The $E A N$ of iron in $F e (C O)_{5}$ and $F e (C_{5} H_{5})_{2}$ is respectively :

36 and 34

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29 and 43

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32 and 33

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None of these

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Solution

The correct option is A 36 and 34
(i) $E A N$ of $F e = 26 + 10$ (from five $C O$ ) $= 36 = K r$ ; (ii) $E A N$ of $F e = 24 + 10$ (from two $C_{5} H_{5}^{-}$ ions) $= 34$

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