StudyQuestionHub

Problem 150

Question

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

Step-by-Step Solution

Verified
Answer
All statements (a), (b), and (c) are correct.
1Step 1: Understanding Statement (a)
Statement (a) says that the bond angle of hybrid bonds increases as \( \mathrm{sp}^{3} < s p^{2} < s p \). This is correct because the bond angle depends on the type of hybridization. In \( \mathrm{sp}^3 \) hybridization, as seen in tetrahedral molecules, the bond angle is approximately 109.5°. For \( \mathrm{sp}^2 \), which is seen in trigonal planar molecules, the bond angle is approximately 120°. For \( \mathrm{sp} \) hybridization, typical in linear molecules, the bond angle is 180°. Hence, \( \mathrm{sp}^{3} < s p^{2} < s p \) is correct.
2Step 2: Analyzing Statement (b)
Statement (b) claims that for hydrogen halides, bond length increases as \( \mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI} \). This is correct because bond length increases with the size of the halogen atom bonded to hydrogen. As we move down Group 17 of the periodic table from fluorine (F) to iodine (I), the size of the halogen atom increases. Therefore, HF (with the smallest F atom) has the shortest bond length and HI (with the largest I atom) has the longest bond length.
3Step 3: Assessing Statement (c)
Statement (c) proposes that the increasing order of bond energy is: Single bond \(<\) Double bond \(<\) Triple bond. This statement is correct because bond energy, which is the amount of energy required to break a bond, is higher in multiple bonds. Single bonds have the least bond energy because they have only one shared pair of electrons. Double bonds, with two shared pairs, and triple bonds, with three shared pairs, progressively have higher bond energies. Therefore, single < double < triple bond energy order holds true.

Key Concepts

HybridizationBond AnglesBond EnergyPeriodic TrendsHydrogen Halides
Hybridization
Hybridization is a concept fundamental to understanding molecular shapes and types of chemical bonds. It's essentially the mixing of atomic orbitals to form new hybrid orbitals that are equal in energy. This idea helps explain how molecules form with certain geometries.
In hybridization, we often encounter three main types:
  • sp: This involves the combination of one s and one p orbital. Hybrid orbitals formed this way have linear shapes with a bond angle of 180°.
  • sp2: Here, one s orbital combines with two p orbitals, resulting in trigonal planar shapes. The bond angles are around 120°.
  • sp3: In this scenario, one s and three p orbitals mix to form a tetrahedral shape with bond angles of approximately 109.5°.
Understanding hybridization is key to explaining why some molecules have certain bond angles. For example, hybrid orbitals dictate that sp3 has a smaller bond angle than sp, due to its more compact, tetrahedral structure.
Bond Angles
Bond angles are crucial for determining the shape and geometry of a molecule. The angles between adjacent lines representing bonds in a molecule directly relate to the hybridization of orbitals involved.
Let's consider some typical bond angles for different hybridizations:
  • A linear configuration, seen in sp hybridization, yields a bond angle of 180°.
  • For a trigonal planar structure, as in sp2 hybridization, the bond angles are 120°.
  • A tetrahedral shape, like in sp3 hybridization, features bond angles of 109.5°.
This understanding helps in visualizing the 3D molecular structure, predicting how molecules might interact, and determining physical properties like melting and boiling points.
Bond Energy
Bond energy refers to the measure of bond strength in a chemical bond. Specifically, it is the energy required to break one mole of bonds in a gaseous substance, indicating the stability of a bond.
Generally, bond energy increases with the number of shared electron pairs:
  • Single Bonds: One pair of electrons is shared; hence, it has the lowest bond energy.
  • Double Bonds: Two pairs of electrons are involved, making it stronger and having higher bond energy than single bonds.
  • Triple Bonds: With three pairs of shared electrons, triple bonds are the strongest, having the highest bond energy.
Understanding bond energies is important for reactions and stability analysis. A higher bond energy implies that more energy is required for a reaction, leading to more stable molecules.
Periodic Trends
Periodic trends are patterns observed in the periodic table, guiding us in predicting properties of elements. Primary periodic trends include atomic size, electronegativity, ionization energy, and metallic character.
These trends explain many chemical behaviors:
  • Atomic Size: Increases down a group due to the addition of electron shells, useful for understanding bond lengths.
  • Electronegativity: Generally increases across a period and decreases down a group, influencing bond polarity.
  • Ionization Energy: Tends to increase across a period and decrease down a group, crucial for understanding element reactivity.
Such trends are particularly helpful when comparing similar groups like halogens, providing a framework to understand variations in properties like bond lengths in hydrogen halides.
Hydrogen Halides
Hydrogen halides are binary compounds consisting of hydrogen and halogens. Their properties often highlight fundamental concepts like bond lengths and strengths, influenced by the size and electronegativity of halogens.
Some key insights include:
  • Bond Length: In hydrogen halides, bond length increases with the size of the halogen. For instance, from HF to HI, the bond length grows as the atomic size of the halogen increases.
  • Bond Strength: Although HF has the strongest bond among hydrogen halides due to high electronegativity of fluorine, bond strength decreases down the group as bond length increases.
These insights into hydrogen halides provide a clear illustration of how periodic trends affect bond properties, aiding in the prediction of their behavior in various chemical reactions.
Previous
Problem 149
Next
Problem 151

Other exercises in this chapter

Problem 148
Which of the pairs have identical values of bond order? (a) \(\mathrm{F}_{2}\) and \(\mathrm{Ne}_{2}\) (b) \(\mathrm{N}_{2}^{+}\)and \(\mathrm{O}_{2}^{+}\) (c)
View solution
Problem 149
Isostructural group of molecule is (a) \(\mathrm{XeO}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{CH}_{4}\) (b) \(\mathrm{CH}_{3}^{-}, \mathrm{NH}_{3}, \mathrm{NF}_{3}\)
View solution
Problem 151
In \(\mathrm{O}_{2}\) molecule which of the following molecular orbitals are lower in energy than \(\sigma 2 \mathrm{p}_{\mathrm{z}}\) orbital (a) \(\sigma 2 \m
View solution
Problem 152
Which combination of the compound and their shapes are correct? (a) \(\mathrm{ClF}_{3}\) - see saw (b) \(\mathrm{ICl}_{4}^{-}-\)square planar (c) \(\mathrm{ICl}
View solution