Using valence bond theory, deduce the shape and hybridisation of [Ni(NH3)6]2+ [Atomic number of Ni = 28]
Ans.
To deduce the shape and hybridization of [Ni(NH3)6]2+, we’ll utilize the valence bond theory.
First, let’s consider the atomic structure of nickel (Ni), with an atomic number of 28. The electron configuration of nickel is [Ar] 3d^8 4s^2. For [Ni(NH3)6]2+, nickel loses two electrons to form Ni^2+, giving it the electron configuration [Ar] 3d^8.
In this complex, nickel forms coordinate bonds with six ammonia (NH3) ligands. Ammonia is a weak field ligand, so it does not cause any d-electron pairing. Therefore, we consider the complex as a high spin complex.
Now, let’s examine the hybridization and shape:
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Hybridization: In the valence bond theory, the central metal ion (nickel) makes available a number of vacant orbitals equal to its coordination number. Here, the coordination number of nickel is 6, as it coordinates with six ammonia ligands. Therefore, the appropriate atomic orbitals (s, p, and d) of nickel hybridize to give a set of equivalent orbitals of definite geometry.
To accommodate six ligands in an octahedral geometry, nickel undergoes dsp^2 hybridization. This means that one s, three p, and two d orbitals hybridize to form six equivalent hybrid orbitals.
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Shape: Since nickel is dsp^2 hybridized and forms six sigma bonds with the ammonia ligands in an octahedral arrangement, the shape of [Ni(NH3)6]2+ is octahedral.
In summary:
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Hybridization: dsp^2
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Shape: Octahedral
