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 linear combinations of atomic orbitals will lead to the formation of molecular orbitals in homonuclear diatomic molecules (internuclear axis in z-direction)?
(1) \( 2p_z \) and \( 2p_x \)
(2) \( 2s \) and \( 2p_x \)
(3) \( 3d_{xy} \) and \( 3d_{x^2-y^2} \)
(4) \( 2s \) and \( 2p_z \)
(5) \( 2p_z \) and \( 3d_{x^2-y^2} \)
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.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32