Consider the following complex ions:
$[\text{Ni(CN)}_4]^{2-}$, $[\text{Fe(CN)}_6]^{3-}$, $[\text{Cu(H}_2\text{O)}_6]^{2+}$, and $[\text{Co(CN)}_6]^{3-}$.
The complex ions that are expected to show diamagnetic behavior at room temperature are:

For $[\text{Ni(CN)}_4]^{2-}$, the oxidation state of Ni is $+2$. The electronic configuration of Ni$^{2+}$ is $3d^8$. Since CN$^{-}$ is a strong field ligand, it causes pairing of electrons, resulting in $dsp^2$ hybridization. All electrons are paired, making it diamagnetic.

For $[\text{Fe(CN)}_6]^{3-}$, the oxidation state of Fe is $+3$. The electronic configuration of Fe$^{3+}$ is $3d^5$. CN$^{-}$ is a strong field ligand, so the configuration is $t_{2g}^5 e_g^0$. It has one unpaired electron, making it paramagnetic.

For $[\text{Cu(H}_2\text{O)}_6]^{2+}$, the oxidation state of Cu is $+2$. The electronic configuration of Cu$^{2+}$ is $3d^9$. It has one unpaired electron, making it paramagnetic.

For $[\text{Co(CN)}_6]^{3-}$, the oxidation state of Co is $+3$. The electronic configuration of Co$^{3+}$ is $3d^6$. CN$^{-}$ is a strong field ligand, so the configuration is $t_{2g}^6 e_g^0$. All electrons are paired, making it diamagnetic.

Therefore, $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Co(CN)}_6]^{3-}$ are diamagnetic.

Answer: $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Co(CN)}_6]^{3-}$