Chemistry, 25.03.2021 18:40 yachtboy1917

Gold is found in nature state but not iron whyon

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a solution is always transparent, light passes through with no scattering from solute particles which are molecule in size. the solution is homogeneous and does not settle out. a colloid is intermediate between asolution and a suspension. while asuspension will separate out a colloid will not.

What’s is a suspension, and how is it different from a solution?

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assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

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Ais stronge

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Electron equations of the given changes are as follow,a) cu → cu⁺² + 2e⁻b) mg⁺² + 2 e⁻ → mgc) cl⁻ → cl + 1e⁻d) s + 2 e⁻ → s²⁻e) o²⁻ → o + 2 e⁻f) p + 3 e⁻ → p³⁻g) fe²⁺ → fe³⁺ + 1 e⁻h) s²⁻ → s⁴⁺ + 6 e⁻

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