1.predict and expalain the order of melting point for propanol, butane, and proanone with refernce to their intermolecular forces.
2. The element sodium, aluminum, silicon, phosphrous, and sulfur are in periods 3 of the periodic table. Decribes the metallic bonding present in alumminium and explain why aluinium has higher melting point that sodium.
I am really connfused.. can anybody help me.??
As the polarity of molecules increase, their melting points increase because bonds between the molecules are stronger.
Metallic bonding is described as a sea of electrons in which the atoms are immersed. Na has one electron from each atom to offer to the sea. Mg has a smaller radius than Na AND it has two electrons to offer; therefore, bonding forces are stronger for Mg and the melting point is a little higher. The same kind of reasoning extends to Al. It is smaller still, it has three electron to offer for each atom, and the melting point is still higher. Here is a site that gives a simplified discussion.
http://www.chemguide.co.uk/atoms/bonding/metallic.html
1. To predict and explain the order of melting point for propanol, butane, and propanone, we need to consider the intermolecular forces present in each compound. Intermolecular forces are the forces of attraction between molecules.
a) Propanol (C3H7OH) is an alcohol that has hydrogen bonding as its strongest intermolecular force. Hydrogen bonding occurs when a hydrogen atom is directly bonded to a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and experiences a strong dipole-dipole interaction with another electronegative atom in a different molecule. Due to hydrogen bonding, propanol has higher melting point compared to compounds without hydrogen bonding.
b) Butane (C4H10) is an alkane and has only weak van der Waals forces (London dispersion forces) as intermolecular forces. Van der Waals forces are temporary dipoles induced by the movement of electrons within molecules. Butane has fewer hydrogen atoms compared to propanol, so the intermolecular forces are weaker. As a result, butane has a lower melting point than propanol.
c) Propanone, also known as acetone (CH3COCH3), is a ketone and has dipole-dipole forces as its intermolecular forces. These forces result from the difference in electronegativity between the carbon and oxygen atoms. While dipole-dipole forces are stronger than van der Waals forces, they are weaker than hydrogen bonding. Therefore, propanone has a lower melting point than propanol, but higher than butane.
So, the order of melting point from highest to lowest would be:
Propanol > Propanone > Butane
2. Metallic bonding describes the bonding that occurs between metal atoms. In metallic bonding, the metal atoms lose their valence electrons and form a lattice of positive metal ions surrounded by a "sea" of delocalized electrons. The positive metal ions are attracted to the negatively charged electrons, creating a strong electrostatic force that holds the metal together.
When comparing sodium and aluminum in terms of metallic bonding and melting points, it is important to consider a few factors:
a) Number of valence electrons: Sodium has one valence electron, while aluminum has three valence electrons. Aluminum's additional valence electrons allow for stronger metallic bonding as there are more delocalized electrons available to create a denser electron "sea." This leads to stronger electrostatic forces and higher melting points compared to sodium.
b) Atomic size: Aluminum has a larger atomic size compared to sodium. Larger atomic size leads to increased distance between positive metal ions in the lattice, which weakens the interatomic attractions. As a result, the metallic bonding in aluminum becomes slightly weaker compared to sodium, but the presence of more valence electrons compensates for this effect, resulting in a higher melting point for aluminum.
Therefore, aluminum has a higher melting point than sodium due to the combination of additional valence electrons and a larger atomic size, which leads to stronger metallic bonding.