The usual Kuratowski ordered pair function does not work on proper classes, because if $A,B$ are proper classes and $\langle A,B\rangle=\{\{A\},\{A,B\}\}$, then since $A\notin\{A\}$ and so on you get $\langle A,B\rangle=\{\emptyset\}$, which is not very useful. Instead, you can use a definition like $(A,B)=(\{0\}\times A)\cup(\{1\}\times B)$ where $0,1$ are any two distinct sets and $A\times B$ is the set of all (Kuratowski) ordered pairs of elements from $A,B$, from which you can extract the elements as $A=\{x:\langle 0,x\rangle\in(A,B)\}$ and similarly for $B$.
Now that process all works perfectly fine, but now I'm interested in defining an unordered pair of the proper classes $A,B$, which is to say a definable term $[A,B]$ which satisfies $$[A,B]=[C,D]\iff(A=C\wedge B=D)\vee(A=D\wedge B=C)$$
for any classes $A,B,C,D$. I don't see any easy way to build this from the above ordered pair definition; does anyone have an idea for such a formula?
Bonus points if you can generalize to an unordered set: given an index set $x$ (not a proper class) and a definable class term $A_i$ with free variable $i$, define a term $[A_i:i\in x]$ which satisfies $$[A_i:i\in x]=[B_j:j\in x]\iff \exists \sigma:x\overset{bij.}{\to} x.\forall i\in x.A_i=B_{\sigma(i)}$$
$[A, B] = \{\{A\cap V_\alpha, B\cap V_\alpha\}: \alpha\in On\}$ will do the trick. To see this, let $[A, B] = [C, D]$ and suppose $A\not = C, D$. Then there is an ordinal $\alpha$ such that $A\cap V_\alpha \not = C\cap V_\alpha, D\cap V_\alpha$. So $\{A\cap V_\alpha, B\cap V_\alpha\} \not = \{C\cap V_\alpha, D\cap V_\alpha\}$ contradicting our assumption that $[A, B] = [C, D]$. It follows from this argument that if $[A, B] = [C, D]$, then $(A = C \wedge B = D) \vee (A = D \wedge B = C)$. The other direction is trivial.