Oxidation number of Cr in $K_{3} {C r O}_{8}$ is
(1) $+ 6$ (2) $+ 5$
(3) $+ 3$ (4) $+ 2$

Open in App

Solution

The correct answer is 2
The structure of $K_{3} C r O_{8}$ is tetragonal
Let oxidation number of Cr=x
$3 + x + 8 \times (- 2)$ =0
$3 + x - 16$ =0
x=+13
x=+5
Sbut since Chromium can have the maximum number +5 oxidation state. This contrary is solved by the bonding by bonding pattern as shown in structure. The Cr is attached to any 5 oxygen through a single bond. hence oxidation state is +5

Suggest Corrections

Similar questions

C r

C r O_{2} C l_{2}, N a_{2} C r_{3} O_{10}, C r_{2} (S O_{4})_{3}

[C r O_{8}]^{3 -}

C r

K_{3} [C r (C_{2} O_{4})_{3}]

C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q .) + 8 H^{+} \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O

C r_{2} O_{7}^{2 -}

S O_{3}^{2 -}

S O_{3}^{2 -}

C r_{2} O_{7}^{2 -}

V, C r, M n

S c

C r

[C r (H_{2} O)_{6}] C l_{3}, [C r (C_{6} H_{6})_{2}]

K_{2} [C r (C N)_{2} (O)_{2} (O_{2}) (N H_{3})]

Join BYJU'S Learning Program