
Give the order of melting points \[KCl,CuCl,CuC{l_2}\]
Answer
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Hint: Softening point (mp): The temperature (or all the more usually temperature range) at which a substance goes through a strong to fluid stage change (i.e., it liquefies) without an expansion in temperature. Then again, the temperature at which a substance exists in harmony between its strong and fluid stages.
Complete step by step answer:
The dissolving purpose of a substance relies upon pressure and is typically determined at a standard weight, for example, 1 environment or 100 kPa. At the point when considered as the temperature of the converse changes from fluid to strong, it is alluded to as the point of solidification or crystallization point.
The more noteworthy the distinction in electronegativity, the more ionic a bond is. The bond with the most covalent character is controlled by electronegativities. The more modest contrast in electronegativities makes a more covalent bond. So you have to choose which particle has ions with the most comparable electronegativities.
\[CuCl{\text{ }} < {\text{ }}KCl{\text{ }} < {\text{ }}CuC{l_2}\]
More covalent character has a high softening point.
Particles have high charge i.e. cation has more prominent positive charge and anion has more noteworthy negative charge have more covalent character.
Cation has little ionic range and has more covalent character.
For example, on the off chance that the softening point depended absolutely on electronegativity, at that point truly, \[N{a_2}O\] would have a higher dissolving point. Yet, it's something other than that. Dissolving point is likewise a factor of particle size. In the two cases, the \[{O^{2 - }}\] particle is a similar size. Be that as it may, the Mg (2+) particle is much more modest than the Na+ particle. Therefore, the \[MgO\] structure is more firmly pressed than the \[N{a_2}O\] structure. Sodium's ionic sweep is 1.02 Angstroms while Magnesium's is 0.72 Angstroms
Note: Offer with the kids that the temperature at which ice liquefies is known as the softening point. The dissolving point is the temperature at which a strong goes to a fluid. The dissolving point where ice — a strong — goes to water — a fluid — is \[32^\circ F{\text{ }}\left( {0^\circ C} \right)\].
Complete step by step answer:
The dissolving purpose of a substance relies upon pressure and is typically determined at a standard weight, for example, 1 environment or 100 kPa. At the point when considered as the temperature of the converse changes from fluid to strong, it is alluded to as the point of solidification or crystallization point.
The more noteworthy the distinction in electronegativity, the more ionic a bond is. The bond with the most covalent character is controlled by electronegativities. The more modest contrast in electronegativities makes a more covalent bond. So you have to choose which particle has ions with the most comparable electronegativities.
\[CuCl{\text{ }} < {\text{ }}KCl{\text{ }} < {\text{ }}CuC{l_2}\]
More covalent character has a high softening point.
Particles have high charge i.e. cation has more prominent positive charge and anion has more noteworthy negative charge have more covalent character.
Cation has little ionic range and has more covalent character.
For example, on the off chance that the softening point depended absolutely on electronegativity, at that point truly, \[N{a_2}O\] would have a higher dissolving point. Yet, it's something other than that. Dissolving point is likewise a factor of particle size. In the two cases, the \[{O^{2 - }}\] particle is a similar size. Be that as it may, the Mg (2+) particle is much more modest than the Na+ particle. Therefore, the \[MgO\] structure is more firmly pressed than the \[N{a_2}O\] structure. Sodium's ionic sweep is 1.02 Angstroms while Magnesium's is 0.72 Angstroms
Note: Offer with the kids that the temperature at which ice liquefies is known as the softening point. The dissolving point is the temperature at which a strong goes to a fluid. The dissolving point where ice — a strong — goes to water — a fluid — is \[32^\circ F{\text{ }}\left( {0^\circ C} \right)\].
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