If for any graph $H$ we define $\alpha(H)$ to be the cardinality of any maximum size indepedent set in $H$. Then under what conditions can any graph $G$ be expressed as a union of $\alpha(G)$ complete graphs?

Or equivalently if $[S]_2$ is the set of all two element subsets of $S$, then for which class of graphs $G$ will there exist $\alpha(G)$ sets $V_{1},V_2,\ldots V_{\alpha(G)}$ that satisfy $V(G)=\bigcup_{n=1}^{\alpha(G)}V_n$ and $E(G)=\bigcup_{n=1}^{\alpha(G)}[V_n]_2$?

Are there any reasonably simple necessary and sufficient conditions for categorizing these graphs?

Lastly also note if there are $n$ complete graphs $C_1,C_2,\ldots C_n$ with $G=\bigcup_{k=1}^n C_k$ and $n<\alpha(G)$ there must exist an independent set $I$ of $G$ such that $|I|=n+1$ however by assumption we have that $I\subseteq V(G)=\bigcup_{k=1}^nV(C_k)$ therefore by the pigeonhole principal there must exists two distinct vertices $u,v\in V(C_j)$ for some integer $1\leq j\leq n$ yet since $V(C_j)$ is a clique of $G$ this means we have $\{u,v\}\in E(G)$ contradicting the fact $I$ is an independent set of $G$, which proves $n\geq \alpha(G)$. Thus a a corollary if we let $\theta(G)$ be the intersection number of $G$ then by definition we have that:

$$G\text{ is a union of }\alpha(G)\text{ complete graphs}\iff \alpha(G)=\theta(G)$$

However I'm unable to simplify these equivalent conditions into anything nice either.