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Assigning Oxidation Numbers

In FeCr2O4,...

Question

In $F e C r_{2} O_{4}$ , the oxidation number of fe and Cr are:

A

+ 2 and + 3

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B

0 and +2

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C

+ 2 and +6

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D

2 and 5

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Solution

The correct option is A + 2 and + 3
Chromium is found in nature principally as the chromite ore FeCr2O4 in which chromium is in the $+ 3$ state.

Oxidation state of

Cr is $3 +$

Oxygen is $2 -$

So, net charge of $C r_{2} O_{4} (3 \times 2 - 4 \times 2) = + 2$

So, Oxidation state of Fe is $2 + .$

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Similar questions

F e C r_{2} O_{4}

the oxidation numbers of Fe and Cr are:

Q. The formal oxidation numbers of Cr and Cl in the ions

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

C l O_{3}^{-}

respectively, are

Q. Oxidation number of

F e

[F e (S C N)]^{+ 2}

[F e (H_{2} O)_{6}]^{+ 3}

would be respectively-

Q. The oxidation states of

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})]

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Q. The d-electron configurations of

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d^{^{7}}

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(Atomic numbers of Cr

= 24, M n = 25, F e = 26, C o = 27)

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