What is the sum of the numbers of neutrons and electrons in the ion 208 pb2+?

For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206

Explanation:

Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.          

The isotope of lead with the mass number 208, has 126 neutrons.

Since, atomic number = number of protons =  number of electrons for neutral atom

Therefore, for ²⁰⁸Pb: number of electrons= 82

So, for ²⁰⁸Pb²⁺ cation: number of electrons= 82 - 2 = 80

Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.

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b) mutualism

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