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Which one of the following pairs of species have the same bond order? (a) \(\mathrm{CN}^{-}\)and \(\mathrm{NO}^{+}\) (b) \(\mathrm{CN}\) - and \(\mathrm{CN}^{+}\) (c) \(\mathrm{O}_{2}^{-}\)and \(\mathrm{CN}^{-}\) (d) \(\mathrm{NO}^{+}\)and \(\mathrm{CN}^{+}\)

Short Answer

Expert verified
The pair \(\mathrm{CN}^-\) and \(\mathrm{NO}^+\) have the same bond order.

Step by step solution

01

Determine bond order formula

The bond order is calculated using the formula: \( \text{Bond Order} = \frac{1}{2} [\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals}] \).
02

Determine number of electrons

Count the total number of electrons for each ion: - \( \text{CN}^- \) has 14 electrons, \( \text{CN}^+ \) has 12, \( \text{NO}^+ \) has 14, \( \text{O}_2^- \) has 17.
03

Calculate bond order for \(\text{CN}^-\) and \(\text{NO}^+\)

Both \(\text{CN}^-\) and \(\text{NO}^+\) have 14 electrons. Their molecular orbital configurations are identical, resulting in the same bond order. Using the formula: \[ \text{Bond Order} = \frac{1}{2} [10 - 4] = 3 \]
04

Calculate bond order for remaining species

For \( \text{CN}^+ \) (12 electrons): \( \text{Bond Order} = \frac{1}{2} [8 - 4] = 2 \).For \( \text{O}_2^- \) (17 electrons): \( \text{Bond Order} = \frac{1}{2} [10 - 7] = 1.5 \).
05

Identify the pair with equal bond order

Compare the bond orders calculated:- \( \text{CN}^- \text{ and } \text{NO}^+ \) both have a bond order of 3.None of the other pairs match this equality in bond orders.

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Key Concepts

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

Bond Order Calculation
Bond order is a fundamental concept in molecular orbital theory that helps us understand the stability of a bond in a molecule. It signifies the number of chemical bonds between a pair of atoms. The formula for calculating bond order is:
  • \[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals} \right) \]
A higher bond order means a stronger bond, while a lower bond order indicates a weaker bond. A bond order of zero suggests that no bond exists between the atoms.
To calculate the bond order accurately, you need to know the molecular orbital configuration of the molecule or ion. This involves placing electrons into molecular orbitals derived from the atomic orbitals of the constituent atoms. The number of electrons in bonding and antibonding orbitals helps determine the bond's characteristics.
Electron Configuration
Electron configuration in molecular orbital theory involves understanding how electrons fill orbitals in a molecule. In molecules, electrons are distributed into molecular orbitals that result from the overlap of atomic orbitals.
The electronic configuration of molecular orbitals is represented by filling electrons into lower energy orbitals first, as per the Aufbau principle, similar to electronic configurations in atoms:
  • Bonding orbitals are filled before antibonding orbitals.
  • Each orbital can hold up to two electrons with opposite spins.
  • Hund's rule applies; electrons fill degenerate orbitals singly before doubly occupying them.
For example, in the molecule \( \text{CN}^- \) and \( \text{NO}^+ \), electrons fill the molecular orbitals up to 14 electrons, filling all lower energy bonding orbitals and resulting in a stable configuration.
This systematic approach helps in predicting the chemical properties and reactivity of molecular ions.
Chemical Bonding
Chemical bonding in the context of molecular orbitals is about understanding how atoms bond to form molecules. The types of chemical bonds include:
  • Ionic bonds - formed through the transfer of electrons from one atom to another.
  • Covalent bonds - formed when atoms share electrons to achieve a full outer shell.
Molecular orbital theory, however, provides a more nuanced view by considering the entire molecule's electron distribution. In this theory, bonded atoms form molecular orbitals, which can be bonding or antibonding:
  • Bonding orbitals arise when the wavefunctions of atomic orbitals constructively interfere, enhancing electron density between atoms and creating a more stable molecule.
  • Antibonding orbitals occur when wavefunctions destructively interfere, leading to a decrease in electron density between the atoms and a less stable molecule.
The occupation of these orbitals by electrons determines the strength and properties of the bond.
Molecular Ions
Molecular ions are species formed by the addition or removal of one or more electrons from a molecule, resulting in a charged species. They are crucial in many chemical processes and reactions.
Molecular ions like \( \text{CN}^- \) and \( \text{NO}^+ \) are important in understanding chemical reactions and bonding properties because their charge alters electron configuration and consequently, their bond order. The extra electron in \( \text{CN}^- \) or the lost electron in \( \text{NO}^+ \) affects the balance between bonding and antibonding electrons:
  • In the case of \( \text{CN}^- \), the extra electron enters a bonding orbital, increasing stability and bond order.
  • In \( \text{NO}^+ \), the loss of an electron typically removes one from an antibonding orbital, also affecting bond stability.
Understanding molecular ions and their configurations is key in areas such as mass spectrometry, where ions are analyzed based on their mass-to-charge ratios.

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Most popular questions from this chapter

Consider the following statements: 1\. the bond order of \(\mathrm{NO}\) is \(2.5\) 2\. the bond order of \(\mathrm{NO}^{+}\)is 3 3\. the bond order of \(\mathrm{O}_{2}\) is \(1.5\) 4\. the bond order of \(\mathrm{CO}\) is 3 Which of these statements are correct? (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,3 and 4 (d) 1,2 and 4

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) \(4<2 \leq 1<3\)

In which of the following molecules/ions all the bonds are unequal? (a) \(\mathrm{SF}_{4}\) (b) \(\mathrm{SiF}_{4}\) (c) \(\mathrm{XeF}_{4}\) (d) \(\mathrm{BF}_{4}^{-}\)

When anions and cations approach each other, the valence shell of anions are pulled towards cation nucleus and thus, shape of anion is deformed. The phenomenon of deformation of anion by a cation is known as polarization and the ability of the cation to polarize the anion is called as polarizing power of cation. Due to polarization, sharing of electrons occurs between two ions to some extent and the bond shows some covalent character. Considering \(\mathrm{BeCl}_{2}, \mathrm{MgCl}_{2}, \mathrm{CaCl}_{2}\) and \(\mathrm{BaCl}_{2}\), predict which of the following statement is true? (a) Covalent character increases as the atomic number of the metal atom increases (b) \(\mathrm{BeCl}_{2}\) is least ionic out of the given chlorides. (c) \(\mathrm{BeCl}_{2}\) has the highest melting point among the given chlorides. (d) All are highly ionic compound.

Which of the following statement is not correct regarding the properties of ionic compounds? (a) ionic compounds have high melting and boiling points (b) their reaction velocity in aqueous medium is very high. (c) ionic compounds in their molten and aqueous solutions do not conduct electricity. (d) they are highly soluble in polar solvents.

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