Since F e has 5 carbonyl around it, it has trigonal bi pyramidal geometry and shape and has hybridization- d s p 3. The splitting of various orbitals of d subshells to different energy is based on the repulsions experienced because of the approaching ligand as, C O is a neutral ligand

What is the shape of F e (C O) 5 molecule and which of the following d-orbitals involved in hybridization ? Tetrahedral d x 2 − y 2; Trigonal bipyramidal, d z 2; Trigonal bipyramidal, d x 2 − y 2; Square pyramidal

Let oxidation no. of F e = x x + (5 × 0) = 0 x = 0. E A N = ( z − x ) + 2 n L ( 1 ) where z = Atomic number of metal x = Oxidation state of metal n = No. of Ligands L = No. of co-ordinate bonds formed by ligand.

Let x be the oxidation state of F e in F e (C O) 5. Since, the overall charge on the complex is 0, the sum of oxidation states of all elements in it should be equal to 0. Therefore, x + 5 (0) = 0 or, x = 0. Hence, the oxidation state of F e in F e (C O) 5 is 0.

Now 5 C O ligands will fill in one 3 d orbital, one 4 s orbital and three 4 p orbitals forming d s p 3 hybridization which results in trigonal bipyramidal shape. Was this answer helpful? 6

Oxidation state of Fe in Fe(CO) 5. Reason. View Solution. Q5. The oxidation state of F e in F e (C O) 5 is ...

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In F e (C O) 5, the F e atom is in d s p 3 hybridised state. Therefore, the shape of molecule is trigonal bipyramidal. Therefore, the shape of molecule is trigonal bipyramidal. The hybridisation is as:

1) The oxidation number of Fe is 0. 2) The charge on a complex is the sum of the oxidation state of the metal center and the charges on the ligands. 3) C O is a neutral ligand. Hence, the charge on the ligand is 0. The total charge on the complex is also 0. 4) Calculate the oxidation number of Fe in the complex, F e (C O) 5 is as follows;

Q5. The zero magnetic moment of octahedral K 2 N i F 6 is due to: View Solution ...