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Chapter 3: Problem 15

(a) Neon molecule is monoatomic but Hydrogen molecule is diatomic. (b) The loss of an electron from a diatomic molecule results in decrease of its bond energy. (c) Bond length of NO is higher than that of \(\mathrm{NO}^{\circ}\) (d) Bond order of \(\mathrm{N}_{2}\) is higher than that of \(\mathrm{N}_{2}^{+}\).

Short Answer

Expert verified
a) Neon is monoatomic and Hydrogen is diatomic. b) Loss of an electron from a molecule decreases its bond energy. c) The bond length of NO is higher than that of \(\mathrm{NO}^{\circ}\). d) The bond order of \(\mathrm{N}_{2}\) is higher than that of \(\mathrm{N}_{2}^{+}\)

Step by step solution

01

Part (a): Atomicity of Neon and Hydrogen

Neon is a noble gas and it exists in a free state as individual atoms. So, it is monoatomic. On the other hand, hydrogen exists as diatomic molecules (\(\mathrm{H}_2\)) in its natural state.
02

Part (b): Loss of Electron and Bond Energy

When an electron is lost from a diatomic molecule, it becomes a positive ion. The loss of electron destabilizes the molecule leading to decrease its bond energy. This is because the bond energy is the energy required to break the bond between two atoms in a molecule.
03

Part (c): Bond Length of NO and \(\mathrm{NO}^{\circ}\)

The bond order of a molecule is inversely proportional to its bond length. The bond order of \(\mathrm{NO}^{\circ}\) is greater than that of NO. Therefore, the bond length of NO is higher than that of \(\mathrm{NO}^{\circ}\).
04

Part (d): Bond Order of \(\mathrm{N}_{2}\) and \(\mathrm{N}_{2}^{+}\)

The bond order of a molecule is the half of the difference between the number of bonding electrons and antibonding electrons. The bond order of \(\mathrm{N}_{2}\) is 3 and the bond order of \(\mathrm{N}_{2}^{+}\) is 2.5. Hence, the bond order of \(\mathrm{N}_{2}\) is higher than that of \(\mathrm{N}_{2}^{+}\).

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