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What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) \(\mathrm{HCN}^{\text {, }}\) (c) \(\mathrm{SO}_{3}\), (d) \(\mathrm{TeCl}_{2}\).

Short Answer

Expert verified
The hybridizations of the central atoms in the given molecules are: (a) SiCl4: sp³, (b) HCN: sp, (c) SO3: sp², and (d) TeCl2: sp³.

Step by step solution

01

(a) Hybridization of the central atom in SiCl4

The central atom is Si (silicon) and it is surrounded by four Cl (chlorine) atoms. Number of valence electrons on Si (silicon) = 4 Number of monovalent atoms surrounding Si (chlorine) = 4 Charge of cation = 0 (As it's a neutral molecule) Charge of anion = 0 (As it's a neutral molecule) Now substitute these values in the formula, Hybridization = 1/2 [4 + 4 - 0 + 0] = 1/2 [8] = 4 Based on hybridization value, we can determine the hybridization type: 4: sp³ hybridization. Thus, the hybridization of the central atom (Si) in SiCl4 is sp³.
02

(b) Hybridization of the central atom in HCN

The central atom is C (carbon), surrounded by H (hydrogen) and N (nitrogen) atoms. Number of valence electrons on C (carbon) = 4 Number of monovalent atoms surrounding C (hydrogen and nitrogen) = 1 + 0 (N is not monovalent, so we don't count it) = 1 Charge of cation = 0 (As it's a neutral molecule) Charge of anion = 0 (As it's a neutral molecule) Now substitute these values in the formula, Hybridization = 1/2 [4 + 1 - 0 + 0] = 1/2 [5] = 2.5 Based on hybridization value, we can determine the hybridization type: 2.5: sp hybridization. Thus, the hybridization of the central atom (C) in HCN is sp.
03

(c) Hybridization of the central atom in SO3

The central atom is S (sulfur) and it is surrounded by three O (oxygen) atoms. Number of valence electrons on S (sulfur) = 6 Number of monovalent atoms surrounding S (oxygen) = 0 (Oxygen is not monovalent) Charge of cation = 0 (As it's a neutral molecule) Charge of anion = 0 (As it's a neutral molecule) Now substitute these values in the formula, Hybridization = 1/2 [6 + 0 - 0 + 0] = 1/2 [6] = 3 Based on hybridization value, we can determine the hybridization type: 3: sp² hybridization. Thus, the hybridization of the central atom (S) in SO3 is sp².
04

(d) Hybridization of the central atom in TeCl2

The central atom is Te (tellurium) and it is surrounded by two Cl (chlorine) atoms. Number of valence electrons on Te (tellurium) = 6 Number of monovalent atoms surrounding Te (chlorine) = 2 Charge of cation = 0 (As it's a neutral molecule) Charge of anion = 0 (As it's a neutral molecule) Now substitute these values in the formula, Hybridization = 1/2 [6 + 2 - 0 + 0] = 1/2 [8] = 4 Based on hybridization value, we can determine the hybridization type: 4: sp³ hybridization. Thus, the hybridization of the central atom (Te) in TeCl2 is sp³.

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Key Concepts

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

Molecular Geometry
Molecular geometry is the 3D arrangement of atoms around a central atom in a molecule. Understanding this helps us predict many physical and chemical properties. For example, the molecule \(\mathrm{SiCl_4}\) has a central silicon atom surrounded by four chlorine atoms. The molecular geometry here is tetrahedral, due to the sp³ hybridization. This shape results because all four bonded pairs are arranged as far apart as possible, minimizing electron pair repulsion.

In contrast, a molecule like \(\mathrm{SO_3}\) has a trigonal planar geometry. Here, the sulfur atom undergoes sp² hybridization and the three oxygen atoms arrange themselves in a triangular plane. The 120-degree separation between the oxygen atoms provides stability to the molecule. Similarly, the \(\mathrm{HCN}\) molecule, with its linear geometry, illustrates how molecular geometry can differ dramatically depending on the types of bonds formed and hybridization of the central atom.

Paying attention to molecular geometry can provide insight into molecules' interaction with light, polarity, and how they participate in chemical reactions.
Valence Electrons
Valence electrons are the outermost electrons found in an atom's shell. They play a crucial role in chemical bonding and reactions. These are the electrons a central atom uses to bond with other atoms in a molecule. For silicon in \(\mathrm{SiCl_4}\), there are four valence electrons. Each of these electrons forms a bond with a chlorine atom, maximizing stability and resulting in a tetrahedral molecular geometry.

The number of valence electrons also determines the hybridization of the central atom, as evident in the case of \(\mathrm{SO_3}\). Sulfur has six valence electrons, leading to a trigonal planar structure with three double bonds.

Valence electrons are essential for predicting reactivity and properties. These electrons are involved in electron sharing or transfer that causes molecules to form. Keeping track of how many valence electrons are involved in chemical bonding helps us predict the type of bonds and the geometry a molecule can form.
Chemical Bonding
Chemical bonding is the process by which atoms connect to form molecules. The type and number of bonds play a huge role in defining the physical and chemical properties of the compound. In \(\mathrm{SiCl_4}\), the bonds between silicon and chlorine are covalent, formed when they share a pair of valence electrons. This sharing is what gives each chlorine the stable electron configuration it desires.

In a molecule like \(\mathrm{HCN}\), where the carbon forms a triple bond with nitrogen and a single bond with hydrogen, the type of bonding is slightly different. The strong triple bond between C and N is a characteristic of sp hybridization, resulting in significant bond strength and a linear molecule.

Chemical bonding not only holds atoms together; it determines molecular shape, stability, and reactivity. Understanding the nature of these bonds helps us predict how a molecule will behave in various environments.
Central Atom
The central atom in a molecule is typically the atom with the highest bonding capacity, often found at the center, surrounded by other atoms. Identifying the central atom helps in determining both the hybridization and the overall molecular geometry. In molecules like \(\mathrm{SiCl_4}\), silicon acts as the central atom because it can form four bonds with chlorine, leading to a tetrahedral shape.

In \(\mathrm{SO_3}\), sulfur serves as the central atom due to its larger number of valence electrons and its ability to form strong covalent bonds with oxygen. Similarly, in \(\mathrm{HCN}\), carbon is the central atom as it connects hydrogen and nitrogen through covalent bonds.

The central atom is crucial in defining the molecule's shape and electron distribution. This influences how the molecule will interact chemically and physically with other molecules. Recognizing the central atom can simplify understanding complex molecular shapes and reactions.

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Most popular questions from this chapter

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{N}_{2}\) . A closely related substance is hydrazobenzene, $\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2}$ . The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the N atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? (c) Predict the \(N-N-C\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

If we assume that the energy-level diagrams for homonuclear diatomic molecules shown in Figure \(9.43\) can be applied to heteronuclear diatomic molecules and ions, predict the bond order and magnetic behavior of (a) \(\mathrm{CO}^{+}\), (b) \(\mathrm{NO}^{-}\), (c) \(\mathrm{OF}^{+}\), (d) \(\mathrm{NeF}^{+}\).

(a) Does \(\mathrm{SCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{BeCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: (a) 3 , (b) 4 , (c) 5, (d) 6 .

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85\) D. (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

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