Answer:
Both are highly reactive.
Explanation:
A has 1 valence electron D has 3
A is sodium D is aluminum
Let us check each statement one by one
a) Sb has a lower ionization energy but a higher electronegativity than I. : As per values given : Definitely Sb has lower ionization energy however the electronegativity of Sb is lower than that of iodine
b) Sb has a higher ionization energy but a lower electronegativity than I. FAlse:
Sb has lower ionization energy than I
c) Sb has a lower ionization energy and a lower electronegativity than I. True
d) Sb has a higher ionization energy and a higher electronegativity than I. False
Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Answer:
The seven SI base units, which are comprised of:
Length - meter (m)
Time - second (s)
Amount of substance - mole (mole)
Electric current - ampere (A)
Temperature - kelvin (K)
Luminous intensity - candela (cd)
Mass - kilogram (kg)
Explanation: