For the following reaction, the particle ' $x$ ' is ${ }_6^{\mathrm{C}^{11}} \longrightarrow{ }_5^{\mathrm{B}^{11}}+\beta^{+}+\mathrm{X}$

The nuclear reaction ${}_6^{11}\mathrm{C} \to {}_5^{11}\mathrm{B} + \beta^+ + X$ requires identifying the unknown particle X by applying conservation laws.

Mass number conservation gives $11 = 11 + 0 + A_X$, yielding $A_X = 0$.

Atomic number conservation gives $6 = 5 + 1 + Z_X$, yielding $Z_X = 0$.

The particle with $A=0$ and $Z=0$ corresponds to a neutrino. In positron emission, a proton converts to a neutron, positron, and electron neutrino: ${}_Z^A X \to {}_{Z-1}^A Y + {}_1^0e + \nu_e$.

Final answer: $\boxed{\text{B}}$