The statement "An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy" is false.
In this context , we will define ionization energy as the minimum energy required to remove a valence electron from a neutral atom in it's gaseous state. In a sense the ionization energy is a measure the amount of 'difficulty' of making something an ion. A high ionization energy means that it takes a lot of energy to remove a valence electron from that atom. A low ionization energy means that it is easy to remove a valence electron from the atom. It is known that group 1 elements generally have a low ionization energy. On the other hand, it is harder for noble gasses and group 7 atoms to loose electrons because they have higher ionization energy.
To form a positive ion, you have to remove an electron. When an electron is removed from an atom, there ion formed has more positive charges than negative charges in it, making it net positive. We have established that atoms with low ionization energy loose elections much more easily. We have also established that atoms with high ionization energy do not loose electrons easily. From this we can gather that the statement is false. An atom with high ionization energy will not form a positive ion more easily that an atom with low ionization energy.
- The chemical symbol of the isotope boron-11 is ¹¹B.
- The atomic mass of the isotope boron-11 is equal to 11.009306.
- The abundance in nature of the isotope boron-11 is equal to 80.1%.
<h3>What is an isotope?</h3>
An isotope can be defined as the atom of a chemical element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of neutrons).
In Chemistry, there are two main isotopes of boron and these include the following:
Boron-11 is the most stable isotope of boron and it is characterized by the following:
- The chemical symbol of the isotope boron-11 is ¹¹B.
- The atomic mass of the isotope boron-11 is equal to 11.009306.
- The abundance in nature of the isotope boron-11 is equal to 80.1%.
Read more on Boron-11 here: brainly.com/question/6283234
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“It can hold two electrons”
The correct option is A.
To calculate the binding energy, you have to find the mass defect first.
Mass defect = [mass of proton and neutron] - Mass of the nucleus
The molar mass of thorium that we are given in the question is 234, the atomic number of thorium is 90, that means the number of neutrons in thorium is
234 - 90 = 144.
The of proton in thourium is 90, same as the atomic number.
Mass defect = {[90 * 1.00728] +[144* 1.00867]} - 234
Note that each proton has a mass of 1.00728 amu and each neutron has the mass of 1.00867 amu.
Mass defect = [90.6552 + 145.24848] - 234 = 1.90368 amu.
Note that the unit of the mass is in amu, it has to be converted to kg
To calculate the mass in kg
Mass [kg] = 1.90368 * [1kg/6.02214 * 10^-26 = 3.161135 * 10^-27
To calculate the binding energy
E = MC^2
C = Speed of light constant = 2.9979245 *10^8 m/s2
E = [3.161135 * 10^-27] * [2.9979245 *10^8]^2
E = 2.84108682069 * 10^-10.
Note that we arrive at this answer because of the number of significant figures that we used.
So, from the option given, Option A is the nearest to the calculated value and is our answer for this problem.
