The bonding and antibonding molecular orbitals which form from liner combinations of the 2 p z atomic orbitals in a homonuclear diatomic molecule. (The 2 p z orbitals are those whose lobes are oriented perpendicular to the bonding axis.) And these molecular orbitals differ from those obtained from linear combinations of the 2 p y atomic orbitals. (The 2 p y orbitals are also oriented perpendicular to the bonding axis, but also perpendicular to the 2 p z orbitals.) Concept Introduction: When two 2 pz orbitals combine in a side-by-side orientation in a homonuclear diatomic molecule, two molecular orbitals are formed—bonding and antibonding.
The bonding and antibonding molecular orbitals which form from liner combinations of the 2 p z atomic orbitals in a homonuclear diatomic molecule. (The 2 p z orbitals are those whose lobes are oriented perpendicular to the bonding axis.) And these molecular orbitals differ from those obtained from linear combinations of the 2 p y atomic orbitals. (The 2 p y orbitals are also oriented perpendicular to the bonding axis, but also perpendicular to the 2 p z orbitals.) Concept Introduction: When two 2 pz orbitals combine in a side-by-side orientation in a homonuclear diatomic molecule, two molecular orbitals are formed—bonding and antibonding.
Interpretation: The bonding and antibonding molecular orbitals which form from liner combinations of the 2pz atomic orbitals in a homonuclear diatomic molecule. (The 2pz orbitals are those whose lobes are oriented perpendicular to the bonding axis.) And these molecular orbitals differ from those obtained from linear combinations of the 2py atomic orbitals. (The 2py orbitals are also oriented perpendicular to the bonding axis, but also perpendicular to the 2pz orbitals.)
Concept Introduction:
When two 2pz orbitals combine in a side-by-side orientation in a homonuclear diatomic molecule, two molecular orbitals are formed—bonding and antibonding.
Consider a solution of 0.00304 moles of 4-nitrobenzoic acid (pKa = 3.442) dissolved in 25 mL water and titrated with 0.0991 M NaOH. Calculate the pH at the equivalence point
Chapter 10 Solutions
Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package
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