A maximum of 8 electrons can share the quantum number n = 2.
Principal Quantum number has a symbol of "n". It tells you the energy level on which an electron resides. Y<span>ou need to determine exactly how many </span>orbitals<span> you have in this energy level before you can determine the number of electrons that can share the value of n.
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The number of orbitals you get per energy level can be found using this formula:
<span>no. of orbitals=<span>n</span></span><span>²</span>
Each orbital can hold a maximum of two electrons, the formula would be:
<span>no. of electrons=2<span>n</span></span><span>²</span>
Using the given formulas:
<span>no. of orbitals = <span>n</span></span><span>² </span><span>= </span><span>2</span><span>² </span><span>= </span><span>4</span>
<span>no. of electrons </span><span>=</span><span>2 *</span><span> </span><span>4 </span><span>= </span><span>8 </span>
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
Answer:
The wavelength the student should use is 700 nm.
Explanation:
Attached below you can find the diagram I found for this question elsewhere.
Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.
At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.
