Chemistry, 09.09.2019 17:30 anitadefrances

Detailed calculations show that the value of z_eff for the outermost electrons m si and cl atoms is 4.29 + and 6.12 +, respectively. what value do you estimate for zeff experienced by the outermost electron in both si and cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant?

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Answer from: alexussaniyah

Assuming those contributions to the screening, the $Z_{eff}$ estimation for Si is 4, and for Cl it is 7.

Explanation:

The general expresion for nuclear efective charge is: $Z_{eff}=Z-S$ , with Z the atomic number of the element and the number of protons in the nucleus, and S the screening constant.

Si and Cl belong to the third row of the periodic table, so their core electrons are 10 (the same as the noble gas before them, Ne). The number of protons in Si nucleus is 14, and in Cl is 17. Each of the core electrons contribute with 1.00, so each one screens one proton's charge, leaving a total $Z_{eff}$ of 4 for Si and 7 for Cl.

For Si, this aproximation underestimate the experimental value, so probably the valence electrons contribute to the screening (contribution bigger than 0.0). For Cl this value is overestimated, and probably the contribution of the core electrons is weaker than 1.0.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

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Detailed calculations show that the value of z_eff for the outermost electrons m si and cl atoms is...

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