The nature of bonding, structure of the coordination compound can be explained to
some extent by valence bond theory. The central metal atom/ion makes available a num
ber of vacant orbitals equal to its coordination number. The appropriate atomic orbitals
(s, p and d) of the metal hybridize to give a set of equivalent orbitals of definite geome
try such as square planar, tetrahedral, octahedral and so on. A strong covalent bond is
formed only when the orbitals overlap to the maximum extent. The d-orbitals involved
in the hybridization may be either inner d-orbitals i.e. (n–1) d or outer d-orbitals i.e. nd.
The complexes formed are called inner orbital complex (low spin complex) and outer
orbital complex (high spin complex) respectively. Further, the complexes can be para
magnetic or diamagnetic in nature. The drawbacks of this theory are that this involves
number of assumptions and also does not explain the colour of the complex.
Answer the following questions:
Question: 1
Predict whether \([CoF_6]^{3-}\) is diamagnetic or paramagnetic and why?
\([Atomic \ \ number : Co = 27]\)
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A complex is paramagnetic if it contains unpaired electrons in its metal's d-orbitals. If all electrons are paired, it is diamagnetic.
Paramagnetic or Diamagnetic Nature of \([CoF_6]^{3-}\):
To predict whether \([CoF_6]^{3-}\) is paramagnetic or diamagnetic, let's look at the electronic configuration of \(Co^{3+}\). The electronic configuration of \(Co\) is \([Ar]\, 3d^7\, 4s^2\). For \(Co^{3+}\), the electrons are removed from the 4s and 3d orbitals, leaving the configuration \(3d^6\). In the case of \([CoF_6]^{3-}\), fluorine is a weak field ligand and does not cause pairing of the electrons. Therefore, the 6 electrons in the \(3d\) orbitals remain unpaired, making \([CoF_6]^{3-}\) paramagnetic.
Step 1: Determine the electron configuration of \(Co^{3+}\)
\[
Co^{3+} : 3d^6
\]
Step 2: The presence of unpaired electrons in the \(3d\) orbitals means that the complex is paramagnetic.
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Question: 2
What is the coordination number of Co in \([Co(en)_2Cl_2]^+\)?
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The coordination number is determined by counting the number of bonds formed between the metal and the ligands. Bidentate ligands contribute two bonds.
The coordination number of a metal in a coordination complex is determined by the number of ligand atoms directly bonded to the metal ion. In \([Co(en)_2Cl_2]^+\), ethylenediamine (en) is a bidentate ligand, meaning each en ligand forms two bonds with the metal. Chloride (Cl) is a monodentate ligand, forming one bond with the metal. Therefore, the total coordination number is:
\[
\text{Coordination number} = 2 \times 2 \, (\text{from two en ligands}) + 2 \, (\text{from two Cl ligands}) = 6
\]
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Question: 3
Write the IUPAC name of the given complex:
\([Pt(NH_3)_2Cl_2]^{2+\)}
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In IUPAC naming, the ligands are named alphabetically before the metal, and the oxidation state of the metal is indicated in parentheses.
In this complex, platinum (Pt) is the central metal ion, and it is surrounded by two ammonia (NH$_3$) ligands and two chloride (Cl$^-$) ligands. According to IUPAC nomenclature:
- Ammonia is a neutral ligand and is named as "ammine."
- Chloride is an anionic ligand and is named as "chloro."
- The metal ion (platinum) is named first, followed by the ligands in alphabetical order.
The IUPAC name of the complex is:
\[
\text{Diamminedichloroplatinum(II)} \, \text{ion}
\]
Step 1: Identify the ligands and their names (ammine and chloro).
Step 2: Name the metal and its oxidation state (Pt(II)).
Step 3: Put the ligands and metal together following IUPAC naming rules.
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Question: 4
Explain \([Co(NH_3)_6]^{3+\) is an inner orbital or outer orbital complex. }
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Inner orbital complexes form when the metal ion uses lower energy \(d\)-orbitals for bonding. Outer orbital complexes use higher energy \(d\)-orbitals.
The complex \([Co(NH_3)_6]^{3+}\) involves the coordination of six ammine (NH$_3$) ligands to a \(Co^{3+}\) ion. For \(Co^{3+}\) (with electron configuration \(3d^6\)), the hybridization occurs using the inner \(3d\), \(4s\), and \(4p\) orbitals, as \(NH_3\) is a weak field ligand and does not cause significant pairing of electrons. Therefore, the complex uses the inner \(d\)-orbitals for hybridization and is considered an inner orbital complex.
Step 1: Consider the electron configuration of \(Co^{3+}\) and the ligands involved.
Step 2: Since \(NH_3\) is a weak field ligand, \(Co^{3+}\) does not undergo pairing, and the metal uses its inner orbitals for hybridization.
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Question: 5
Using valence bond theory, deduce the shape and hybridisation of \([Ni(NH_3)_6]^{2+\)
\([Atomic \ \ number \ of \ Ni = 28]\) }
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For a coordination number of 6, the complex adopts an octahedral shape with sp$^3$d$^2$ hybridization.
For \([Ni(NH_3)_6]^{2+}\), the coordination number is 6, and the ligand is ammine (NH$_3$), which is a weak field ligand. Thus, the metal ion, \(Ni^{2+}\), will undergo octahedral hybridization using \(3d\), \(4s\), and \(4p\) orbitals. Since the coordination number is 6, the complex must have an octahedral shape.
Step 1: The \(Ni^{2+}\) ion has the electron configuration \(3d^8 4s^2\), and upon losing two electrons, it forms \(Ni^{2+}\) with \(3d^8\).
Step 2: The \(Ni^{2+}\) ion uses its \(3d\), \(4s\), and \(4p\) orbitals to form six hybrid orbitals, leading to an octahedral shape.
Step 3: The hybridization of the \(Ni^{2+}\) ion is sp$^3$d$^2$, and the complex adopts an octahedral shape.
