The metal complex that is diamagnetic is (Atomic number: Fe, 26; Cu, 29)

We need to identify the diamagnetic metal complex from the given options.

A complex is diamagnetic if it has no unpaired electrons.

Option A: $$K_3[Cu(CN)_4]$$

Here, $$Cu$$ has a charge of $$+1$$ (since $$3 \times (+1) + x + 4 \times (-1) = 0$$, so $$x = +1$$).

$$Cu^+$$ has electronic configuration: $$[Ar] 3d^{10}$$

With 10 d-electrons, all orbitals are completely filled. Therefore, there are no unpaired electrons.

$$CN^-$$ is a strong field ligand, but regardless of the field strength, $$d^{10}$$ always has zero unpaired electrons.

This complex is diamagnetic.

Option B: $$K_2[Cu(CN)_4]$$

Here, $$Cu$$ has a charge of $$+2$$ (since $$2 \times (+1) + x + 4 \times (-1) = 0$$, so $$x = +2$$).

$$Cu^{2+}$$ has electronic configuration: $$[Ar] 3d^9$$

With 9 d-electrons, there is 1 unpaired electron regardless of the crystal field. This complex is paramagnetic.

Option C: $$K_3[Fe(CN)_4]$$

Here, $$Fe$$ has a charge of $$+1$$ (since $$3 \times (+1) + x + 4 \times (-1) = 0$$, so $$x = +1$$).

$$Fe^+$$ has electronic configuration: $$[Ar] 3d^6 4s^1$$ or $$[Ar] 3d^7$$

With $$CN^-$$ as strong field ligand, $$3d^7$$: $$t_{2g}^6 e_g^1$$ gives 1 unpaired electron. This complex is paramagnetic.

Option D: $$K_4[FeCl_6]$$

Here, $$Fe$$ has a charge of $$+2$$ (since $$4 \times (+1) + x + 6 \times (-1) = 0$$, so $$x = +2$$).

$$Fe^{2+}$$ has electronic configuration: $$[Ar] 3d^6$$

$$Cl^-$$ is a weak field ligand (high spin): $$t_{2g}^4 e_g^2$$ gives 4 unpaired electrons. This complex is paramagnetic.

Therefore, the correct answer is Option A: $$K_3[Cu(CN)_4]$$.