Why is black phosphorus semiconductor with direct bandgap?

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Black phosphorus (BP) is a semiconductor characterized by a direct bandgap of 2 eV at the G point of the first Brillouin zone. This property arises from phosphorus having three valence electrons, allowing each atom to form three covalent bonds with neighboring atoms, unlike carbon which has four. The bonding orbitals create a full valence band, while the anti-bonding orbitals result in an empty conduction band. This unique electronic structure differentiates black phosphorus from materials like graphene.

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Douasing
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Why is black phosphorus a semiconductor with a direct bandgap?

The problem is mentioned by the two following references:
"The three bonds take up all three valence electrons of phosphorus,so, unlike graphene, monolayer black phosphorus is a semiconductor with a predicted direct bandgap of 2 eV at the G point of the first Brillouin zone."
(see http://www.nature.com/nnano/journal/v9/n5/full/nnano.2014.35.html)

"Unlike carbon, phosphorus has only three valance electrons which leads to BP being semiconducting since each atom is bonded to three neighboring atoms."
(see http://scitation.aip.org/content/aip/journal/apl/104/10/10.1063/1.4868132)

In order to check the point,I input the keyword "valence" in wiki:
http://en.wikipedia.org/wiki/Valence_(chemistry)

I found that typical valencies are three and five for phosphorus,but only four for carbon.I somehow doubt the explanation in the references and some meticulous and clear details shoud be supplied.
 
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p has a full s orbital and 3 additional electrons in the remaining three p orbitals. Hence these three orbitals can form 3 covalent bonds with neighbouring atoms. There are also 3 anti-bonding orbitals at higher energy. The bonding orbitals will form a full valence band and the anti-bonding orbitals an empty conduction band.
 

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