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Rank the following atoms/ions in order of atomic radius from smallest to largest: Br / Cl / I- / I

Question

Rank the following atoms/ions in order of atomic radius from smallest to largest: Br / Cl / I- / I

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Solution 1

The atomic radius generally increases as we move down a group in the periodic table because additional electron shells are added. Also, anions (negative ions) are larger than their parent atoms because the addition of electrons increases electron-electron repulsion, causing the electron cloud to expand.

  1. Cl: Chlorine is in the 3rd period of the periodic table, so it has three electron shells.

  2. Br: Bromine is in the 4th period, so it has four electron shells, making it larger than chlorine.

  3. I: Iodine is in the 5th period, so it has five electron shells, making it larger than bromine.

  4. I-: Iodine ion is larger than the iodine atom because the addition of an electron increases electron-electron repulsion, causing the electron cloud to expand.

So, the order from smallest to largest atomic radius is: Cl < Br < I < I-.

This problem has been solved

Solution 2

The atomic radius generally increases as we move down a group in the periodic table. This is because with each step down a group, an additional electron shell is added. Therefore, the atomic radius increases due to the increased number of electron shells.

Also, anions (negatively charged ions) are larger than their parent atoms because the addition of an electron increases electron-electron repulsion, causing the electron cloud to expand.

Given these rules, we can rank the atoms/ions as follows:

  1. Cl (Chlorine) - It is in the 3rd period of the periodic table, so it has fewer electron shells than bromine or iodine.
  2. Br (Bromine) - It is in the 4th period of the periodic table, so it has more electron shells than chlorine but fewer than iodine.
  3. I (Iodine) - It is in the 5th period of the periodic table, so it has more electron shells than both chlorine and bromine.
  4. I- (Iodide ion) - It is an iodine atom that has gained an extra electron, making it larger than a neutral iodine atom due to increased electron-electron repulsion.

So, the order from smallest to largest atomic radius is: Cl < Br < I < I-.

This problem has been solved

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