Question

H₂Se is structurally analogous to water, but containing an atom much lower in the oxygen family, the bonding between the H’s and the Se may involve d orbitals.    Using the s, p, and d orbitals of Se as your basis set, determine which orbitals would be involved in creating MO’s for the molecule.    Draw the MO energy level diagram for the molecule and show the arrangement of electron in the MO’s.

Answer 1

Atomic Number of Selinium = 34,

Hence the elctronic configuration is [Ar]3d¹⁰4s²4p⁴ (EC of Ar = 1s²2s²2p⁶3s²3p⁶)

Atomic number of Hydrogen = 1 and hence the electronic congfiguration of H is 1s¹

Here the Se is the central atom and it's outer most shell (4) having 6 electrons, in which 4 electrons are paired (2 in 4s and 2 in 4p) (Remember that, the d orbitals of Se will not involved in the hybridization because it is fully filled orbital). The remaining two electrons are unpaired. See the figure.

Now these 4 orbitals (2 paired atomic orbital 2 unpaired atomic orbitals) mix up (hybridization) to get 4 molecular orbitals and the hybridization is sp³ hybridization (1 s and 3 p orbitals), in which 2 orbitals are paired and 2 are unpaired. Two Hydrogen atoms can share their electrons to the unpaired molecular orbitals to complete the pairing as shown in the figure.

Additional Note: As mentioned above,4 sp³ hybridized orbitals of Se contain 2 paired orbitals. There will be strong repulsion between two paired orbitals. Hence structure of H₂Se is bend structure or shape (as in H₂O) and the bond angle between hydrogen and selinium is 97.7^o.