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mars1129 [50]
3 years ago
15

Which pair of atoms has the highest electronegativity difference?

Chemistry
1 answer:
zavuch27 [327]3 years ago
4 0
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:

A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4

As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.
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Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

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