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.
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-
Br is an atom.
Yes, it's possible. Cis: Br/H=Br/H or Trans: Br/H=H/Br
Bromine forms the Br- anion
Fluorine.
Among halogen acids, HX (X = F, Cl, Br, I), HI is the strongest acid.
2 NaBr (aq) + Cl2 (g) β 2 NaCl (aq) + Br2 (aq) Oxidation reduction reaction 2 Br-I - 2 e- β 2 Br0 (oxidation) 2 Cl0 + 2 e- β 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.
Br, not Br-
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
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Br-I
Br-I