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IgorLugansk [536]
2 years ago
5

When neptunium (Np) and plutonium (Pu) were discovered, the periodic table did not include the actinides, so these elements were

placed in Groups 7B(7) and 8B(8). When americium(Am) and curium (Cm) were synthesized, they were placed in Groups 8B(9) and 8B(10). However, during chemical isolation procedures, Glenn Seaborg and his colleagues, who had synthesized these elements, could not find their compounds among other compounds of members of the same groups, which led Seaborg tosuggest they were part of a new inner transition series.(b) The highest fluorides of Np and Pu are hexafluorides, as is the highest fluoride of uranium. How does this chemical evidence support the placement of Np and Pu as inner transition elements rather than transition elements?
Chemistry
1 answer:
lorasvet [3.4K]2 years ago
3 0

5f− inner transition series of elements are placed separately at the bottom of the periodic table. Reason: (i) This prevents the undue expansion of the periodic table i.e., maintains its structure. (ii) This preserves the principle of classification by keeping elements with similar properties in a single column.

<h3>What is the difference between transition and inner transition elements?</h3>

The main difference between transition metals and inner transition metals is that transition metal atoms have their valence electrons in the outermost d orbital whereas inner transition metal atoms have their valence electrons in the f orbital of the inner penultimate electron shell.

<h3>How many inner transition elements are there?</h3>

Total number of inner transition elements in periodic table is 28, 14 each in lanthanoid and actinioid series.

Learn more about inner transition metals here:

<h3>brainly.com/question/13632319</h3><h3 /><h3>#SPJ4</h3>
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(aq) + 2I⁻

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²⁻(aq)

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<u>Explanation:</u>

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(aq) + 2I⁻

(aq) → I₂(aq) + 2SO₄

²⁻(aq)

To measure the rate of this reaction we must measure the rate of concentration change of one of  the reactants or products. To do this, we will include (to the reacting S₂O₈

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ii) some starch indicator.

The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂  as soon as it is formed.

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(aq) + 2I⁻

(aq)

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