The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C
1 answer:
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Explanation:
More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.
(a) Electronegativity value of P = 2.19
Electronegativity value of Cl = 3.16
Electronegativity value of Br = 2.96
Electronegativity value of F = 3.98
Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97
Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77
Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79
Since, a P-F bond has the highest electronegativity difference. Therefore, is the most ionic compound and
is the least ionic compound.
(b) Electronegativity value of B = 2.04
Electronegativity value of N = 3.04
Electronegativity value of C = 2.55
Electronegativity value of F = 3.98
Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94
Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94
Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43
Since, a B-F bond has the highest electronegativity difference. Therefore, is the most ionic compound and
is the least ionic compound.
(c) Electronegativity value of Se = 2.55
Electronegativity value of Te = 2.1
Electronegativity value of Br = 2.96
Electronegativity value of F = 3.98
Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43
Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88
Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02
Since, a Te-F bond has the highest electronegativity difference. Therefore, is the most ionic compound and
is the least ionic compound.