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juin [17]
3 years ago
6

Explain why the lattice energy and the energy of crystallization of a substance differ in sign but are equal in magnitude.

Chemistry
1 answer:
Ne4ueva [31]3 years ago
3 0

Answer:

Lattice energy is energy evolved when a crystal lattice is formed. Crystalization energy refers to energy supplied to collapse the lattice structure.

Explanation:

The lattice energy of an ionic crystal is defined in as the change in energy of the system of atoms and molecules when a crystal is formed from the free ionic forms of the components in the gas phase. It can also be defined as the necessary energy to separate ions of a mole of an ionic crystal as gaseous ions from each other by infinite distances.

The strength of ionic compounds is measured by the magnitude of lattice energy. It is the energy released when one mole of an ionic compound is formed. This means when the individual ions of the compound come together to form the crystal lattice, they need a lesser energy to remain together, hence energy is released, the energy released is called the lattice energy. The bond force between ions of opposite charge is strongest when the ions are small.

Similarly, the crystalization energy is the energy required maintain the crystal structure of the compound. It is also the energy that must be supplied for the crystal structure structure to collapse. Hence its sign is positive.

While lattice energy deals with energy evolved when ions come together to form a crystal, crystallization energy deals with energy absorbed to separate the ions and collapse the crystal structure

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<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

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Explanation:

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