The oxidation number of chromium (Cr) in CrO₂, Cr₂O₇²⁻ and Cr₂(SO₄)₃³⁻ is +4, +6 and +1.5 respectively.
<h3>What is oxidation number?</h3>
Oxidation number of any compound gives idea about the number of exchangeable electrons from the outer most shell of an atom.
Let the oxidation state of chromium is x and charge on the whole molecule is zero, so it will be calculated as:
x + 2(-2) = 0
x - 4 = 0
x = +4
Let the oxidation state of chromium is x and charge on the whole molecule is -2, so it will be calculated as:
2x + 7(-2) = -2
2x - 14 = -2
2x = 12
x = +6
Let the oxidation state of chromium is x and charge on the whole molecule is -3, so it will be calculated as:
2x + 3(-2) = -3
2x - 6 = -3
2x = 3
x = +1.5 (which is impossible)
Hence oxidation state of chromium atom in CrO₂, Cr₂O₇²⁻ and Cr₂(SO₄)₃³⁻ is +4, +6 and +1.5 respectively.
To know more about oxidation state, visit the below link:
brainly.com/question/8990767
#SPJ4
Answer:
Ionization energy is the energy required to remove an electron away from an atom, which can be written as IE. ... Electron affinity, on the other hand, is the energy released when an electron is added to the atom. On the periodic table, electron affinity usually decreases down a group, and increases across a period.
Explanation:
I can't see the full question and answers on your picture