Why is the singlet state higher in energy than the corresponding triplet state?

Because the singlet state has electron repulsion and no exchange energy, i.e. it has paired antiparallel electrons instead of unpaired parallel electrons.

Singlet and triplet are just terminology that mean the following:

• Singlet: zero unpaired electrons
• Triplet: two unpaired electrons

Their names come from the notion of spin multiplicity, #2S + 1#. It is the degeneracy of spin angular momentum.

#S# is basically the total spin of all the electrons that aren't paired. In other words,

#S = |sum_(i=1)^(2l+1) m_(s,i)|#

where #l# is the angular momentum quantum number of the set of orbitals with unpaired electrons, and #m_s = pm1/2# is the spin quantum number.

• If you have two paired electrons,

#2S + 1 = 2(-1/2 + 1/2) + 1 = 1#, a singlet state.

#ul(uarr darr)" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#

• If you have two unpaired electrons,

#2S + 1 = 2(1/2 + 1/2) + 1 = 3#, a triplet state.

#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))#

We call this the Russell-Saunders-uncoupled degeneracy. Singlet states don't split in spin-orbit coupling, and triplet states split into three levels in spin-orbit coupling.

Just from Hund's rule, we expect the triplet configuration to be lower in energy.

He remarks (and I'm paraphrasing here) that a state with electrons unpaired is energetically favored over a state with those same electrons paired up.

The reasoning behind that is that there is less electron repulsion energy (a destabilizing factor) and more exchange energy (a stabilizing factor). That leads to a more stable electron configuration.