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Zeff = Z - S
Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.
For boron, the electronic configuration is 1s₂ 2s₂ 2p₄
Z = 5, S = 2
Zeff = 5-2 = +3
For O, electronic configuration is 1s₂ 2s₂ 2p₄
Z = 8, S = 2
Zeff = 8-2 = +6
Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.
Answer:
This is an oxidation-reduction (redox) reaction:
2 Ni0 - 4 e- → 2 NiII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Ni is a reducing agent, O2 is an oxidizing agent.
Answer:
Elements in the same period have the same (number of electron shells)