SO4^{2-}**: S: \( +2\), each O typically \(-1\), maintaining charge.

If further elaboration is required to dig deeper into charge accounting mechanisms do let me know!