The element of the group 17 that is most active non metal is fluorine.
The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).
Among all the elements of the group 17. Fluorine is the smallest in size.
Because of the small size of fluorine it has the highest electronegativity in group 17.
This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.
Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.
It reacts readily to form oxides and hydroxides.
So, we can conclude here that fluorine is the most active non metal of group 17.
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To find the number of neutrons, subtract the number of protons from the mass number. number of neutrons=40−19=21.
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
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What is base dissociation constant?
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The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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Its charge would be the amount of electrons that are lost in total, which the information is not stated
