Home >
Community >
Why would the presence of bromide or iodide show up as a false positive in the test for chloride?
Upvote
14
Downvote
+ Inorganic chemistry
+ Chemistry
Posted by
Marinel Subu
Why would the presence of bromide or iodide show up as a false positive in the test for chloride?
It is larger, therefore its outer electrons experience a weaker force of attraction from the nucleus (although the nuclear charge is higher, this is outweighed by the increased distance from the nucleus of outer electrons and the increased shielding of charge by inner shells of electrons). The weaker attraction of the nucleus on the electrons means that the electron cloud can more easily be distorted (as an analogy, it’s much easier to change the shape of a hot air balloon than a party balloon). It is the distortion of the electron cloud that allows ions to be polarised.
It is larger, therefore its outer electrons experience a weaker force of attraction from the nucleus (although the nuclear charge is higher, this is outweighed by the increased distance from the nucleus of outer electrons and the increased shielding of charge by inner shells of electrons). The weaker attraction of the nucleus on the electrons means that the electron cloud can more easily be distorted (as an analogy, it’s much easier to change the shape of a hot air balloon than a party balloon). It is the distortion of the electron cloud that allows ions to be polarised.
It depends on the test you are using and the relative concentrations. You can test for halogens, in general, by adding silver nitrate (AgNO3), and you will get precipitates with Cl-, Br-, and I- (and maybe F-, depends on the concentration). So, if you were testing fro chloride with AgNO3 and any Br- or I- was around, you would get a false positive for Cl-.
It depends on the test you are using and the relative concentrations. You can test for halogens, in general, by adding silver nitrate (AgNO3), and you will get precipitates with Cl-, Br-, and I- (and maybe F-, depends on the concentration). So, if you were testing fro chloride with AgNO3 and any Br- or I- was around, you would get a false positive for Cl-.
I am not sure who discovered or first made bromides, but bromine was discovered independently by Carl Jacob Loewig in 1825 and Antoine Balard in 1826. Loewig’s discovery preceeded that of Balard, although Balard’s paper was published first.
I am not sure who discovered or first made bromides, but bromine was discovered independently by Carl Jacob Loewig in 1825 and Antoine Balard in 1826. Loewig’s discovery preceeded that of Balard, although Balard’s paper was published first.
It is a very interesting redox reaction. Copper iodide cannot exist. The iodide ions reduce the Cu(II) to Cu(I) and the iodide is oxidised to iodine.
The iodine formed reacts with potassium ions in the reaction mixture to give potassium triiodide. So, you will get a precipitate of white copper (I) iodide in a brown solution of potassium triiodide.
It is a very interesting redox reaction. Copper iodide cannot exist. The iodide ions reduce the Cu(II) to Cu(I) and the iodide is oxidised to iodine.
The iodine formed reacts with potassium ions in the reaction mixture to give potassium triiodide. So, you will get a precipitate of white copper (I) iodide in a brown solution of potassium triiodide.
The chromyl chloride test entails heating a sample suspected of containing chloride with potassium dichromate and concentrated sulfuric acid. If chloride is present, chromyl chloride is formed and red fumes of CrO2Cl2 are evident. If there is no chloride present, no red fumes are produced.
No analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides.
The chromyl chloride test entails heating a sample suspected of containing chloride with potassium dichromate and concentrated sulfuric acid. If chloride is present, chromyl chloride is formed and red fumes of CrO2Cl2 are evident. If there is no chloride present, no red fumes are produced.
No analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides.
Iodide is a larger anion and hence gets polarised easily by methyl carbocation. This leads to greater difference in electronegativity between C and I than in methyl chloride and methyl bromide. Therefore dipolemoment of CH3I is greater than CH3Cl and CH3Br.
Iodide is a larger anion and hence gets polarised easily by methyl carbocation. This leads to greater difference in electronegativity between C and I than in methyl chloride and methyl bromide. Therefore dipolemoment of CH3I is greater than CH3Cl and CH3Br.
It is larger, therefore its outer electrons experience a weaker force of attraction from the nucleus (although the nuclear charge is higher, this is outweighed by the increased distance from the nucleus of outer electrons and the increased shielding of charge by inner shells of electrons). The weaker attraction of the nucleus on the electrons means that the electron cloud can more easily be distorted (as an analogy, it’s much easier to change the shape of a hot air balloon than a party balloon). It is the distortion of the electron cloud that allows ions to be polarised.
