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Hybridization of the underlined atom changes in which of the following transitions? (a) \(\mathrm{AIH}_{3}\) changes to \(\mathrm{AlH}_{4}^{-}\) (b) \(\mathrm{H}_{2} \underline{\mathrm{O}}\) changes to \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{NH}_{3}\) changes to \(\mathrm{NH}_{4}^{+}\) (d) in all cases

Short Answer

Expert verified
Hybridization changes in transition (a) \(\mathrm{AIH}_{3}\) to \(\mathrm{AlH}_{4}^{-}\).

Step by step solution

01

Identify hybridization in initial molecules

Determine the hybridization state of the underlined atom in the initial given molecules: (a) In \(\mathrm{AIH}_{3}\), the Al atom is \(sp^2\) hybridized.(b) In \(\mathrm{H}_{2}O\), the O atom is \(sp^3\) hybridized.(c) In \(\mathrm{NH}_{3}\), the N atom is \(sp^3\) hybridized.
02

Identify hybridization in resulting molecules

Determine the hybridization state of the underlined atom in the resulting molecules:(a) In \(\mathrm{AlH}_{4}^{-}\), the Al atom is \(sp^3\) hybridized.(b) In \(\mathrm{H}_{3}O^{+}\), the O atom remains \(sp^3\) hybridized.(c) In \(\mathrm{NH}_{4}^{+}\), the N atom is \(sp^3\) hybridized.
03

Compare hybridization states before and after

Compare the hybridization states of the underlined atoms in the starting and resulting molecules to find any changes:(a) In \(\mathrm{AIH}_{3}\) to \(\mathrm{AlH}_{4}^{-}\), Al changes from \(sp^2\) to \(sp^3\) hybridization.(b) In \(\mathrm{H}_{2}O\) to \(\mathrm{H}_{3}O^{+}\), the O atom remains \(sp^3\).(c) In \(\mathrm{NH}_{3}\) to \(\mathrm{NH}_{4}^{+}\), the N atom remains \(sp^3\).

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Key Concepts

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

Hybridization change
Hybridization change refers to the process where an atom adopts a new hybridization state due to changes in its bonding situation. This is often seen in chemical reactions where the bonding environment of an atom changes. For example, in the transition from \(\mathrm{AIH}_{3}\) to \(\mathrm{AlH}_{4}^{-}\), the aluminum atom changes its hybridization from \(sp^2\) to \(sp^3\) to accommodate an extra hydrogen atom. This transition is due to the addition of an electron, allowing the Al atom to use all four of its orbitals to form bonds.
In the initial state, the Al atom in \(\mathrm{AIH}_{3}\) is bonded in a planar trigonal fashion, needing only three hybrid orbitals, hence \(sp^2\). When it transitions to \(\mathrm{AlH}_{4}^{-}\), it requires four equivalent tetrahedral hybrid orbitals, hence an \(sp^3\) hybridization.
Understanding such changes is crucial in predicting molecular structure and reactivity because hybridization directly affects bond angles and molecular geometry.
Chemical bonding
Chemical bonding is a fundamental concept that describes how atoms connect and interact in molecules. There are several types of chemical bonds that affect the hybridization of the involved atoms.
  • **Covalent Bonds**: Atoms share electrons to fill their outer shells, as seen in water (\(\mathrm{H}_2\mathrm{O}\)) where each hydrogen shares its electron with oxygen, forming polar covalent bonds.
  • **Ionic Bonds**: Electrons are transferred between atoms, resulting in positively and negatively charged ions that attract each other.
  • **Metallic Bonds**: Involves the pooling of electrons that are free to move around, giving metals their conductivity.
In \(\mathrm{NH}_3\) changing to \(\mathrm{NH}_4^{+}\), nitrogen initially has three hydrogen bonds, all involving shared electrons, resulting in a covalent nature. As it gains a hydrogen ion (\(H^+\)), an additional covalent bond forms, although the hybridization type remains \(sp^3\).
Chemical bonding is key to understanding molecular structures and transition states because it governs the angles and arrangement of atoms in molecules.
Transition states
Transition states are high-energy states during reactions where bonds are in the process of breaking and forming. They are crucial for understanding reaction mechanisms as they provide a glimpse into the "middle step" of a reaction that cannot be isolated.
The transition from \(\mathrm{AIH}_3\) to \(\mathrm{AlH}_4^{-}\) involves a transition state where an additional hydrogen atom is approaching the aluminum, causing it to begin shifting its electron occupancy and hybridization. Identifying these states helps chemists understand how easily reactions might proceed and predicts reaction outcomes.
In our examples, while neither \(\mathrm{H}_2\mathrm{O}\) to \(\mathrm{H}_3\mathrm{O}^{+}\) nor \(\mathrm{NH}_3\) to \(\mathrm{NH}_4^{+}\) changes hybridization state, understanding their transition states is still critical for understanding the speed and behavior of these reactions.
Molecular structure
Molecular structure refers to the three-dimensional arrangement of atoms in a molecule, dictated by chemical bonding and hybridization. This structure defines many physical and chemical properties of the substance.
  • **Linear and Angular Structures**: Molecules like \(\mathrm{H}_2\mathrm{O}\) exhibit bent structures due to \(sp^3\) hybridization, giving water its distinct properties.
  • **Trigonal Planar and Tetrahedral Structures**: Molecules like \(\mathrm{NH}_3\) generally have a pyramidal structure due to their \(sp^3\) hybridization; when forming \(\mathrm{NH}_4^{+}\), it converts into a tetrahedral shape.
Molecular structure is fundamental for the functionality and reactivity in biological, chemical, and material sciences. Understanding hybridization and bonding allows us to predict and explain molecular shapes, which are critical for applications like drug development, where molecular fit into binding sites can mean the difference between efficacy and inactivity.

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Most popular questions from this chapter

Which of the following species have undistorted octahedral structures? (1) \(\mathrm{SF}_{6}\) (2) \(\mathrm{PF}_{6}^{-}\) (3) \(\mathrm{SiF}_{6}^{2}\) (4) \(\mathrm{XeF}_{6}\) Select the correct answer using the codes given below: (a) 1,3 and 4 (b) 1,2 and 3 (c) 1,2 and 4 (d) 2,3 and 4

Among the following species, identify the isostructural pairs \(\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{HN}_{3}\) (a) \(\left[\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}\right]\)and \(\left[\mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\) (b) \(\left[\mathrm{NF}_{3}, \mathrm{HN}_{3}\right]\) and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (c) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (d) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{HN}_{3}, \mathrm{BF}_{3}\right]\)

The correct statements if the following are: (a) The bond angle of hybrid bonds increases as \(\mathrm{sp}^{3}<\) \(s p^{2}

\(\mathrm{H}_{2} \mathrm{O}\) is dipolar, whereas \(\mathrm{BeF}_{2}\) is not. It is because (a) the electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{O}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) involves hydrogen bonding where as \(\mathrm{BeF}_{2}\) is a discrete molecule (c) \(\mathrm{H}_{2} \mathrm{O}\) is linear and \(\mathrm{BeF}_{2}\) is angular (d) \(\mathrm{H}_{2} \mathrm{O}\) is angular and \(\mathrm{BeF}_{2}\) is linear

Arrange the following compounds in order of increasing dipole moment 1\. Toluene, 2\. m-dichlorobenzene 3\. o-dichlorobenzene, 4\. p-dichlorobenzene (a) \(1<4<2<3\) (b) \(4<1<2<3\) (c) \(4<1<3<2\) (d) \(4<2 \leq 1<3\)

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