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8_murik_8 [283]
3 years ago
10

What is the correct unabbreviated electron configuration of the following?:

Chemistry
1 answer:
Rudik [331]3 years ago
5 0

The correct  unabbreviated electron configuration  is as below


Vanadium - 1S2 2S2 2P6 3S2 3p6 3d3 4s2

Strontium -  1s2 2S2 2P6 3S2 3P6 3d10 4S2 4P6 4S2

Carbon =1S2 2S2 2P2


<u><em> Explanation</em></u>

vanadium  is in atomic number  23  in the periodic table  hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2

Strontium is in atomic number 38  in periodic table  hence  its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d10  4s2 4p6 4s2

Carbon is in atomic number 6 in periodic  table  therefore its electron configuration is 1s2 2s2 2p2

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0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
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