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The molecular shapes of \(\mathrm{SF}_{4}, \mathrm{CF}_{4}\) and \(\mathrm{XeF}_{4}\) are (a) the same with 2,0 and 1 lone pairs of electrons on the central atom, respectively (b) the same with 1,1 and 1 lone pair of electrons on the central atom, respectively (c) different with 0,1 and 2 lone pairs of electrons on the central atom, respectively (d) different with 1,0 and 2 lone pairs of electrons on the central atom, respectively

Short Answer

Expert verified
The correct answer is (d) different with 1,0, and 2 lone pairs on the central atom, respectively.

Step by step solution

01

SF4 Analysis

The structure of \(\text{SF}_4\) includes a central sulfur atom bonded to four fluorine atoms. Sulfur has 6 valence electrons; 4 are involved in bonding with fluorine atoms, leaving 2 electrons as a lone pair. The lone pair causes a seesaw shape due to the repulsion on the bonded pairs.
02

CF4 Analysis

The molecule \(\text{CF}_4\) consists of a carbon atom bonded to four fluorine atoms. Carbon has 4 valence electrons, all of which are utilized in forming four bonds with fluorine. This results in no lone pairs on the central atom, giving \(\text{CF}_4\) a tetrahedral shape.
03

XeF4 Analysis

For \(\text{XeF}_4\), xenon is the central atom, with 8 valence electrons. Four of these electrons form bonds with four fluorine atoms, while the remaining 4 electrons form 2 lone pairs. This gives \(\text{XeF}_4\) a square planar shape due to the lone pairs.
04

Compare Molecular Shapes and Lone Pairs

\(\text{SF}_4\) has a seesaw shape with 1 lone pair, \(\text{CF}_4\) is tetrahedral with 0 lone pairs, and \(\text{XeF}_4\) is square planar with 2 lone pairs. All three have different shapes and lone pair configurations.

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Key Concepts

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

Molecular Geometry
Understanding the molecular geometry of a compound is key to predicting its chemical behavior and physical properties. Molecular geometry is determined by the number of bonding pairs and lone pairs on the central atom in a molecule. The Valence Shell Electron Pair Repulsion (VSEPR) Theory is a handy model to predict the shape. This theory states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.
  • Lone pairs repel more strongly than bonding pairs, altering the geometry.
  • Molecular geometries include shapes like linear, trigonal planar, tetrahedral, and octahedral.
The shape significantly influences molecular polarity and intermolecular interactions.
Lone Pairs
Lone pairs are non-bonding pairs of electrons in the valence shell of an atom. They are crucial as they affect both the shape and reactivity of a molecule. While they do not participate in bonding, lone pairs repel nearby electrons more strongly. This altered repulsion changes the angles between bonds and thus the overall shape.
  • The presence of lone pairs usually leads to deviations from ideal geometric angles.
  • Lone pairs are often responsible for a molecule's polarity.
In cases like \( ext{SF}_4\), \( ext{CF}_4\), and \( ext{XeF}_4\), the number and orientation of lone pairs are key to the final geometry.
SF4 Structure
The molecule \(\text{SF}_4\) offers a fascinating look at how lone pairs affect molecular shape. In this molecule, sulfur is the central atom, bonded to four fluorine atoms. Sulfur starts with six valence electrons. Four electrons are used in bonding, leaving one lone pair. The presence of this lone pair distorts the perfect arrangement of the bonded pairs. The result is a seesaw shape.
  • VSEPR theory helps predict this by suggesting that the lone pair will push down, altering bond angles.
  • The final seesaw shape is less symmetric than other geometries like tetrahedral.
This shape is significant in determining the physical properties of the molecule.
CF4 Structure
In \(\text{CF}_4\), the central carbon atom is bonded to four fluorine atoms. With four valence electrons, carbon uses all of them in forming bonds with fluorine. Since there are no lone pairs, the molecule achieves a perfect tetrahedral shape, which is highly symmetrical.
  • This symmetry influences the non-polar nature of \(\text{CF}_4\).
  • All bond angles are 109.5 degrees, typical for tetrahedral arrangements.
The absence of lone pairs simplifies predicting the geometry and properties of such molecules.
XeF4 Structure
The structure of \(\text{XeF}_4\) provides insight into the effect of multiple lone pairs on geometry. Xenon, having eight valence electrons, forms four bonds with fluorine atoms, leaving two lone pairs behind. The VSEPR theory predicts the shape based on the arrangement of both bonding pairs and lone pairs. Due to these two lone pairs, \(\text{XeF}_4\) assumes a square planar shape.
  • The lone pairs are placed opposite each other, which cancels their repulsive forces to some extent.
  • This configuration allows the bonded pairs to remain in a flat square arrangement.
Understanding this structure showcases how lone pairs can lead to unique geometries even with the same number of bonded atoms.

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Most popular questions from this chapter

Match the following: List I (Hydridization) 1\. \(\mathrm{sp}^{2}\) 2\. \(\mathrm{sp}^{3}\) 3\. \(\mathrm{sp}\) \(4 .\) List II (Geometry of the molecule) (i) trigonal bipyramidal (ii) planar trigonal (iii) octahedral (iv) tetrahedral (v) linear

In which one of the following pairs, molecules/ions have similar shape? (a) \(\mathrm{CCl}_{4}\) and \(\mathrm{PtCl}_{4}\) (b) \(\mathrm{NH}_{3}\) and \(\mathrm{BF}_{3}\) (c) \(\mathrm{BF}_{3}\) and \(\mathrm{t}\)-butyl carbonium ion (d) \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\)

Which of the following are non-polar? (1) \(\mathrm{SiF}_{4}\) (2) \(\mathrm{XeF}_{4}\) (3) \(\mathrm{SF}_{4}\) (4) \(\mathrm{BF}_{3}\) (5) \(\mathrm{NF}_{3}\) Select the correct answer using the code given below: (a) 1,2 and 4 (b) 3,4 and 5 (c) 2,3 and 4 (d) 1,3 and 4

Consider the following halogen containing compounds (i) \(\mathrm{CHCl}_{3}\) (ii) \(\mathrm{CCl}_{4}\) (iii) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (iv) \(\mathrm{CH}_{3}^{+} \mathrm{Cl}\) (v) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}_{2}\) The compounds with a net zero dipole moment are (a) (ii) and (v) only (b) (iii) only (c) (iii) and (iv) only (d) (iv) only

Match the following: \(\begin{array}{ll}\text { List I } & \text { List II }\end{array}\) (Species) (Hybridization) 1\. \(\mathrm{BCI}_{3}\) (i) \(\mathrm{sp}^{3}\) 2\. \(\mathrm{NH}_{3}\) (ii) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) 3\. \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (iii) \(\mathrm{sp}^{2}\) 4\. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (iv) \(\mathrm{sp}^{2} \mathrm{~d}\)

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