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A square planar complex is formed by hybridization of the following atomic orbitals (a) s, \(\mathrm{p}_{\mathrm{x}}, \mathrm{p}_{y}, \mathrm{p}_{z}\) (b) \(s, p_{x}, p_{y}, p_{z}, d\) (c) \(\mathrm{d}_{x^{2}-y^{2}}, \mathrm{~s}, \mathrm{p}_{x}, \mathrm{p}_{\mathrm{y}}\) (d) \(\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{y}, \mathrm{p}_{2}, \mathrm{~d}_{z^{2}}\)

Short Answer

Expert verified
Option (c) forms a square planar complex: d_{x^2-y^2}, s, p_x, p_y.

Step by step solution

01

Identify the Hybridization Required for Square Planar

Square planar complexes typically involve dsp^2 hybridization. This means one d, one s, and two p orbitals are required for the square planar arrangement.
02

Analyze Option (a)

Option (a) involves the orbitals: s, p_x, p_y, p_z. This does not include a d orbital, which is necessary for square planar dsp^2 hybridization. Hence, option (a) is not suitable.
03

Analyze Option (b)

Option (b) involves the orbitals: s, p_x, p_y, p_z, d. At a glance, it appears possible; however, for square planar hybridization, the specific d orbital needed is d_{x^2-y^2}, which is not specified here. Therefore, option (b) may not be ideal.
04

Analyze Option (c)

Option (c) includes the orbitals: d_{x^2-y^2}, s, p_x, p_y. These orbitals perfectly match the dsp^2 hybridization required for a square planar complex. Hence, this is a viable choice.
05

Analyze Option (d)

Option (d) lists: s, p_x, p_y, p_2, d_{z^2}. The use of p_2 is unusual, and more importantly, d_{x^2-y^2} is necessary for square planar, which d_{z^2} does not fulfill. Thus, option (d) is not suitable.
06

Conclusion

Based on the analysis, the only set of orbitals that forms a square planar complex through dsp^2 hybridization is found in option (c): d_{x^2-y^2}, s, p_x, and p_y.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Planar Complex
In chemistry, a square planar complex is a type of coordination compound where the central atom is surrounded by ligand atoms in a square planar geometry. This specific arrangement involves the atoms being positioned at the corners of an imaginary square on the same plane centered around the metal atom. Square planar complexes are often seen in transition metal chemistry, particularly involving metals such as platinum and palladium.

Key points include:
  • Square planar geometry is frequently observed in complexes with a coordination number of four. This means there are four ligand atoms attached to the central metal atom.
  • The orientation of the ligands influences the chemical properties and reactivity of the complex significantly, making these structures particularly interesting in catalysis.
  • This arrangement is less common than the alternative tetrahedral configuration for four-coordinate complexes.
Understanding square planar complexes is crucial for unraveling the behavior of certain catalysts, especially those used in organic reactions like hydrogenation or carbon-carbon coupling reactions.
Orbital Hybridization
Orbital hybridization is a vital concept in chemistry that helps us understand how atoms bond in molecules. It refers to the process where atomic orbitals mix to form new hybrid orbitals suitable for pairing with electrons from other atoms to form chemical bonds.

For square planar complexes, the dsp^2 hybridization is crucial. Here's a breakdown of this process:
  • The **d** orbital involved is typically the **d_{x^2-y^2}** orbital. This orbital is oriented in such a way that it can form strong bonds with ligands lying in the xy-plane.
  • The **s** and **p** orbitals used are generally the **s, p_x, and p_y**. These orbitals help form the planar structure by interacting with the ligands placed along their axes.
  • Hybridization allows for the formation of equivalent hybrid orbitals which can form more stable, stronger covalent bonds.
The concept of hybridization and its types, like sp, sp^2, sp^3, dsp^2, etc., help explain the varied geometry and reactivity patterns observed in chemical bonding.
Chemical Bonding Concepts
Chemical bonding is the foundation of molecular structure and properties. Understanding it requires a grasp of how atoms interact to form more stable configurations.

In the context of square planar complexes, bonding involves:
  • **Covalent Bonding**: Predominantly, the bonds in these complexes are covalent, involving the sharing of electrons between the central metal atom and ligands.
  • **Coordination Bonds**: These are a unique type of covalent bond formed between a metal atom and ligands, essential in the formation of coordination complexes.
  • **Geometry and Stability**: The square planar structure is influenced by electronic factors, with the dsp^2 hybridization playing a crucial role in minimizing electron repulsion and maximizing bonding stability.
Comprehending these bonding concepts helps predict the properties and behaviors of complex molecules. For students and professionals alike, mastering these foundational ideas is key to understanding more advanced topics in inorganic chemistry and beyond.

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Most popular questions from this chapter

Of the following the number of species having unpaired electron are \(\mathrm{B}_{2}, \mathrm{KO}_{2}, \mathrm{BaO}_{2}, \mathrm{NO}_{2}, \mathrm{O}_{2}, \mathrm{ClO}_{2}, \mathrm{O}_{2}\left[\mathrm{AsF}_{6}\right]\)

In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed? (a) \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\) (b) \(\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}\) (c) \(\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}\) (d) \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}^{+}\)

Which of the following is correct? (a) The rate of ionic reactions are very slow (b) The number of electrons present in the valence shell of \(\mathrm{S}\) in \(\mathrm{SF}_{6}\) is 12 (c) According to VSEPR theory \(\mathrm{SnCl}_{2}\) is a linear molecule (d) The correct order of stability to form ionic compounds among \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) is \(\mathrm{Al}^{3+}>\) \(\mathrm{Mg}^{2+}>\mathrm{Na}^{+}\)

Consider the given statements about the molecule \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} .\) 1\. Three carbon atoms are sp \(^{3}\) hybridized 2\. Three carbon atoms are sp \(^{2}\) hybridized 3\. Two carbon atoms are sp hybridized Of three statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct

Identify the correct sequence of increasing number of \(\pi\) bonds in the structures of the following molecules 1\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\) 2\. \(\mathrm{H}_{2} \mathrm{SO}_{3}\) 3\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{5}\) (a) \(1,2,3\) (b) \(2,3,1\) (c) \(2,1,3\) (d) \(1,3,2\)

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