Let and be two unit vectors andθ is the angle between them.

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ be the angle between them.

Then, $|\vec{a}|=|\vec{b}|=1$

Now, $\vec{a}+\vec{b}$ is a unit vector if $|\vec{a}+\vec{b}|=1$.

$|\vec{a}+\vec{b}|=1$

$\Rightarrow(\vec{a}+\vec{b})^{2}=1$

$\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1$

$\Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=1$

$\Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1$

$\Rightarrow 1^{2}+2|\vec{a}||\vec{b}| \cos \theta+1^{2}=1$

$\Rightarrow 1+2 \cdot 1 \cdot 1 \cos \theta+1=1$

$\Rightarrow \cos \theta=-\frac{1}{2}$

$\Rightarrow \theta=\frac{2 \pi}{3}$

Hence, $\vec{a}+\vec{b}$ is a unit vector if $\theta=\frac{2 \pi}{3}$.

The correct answer is D.