The relative order of stability of the following species: \(\mathrm{O}_2,\mathrm{O}^+_2,\mathrm{O}^-_2~\mathrm{and}~\mathrm{O}^{2-}_2\) is -
1. | \(\mathrm{O}^+_2>\mathrm{O}_2>\mathrm{O}^-_2>\mathrm{O}^{2-}_2\) | 2. | \(\mathrm{O}^{2-}_2>\mathrm{O}_2>\mathrm{O}^-_2>\mathrm{O}^{+}_2\) |
3. | \(\mathrm{O}_2>\mathrm{O}^+_2>\mathrm{O}^-_2>\mathrm{O}^{2-}_2\) | 4. | \(\mathrm{O}_2>\mathrm{O}^{2-}_2>\mathrm{O}^-_2>\mathrm{O}^{+}_2\) |
The plus and negative sign of the orbitals mean:
1. Wave function.
2. Probability density.
3. Quantum number of orbitals.
4. Frequency of orbitals.
In the option are given sequences of increasing bond order for the following species. The correct option is :
\(\mathrm{O}_2,\mathrm{N}_2,\mathrm{O}^+_2 ~\mathrm{and}~\mathrm{O}^-_2\)
1. | \(\mathrm{O}_2>\mathrm{N}_2>\mathrm{O}^+_2>\mathrm{O}^-_2\) | 2. | \(\mathrm{O}^-_2>\mathrm{O}_2>\mathrm{O}^+_2>\mathrm{N}_2\) |
3. | \(\mathrm{N}_2>\mathrm{O}^+_2>\mathrm{O}_2>\mathrm{O}^-_2\) | 4. | \(\mathrm{O}^-_2>\mathrm{O}^+_2>\mathrm{O}_2>\mathrm{N}_2\) |
Be2 molecule does not exist. The best explanation to explain this on the basis of the molecular orbital theory is -
1. The bond order of Be2 is one.
2. The bond order of Be2 is negative.
3. The bond order of Be2 is zero.
4. The bond order of Be2 is two.
The condition(s) required for the linear combination of atomic orbital is :
1. | Combining atomic orbitals must have the same or nearly the same energy. |
2. | Combining atomic orbitals must have proper orientations to ensure that the overlap is maximum |
3. | Both 1 and 2 |
4. | None of the above |