JEE MAIN - Chemistry (2021 - 26th February Morning Shift - No. 18)
Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ___________.
Answer
6
Explanation
$$C{r_2}O_7^{2 - } + 2O{H^ - }$$ ⇌ $$2CrO_4^{2 - } + {H_2}O$$
Let Oxidation state of Cr in $$CrO_4^{2 - }$$ is = x.
$$ \therefore $$ x + (–2 × 4) = –2
$$ \Rightarrow $$ x = 6
Let Oxidation state of Cr in $$CrO_4^{2 - }$$ is = x.
$$ \therefore $$ x + (–2 × 4) = –2
$$ \Rightarrow $$ x = 6
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