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.	\(\mathrm{O}_2>\mathrm{O}^{2-}_2>\mathrm{O}^-_2>\mathrm{O}^{+}_2\)

$\mathrm{}$

The plus and negative sign of the orbitals mean:

1. Wave function.

2. Probability density.

3. Quantum number of orbitals.

4. Frequency of orbitals.

${\mathrm{AlCl}}_3$ $+$ ${\mathrm{Cl}}^{-}$ $\to$ ${\mathrm{AlCl}}_4^{-}$

The change in the hybridization of the Al atom in the above reaction is:

Considering the X-axis as the internuclear axis, a sigma bond will not be formed in:

$1.$ $1\mathrm{s}$ $\mathrm{and}$ $2\mathrm{s}$
$2.$ $1\mathrm{s}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{x}}$
$3.$ $2{\mathrm{p}}_{\mathrm{y}}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{y}}$
$4.$ $1\mathrm{s}$ $\mathrm{and}$ $1\mathrm{s}$

The set of oxides of nitrogen that are paramagnetic in nature is:

1. NO, N₂O

2. N₂O₃, NO

3. NO, NO₂

4. N₂O, NO₂

Which of the compounds depicted below will exhibit a linear structure?

1. NO₂

2. HOCl

3. O₃

4. N₂O

Match the compounds of Xe in Column I with the molecular structure in Column II.

Match the coordination number and type of hybridization with the distribution of hybrid orbitals in space based on Valence bond theory.

The potential energy (y) curve for H₂ formation as a function of internuclear distance (x) of the H atoms is shown below.

The bond energy of H₂ is:

1. (b – a)
2. \(\frac{\left(\right. c - a \left.\right)}{2}\)
3. \(\frac{\left(\right. b - a \left.\right)}{2}\)
4. (c – a)