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Magnetic Moment Examples

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Question

Write the formula for a sulphate of iron with $F e$ in the oxidation state $+ 3$ .

{F e}_{2} {({S O}_{4})}_{3}

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{F e}_{2} {({S O}_{4})}_{2}

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{F e}_{3} {({S O}_{4})}_{2}

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{F e}_{3} {({S O}_{4})}_{3}

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Solution

The correct option is A ${F e}_{2} {({S O}_{4})}_{3}$
The formula for the sulphate of iron with Fe in the oxidation state +3 is ${F e}_{2} {({S O}_{4})}_{3}$ .
Each ferric ion has +3 charge. Two ferric ions have total +6 charge.
Each sulphate ion has -2 charge.
Three sulphate ions have -6 charge.
The total positive charge is equal to total negative charge.
Thus, electrical neutrality is maintained.

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Similar questions

F e | F e^{2 +} | | F e^{3 +} | F e

F e | F e^{2 +} | | F e^{3 +}, F e^{2 +} | P t

F e | F e^{3 +} | | F e^{3 +}, F e^{2 +} | P t

F e

F e^{2 +}

F e^{3 +}

E_{1}^{o}, E_{2}^{o}

E_{3}^{o}

F e | F e^{2 +}, F e^{2 +} | F e^{3 +}

F e | F e^{3 +}

E_{3}^{o} = \frac{E_{2}^{o} + 2 E_{1}^{o}}{n}

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e 6 - \to F e; E_{F}^{o} e_{F e^{2 +} / F e} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3 +} / F e^{2 +}}^{o} = + 0.77 V

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

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