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Q.

The hybridization and unpaired electrons in $[Fe {(H_{2} O)}_{6}]^{2 +}$ ion are:

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a

$d^{2} {sp}^{3}; 3$

b

${sp}^{3} d^{2}; 2$

c

${sp}^{3} d; 4$

d

${sp}^{3} d^{2}; 4$

answer is A.

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Detailed Solution

When two atomic orbitals join to generate a hybrid orbital in a molecule, the energy of the individual atoms' orbitals is redistributed to give orbitals of equivalent energy.
In the third orbital of [Fe(H₂O)₆]²⁺, there are 4 unpaired electrons. The hybridization and unpaired electrons in it are ${sp}^{3} d^{2}$ and 4 respectively.

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