When an atom forms a positive ion (by losing electrons). will the radius of an ion change from the radius of an atom? explain with the example of sodium atom (atomic no. 11) and sodium ion.

Radius of the positive ion will be smaller than the radius of its parent atom.

Explanation

Sodium Na has atomic number 11. A neutral Na atom has 11 electrons. It has

2 electrons in the first main shell,8 electrons in the second main shell, and1 electron in the third main shell.

Na tends to lose one electron. It would form an Na⁺ ion. The ion is positive and has ten electrons. It has

2 electrons in the first main shell, and8 electrons in the second main shell.

Three of the main shells of Na contain electrons. Only two main shells contain electrons in Na⁺. The energy of the second main shell is higher than the first. It is more distant from the nucleus than the first electron shell. Similarly, the third main shell is further away from the nucleus than the second.

The outermost shell in Na is the third main shell. The outermost shell in Na⁺ is the second main shell. As a result, the outermost electrons in Na are further away from the nucleus that electrons in Na⁺. As a result, the radius of Na appears to be larger than that of Na⁺.

When an atom loses electrons to form a positive cation, it tends to lose all electron in its outermost shell (the valence shell). That will give the ion an octet, which is more stable than having one or two electrons left in the valence shell. The rule on Na shall still apply. The positive ion will have one less main shell; its radius shall be smaller than that of the parent atom.

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.

Answer:

5.6⋅10−^19  j

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