Question

1. identifie latome qui aura une charge partielle négative dans les liaisons suivantes :
2. \\(\text{o}-\text{h}\\)
3. \\(\text{c}-\text{h}\\)

1. la phosphine (phosphane, nom selon luicpa) est un hydrure de phosphore, dont la formule est \\(\text{ph}_3\\).

a) dessine une structure de lewis (électrons représentés par des points) de la phosphine.
b) résume si tu tattends à ce que les liaisons dans la phosphine soient polaires ou non polaires, en donnant une brève justification.

Explanation:

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<reasoning>

Identify electronegativity values

To determine partial charges and bond polarity, we compare the electronegativity values of the atoms involved.

• Oxygen (\(\text{O}\)): \(3.44\)
• Hydrogen (\(\text{H}\)): \(2.20\)
• Carbon (\(\text{C}\)): \(2.55\)
• Phosphorus (\(\text{P}\)): \(2.19\)

Determine partial charges for Question 2

The atom with the higher electronegativity attracts the shared electrons more strongly, acquiring a partial negative charge (\(\delta^-\)).

• For the \(\text{O}-\text{H}\) bond: Oxygen (\(3.44\)) is more electronegative than Hydrogen (\(2.20\)), so Oxygen carries the partial negative charge.
• For the \(\text{C}-\text{H}\) bond: Carbon (\(2.55\)) is more electronegative than Hydrogen (\(2.20\)), so Carbon carries the partial negative charge.

Analyze Lewis structure for Question 3a

Phosphorus (\(\text{P}\)) is in Group 15 and has 5 valence electrons. Each Hydrogen (\(\text{H}\)) has 1 valence electron.

• The central \(\text{P}\) atom forms 3 single covalent bonds with the 3 \(\text{H}\) atoms.
• This leaves one lone pair of electrons on the \(\text{P}\) atom, completing its octet.

Evaluate bond polarity for Question 3b

We calculate the electronegativity difference (\(\Delta\text{EN}\)) between Phosphorus and Hydrogen:
\[
\Delta\text{EN} = |\text{EN}_{\text{P}} - \text{EN}_{\text{H}}| = |2.19 - 2.20| = 0.01
\]
A difference of \(0.01\) is extremely close to \(0\). Bonds with \(\Delta\text{EN} < 0.4\) are considered non-polar covalent. Therefore, the \(\text{P}-\text{H}\) bonds in phosphine are non-polar.
</reasoning>

<answer>

Question 2

1. \(\text{O}-\text{H}\): Oxygen (\(\text{O}\))
2. \(\text{C}-\text{H}\): Carbon (\(\text{C}\))

Question 3

a) The Lewis structure of phosphine (\(\text{PH}_3\)) consists of a central Phosphorus (\(\text{P}\)) atom bonded to three Hydrogen (\(\text{H}\)) atoms via single covalent bonds, with one lone pair of electrons remaining on the Phosphorus atom:
\[

$$\begin{array}{ccc} & \ddot{\text{P}} & \\ \text{H} & \cdot\ \cdot & \text{H} \\ & \text{H} & \end{array}$$

\]

b) The bonds in phosphine (\(\text{PH}_3\)) are expected to be non-polar.

Justification: The electronegativity of Phosphorus (\(2.19\)) and Hydrogen (\(2.20\)) are nearly identical. The electronegativity difference is extremely small (\(\Delta\text{EN} = 0.01\)), which is well below the threshold of \(0.4\) required for a bond to be considered polar.
</answer>

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Answer:

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<reasoning>

Identify electronegativity values

To determine partial charges and bond polarity, we compare the electronegativity values of the atoms involved.

• Oxygen (\(\text{O}\)): \(3.44\)
• Hydrogen (\(\text{H}\)): \(2.20\)
• Carbon (\(\text{C}\)): \(2.55\)
• Phosphorus (\(\text{P}\)): \(2.19\)

Determine partial charges for Question 2

The atom with the higher electronegativity attracts the shared electrons more strongly, acquiring a partial negative charge (\(\delta^-\)).

• For the \(\text{O}-\text{H}\) bond: Oxygen (\(3.44\)) is more electronegative than Hydrogen (\(2.20\)), so Oxygen carries the partial negative charge.
• For the \(\text{C}-\text{H}\) bond: Carbon (\(2.55\)) is more electronegative than Hydrogen (\(2.20\)), so Carbon carries the partial negative charge.

Analyze Lewis structure for Question 3a

Phosphorus (\(\text{P}\)) is in Group 15 and has 5 valence electrons. Each Hydrogen (\(\text{H}\)) has 1 valence electron.

• The central \(\text{P}\) atom forms 3 single covalent bonds with the 3 \(\text{H}\) atoms.
• This leaves one lone pair of electrons on the \(\text{P}\) atom, completing its octet.

Evaluate bond polarity for Question 3b

We calculate the electronegativity difference (\(\Delta\text{EN}\)) between Phosphorus and Hydrogen:
\[
\Delta\text{EN} = |\text{EN}_{\text{P}} - \text{EN}_{\text{H}}| = |2.19 - 2.20| = 0.01
\]
A difference of \(0.01\) is extremely close to \(0\). Bonds with \(\Delta\text{EN} < 0.4\) are considered non-polar covalent. Therefore, the \(\text{P}-\text{H}\) bonds in phosphine are non-polar.
</reasoning>

<answer>

Question 2

1. \(\text{O}-\text{H}\): Oxygen (\(\text{O}\))
2. \(\text{C}-\text{H}\): Carbon (\(\text{C}\))

Question 3

a) The Lewis structure of phosphine (\(\text{PH}_3\)) consists of a central Phosphorus (\(\text{P}\)) atom bonded to three Hydrogen (\(\text{H}\)) atoms via single covalent bonds, with one lone pair of electrons remaining on the Phosphorus atom:
\[

$$\begin{array}{ccc} & \ddot{\text{P}} & \\ \text{H} & \cdot\ \cdot & \text{H} \\ & \text{H} & \end{array}$$

\]

b) The bonds in phosphine (\(\text{PH}_3\)) are expected to be non-polar.

Justification: The electronegativity of Phosphorus (\(2.19\)) and Hydrogen (\(2.20\)) are nearly identical. The electronegativity difference is extremely small (\(\Delta\text{EN} = 0.01\)), which is well below the threshold of \(0.4\) required for a bond to be considered polar.
</answer>

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