Theorem 2.2.18 in Chang and Kiesler uses omitting types to show that any countable model of ZF has an elementary end extension.
Can we control the countable order type of such a model? for example, if $ X \prec H\_ {\omega\_2}$ 
can we have an elementary extension $Y \prec H\_{\omega\_2}$ such the $order type(Y)$ is bounded by some ordinal?


Any help or reference would be appreciated.



Elementary end extension of a countable model for ZF
There are several interesting issues arising in your question.


First, you ask about models of ZF, but then mention elementary
substructures of $H\_{\omega\_2}$, which of course is not a model of ZF,
because it does not satisfy the power set axiom. In the context of
$H\_{\omega\_2}$, you probably intend to discuss the theory
$\text{ZFC}^-$, meaning the theory of ZFC without the power set
axiom, and this makes for a very interesting question. (Meanwhile,
please see my recent paper with Gitman and Johnstone [What is the
theory of ZFC without power set?](http://jdh.hamkins.org/what-is-the-theory-zfc-without-power-set/) for
reasons to be careful about how this theory is described---in
particular, one should be sure to include collection rather than
merely replacement; they are no longer equivalent without power
set.)


Second, in the context of set theory, there are distinct concepts
of end-extension.


* One model $(M,E)$ *end-extends* to another $(N,F)$ if $(M,E)$
is a substructure of $(N,F)$, meaning that $M\subset N$ and
$E=F\cap M\times M$, and also $a\mathrel{F}b\in M$ implies $a\in
 M$. That is, the sets of $M$ do not gain new elements in $N$. In
other words, $M$ is a transitive subclass of $N$.
* One model $M$ *top-extends* to another model $N$, if $M$ is a
rank initial segment of $N$. In other words, $M$ end-extends to
$N$ and also all new objects of $N$ have ordinal rank
larger than any ordinal of $M$. For example, this is the situation
when $V\_\alpha\prec V\_\beta$, but in the general case,
top-extensions needn't have that the height of $M$ is an ordinal
of $N$.


Set theorists often use the term "end-extension" to refer to
top-extensions, so one must take care, as the concepts are distinct
for models of set theory. For example, a nontrivial forcing
extension is an end-extension but not a top-extension. When it
comes to *elementary* extensions of models of ZF, however, even in
the case of nonstandard models, then the two notions coincide. This
is because if $M\prec N$ is an elementary end-extension, then by
elementarity, the $V\_\alpha^M$ remains the $V\_\alpha$ of $N$ for
any ordinal $\alpha$ in $M$, and furthermore $N$ cannot add any new
ordinals below an ordinal of $M$. Thus, all new sets of $N$ must
have rank larger than any ordinal of $M$ and so $N$ is a
top-extension. So for models of ZF, the order type of an elementary
end-extension must be strictly larger.


The corresponding fact is not true for models of $\text{ZFC}^-$,
where the rank initial segments $V\_\alpha$ of the von Neumann hierarchy do not exist as sets. To see this,
start with GCH in $V$ and then force to add $\aleph\_2$ many Cohen
reals to form the extension $V[G]$, and consider the model
$M=H\_{\omega\_2}^{V[G]}$. Now add one more Cohen real $c$ over
$V[G]$, and consider $N=M[c]=H\_{\omega\_2}^{V[G][c]}$. I claim that
$N$ is an elementary end-extension of $M$. First, it is clearly an
end-extension, since both are transitive sets in $V[G][c]$. Second,
for elementarity, observe that the two-step forcing
$\text{Add}(\omega,\omega\_2)\*\text{Add}(\omega,1)$ is isomorphic to
$\text{Add}(\omega,\omega\_2)$ by absorbing the final factor into an
earlier stage. Furthermore, this isomorphism can be arranged to fix
any $\omega\_1$ sized initial segment of the forcing. Now, any fact
true in $N$ about some parameters in $V$ and an $\aleph\_1$ sized
piece of $G$, but not involving $c$, is forced by a condition, and
thus will be true whether as computed in $H\_{\omega\_2}^{V[G]}$ or
in $H\_{\omega\_2}^{V[G][c]}$, by using that isomorphism. And so the
inclusion $M\subset N$ is elementary.


The following theorem shows that the question of whether one must
go to strictly larger order types is actually independent of ZFC.


**Theorem.**


1. It is relatively consistent with ZFC that every countable $X\prec
 H\_{\omega\_2}$ has a nontrivial end-extension $X\prec Y\prec H\_{\omega\_2}$
with exactly the same ordinals.
2. It is also relatively consistent with ZFC that whenever $X\prec
 Y\prec H\_{\omega\_2}$ and $X\neq Y$, then $Y$ has strictly higher order type than $X$.


Proof. For statement 1, use the model $V[G]$ as described above.
The point is that for any countable $X\prec H\_{\omega\_2}^{V[G]}$,
we can find an $X$-generic Cohen real $c$ in $V$, and it will
follow that $X\prec X[c]$ by the argument given above.


For statement 2, use $L$ and the fact that
$H\_{\omega\_2}^L=L\_{\omega\_2}$. If $X\prec Y\prec L\_{\omega\_2}$ and
these are end-extensions, then because there is a definable
well-ordering, each new object of $Y$ not in $X$ has an ordinal
position $\alpha$ in the canonical $L$ order, and this ordinal is
in $Y$ but not in $X$. So by the end-extension property, it cannot
be smaller than any ordinal of $X$, and so the order type of $Y$ is
strictly higher than the order type of $X$. QED


I've written too much already, but one can say much more about
bounding the order-type of $Y$ over $X$, by fixing suitable Skolem
functions and then arguing that there is a club of closure points. And there are also interesting things to say for the case of extensions of models of full ZF rather than $\text{ZF}^-$. 


