Answer
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Hint: Think about the position of all the elements in the periodic table, and how the trend of properties affects the losing or gaining of electrons to form octets.
Complete answer:
- As we go along a period from left to right, the atomic orbital size is said to decrease. This happens since the number of protons in the nucleus keep on increasing but no new shells are added. The increase in the distance between the farthest electron and the nucleus does not happen and the nucleus tends to hold on to the electrons more strongly as we go from left to right along a period.
Hence, we can infer that electronegative character increases from left to right along a period as the atoms are less likely to let go of their electrons.
- As we move down a group, the number of orbitals increases and the distance between the nucleus and the farthest atom increases. The increase in the number of protons cannot compensate for the increased distance as it is very large and also due to the shielding effect. Thus, atoms are more likely to let go of electrons as we go down a group.
Hence, we can infer that the electronegative character decreases as we go down a group as the size of the atom increases and it is more likely to let go of the electrons.
Thus, from both these inferences, we can conclude that the most electronegative element will be present at the top right corner of the periodic table. The group of noble gases can be excluded since they do not react. Therefore, the most electronegative element is present in group 7A of the ${{2}^{nd}}$ period. This place is occupied by Fluorine which has the value of 4.0 on Pauling’s scale of electronegativity. Where, 4.0 is the most electronegative and 0.0 is the least electronegative.
So, the correct answer is “Option B”.
Note: Even though fluorine is the most electronegative element, oxygen is a close second with a value of 3.5 on Pauling’s scale of electronegativity. Oxygen takes electrons from all other atoms but when it is bonded with fluorine, it has to give up electrons, this proves that fluorine is the most electronegative.
Complete answer:
- As we go along a period from left to right, the atomic orbital size is said to decrease. This happens since the number of protons in the nucleus keep on increasing but no new shells are added. The increase in the distance between the farthest electron and the nucleus does not happen and the nucleus tends to hold on to the electrons more strongly as we go from left to right along a period.
Hence, we can infer that electronegative character increases from left to right along a period as the atoms are less likely to let go of their electrons.
- As we move down a group, the number of orbitals increases and the distance between the nucleus and the farthest atom increases. The increase in the number of protons cannot compensate for the increased distance as it is very large and also due to the shielding effect. Thus, atoms are more likely to let go of electrons as we go down a group.
Hence, we can infer that the electronegative character decreases as we go down a group as the size of the atom increases and it is more likely to let go of the electrons.
Thus, from both these inferences, we can conclude that the most electronegative element will be present at the top right corner of the periodic table. The group of noble gases can be excluded since they do not react. Therefore, the most electronegative element is present in group 7A of the ${{2}^{nd}}$ period. This place is occupied by Fluorine which has the value of 4.0 on Pauling’s scale of electronegativity. Where, 4.0 is the most electronegative and 0.0 is the least electronegative.
So, the correct answer is “Option B”.
Note: Even though fluorine is the most electronegative element, oxygen is a close second with a value of 3.5 on Pauling’s scale of electronegativity. Oxygen takes electrons from all other atoms but when it is bonded with fluorine, it has to give up electrons, this proves that fluorine is the most electronegative.
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