Can antimatter exists in the form of compound? I just found an article describing the cooling of the antimatter, "Laser cooling of antihydrogen atoms". Naturally, if there's something, one turned to be like to hold it.
Once the antihydrogen had formed with $e^+ e^-$ pair, the $\bar p$ became the "nucleus" of a charged particle, like an ion with a negative charge. Then, since electromagnetic force acted in a much longer distance than the strong force, is it possible to make a $\bar H^-$ to form a compound with, say $Ni^+$ and form $\bar HN_i$ by annihilating or getting rid of a pair of $e^+e^-$?
Naively thinking, once that's done, they were essentially trapped in the electron cloud, and, then, can they be picked up just like any other materials?
 A: Your proposal of binding antihydrogen with an atom  made of particles ignores quantum mechanics.
Here is a discussion of  the atomic orbitals, the probable locations of the electrons about the nucleus. In the molecule you envisage, the orbital of the antihydrogen positron has a high probability to  overlap with an electron orbital , and annihilating into two photons. The antiproton, because of its large mass will have an orbital within  the nucleus, with high probability of annihilating with a proton or neutron.
Muonic atoms) give an experimental confirmation of what happens when heavier particles take over the(n,l,m) of an electron orbital.


Since the orbital of the muon is very near the atomic nucleus, that muon can be considered as a part of the nucleus.

....

Since there is only one electron outside the nucleus, the hydrogen-4.1 atom can react with other atoms. Its chemical behavior is that of a hydrogen atom and not a noble helium atom.

So the nuclear structure changes , when a muon occupies one of the two  electrons location, ( n, l,m) quantum numbers.
The antiproton in your proposal  will have an even tighter orbital and will be within the positive nucleus,  an orbital with a good probability of overlap and annihilation with another nucleon .
So the answer is, antimatter cannot exist in a compound with matter in stable form.
A: Topics involved:
Notice that the process meant for $e^-e$ pair to annihilate each other in the very beginning to create ion like structure.

*

*Pauli's exclusion principle: while Pauli's exclusion principle worked for fermions, the electron and anti proton were two types of particles for the interests of the situation. Thus it's rather most useful to think $\bar p^-$ and $e^-$ to be two independent systems from the beginning, labeled by two quantum number $n_{\bar p^-}$ and $n_{e}$.


*Though the usual question was to ask what's the wave function of the Exotic atom or Muonic atoms or "Hydrogen 4.1", the configuration of those were known and it didn't work because $\bar p^-$ when treated like hydrogen atom's electron had 2000 times mass than electron, and it's interaction were around the nuclei nuclei range.(see anna v's answer. ) At $n_{\bar p^-}=1,2,...$ , the overlap between the wave function of the $\bar p^-$ and nuclei was significant, thus the annihilation process was expected to happen with a high probability. (a solution to the hydrogen atom could be of useful, see anna v's answer.)
However, question regarded was to send  $\bar H^-$ to ions such as $Ni^+$ from the outside. It's a dynamic process rather than a static one. The question of interest here was for $n_{\bar p^-}$ to be a larger number, i.e. 2000000, and then transit to the state of $n_{\bar p^-}=1,2,...$.


*For the interest of the situation, during the scale of nuclei one could assume that the probability distribution of the electron mass coincide with the probability distribution of the electron charge. In face, that's how Lagrangian was written.

Change in electron cloud lagrangian
Once that's done, take for one instance to think the condition of when $\bar p^-$ was inside the nuclei with $n_{\bar p^-}=1,2,...$, the effective Lagrangian for electrons were no longer $\frac{Ze^2}{2\pi\epsilon_0 r}$ but rather $\frac{(Z-1)e^2}{2\pi\epsilon_0 r}$. The energy of the electron cloud had already been drastically changed.
Let $a+1=Z$ be the number of the original nuclei, this lead to a percentage change of $\frac{2a+1}{(a+1)^2}$ per electron's original energy, or a percentage change of $\frac{2a^2 -a-1}{(a+1)^2}$ of the combined energy of the electron could. Notice that in the Muonic helium this never mattered, because for $a=1$, $\frac{2a^2 -a-1}{(a+1)^2}=0$(corrected) consistent with the fact anna v had mentioned, that in Muonic helium the other conditions simply didn't matter, where $n_{\bar p^-}$ were free to decay and annihilate in the overlap of the wave functions.
However, that's not the same case for Nikel, for Nikel, $a=27$, and $\frac{2a^2 -a-1}{(a+1)^2}=1.8239$. This drastic change had to be considered, other wise the charge conservation and on shell condition wouldn't hold. This created a sort of "repulsion" of electron cloud with high atomic number unseen from that of the Muonic helium, or the "shelding" of the nuclei from the electron cloud.
Transition probability(incomplete)
Thus, the more relevant value was to calculate A. the transition probability of the process $$|n_{\bar p^-}(high)\rangle \otimes|a+1,n_e\rangle\Rightarrow |n_{\bar p^-}(low)\rangle \otimes|a,n_e\rangle$$. However, one had to consider B. what if $\bar p^-$ started to push the atom one states by another and thus "fuse" through one layer of the electron cloud to another. This would increase the probability and had to be discussed. Further, though the fact that $m_{\bar p^-}$ was much larger to be comparable, it also lead to $\mu_{\bar p^-}=2000 \mu_e$, and this increased the probability of the transition to happen by composite the energy of the $\bar \mu^-$.
Change in $\bar p^-$ energy
The above discussion only talked about the electron system, notice that the sheilding also meant a change in the $\bar p^-$ energy as well, i.e.
$$E_{\bar p^-}\propto -\frac{2000 \mu_e 1^2}{(n_{\bar p^-}large)^2}\Rightarrow  -\frac{2000\mu_e Z^2}{(1,2...)^2}$$ of $\Delta E_{\bar p^-}\approx -\frac{2000\mu_e Z^2}{(1,2...)^2}$. But $\Delta E_{total_e}<\frac{(2Z^2-Z)\mu_e}{1^2}$. That had already brought suspension.
Calculating
$$\Delta E_{p,\alpha,a}= \frac{2000\mu_e (a+1)^2}{(n_p+\alpha)^2}- \frac{2000\mu_e (a)^2}{(n_p)^2}  \approx \frac{2000\mu_e}{n_p+\alpha} (\frac{\alpha(a+1)+2a n_p  }{n_p^2}) \approx \frac{2000\mu_e}{n_p+\alpha} (\frac{c*a}{n_p})$$ where $n_p\sim \alpha>>c,a$, while
$$\Delta E_{e,a}= -\frac{\mu_e (a+1)^2}{n_e^2}+ \frac{\mu_e (a)^2}{n_e^2}= -\frac{\mu_e (2a+1)}{n_e^2}$$.
Notice that in the above notation, the $a$ for $e$ and $a$ for $p$ was different and with a sum of the atomic number $Z$.
For $Ni$ (atomic number $28$), $a=27$, and if filled to the out shell of $n_e=3$.
Hypothesis (B.)
Thus, though the "lowest" energy states was for $n_p$ reduce to $1,2,...$, the effect of the electron cloud provided some sort of repulsion, where at $n_p$  large, the reduction of $n_p$ would eventually reach to a point where it required the absorption of the energy to enable the further quantized orbital decay to the nearby lower energy states $n_p-\alpha$.
(A.)For at least three energetic photon to mediate the process and taking into the consideration of spin, it took some probability to happen. But it doesn't just happen.
When the atomic number $Z$ was small, anna v's answer pretty much summarized it in QM.
