\(2 K _2 CrO _4+2 HNO _3(\text { dil }) \rightarrow K _2 Cr _2 O _7 2 KNO _3+ H _2 O\)

Therefore \(; CrO _4^{2-}+2 HNO _3(\) dil \() \rightarrow Cr _2 O _7^{2-}+2 NO _3^{-}+ H _2 O\)

Dilute \(HNO _3\) is an oxidising agent. In this reaction oxidation no. of \(Cr\) changes from \(+3\) to \(+6\). So, dilute \(HNO _3\) oxidises \(K _2 CrO _4\) to \(K _2 Cr _2 O _7\)