To determine the truth of the given statements about molecular orbitals, we need to understand two concepts: \(\pi\) bonding molecular orbitals (MOs) and \(\pi^*\) antibonding molecular orbitals.
Based on the reasoning above:
Thus, the correct answer is: Statement I is false but Statement II is true.
Statement (I) Analysis:
A $\pi$ bonding molecular orbital (MO) is formed by the sideways overlap of $p$-orbitals.
This type of bonding MO has higher electron density above and below the internuclear axis, leading to a bonding interaction between atoms.
Therefore, Statement (I) is false.
Statement (II) Analysis:
The $\pi^*$ antibonding molecular orbital is formed when $p$-orbitals combine in such a way that destructive interference occurs between the wave functions of the atomic orbitals.
This creates a node (a region of zero electron density) between the nuclei.
Therefore, Statement (II) is true.
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.
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 \( 3d_{x^2-y^2} \)

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.