Question:
Answer the following questions about the complexes
\[
[NiCl_4]^{2-}
\quad \text{and} \quad
[Ni(CN)_4]^{2-}
\]
(i) Write the hybridization involved in each case.
(ii) Which of them is the inner orbital complex and which one is the outer orbital complex ?
(iii) Compare their magnetic behaviour.
(Atomic Number of Ni = 28)
Answer the following questions about the complexes
\[
[NiCl_4]^{2-}
\quad \text{and} \quad
[Ni(CN)_4]^{2-}
\]
(i) Write the hybridization involved in each case.
(ii) Which of them is the inner orbital complex and which one is the outer orbital complex ?
(iii) Compare their magnetic behaviour.
(Atomic Number of Ni = 28)
Show Hint
CN$^-$ is a strong field ligand and usually causes electron pairing, whereas Cl$^-$ is a weak field ligand and generally does not cause pairing.
Updated On: Jun 29, 2026
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Solution and Explanation
Concept:
The geometry, hybridization and magnetic properties of coordination compounds depend upon:
• Oxidation state of the central metal ion.
• Electronic configuration of the metal ion.
• Strength of the ligand present.
• Crystal field splitting produced by the ligand.
According to Crystal Field Theory, strong field ligands pair the electrons present in d-orbitals whereas weak field ligands generally do not cause pairing. Consequently, the type of hybridization and magnetic behaviour of the complex changes significantly.
Step 1: Determine the oxidation state of nickel in both complexes. For \[ [NiCl_4]^{2-} \] let the oxidation state of nickel be \(x\). \[ x+4(-1)=-2 \] \[ x=+2 \] Similarly, for \[ [Ni(CN)_4]^{2-} \] \[ x+4(-1)=-2 \] \[ x=+2 \] Thus, in both complexes: \[ Ni^{2+} \] is present.
Step 2: Write the electronic configuration of \(Ni^{2+}\). Atomic number of nickel: \[ Z=28 \] Electronic configuration of Ni: \[ [Ar]\,3d^8\,4s^2 \] Electronic configuration of \[ Ni^{2+} \] is: \[ [Ar]\,3d^8 \]
Step 3: Analysis of \([NiCl_4]^{2-}\). Chloride ion is a weak field ligand. It is unable to pair the electrons present in the \(3d\)-orbitals. Hence electron pairing does not occur. The complex therefore uses outer orbitals for hybridization. Hybridization: \[ sp^3 \] Geometry: Tetrahedral Since two unpaired electrons remain present, the complex is paramagnetic. \[ \boxed{ [NiCl_4]^{2-} \rightarrow sp^3 \text{ hybridization} } \]
Step 4: Analysis of \([Ni(CN)_4]^{2-}\). Cyanide ion is a strong field ligand. It causes pairing of the \(3d\)-electrons. After pairing, one \(3d\)-orbital becomes vacant and participates in hybridization. Hybridization: \[ dsp^2 \] Geometry: Square planar Since all electrons become paired, the complex is diamagnetic. \[ \boxed{ [Ni(CN)_4]^{2-} \rightarrow dsp^2 \text{ hybridization} } \]
Step 5: Identify inner and outer orbital complexes. In \[ [Ni(CN)_4]^{2-} \] the inner \(3d\)-orbital participates in hybridization. Therefore it is an inner orbital complex. In \[ [NiCl_4]^{2-} \] the outer orbitals participate in hybridization. Therefore it is an outer orbital complex.
Final Answers: \[ \boxed{ [NiCl_4]^{2-} : sp^3 \text{ hybridization} } \] \[ \boxed{ [Ni(CN)_4]^{2-} : dsp^2 \text{ hybridization} } \] \[ \boxed{ [NiCl_4]^{2-} \text{ is an outer orbital complex} } \] \[ \boxed{ [Ni(CN)_4]^{2-} \text{ is an inner orbital complex} } \] \[ \boxed{ [NiCl_4]^{2-} \text{ is paramagnetic} } \] \[ \boxed{ [Ni(CN)_4]^{2-} \text{ is diamagnetic} } \]
• Oxidation state of the central metal ion.
• Electronic configuration of the metal ion.
• Strength of the ligand present.
• Crystal field splitting produced by the ligand.
According to Crystal Field Theory, strong field ligands pair the electrons present in d-orbitals whereas weak field ligands generally do not cause pairing. Consequently, the type of hybridization and magnetic behaviour of the complex changes significantly.
Step 1: Determine the oxidation state of nickel in both complexes. For \[ [NiCl_4]^{2-} \] let the oxidation state of nickel be \(x\). \[ x+4(-1)=-2 \] \[ x=+2 \] Similarly, for \[ [Ni(CN)_4]^{2-} \] \[ x+4(-1)=-2 \] \[ x=+2 \] Thus, in both complexes: \[ Ni^{2+} \] is present.
Step 2: Write the electronic configuration of \(Ni^{2+}\). Atomic number of nickel: \[ Z=28 \] Electronic configuration of Ni: \[ [Ar]\,3d^8\,4s^2 \] Electronic configuration of \[ Ni^{2+} \] is: \[ [Ar]\,3d^8 \]
Step 3: Analysis of \([NiCl_4]^{2-}\). Chloride ion is a weak field ligand. It is unable to pair the electrons present in the \(3d\)-orbitals. Hence electron pairing does not occur. The complex therefore uses outer orbitals for hybridization. Hybridization: \[ sp^3 \] Geometry: Tetrahedral Since two unpaired electrons remain present, the complex is paramagnetic. \[ \boxed{ [NiCl_4]^{2-} \rightarrow sp^3 \text{ hybridization} } \]
Step 4: Analysis of \([Ni(CN)_4]^{2-}\). Cyanide ion is a strong field ligand. It causes pairing of the \(3d\)-electrons. After pairing, one \(3d\)-orbital becomes vacant and participates in hybridization. Hybridization: \[ dsp^2 \] Geometry: Square planar Since all electrons become paired, the complex is diamagnetic. \[ \boxed{ [Ni(CN)_4]^{2-} \rightarrow dsp^2 \text{ hybridization} } \]
Step 5: Identify inner and outer orbital complexes. In \[ [Ni(CN)_4]^{2-} \] the inner \(3d\)-orbital participates in hybridization. Therefore it is an inner orbital complex. In \[ [NiCl_4]^{2-} \] the outer orbitals participate in hybridization. Therefore it is an outer orbital complex.
Final Answers: \[ \boxed{ [NiCl_4]^{2-} : sp^3 \text{ hybridization} } \] \[ \boxed{ [Ni(CN)_4]^{2-} : dsp^2 \text{ hybridization} } \] \[ \boxed{ [NiCl_4]^{2-} \text{ is an outer orbital complex} } \] \[ \boxed{ [Ni(CN)_4]^{2-} \text{ is an inner orbital complex} } \] \[ \boxed{ [NiCl_4]^{2-} \text{ is paramagnetic} } \] \[ \boxed{ [Ni(CN)_4]^{2-} \text{ is diamagnetic} } \]
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