The oxidation states of transition metal atoms in $$K_2Cr_2O_7$$, $$KMnO_4$$ and $$K_2FeO_4$$, respectively, are x, y and z. The sum of x, y and z is ___________.

First, we examine the compound $$K_2Cr_2O_7$$ (potassium dichromate). According to the common oxidation-state rules, each potassium atom carries an oxidation state of $$+1$$ and each oxygen atom carries an oxidation state of $$-2$$. Let the oxidation state of each chromium atom be $$x$$.

Because the molecule is electrically neutral, the algebraic sum of all oxidation states must be zero. We therefore write

$$2(+1) \;+\; 2(x) \;+\; 7(-2) \;=\; 0.$$

Simplifying term by term gives

$$+2 \;+\; 2x \;-\; 14 \;=\; 0.$$

Combining the constant numbers, we get

$$2x \;-\; 12 \;=\; 0.$$

Adding $$12$$ to both sides yields

$$2x \;=\; 12,$$

and dividing by $$2$$ finally gives

$$x \;=\; +6.$$

Next, consider $$KMnO_4$$ (potassium permanganate). Potassium again is $$+1$$ and oxygen is $$-2$$. Let the oxidation state of manganese be $$y$$. Setting up the neutrality condition, we have

$$1(+1) \;+\; y \;+\; 4(-2) \;=\; 0.$$

This expands to

$$+1 \;+\; y \;-\; 8 \;=\; 0.$$

Combining the constants gives

$$y \;-\; 7 \;=\; 0.$$

Adding $$7$$ to both sides, we get

$$y \;=\; +7.$$

Finally, we analyze $$K_2FeO_4$$ (potassium ferrate). As before, potassium is $$+1$$ and oxygen is $$-2$$. Let the oxidation state of iron be $$z$$. Writing the charge-balance equation, we have

$$2(+1) \;+\; z \;+\; 4(-2) \;=\; 0.$$

This simplifies to

$$+2 \;+\; z \;-\; 8 \;=\; 0.$$

Combining constants gives

$$z \;-\; 6 \;=\; 0.$$

Adding $$6$$ to both sides yields

$$z \;=\; +6.$$

We now have the individual oxidation states:

$$x = +6,\;\; y = +7,\;\; z = +6.$$

The question asks for the sum $$x + y + z$$. Substituting the values we found,

$$x + y + z = 6 + 7 + 6 = 19.$$

So, the answer is $$19$$.