Select one donor impurity among the following: Boron (B), Aluminium (Al), and Arsenic (As).

Hint: Pentavalent elements can donate one electron to the doped material. So, these group $V$ elements are called donor impurity. Here, the elements having $5$ valence electrons in its outermost shell will be the donor impurity among the given elements.

Complete solution:
A dopant that has $5$ electrons in its valence shell, when doped with a semiconductor to increase its conductivity, is called donor impurity. This is also called pentavalent impurity for having $5$ electrons in the outermost shell. It can donate the extra electron present in its valence shell to the neighboring atom.
A dopant which has $3$ electrons in its valence shell, when doped with a semiconductor to raise its conductivity, is called acceptor impurity. This is also called trivalent impurity for having $3$ electrons in the outermost shell. It can accept one electron from a neighboring atom as it has a vacancy of the electron.
Here, Boron (B) and Aluminium (Al) are the group $III$ elements and have $3$ electrons in the valence shell. So, they are acceptor impurities.
But, Arsenic (As) is the group $V$ element and has $5$ electrons in the valence shell. So, Arsenic (As) is the donor impurity.

Additional information:
Both the p-type and n-type crystals, formed by doping acceptor and donor impurities, respectively, are electrically neutral.

Note: By doping donor impurities in the crystal of an intrinsic semiconductor in a controlled manner, an n-type semiconductor is formed, with electrons as majority carriers. And acceptor impurities are doped in the crystal of an intrinsic semiconductor in a controlled manner to form a p-type semiconductor, with holes as the majority carriers.