For $B$, $Z=5$, i.e. its nucleus contains 5 protons, 5 positively charged, massive particles. And if there 5 protons, the neutral atom has 5 electrons conceived to whiz about the nuclear core, with a formal $1s^(2)2s^(2)2p^1$ configuration.
There are 2 common of boron: (i) $""^11B$ contains 6 neutrons, and is approx. 80% naturally abundant; and (ii) $""^10B$ with 5 neutrons, makes up the balance. The atomic mass of boron quoted on the Periodic Table, $10.81*g$, reflects this isotopic abundance.
Most of the time, boron forms $BR_3$ and $BX_3$ , i.e. $B^(III+)$..........it can expand its valence shell to form $BX_4^-$ and $BR_4^(-)$ derivatives.
