The table below shows the first three ionization energies for atoms of four elem

Question

The table below shows the first three ionization energies for atoms of four elements from the third period of the periodic table.

The elements are numbered randomly. Use the information in the table to answer the following questions.

(a) Which element is most metallic in character? Explain your reasoning.

(b) Identify element 3. Explain your reasoning.

(c) Write the complete electron configuration for an atom of element 3.

(d) What is the expected oxidation state for the most common ion of element 2?

(e) What is the chemical symbol for element 2?

(f) A neutral atom of which of the four elements has the smallest radius? Explain your reasoning.

Explanation / Answer

a) Metallic character decreases from left to right in each Period.Where as ionisation energy increases from left to right. so the one having having more metallic character have less ionisation energy.More over after removing one electron from element 2 the second IE is very high it means it may have acquired inert gas configuration.
therefore element 2 will have most metallic character and it may be sodium.
b) element 3 is magnesium because after removing 2 electrons it achieves inert gas configuration.
we can conclude it because the third IE is very high for element 3.
c) Mg has 12 electrons: 1s2 2s2 2p6 3s2

d) expected oxidation state for the most common ion of Na = +1

e)Na
f) Element 1
if IE increses from left to right Atomic radius decreases from left to right .
since more energy is required to pull the outer most electron.to the right most we have inert gases,the electrons are tightly bound and difficult to remove them so size decreses.

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