What is the electron configuration for $F e_{2}$ $Fe_2$ ?

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What is the electron configuration for $F e_{2}$ $Fe_2$ ?

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Answer 1 · 2017-06-16T06:54:17

The ion $F e^{2 +}$ $Fe^(2+)$ electron configuration is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$ $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$

Explanation:

The electron configuration for iron, $F e$ $Fe$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}$ $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$

The orbital filling order for $F e$ $Fe$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$

Iron does not occur as $F e_{2}$ $Fe_2$ in nature. It is not a diatomic element.

The ion $F e^{2 +}$ $Fe^(2+)$ electron configuration is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$ $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$ .

It is missing the two valence electrons in the $4 s^{2}$ $4s^2$ orbital. This makes the ion a cation, an ion that is positively charged because it has lost valence electrons.

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What is the electron configuration for $F e_{2}$ $Fe_2$ ?

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