In a diatomic molecule, when combining atomic orbitals, both bonding and anti-bonding molecular orbitals are formed. For atomic orbitals 2s and 2p:
Total number of anti-bonding orbitals:
Total anti-bonding orbitals = 1 (from 2s) + 1 (from σ2p*) + 2 (from 2 π2p*) = 4.
This total, 4, falls within the given range (4,4).
In molecular orbital theory, the anti-bonding molecular orbitals formed from atomic orbitals are as follows:
- Anti-bonding molecular orbital from 2s: 1
- Anti-bonding molecular orbitals from 2p: 3
Thus, the total number of anti-bonding molecular orbitals is:
\(1 + 3 = 4\)
The Correct Answer is: 4
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 \( 3d_{x^2-y^2} \)
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g H₂O. Molar mass of (X) is ___________g mol\(^{-1}\).
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: