How is the dichromate anion generated from potassium dichromate? [closed]

Question

How does $\ce{K2Cr2O7}$ become $\ce{Cr2O7^2-}$ when dissolved in water?

Answer 1

You don't seem to be overly familiar with the concepts of salts (made of cations and anions), ionic bonds and solvation of salts in water.

Lets's walk through some examples in which a solid material made from cations and anions is dissolved in water in such a way that the anions and cations are each surrounded and stabilized by water molecules (i.e. they are in an aqueous state, denoted as aq):

$$\begin{align} \ce{NaCl (s) &->[H2O] Na+ (aq) + Cl- (aq)} \\ \ce{MgSO4 (s) &->[H2O] Mg^2+ (aq) + SO4^2- (aq)} \\ \ce{Na2SO4 (s) &->[H2O] 2Na+ (aq) + SO4^2- (aq)} \\ \ce{K3PO4 (s) &->[H2O] 3K+ (aq) + PO4^3- (aq)} \\ \ce{Mg3(PO4)2 (s) &->[H2O] 3Mg^2+ (aq) + 2PO4^3- (aq)} \end{align}$$

As a general rule for the solvation of the salts, we can derive:

$$\ce{M_mX_n (s) ->[H2O] mM^n+ (aq) + nX^m- (aq)}$$

Applying this to potassium dichromate, we recognise that the cation is $\ce{K+}$ and the anion is $\ce{Cr2O7^2-}$. Therefore, it is the dissolution of potassium dichromate in water which generates the dichromate anion:

$$\ce{K2Cr2O7 (s) ->[H2O] 2K+ (aq) + Cr2O7^2- (aq)}$$

As for the physical reason behind why the salts dissolve, you may find a more in-depth discussion here: Why do salts such as NaCl dissolve?