Question

The total number of unpaired electrons in the two complexes [cr(h2o)6]2+ and [cr(cn)6]4- having octahedral geometry are

Answer 1

5 + 1 = 6 .

[cr(h2o)6]2+ = 5

[cr(cn)6]4- = 1

Answer 2

Answer:

Two complexes having octahedral geometry, [cr(h2o)6]2+ and [cr(cn)6]4- are 6.

Explanation:

• Because of its tiny size and high z-effect, chromium forms inner orbital complexes, and according to crystal field theory, it is d2sp3.
• The oxidation state of Cr in [Cr(CN)6]4 is +2, making it a d4 metal. To see this, look at a periodic table.
• It has four d electrons to distribute in its eg and t2g orbitals because it is a d4 metal.
• [Cr(H2O)6]2+ chromium has a d4 configuration, and H2O is a weak field ligand d4 (weak field)chromium has a d4 configuration, and CN-ion is a strong field ligand in [Cr(CN)6]4-chromium.
• Weak field ligands build high spin complexes by causing tiny splittings between the d orbitals.
• The halogens, OH-, and H2O are examples of weak field ligands.
• These ions or molecules all have full orbitals that can connect with the metal d orbitals in the ligand field.
• As a result, the CN-ion causes electron pairing. As a result, the total number of unpaired electrons in both complexes is 6.

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