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Postby NRodgers_1C » Sun Nov 17, 2019 8:49 pm

2D.3: "Which of these compounds has bonds that are primarily ionic? (a) BBr3 ; (b) BaBr2 ; (c) BeBr2."

I understand why BBr3 is not a good option, but how are we able to establish that "the electronegativity difference is greater between Ba and Br than between Be and Br, making the Ba---Br bond more ionic" if Be and Ba are the same distance away from Br on the periodic table?

Jessica Chen 2C
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Re: 2D.3

Postby Jessica Chen 2C » Sun Nov 17, 2019 8:59 pm

Keep the electronegativity trends in mind: F is the most electronegative atom, so as you move away from F on the periodic table, the atoms become less electronegative. Since Ba is further from F than Be is, Ba is less electronegative than Be. This means that the electronegativity difference between Ba and Br is bigger than the electronegativity difference between Be and Br.

Cynthia Rodas 4H
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Re: 2D.3

Postby Cynthia Rodas 4H » Sun Nov 17, 2019 10:04 pm

The periodic trend for electronegativity is that electronegativity decreases as you go down the periodic table and electronegativity increases as you go from left to right of the periodic table (excluding noble gases). With that being said, although Ba and Be are the same distance from Br in terms of groups, the electrongegativity of Be is higher than that of Ba due to their period trend. Therefore, the electronegativity distance is greater between Ba and Br than it is between Be and Br.

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Re: 2D.3

Postby 005321227 » Fri Nov 22, 2019 7:30 pm

You can determine ionic bonds by their electronegativity differences: if the difference is over 2, it is considered ionic.

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