Question

Based on the octet rule, boron will most likely form:

• A. $B^{3-}$
• B. $B^{+}$
• C. $B^{3+}$
• D. $B^{2+}$

Answer 1

The correct option is C $B^{3+}$

Atomic number of Boron = 5
Electronic configuration of Boron is $\left[He\right]2s^{2}2p^1$
Configuration of the valence shell: $2s^{2}2p^1$
It does not have a complete octet.

(a) $B^{3-}$
Electronic configuration of $B^{3-}$ is $\left[He\right]2s^{2}2p^4$
Configuration of the valence shell: $2s^{2}2p^4$
It does not have a complete octet.

(b) $B^{+}$
Electronic configuration of $B^{1+}$ is $\left[He\right]2s^2$
Configuration of the valence shell: $2s^2$
It does not have a complete octet.

(c) $B^{3+}$
Electronic configuration of $B^{3+}$ is $\left[He\right]$
Configuration of the valence shell: $1s^2$
Boron is most likely to form $B^{3+}$ as then it achieves the noble gas electronic configuration of Helium (complete duplet).

(d) $B^{2+}$
Electronic configuration of $B^{2+}$ is $\left[He\right]2s^1$
Configuration of the valence shell: $2s^1$
It does not have a complete octet.