The geometry of [NiCl4]2- and [Ni(CN)4]2- ions are:
The geometry of [NiCl4]2- and [Ni(CN)4]2- ions are:
Both tetrahedral
Both square planar
Both octahedral
Square planar and tetrahedral,respectively
Tetrahedral and square planar,respectively
The Correct Option is
Approach Solution - 1
Coordination complexes analysis:
- [NiCl4]2-:
- Weak field ligand (Cl-)
- sp3 hybridization → Tetrahedral
- [Ni(CN)4]2-:
- Strong field ligand (CN-)
- dsp2 hybridization → Square planar
Thus, the correct option is (E): Tetrahedral and square planar, respectively.
Approach Solution -2
This question is about the geometry of two ions, \([NiCl_4]^{2-}\) and \([Ni(CN)_4]^{2-}\).
Let's go through the options:
- Option (A) Both tetrahedral: This is incorrect because \([NiCl_4]^{2-}\) has a tetrahedral geometry, but \([Ni(CN)_4]^{2-}\) generally has a square planar geometry.
- Option (B) Both square planar: This is incorrect, as \([NiCl_4]^{2-}\) is tetrahedral, not square planar.
- Option (C) Both octahedral: This is incorrect. Neither of the ions has an octahedral geometry.
- Option (D) Square planar and tetrahedral, respectively: This is incorrect because the order is reversed. \([Ni(CN)_4]^{2-}\) has a square planar geometry, while \([NiCl_4]^{2-}\) is tetrahedral.
- Option (E) Tetrahedral and square planar, respectively:
This is the correct answer. \([NiCl_4]^{2-}\) has a tetrahedral geometry, and \([Ni(CN)_4]^{2-}\) has a square planar geometry.
So, the correct answer is (E) Tetrahedral and square planar, respectively.
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Concepts Used:
VSEPR theory
The Valence Shell Electron Pair Repulsion Theory abbreviated as VSEPR theory is based on the premise that there is a repulsion between the pairs of valence electrons in all atoms, and the atoms will always tend to arrange themselves in a manner in which this electron pair repulsion is minimalized.
Postulates of Valence Bond Theory
- The overlapping of two half-filled valence orbitals of two different atoms forms the covalent bond. The electron density between two linked atoms increases due to the overlapping. The molecule thus gains the property of stability.
- More than one bond can be created if the atomic orbitals contain more than one unpaired electron. As per the valence bond theory, the paired electrons in the valence shell cannot participate in such bond formation.
- A covalent bond has a specific direction. The region of overlapping atomic orbitals is also parallel to such a bond.
- There are two types of covalent bonds based on the overlapping pattern: sigma bonds and pi bonds. The pi bond is formed by the overlapping of atomic orbitals sideways, whereas the sigma bond is created by the overlapping of atomic orbitals along the axis of the nucleus.
Types of Hybrid Orbitals:
Applications of Valence Bond Theory
- The maximum overlap conditions which is given by the valence bond theory can explain the formation of covalent bonds in many compounds.
- One of its most important applications is the variation in the length and strength of chemical bonds in H2 and F2 molecules can be explained by the difference in their overlapping orbitals.
- The overlap of the 1s orbital of the hydrogen atom and the 2p orbital of the fluorine atom makes the covalent bond in an HF molecule which is explained by the valence bond theory.
Limitations of the VSEPR theory
- The VSEPR model is not a theory. It does not explain or attempt to explain any observations or predictions. Rather, it is an algorithm that accurately predicts the structures of a large number of compounds.
- VSEPR is simple and useful but does not work for all chemical species.
- First, the idealized bond angles do not always match the measured values. For example, VSEPR predicts that and will have the same bond angles, but structural studies have shown the bonds in the two molecules are different by 12 degrees.
- VSEPR also predicts that group-2 halides such as will be linear when they are actually bent. Quantum mechanics and atomic orbitals can give more sophisticated predictions when VSEPR is inadequate.