The rotation operator of spin-1/2 system is
$$ \hat{R}(\hat{a},\theta) =\exp\left(-\frac{i}{2} \vec{a}\cdot\vec{\sigma}\theta\right) = \begin{bmatrix} \cos\frac{\theta}{2} +ia_z\sin\frac{\theta}{2} && (-ia_x+a_y)\sin\frac{\theta}{2}\\ (-ia_x-a_y) \sin\frac{\theta}{2}&& \cos\frac{\theta}{2} - i a_z \sin\frac{\theta}{2}\end{bmatrix} $$
$$ =\hat{I} \cos\frac{\theta}{2} -i\hat{a}\cdot\vec{\sigma}\sin\frac{\theta}{2} = \hat{I} \cos\frac{\theta}{2} -i (a_x\hat{\sigma}^x+a_y\hat{\sigma}^y+a_z\hat{\sigma}^z)\sin\frac{\theta}{2} $$
If the rotational angle is $\theta=\pi$
$$ \hat{R}(\hat{a},\pi) = -i (a_x\hat{\sigma}^x+a_y\hat{\sigma}^y+a_z\hat{\sigma}^z) = \begin{bmatrix} ia_z && (-ia_x+a_y)\\ (-ia_x-a_y) && - i a_z \end{bmatrix} $$
If the rotational axis is $\vec{a} = \hat{\sigma}^x$
$$ \hat{R}(\hat{\sigma}^x,\pi) = -i \hat{\sigma}^x = \begin{bmatrix} 0 && -i\\ -i&& 0\end{bmatrix} $$
$$ \hat{R}(\hat{\sigma}^y,\pi) = -i \hat{\sigma}^y = \begin{bmatrix} 0 && 1\\ -1&& 0\end{bmatrix} $$
$$ \hat{R}(\hat{\sigma}^z,\pi) = -i \hat{\sigma}^z = \begin{bmatrix} i && 0\\ 0&& -i\end{bmatrix} $$
Similarly, the ‘2pi-pulse’ becomes
$$ \hat{R}(\hat{a},2\pi) = -\hat{I}= \begin{bmatrix} -1&& 0\\ 0 && -1 \end{bmatrix} $$
Therefore,
$$ \hat{R}(\vec{a},2\pi)\left|\psi\right> = -\left|\psi\right> $$
It is related with spin-1/2.