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enot [183]
3 years ago
10

Explain in terms of electronegativity whether the bonds in CHIFBr (above) are polar or not.

Chemistry
2 answers:
SOVA2 [1]3 years ago
8 0
This is non polar because the elements are relatively close on the periodic table
kirill115 [55]3 years ago
5 0

A bond between two atoms is considered to be polar, if the electronegativity difference between the atoms bonded is greater than 0.4

The electro-negativity values of C, H, F, Br and I are,

C = 2.55, H = 2.20, F = 3.98, Br = 2.96, I = 2.66

The electronegativity difference between C and H bond = 2.55 - 2.20 = 0.35

The electronegativity difference between C and F bond = 3.98-2.55 = 1.43

The electronegativity difference between C and Br bond = 2.96 - 2.55 = 0.41

The electronegativity difference between C and I bond = 2.66 - 2.55= 0.11

All the bonds have electro-negativity difference less than or equal to 0.4 except C-Br and C - F bond. So, the C-Br and C-F bond will be polar, and all the other bonds will be non-polar.

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\begin{array}{cccccc}\text{2NO} & + & \text{I}_{2} &\, \rightleftharpoons \, & \text{2NOI} & & \\ 2.0 &   &  4.0 &   & 1.0 &  & \\ -2x &   & -x &   & +2x &  &  \\ 2.0-2x &   & 4.0-x &   & 1.0+2x &  & \\\end{array}

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