Sulfur has two filled energy levels and six electrons on the third energy level. The corresponding electron configuration is A.
B is incorrect because there are no p orbitals at the first energy level, ie, no 1p orbitals. C is incorrect because the 4s1 electron would spontaneously drop into the 3p orbitals. D is incorrect because the 3d electrons would spontaneously drop into the 3p orbitals.
Answer:
Silicon is the basic material used to make computer chips, transistors, silicon diodes and other electronic circuits and switching devices because its atomic structure makes the element an ideal semiconductor.
Barium nitrate and methane (CH4) are both soluble. They both will dissolve in water, however, barium nitrate will dissociate becoming barium 2+ ions and nitrate becoming NO3 1- ions. All nitrates are soluble and dissociate. CH4 is a weak base and does dissolves but doesn't dissociate. So in solubility terms.... they are both equally soluble just one happens to dissociate into its cations and anions. Hope this helps!
Answer:
The effective nuclear charge for a valence electron in oxygen atom: 
Explanation:
Effective nuclear charge ![[Z_{eff}]](https://tex.z-dn.net/?f=%5BZ_%7Beff%7D%5D)
is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s). </em>
<u>For an oxygen atom</u>-
Electron configuration: (1s²) (2s² 2p⁴)
<em>The atomic number (actual nuclear charge): </em>Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>
<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u>
