Question

In: Physics

Which defects will be formed when compound B is doped with compound A? Display the defects...

Which defects will be formed when compound B is doped with compound A? Display the defects using the Kröger Vink notation and the effective charge.

A B
Lithium bromide Calcium bromide
Magnesium oxide Iron (III) oxide
Cadmium chloride Sodium chloride

Solutions

Expert Solution

Hi,

hope you are doing well.

CASE 1:

When calcium bromide (CaBr2) is doped with Lithium bromide(LiBr),

2 Li+ ions will occupy the lattice, one at cation vacancy (of Ca2+) and the other at the interstitial position of lattice.

This type of defect is called Metal excess defect, in which the no.of cations will be more than no. of anions.

In this type of defect, the density of the lattice will increase (due to increase in mass) and the effective charge remains the same.

CASE 2:

When Iron oxide(Fe2O3) is doped with Magnesium oxide(MgO),

2 Fe3+ ions will be released from the lattice, and to maintain electrical neutrality, three Mg2+ ions will occupy the lattice, two at cation vacancy (made by 2 Fe2+) and the other at the interstitial position of lattice.

This type of defect is called Metal excess defect, in which the no.of cations will be more than no. of anions.

In this type of defect, the density of the lattice will increase (due to increase in mass) and the effective charge remains the same.

CASE 3:

When Sodium chloride(NaCl) is doped with Cadmium chloride(CdCl2),

2 Na+ ions will be released from the lattice, and to maintain electrical neutrality, one Cd2+ ion will occupy the lattice, at cation vacancy (made by Na+) and there will be a cation vacancy.

This type of defect is called Impurity defect, in which the no.of anions will be more than no. of cations.

In this type of defect, the density of the lattice will decrease (due to cation vacancy) and the effective charge remains the same.

Hope this helped for your studies. Keep learning. Have a good day.

Feel free to clear any doubts at the comment section.


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Thank you. :)


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