The electronic configuration of O$_2$ and its ions in terms of molecular orbitals is as follows:
$\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 = \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1$
For O$_2$: There are 2 electrons in $\pi^*$ orbitals.
For O$_2^+$: 1 electron is removed from the $\pi^*$ orbitals, leaving 1 electron.
For O$_2^-$: 1 electron is added to the $\pi^*$ orbitals, making it 3 electrons.
Total electrons in ($\pi^*$) molecular orbitals $= 2 + 1 + 3 = 6$
Final Answer: (6)
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to:
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to: