In semiconductors when electrons jump from VB to CB, do they leave behind their parent atom's nuclei? In semiconductors/conductors when electrons jump from VB to CB, do they leave behind their parent atom's nuclei?
If yes,  when this happens in Si (electron jump from VB to CB) why don't they ionize Si to Si +?
If no, why do they ionize doped As atom when the unbonded free electron from VB goes to CB?
 A: First, the electrons in the valence or conduction band are not localized to a single nucleus. They move about in the crystal lattice much like atoms in a gas, so we call them an "electron gas". 
Second, the conduction and valence bands are bands of states of electrons, not of whole molecules. Finding an electron in one or the other band doesn't tell you anything about the energy state of the nuclei nearby (and the energy states of the nuclei don't much affect semiconductor electrical behavior).

If yes, when this happens in Si (electron jump from VB to CB) why don't they ionize Si to Si +?

Because the electron has only changed energy state, it hasn't moved to a different location far away from where it was before. So any positively charged nuclei nearby are still balanced by the negative charge of the electron.
Note, though, that ionized donor impurities do contribute an electron to the conduction band while leaving a localized positively charged nuclear site behind.
A: When an electron with its negative elementary charge moves from the VB to the CB it leaves behind a positive elementary charge, a hole. The positive charge of the hole is, of course, related to the positive Si+ ions in the crystal lattice. But the hole is not localized at a specific Si atom. Its wave function is spread out over the crystal. Thus it can be found near any Si atom of the lattice just like an electron in the CB.
The case is different for an As atom occupying a lattice point. If its electron moves to the conduction band, it leaves behind a positive As ion fixed at that location.  
