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Step-by-Step Solution
Step 1: Determine the oxidation state of Ni in NiCl2
In NiCl2, nickel is typically in the +2 oxidation state. Hence Ni is Ni(II).
The electronic configuration of Ni (atomic number = 28) in the ground state is
[Ar]\,3d^8\,4s^2 . For Ni2+, the configuration becomes 3d^8 .
Step 2: Identify the number of unpaired electrons in Ni(II) with Cl-
Chloride (Cl-) is a weak field ligand, so in an octahedral environment
with Ni2+ ( 3d^8 ), there are two unpaired electrons.
Therefore, in NiCl2, nickel has 2 unpaired electrons.
Step 3: Determine the oxidation state of Ni in [Ni(CN)6]2−
When NiCl2 is heated with excess sodium cyanide in the presence of a strong
oxidizing agent, the nickel is oxidized to Ni(IV) and forms the complex
[Ni(CN)6]2−. In this complex, Ni is Ni(IV), i.e., Ni4+.
Step 4: Determine the electronic configuration of Ni(IV)
In Ni4+, nickel loses 4 electrons. So from the ground state ( 3d^8\, 4s^2 ),
four electrons are removed (2 from the 4s and 2 from the 3d), leaving
3d^6 for Ni4+.
Step 5: Identify the number of unpaired electrons in [Ni(CN)6]2−
Cyanide (CN−) is a strong field ligand. For a d6 metal ion
in an octahedral field of strong ligands, pairing of electrons occurs, typically leading to
a low-spin complex with 0 unpaired electrons.
Therefore, in [Ni(CN)6]2−, nickel has
0 unpaired electrons.
Step 6: Calculate the total change in the number of unpaired electrons
• Unpaired electrons in NiCl2 (Ni2+): 2
• Unpaired electrons in [Ni(CN)6]2− (Ni4+): 0
Change in unpaired electrons
= (Unpaired in NiCl2) − (Unpaired in [Ni(CN)6]2−)
= 2 − 0
= 2.
Final Answer
The total change in the number of unpaired electrons on the nickel center is
2.
