Which one of the following compounds is not a yellow colored compound?

First, we recall that the colour shown by an ionic or coordination compound arises from either (i) charge-transfer transitions as in $$CrO_4^{2-}, MoO_4^{2-}$$ etc., or (ii) d-d electronic transitions inside the partially filled d-sub-shell of a transition-metal ion. Compounds that do not possess such transitions generally appear white.

We examine every option one by one and relate its observed colour to the presence or absence of these electronic transitions.

We have the compound in Option A:

$$BaCrO_4$$

The anion $$CrO_4^{2-}$$ contains chromium in the +6 oxidation state. The intense yellow colour of chromate comes from an oxygen (2p) → chromium (3d) charge-transfer transition. Hence barium chromate is a characteristic yellow precipitate in qualitative analysis.

Now we consider Option B:

$$Zn_2[Fe(CN)_6]$$

Inside the complex anion $$[Fe(CN)_6]^{4-}$$ iron is in the +2 oxidation state with electronic configuration $$3d^{6}$$. However, the strong field ligand $$CN^-$$ causes complete pairing of the d electrons, giving a t2g6 configuration. This leaves no unpaired d electron, so d-d transitions are either absent or extremely weak. Further, zinc(II) is a d10 ion with a completely filled d-sub-shell, also incapable of d-d transitions. There is likewise no charge-transfer transition of low energy in this lattice. Because of the absence of any significant absorption in the visible region, the solid reflects all wavelengths equally and therefore appears white.

Next, Option C:

$$K_3[Co(NO_2)_6]$$

In the anion $$[Co(NO_2)_6]^{3-}$$ cobalt is in the +3 oxidation state with configuration $$3d^6$$. The nitrito ligand $$NO_2^-$$ is a fairly strong field ligand, so the complex becomes low-spin, but there are still allowed d-d transitions within the visible region. Consequently the salt shows a yellow colour that deepens to brown on standing.

Finally, Option D:

$$(NH_4)_3[As(Mo_3O_{10})_4]$$

This is ammonium molybdoarsenate, a typical heteropoly-molybdate anion. The intense yellow colour is again due to ligand-to-metal charge-transfer transitions involving $$Mo^{VI}$$ in the $$MoO_4^{2-}$$ units that are linked together in the poly-oxometalate framework.

Summarising our observations:

$$BaCrO_4$$  →  yellow

$$K_3[Co(NO_2)_6]$$  →  yellow

$$(NH_4)_3[As(Mo_3O_{10})_4]$$  →  yellow

$$Zn_2[Fe(CN)_6]$$  →  white

Only the compound in Option B lacks a yellow colour; it is white. Hence, the correct answer is Option 2.