Problem 214 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following pairs of molecules/ions, both the species are not likely to exist? (a) \(\mathrm{H}_{2}^{2+}, \mathrm{He}_{2}\) (b) \(\mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2+}\) (c) \(\mathrm{H}_{2}^{+}, \mathrm{He}_{7}^{2-}\) (d) \(\mathrm{H}_{2}^{-}, \mathrm{He}_{2}^{2-}\)

Step-by-Step Solution

Answer

Both species in pairs (a) and (c) are unlikely to exist.

Step 1: Understand Molecular Orbital (MO) Theory

To determine the stability of molecules or ions, we use Molecular Orbital Theory. According to this theory, molecules are stabilized when electrons are added to bonding molecular orbitals and destabilized when electrons occupy antibonding molecular orbitals. We calculate the bond order to predict the stability; a bond order greater than zero indicates a likely existing species.

Step 2: Determine Bond Order for Given Molecules

The bond order (BO) is calculated using the formula \( BO = \frac{1}{2} (n_b - n_a) \), where \( n_b \) is the number of electrons in bonding orbitals and \( n_a \) is the number in antibonding orbitals. We calculate this for both species in each pair to assess their existence.

Step 3: Calculate for Pair (a) \( \mathrm{H}_{2}^{2+}, \mathrm{He}_{2} \)

- \( \mathrm{H}_{2}^{2+} \): It has 0 electrons (both protons are ionized), leading to a bond order of 0.- \( \mathrm{He}_{2} \): Has 4 electrons (2 in bonding, 2 in antibonding), leading to a bond order of 0.Both have bond order of 0, suggesting neither species exist.

Step 4: Calculate for Pair (b) \( \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2+} \)

- \( \mathrm{H}_{2}^{-} \): Has 3 electrons (2 in bonding, 1 in antibonding), giving a bond order of \( \frac{1}{2}(2-1) = 0.5 \).- \( \mathrm{H}_{2}^{2+} \): As seen, bond order is 0.\( \mathrm{H}_{2}^{2+} \) is unlikely to exist due to bond order 0.

Step 5: Calculate for Pair (c) \( \mathrm{H}_{2}^{+}, \mathrm{He}_{7}^{2-} \)

- \( \mathrm{H}_{2}^{+} \): Has 1 electron, all in bonding, giving a bond order of \( \frac{1}{2}(1-0) = 0.5 \).- \( \mathrm{He}_{7}^{2-} \): Highly unstable and not known molecule/ion in chemistry.\( \mathrm{He}_{7}^{2-} \) likely does not exist.

Step 6: Calculate for Pair (d) \( \mathrm{H}_{2}^{-}, \mathrm{He}_{2}^{2-} \)

- \( \mathrm{H}_{2}^{-} \): As before, bond order is 0.5.- \( \mathrm{He}_{2}^{2-} \): Has 6 electrons (2 in bonding, 4 in antibonding), giving bond order of \( \frac{1}{2}(2-4) = -1 \).\( \mathrm{He}_{2}^{2-} \) has negative bond order and is unstable.

Key Concepts

Bond Order

In Molecular Orbital Theory (MO theory), bond order plays a crucial role in determining the stability of molecules or ions. The bond order is the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals, divided by two. The formula is: \[ BO = \frac{1}{2} (n_b - n_a) \] Here, \( n_b \) is the number of bonding electrons, and \( n_a \) is the number of antibonding electrons.

• A bond order greater than zero usually indicates a stable molecule or ion, which means it is likely to exist under normal conditions.
• A bond order of zero or less signifies instability; hence, the molecule or ion might not exist naturally or readily.

For example, in the case of \( \mathrm{H}_2^{2+} \), both protons are ionized, and no electrons are left to form a bond, resulting in a bond order of zero, indicating that \( \mathrm{H}_2^{2+} \) doesn't likely exist.

Stability of Molecules

The stability of molecules is correlated with their bond order. Higher bond orders typically suggest more stable molecules due to stronger bonding interactions. When electrons occupy more bonding orbitals than antibonding ones, it leads to a lower energy state and increased stability. Several factors contribute to this stability:

• Electron Arrangement: Electrons favor lower energy levels, so a stable configuration usually results from a completed valence shell.
• Energy States: Stability is enhanced when electrons fill bonding molecular orbitals while avoiding or minimizing occupation in antibonding orbitals.
• Energy Gaps: Wider gaps between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) can indicate stability, as they reduce the potential for electronic transitions that can destabilize the molecule.

In the molecular pair \( \mathrm{He}_2 \), each of its electrons occupies both bonding and antibonding orbitals equally, leading to a bond order of zero. This factor decreases its likelihood of being a stable and existing species.

Ions in Molecular Orbital Theory

Understanding ions through Molecular Orbital Theory involves evaluating their electron arrangement in bonding and antibonding orbitals. Ions often result in a net change in electron count, affecting the bond order and consequently their stability.

• Positive ions, or cations, usually involve a loss of electrons, potentially reducing bonding interactions and stability if it leads to electron deficiencies.
• Negative ions, or anions, involve electron gain, which can either stabilize the molecule by filling bonding orbitals or destabilize if it leads to increased occupancy in antibonding orbitals.

For example, the odd number of electrons in \( \mathrm{H}_2^{-} \) results in filling a bonding orbital partly, with one electron occupying an antibonding orbital, giving a fractional bond order. This reflects partial stability compared to its neutral counterpart, \( \mathrm{H}_2 \), but maintains the overall existence of the ion. While \( \mathrm{He}_2^{2-} \) has an excess of antibonding electrons, resulting in a negative bond order and indicating instability and non-existence.