Bond Energies

To determine the correctness of the given statements about the magnetic properties of substances, let's explore the relevant concepts:

1. Statement I: are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.
   This statement describes the characteristics of paramagnetic substances. Paramagnetic materials are attracted by magnetic fields and their internal magnetic field aligns with the direction of the external magnetic field. Both (oxygen molecule) and transition metal ions like and have unpaired electrons, which give them their paramagnetic properties. Therefore, Statement I is correct.
2. Statement II: and are weakly magnetized in opposite directions to the magnetic field.
   This describes diamagnetic materials, which are characterized by a very weak magnetic moment that opposes the direction of the applied magnetic field. Both sodium chloride (an ionic compound) and water (a simple molecular compound) are diamagnetic because all their electrons are paired and thus they create a small magnetic field in the opposite direction. Hence, Statement II is correct.

Based on the explanations above, the correct answer is:

Both Statement I and Statement II are correct.

Step 1: Determine the Electron Configuration of the Acetylide Ion (C2^2-)
The acetylide ion (C2^2-) consists of two carbon atoms, each contributing 6 electrons, and it has a -2 charge. Therefore, the total number of electrons is:
- Total electrons = 2 (from C) + 2 (from charge) = 14 electrons.
Step 2: Write the Molecular Orbital Diagram for C2^2-
The molecular orbital configuration for C2^2- can be derived from the molecular orbital theory. The order of filling for the molecular orbitals for diatomic carbon species is:
- σ(1s), σ*(1s), σ(2s), σ*(2s), σ(2p_z), π(2p_x), π(2p_y), π*(2p_x), π*(2p_y)
For C2^2-, the electron configuration will be:
- σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p_z)² π(2p_x)² π(2p_y)²
Step 3: Calculate the Bond Order
The bond order can be calculated using the formula:
Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2
From the configuration:
- Bonding electrons = 10 (σ(1s)² + σ(2s)² + σ(2p_z)² + π(2p_x)² + π(2p_y)²)
- Antibonding electrons = 4 (σ*(1s)² + σ*(2s)²)
Now, substituting into the formula:
Bond Order = (10 - 4) / 2 = 3
Step 4: Determine the Magnetic Property
To determine the magnetic property, we look at the number of unpaired electrons. In the case of C2^2-, all electrons are paired in the molecular orbital configuration. Therefore, C2^2- is diamagnetic.
Step 5: Compare with Other Species
We need to find another species with the same bond order (3) and magnetic property (diamagnetic).
- The species that fits this description is the nitrosyl cation (NO⁺).
- The electron configuration for NO⁺ (which has 14 electrons) is similar, leading to a bond order of 3 and being diamagnetic.
Conclusion
The bond order and magnetic property of the acetylide ion (C2^2-) are the same as that of the nitrosyl cation (NO⁺).
Final Answer
The correct answer is: NO⁺ (Nitrosyl cation).

To determine the stability order of the resonance structures of , we can evaluate each structure based on several key factors affecting resonance stability:

1. **Octet Rule Compliance**: The most stable resonance structure usually has atoms with completed octets.
2. **Charge Distribution**: Resonance structures with minimal formal charges are typically more stable.
3. **Electronegativity**: Negative charges are more stable on more electronegative atoms (e.g., oxygen).

Let's analyze each structure:

1. Structure I: The carbon atom in the chain has a negative charge. The electronegative oxygen has a positive charge, which is not favorable.
2. Structure II: The positive charge is on a carbon atom, while the negative charge is on oxygen. This is more favorable compared to Structure I.
3. Structure III: There are no charges, and all atoms comply with the octet rule. The double bond between carbon and oxygen is highly stable due to resonance.

Based on these observations, the stability order is:

• Most Stable: Structure III – All atoms have full octets with no charges.
• Intermediate Stability: Structure II – Favorable charge distribution with negative charge on oxygen.
• Least Stable: Structure I – Unfavorable charge distribution. Negative charge on carbon and positive on oxygen.

Thus, the correct stability order is: III > II > I.