In one of his experiments, Lavoisier placed 10.0 grams of mercury (II) oxide into a sealed container and heated it. The mercury
1 answer:
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Answer: Option (d) is the correct answer.
Explanation:
Electronegativity value of hydrogen is 2.2.
Electronegativity value of chlorine is 3.16.
Electronegativity value of carbon is 2.55.
Electronegativity value of oxygen is 3.44.
Electronegativity value of nitrogen is 3.04.
Electronegativity value of sodium is 0.93.
Electronegativity value of iodine is 2.66.
Therefore, calculate the electronegativity difference between the bonded atoms as follows.
- Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen
= 3.16 - 2.2
= 0.96
- Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon
= 3.44 - 2.55
= 0.89
- Electronegativity difference of
= Electronegativity value of nitrogen - electronegativity value of nitrogen
= 3.04 - 3.04
= 0
- Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium
= 2.66 - 0.93
= 1.73
So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.
Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.
