Compare and contrast the bonding found in the \(\mathrm{H}_{2}(g)\) and HF \((g)\) molecules with that found in NaF(s).

Short Answer

Expert verified
In H2, there is a non-polar covalent bond with a bond strength of 436 kJ/mol and a molecular orbital configuration of \(\sigma_{1s}^2\). HF has a polar covalent bond with a bond strength of 565 kJ/mol, formed by the overlap of hydrogen's 1s orbital and fluorine's 2p orbital. NaF has an ionic bond, with a bond strength of 924 kJ/mol (lattice energy), formed by the transfer of an electron from sodium to fluorine and the electrostatic attraction between the resulting Na+ and F- ions.

Step by step solution

01

Identify the type of bonding in each compound

In H2, the bonding is a covalent bond between two non-metal hydrogen atoms. In HF, the bonding is a polar covalent bond between a non-metal hydrogen atom and a non-metal fluorine atom. In NaF, the bonding is an ionic bond between a metal sodium atom and a non-metal fluorine atom.
02

Compare bond strength and polarity

H2 has a non-polar covalent bond with a bond strength of about 436 kJ/mol. HF has a polar covalent bond with a bond strength of about 565 kJ/mol, which is greater than that of H2 due to the higher electronegativity of fluorine. NaF has ionic bonding with a bond strength of about 924 kJ/mol (lattice energy), which is greater than both H2 and HF due to the strong electrostatic attraction between the positive sodium ion and the negative fluoride ion.
03

Examine molecular orbital configurations

The molecular orbital configuration of H2 is represented by the following notation: \(\sigma_{1s}^2\). This means that there are two electrons in the \(\sigma_{1s}\) orbital, which is a bonding orbital that is formed by the overlap of the hydrogen 1s atomic orbitals. The molecular orbital configuration of HF is also represented by the overlap of atomic orbitals: the hydrogen 1s orbital with the fluorine 2p orbital. The resulting molecular orbitals consist of a bonding \(\sigma\) orbital and an antibonding \(\sigma^{*}\) orbital. The higher electronegativity of fluorine results in the electron density being more concentrated around the fluorine atom and gives rise to the polar nature of the bond. As for NaF, there are no molecular orbitals since it is an ionic compound. Instead, the sodium atom loses an electron from its 3s orbital to form a positively charged Na+ ion, and the fluorine atom gains an electron in its 2p orbital to form a negatively charged F- ion. The attraction between these oppositely charged ions is responsible for the strong ionic bond in NaF.
04

Summarize and contrast the bonding in H2, HF, and NaF

In summary, the bonding in H2 is a non-polar covalent bond with a molecular orbital configuration of \(\sigma_{1s}^2\) and has a bond strength of 436 kJ/mol. The bonding in HF is a polar covalent bond where the fluorine atom has a higher electronegativity, resulting in a stronger bond strength of 565 kJ/mol. The molecular orbitals in HF are formed by the overlap of the hydrogen 1s orbital and the fluorine 2p orbital. NaF has an ionic bond with a bond strength of 924 kJ/mol (lattice energy), and no molecular orbitals are present. Instead, NaF's bonding arises from the transfer of an electron from the sodium atom to the fluorine atom and the subsequent electrostatic attraction between the resulting Na+ and F- ions.

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