Problem 214
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) \(\mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{2-}\) (d) \(\mathrm{H}_{2}^{-}, \mathrm{He}_{2}^{2-}\)
Step-by-Step Solution
Verified Answer
Both species in (a) \( \mathrm{H}_{2}^{2+} \) and \( \mathrm{He}_{2} \) are unlikely to exist.
1Step 1: Understanding the problem
We need to determine which pair of species is unlikely to exist. For this, we will use molecular orbital theory and bond order calculations to evaluate the stability of each species.
2Step 1: Calculate bond order
Bond order is calculated as \( \text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \). A bond order of zero or negative indicates that the molecule is unlikely to exist.
3Step 2: Evaluate H2^2+
\( \mathrm{H}_{2}^{2+} \) has two protons and no electrons. Its bond order is \( \frac{0 - 0}{2} = 0 \), indicating it is unlikely to exist.
4Step 3: Evaluate He2
\( \mathrm{He}_{2} \) has four electrons (two in \(\sigma_{1s}\) and two in \(\sigma^{*}_{1s}\)). Its bond order is \( \frac{(2-2)}{2} = 0 \), indicating it is unlikely to exist.
5Step 4: Check other species in given pairs
We cross-check all other given species to verify if any pair contains two unlikely-to-exist species. As calculated, \( \mathrm{H}_{2}^{2+} \) and \( \mathrm{He}_{2} \) both do not exist, covering option (a). All other options have at least one existing species, confirming (a) as the correct answer.
Key Concepts
Bond OrderStability of MoleculesExistence of Species
Bond Order
The concept of bond order is a fundamental part of molecular orbital theory. It provides insight into the strength and stability of a bond within a molecule. Bond order is calculated using the formula:
- Find the number of bonding electrons, which are electrons that participate in bond formation.
- Find the number of antibonding electrons, which are electrons that destabilize the bond.
- Use the formula: \( \text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \).
Stability of Molecules
When determining the stability of a molecule, bond order plays a crucial role. A higher bond order usually correlates with stronger and more stable bonds.
The stability of a molecule can also be inferred from:
The stability of a molecule can also be inferred from:
- Bond Length: Shorter bonds are typically stronger and correspond to a higher bond order.
- Bond Energy: Higher bond energy indicates a more stable molecule as more energy is required to break the bond.
Existence of Species
Molecules or ions are considered viable species if their structural makeup allows them to exist under normal conditions. The likelihood of existence is primarily determined by stability factors derived from molecular orbital theory.
In many cases, species with a bond order of 0 or less are deemed non-existent or highly unstable. For instance:
In many cases, species with a bond order of 0 or less are deemed non-existent or highly unstable. For instance:
- \( \mathrm{H}_{2}^{2+} \) lacks electrons to form any bond, leading to its non-existence.
- \( \mathrm{He}_2 \) has equal bonding and antibonding electrons, leading to a bond order of 0, making it unlikely to exist.
Other exercises in this chapter
Problem 211
The struture of \(\mathrm{IF}_{7}\) is: (a) Octahedral (b) Trigonal bipyramid (c) Square pyramid (d) Pentagonal bipyramid
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The speces in which the atom is in a state of sphybridization is? (a) \(\mathrm{NO}_{2}^{-}\) (b) \(\mathrm{NO}_{3}^{-}\) (c) \(\mathrm{NO}_{2}\) (d) \(\mathrm{
View solution Problem 210
The hybridization of orbitals of \(\mathrm{N}\) atom in \(\mathrm{NO}_{3}^{-}, \mathrm{NO}_{2}^{+}\) and \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}\)are respectively: \([2011]\)
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