Chemistry, 04.07.2019 03:30 paytonevans696

Which element has three half-filled orbitals in the 3p level?

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Phosphorus Phosphorus is in group 15 period 3 in the periodic table hence it is in atomic number 15. Phosphorus electronic configuration is 1S2 2S2 2P6 3S2 3P3 or [Ne] 3S2 3P3

From the electronic configuration of phosphorus the 3p3 sub level is unfilled since the p orbital can hold a maximum of 6 electrons.

Answer from: hannahmxx2

Phosphorus is the element which has 3 half-filled orbitals in 3p level.

Explanation:

Some periodic table characterization of phosphorus is given below:

Periodic table representation:P Atomic mass: 30.9Atomic number: 15 Electron configuration: 1s2 2s2 2p6 3s2 3p3

As the 3p orbital can contain 6 electron and according to above given electronic configuration of phosphorus, 3p has 3 electrons, means half filled.

Answer from: Quest

they are all solids.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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