I is the strongest reducing agent since I is the weakest oxidizing agent among the halogens. This is because the strength of oxidizing agent increases down the group.
An aqueous solution of iodine (I₂) will not react with an aqueous solution of potassium bromide (KBr) because iodine is less reactive than bromine in terms of oxidation potential. In this case, iodine cannot oxidize bromide ions (Br⁻) to bromine (Br₂), as bromide is a stronger reducing agent. Therefore, no reaction occurs between the two solutions.
Neutral atoms are smaller than negatively charged ions of the same element.Bromine is smaller than iodine.So neutral bromine would have the smallest radius of the species listed.
Options are not given in the question. But the following species will have the same electronic configuration as Br- ion: Neutral atom: Kr Cations: Rb+, Sr2+ Anion: Se2-
Yes, it's possible. Cis: Br/H=Br/H or Trans: Br/H=H/Br
Br is an anion. It is called bromide and is the ion form of the element bromine.
In this reaction, Br is the reducing agent because it is being oxidized from Br- to Br2, thereby causing the reduction of Ca from Ca to Ca2+.
Fluorine is the strongest oxidizing agent among the elements chlorine, fluorine, iodine, and bromine. It has the highest electronegativity and is most effective at accepting electrons in a redox reaction.
Among halogen acids, HX (X = F, Cl, Br, I), HI is the strongest acid.
2 NaBr (aq) + Cl2 (g) → 2 NaCl (aq) + Br2 (aq) Oxidation reduction reaction 2 Br-I - 2 e- → 2 Br0 (oxidation) 2 Cl0 + 2 e- → 2 Cl-I (reduction) NaBr is a reducing agent, Cl2 is an oxidizing agent.
Neutral atoms are smaller than negatively charged ions of the same element.Bromine is smaller than iodine.So neutral bromine would have the smallest radius of the species listed.
1. H-H 2. H-I 3. H-Br 4. H-Cl 5. H-F
Options are not given in the question. But the following species will have the same electronic configuration as Br- ion: Neutral atom: Kr Cations: Rb+, Sr2+ Anion: Se2-
brick brim broth brat brake brawl.............. oh and br br br br br
Br-101 br-116 br-163 br-158 br-153 br-280 br-282
S in (SO4)-2 has a charge of +6, in SO2 S has a charge of +4. To go from +6 to +4 you must gain two electrons. The (SO4)-2 is reduced (reduction is gaining electrons). Therefore it reduces the 2br-, so (SO4)-2 is the reducing agent
You do this; <br> You do this; <br> You do this; <br>
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