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Grade: upto college level
        Atomic radii of fluorine and neon in Angstrom units are respectively given by

a. 0.72, 1.60
b. 1.60, 1.60
c. 0.72, 0.72
d. None of these
5 years ago

Answers : (6)

Jitender Pal
askIITians Faculty
365 Points
							Sol.  First ionization energy of oxygen is less than that of nitrogen on the ground of stability of valence shell configuration, hence, (a) is the correct answer.
5 years ago
Hitesh munot
12 Points
							According to general trend in a period atomic radius decreases but neon`s numeric value of atomic radius is greater than fluorine because neon`s Vander Waal radius is calculated and fluorine`s covalent radius is calculated.So correct answer is `a`
2 years ago
14 Points
							Ans is option A.  Fluorine is calculated covelent radii neon is calculated vanderwal radii   vanderwal greater than covalent radii
one year ago
Shallini Maheshwari
15 Points
The radii of noble gas elements are "Van Der wal's radii". Hence, Neon's atomic radius must be much more than that of fluorine. And it is not possible to get covalent and metallic radii for noble gases. Van Der wal's radii is greater then covalent radii. Where fluorine is covalent radii and neon is van der Wal radii. So, the answer is (A).
one year ago
Tarun Mahajan
19 Points
Since atomic radii of Noble gases are measured by vanderwaal radii so the atomic radii of neon is greater than fluorine 
one year ago
Attin singh
13 Points
Neon belongs to vanded waal radius and flourine to covalent radius r vander is greater than r covalent ans. Is a
one year ago
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