A sample of wustite ${F e}_{x} O$ contains one $F e (I I I)$ for every three $F e (I I)$ . The value of $x$ is:

9 / 8

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8 / 9

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3 / 4

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4 / 3

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Solution

The correct option is A $9 / 8$
Let there be 1 mol of iron atom.
Amount of Fe(III)=1/4 mol
Amount of Fe(II)=3/4 mol
Total positive charges =1/4(3)+3/4(2)
= 9/4 mol (positive charge because iron have positive have in bonded state)
Let x be the amount of oxygen atoms,we have
Total negative charges =x(2) (oxygen have negative charge)
To satisfy electrical neutrality,we have
Total positive charges =Total negative charges
9/4 mol = 2x
x=9/8 mol

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\begin{matrix} F e^{I I} {[F e^{I I} {(C N)}_{6}]}^{2 -} & F e^{I I I} [F e^{I I I} {(C N)}_{6}] I & I I F e^{I I} {[F e^{I I I} {(C N)}_{6}]}^{3 -} & F e^{I I I} {[F e^{I I} {(C N)}_{6}]}^{4 -} I I I & I V \end{matrix}

Factorise $9 x^{2} - 64$

24 \div 3 - 9 \times 3 + (52 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 8 - 4) \div 4 + 9

F e_{0.93} O_{1.0}

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