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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

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

Grade:upto college level

7 Answers

Jitender Pal
askIITians Faculty 365 Points
7 years ago
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.
Hitesh munot
12 Points
3 years ago
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`
Ramakrishna
14 Points
3 years ago
Ans is option A. Fluorine is calculated covelent radii neon is calculated vanderwal radii vanderwal greater than covalent radii
Shallini Maheshwari
15 Points
2 years ago
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).
Tarun Mahajan
19 Points
2 years ago
Since atomic radii of Noble gases are measured by vanderwaal radii so the atomic radii of neon is greater than fluorine 
Attin singh
13 Points
2 years ago
Neon belongs to vanded waal radius and flourine to covalent radius r vander is greater than r covalent ans. Is a
 
ankit singh
askIITians Faculty 614 Points
10 months ago
Atomic radii of fluorine and neon in angstrom units are respectively 0.42 and 0.38 amstrong. Atomic radius of fluorine is higher than neon because if we move from fluorine to neon one electron increases in the same valence shell. Hence the nuclear attraction towards valence shell increases.

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