It is larger, therefore its outer electrons experience a weaker force of attraction from the nucleus (although the nuclear charge is higher, this is outweighed by the increased distance from the nucleus of outer electrons and the increased shielding of charge by inner shells of electrons). The weaker attraction of the nucleus on the electrons means that the electron cloud can more easily be distorted (as an analogy, it’s much easier to change the shape of a hot air balloon than a party balloon). It is the distortion of the electron cloud that allows ions to be polarised.
More
VOTE
It depends on the test you are using and the relative concentrations. You can test for halogens, in general, by adding silver nitrate (AgNO3), and you will get precipitates with Cl-, Br-, and I- (and maybe F-, depends on the concentration). So, if you were testing fro chloride with AgNO3 and any Br- or I- was around, you would get a false positive for Cl-.
It depends on the test you are using and the relative concentrations. You can test for halogens, in general, by adding silver nitrate (AgNO3), and you will get precipitates with Cl-, Br-, and I- (and maybe F-, depends on the concentration). So, if you were testing fro chloride with AgNO3 and any Br- or I- was around, you would get a false positive for Cl-.
More
VOTE
Ethyl bromide can be converted to ethyl iodide by treatment with sodium iodide in acetone.
CH3CH2Br +NaI ⇌ CH3CH2I + NaBr
It is called Finkelstein reaction.
Ethyl bromide can be converted to ethyl iodide by treatment with sodium iodide in acetone.
CH3CH2Br +NaI ⇌ CH3CH2I + NaBr
It is called Finkelstein reaction.
More
VOTE
Only happens if the test is not specific enough for chloride.
Removing the effect of interfering species is always a n issue in analytical chemistry.
Only happens if the test is not specific enough for chloride.
Removing the effect of interfering species is always a n issue in analytical chemistry.
More
VOTE
I am not sure who discovered or first made bromides, but bromine was discovered independently by Carl Jacob Loewig in 1825 and Antoine Balard in 1826. Loewig’s discovery preceeded that of Balard, although Balard’s paper was published first.
I am not sure who discovered or first made bromides, but bromine was discovered independently by Carl Jacob Loewig in 1825 and Antoine Balard in 1826. Loewig’s discovery preceeded that of Balard, although Balard’s paper was published first.
More
VOTE
It is a very interesting redox reaction. Copper iodide cannot exist. The iodide ions reduce the Cu(II) to Cu(I) and the iodide is oxidised to iodine.
The iodine formed reacts with potassium ions in the reaction mixture to give potassium triiodide. So, you will get a precipitate of white copper (I) iodide in a brown solution of potassium triiodide.
It is a very interesting redox reaction. Copper iodide cannot exist. The iodide ions reduce the Cu(II) to Cu(I) and the iodide is oxidised to iodine.
The iodine formed reacts with potassium ions in the reaction mixture to give potassium triiodide. So, you will get a precipitate of white copper (I) iodide in a brown solution of potassium triiodide.
More
VOTE
The chromyl chloride test entails heating a sample suspected of containing chloride with potassium dichromate and concentrated sulfuric acid. If chloride is present, chromyl chloride is formed and red fumes of CrO2Cl2 are evident. If there is no chloride present, no red fumes are produced.
No analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides.
The chromyl chloride test entails heating a sample suspected of containing chloride with potassium dichromate and concentrated sulfuric acid. If chloride is present, chromyl chloride is formed and red fumes of CrO2Cl2 are evident. If there is no chloride present, no red fumes are produced.
No analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides.
More
VOTE
Iodide is a larger anion and hence gets polarised easily by methyl carbocation. This leads to greater difference in electronegativity between C and I than in methyl chloride and methyl bromide. Therefore dipolemoment of CH3I is greater than CH3Cl and CH3Br.
Iodide is a larger anion and hence gets polarised easily by methyl carbocation. This leads to greater difference in electronegativity between C and I than in methyl chloride and methyl bromide. Therefore dipolemoment of CH3I is greater than CH3Cl and CH3Br.
More
VOTE
In Chain isomers you re-order the chain to create a different structure. Hence the chain isomers of C4H9Br are:
In Chain isomers you re-order the chain to create a different structure. Hence the chain isomers of C4H9Br are:
More
VOTE