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Topic: Chemistry Electronic Configurations and Bond Polarity  (Read 3331 times)

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Offline Janderson

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Chemistry Electronic Configurations and Bond Polarity
« on: May 18, 2013, 05:53:47 AM »
Hi,
I've got two questions from some chemistry homework that I'm not too sure about.

Q1) The following are hypothetical configurations for a phosphorus atom.
1. 1s2, 2s2, 2p6, 3s1, 3p4.
2. 1s3, 2s3, 2p6, 3s2, 3p1.
3. 1s2, 2s2, 2p6, 3s2, 3p2, 3d1.
4. 1s2, 2s2, 2p6, 3s2, 2d1, 3p2.
5. 1s2, 1p6, 2s2, 2p5.
6. 1s2, 2s2, 2p6, 3s2, 3p3.

Which use nonexistent orbitals? Select all possible answers.

My answer: I thought that seeing as 2. contained 1s3 and 2s3 orbitals, 4. contained a 2d1 orbital, and 5. contained a 1p6 orbital that these 3 options were non-existent (i.e. options 2, 4 and 5). But, apparently this is wrong. Do you know where I went wrong? I have no idea.

Q2) Arrange the following molecules in order of increasing bond polarity.
1. BH3
2. NH3
3. PH3
4. AlH3

My answer: I worked out the electronegativity differences to be:
1. B-H = 0.1
2. N-H = 0.9
3. P-H = 0
4. Al-H = 0.6
So, increasing bond polarity order is 3<1<4<2
But, everywhere I've looked on the internet says the proper order is 4=1<3<2 and I don't understand why. Any ideas?

Thank you in advance.

Offline Schrödinger

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Re: Chemistry Electronic Configurations and Bond Polarity
« Reply #1 on: May 23, 2013, 05:37:17 AM »
Q2 : Hint : Bond polarity is different from overall molecule's polarity.

Q1 : Going strictly by the words 'non-existent orbitals', your answer seems to be correct. However, what according to the book is the right answer? Maybe we could make some sense out of that?
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