Coordination Compounds (Chapter Test - Medical) - Class 12 MCQ

In [Cu(NH3)5Cl]Cl2, there are a total of 3 chloride ions present.

- Out of these 3 chloride ions, one chloride ion present inside the coordination sphere cannot be ionized.

- The remaining 2 chloride ions present outside the coordination sphere can be ionized.

- So, only these 2 chloride ions can be precipitated out by AgNO3 as AgCl.

- So, the fraction of chloride precipitated is 2/3.

As 1/3 of total chlorine of the compound is precipitated by adding AgNO₃​ to its aqueous solution so only one chloride ion is available for ionization, so compound is [CrCl₂​(H₂​O)₄​]Cl.2H₂​O.

• Cr(CO)₆: The electronic configuration for Cr is 4s^​13d⁵. CO is neutral and is a strong field ligand ( it will pair up the electrons). therefor; the hybridisation is d²sp³ and it is diamagnetic in nature. The complex is octahedral with Cr–C and C–O distances of 1.91 and 1.14 Å, respectively.
   
• Fe(CO)₅: The oxidation state of Fe in this complex is Zero and it has outer electronic configuration 3d⁶4s².
  

  Now there are 4 unpaired electrons in 3d orbital and complex with C.N.-5, central atom may involve dsp³ hybridisation, thus 4s must be empty. Two electrons of 4s orbital and one electron of 3d orbital are pushed into 3d orbitals to pair up with the three unpaired 3d electrons. since the complex has no unpaired electron, it will be diamagnetic and has trigonal bipyramidal geometry.

• [CoF_6​]³⁻: The central Co atom is sp³d² hybridised. It forms an octahedral complex with paramagnetic behaviour due to the presence of 4 unpaired electrons. It forms outer orbital or high spin or spin-free complex as it utilizes outer 4d orbitals in sp³d² hybridisation.
  

• Ni(CO)₄: The valence shell electronic configuration of the ground state Ni atom is 3d⁸ 4s². * All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three 4p orbitals undergo sp³ hybridization and form bonds with CO ligands to give Ni(CO)₄. 
  In Ni(CO)4, Ni is sp³ hybridised and its geometry is tetrahedral (as shown in the figure).
  Carbonyl, CO being a strong field ligand causes the pairing of up valence electrons in the Ni atom against the Hund's Rule of Maximum Multiplicity. This results in the formation of an inner orbital complex, [Ni(CO)₄] having diamagnetic character. [Ni(CO)₄] has sp³ hybridization.