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When the both the cations and anions are absent from the crystal lattice it is called Schottky defect. This defect is shown when the anions and cations have comparable size. The Frenkel defect is shown by ionic molecules when their is a large difference in the size of anions and cations. The smaller anions are very much mobile and they occupy interstitial site. AgBr has cations and anions with comparable size and hence it shows Schottky defect but the Ag+ ion is very much mobile and it easily occupies interstitial place getting dislocated from its original place, that's why the ionic crystal AgBr shows both Schottky and Frenkel defects.
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Why does AgBr show both Frenkel and Schottky defects?
The radius ratio for AgBr is intermediate. Thus it shows both frenkel and schottky defects. The major defect in AgBr is the Frenkel defect. It has a rocksalt structure i.e. CCP lattice of of Br with atoms of Ag occupying all octahedral holes.Ag moves from octahedral to tetrahedral sites causing only cations to precipitate. Schottky defect arise due to missing of ions from their lacttice point and frenkel arise when the mmissing ions occupy interstitial sites. in AgBr, ag+ ion is small in size and when removed from lacttice point they can occupy interstitial site and therefore show both frenkel and schottky defect. SCHOTTKY Defect in AgBr is exhibited due to precipitation of both Cations and Anions.
What is the chemical formula AgBr?
AgBr is the chemical formula of silver bromide.
What is the chemical formula for silver bromide?
The chemical formula for silver bromide is AgBr.
What color is AgBr as a precipitate?
Silver bromide (AgBr) is a light yellowish precipitate.
What is the mass of NaBr that will be produced from 42.7 g of AgbR?
Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)
Why AgBr shows both frenkel and schottky defect?
AgBr can exhibit both Frenkel and Schottky defects due to the presence of both cation (Ag+) and anion (Br-) vacancies in its crystal structure. Frenkel defect occurs when a cation occupies an interstitial site, while a Schottky defect involves the simultaneous absence of a cation and anion from their respective lattice sites. The relative sizes of the cation and anion in AgBr make both types of defects possible.
Why does AgBr show both Frenkel and Schottky defects?
The radius ratio for AgBr is intermediate. Thus it shows both frenkel and schottky defects. The major defect in AgBr is the Frenkel defect. It has a rocksalt structure i.e. CCP lattice of of Br with atoms of Ag occupying all octahedral holes.Ag moves from octahedral to tetrahedral sites causing only cations to precipitate. Schottky defect arise due to missing of ions from their lacttice point and frenkel arise when the mmissing ions occupy interstitial sites. in AgBr, ag+ ion is small in size and when removed from lacttice point they can occupy interstitial site and therefore show both frenkel and schottky defect. SCHOTTKY Defect in AgBr is exhibited due to precipitation of both Cations and Anions.
Why silver halides show frenkel defect?
Silver halides, such as AgCl, AgBr, and AgI, exhibit Frenkel defects due to the small size of silver ions compared to the halide ions. In a Frenkel defect, a cation (in this case, Ag+) is displaced from its lattice position and occupies an interstitial site, creating a vacancy in its original position. This defect is favored because it allows for the accommodation of the smaller silver ions without significantly distorting the crystal lattice. As a result, Frenkel defects enhance the ionic conductivity of silver halides, making them useful in various applications, including photography and as semiconductors.
Why AgBr shows both types of defects?
Silver bromide (AgBr) exhibits both Schottky and Frenkel defects due to its ionic crystal structure and the presence of both cations (Ag⁺) and anions (Br⁻). Schottky defects occur when an equal number of cations and anions are missing from the lattice, creating vacancies. In contrast, Frenkel defects arise when an ion (usually a cation like Ag⁺) is displaced from its lattice site to an interstitial position, leaving a vacancy behind. The ability of AgBr to accommodate these defects is influenced by its relatively high ionic mobility and the nature of ionic bonding.
What is the chemical formula AgBr?
AgBr is the chemical formula of silver bromide.
What is the chemical formula for silver bromide?
The chemical formula for silver bromide is AgBr.
What does the symbol AgBr stand for in chemistry?
AgBr is the chemical formula (not symbol) of silver bromide.
What is the product when AgBr reacts with NH4OH?
Образуется растворимое комплексное соединение: AgBr + 2 NH4OH -----> [Ag(NH3)2]+ + Cl- + 2 H2O.
What color is AgBr as a precipitate?
Silver bromide (AgBr) is a light yellowish precipitate.
What is the mass of NaBr that will be produced from 42.7 g of AgbR?
Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)
What is the name of the ionic compound AgBr?
Silver bromide.
Is AgBr a element or a compound?
AgBr is a compound composed of silver (Ag) and bromine (Br) elements.
