Chemistry mcq help please
Hi,
Please could I have some help on these questions?
5) I thought there would be 4 lone pairs of electrons, carbon would have 2 and oxygen would have 2 for a full outer shell but that is incorrect?
Here is the question: https://ibb.co/g6kPVZb
12) i thought the answer would be d but the answer is A. I worked out the electron configuration of sodium being 1s2 2s2 2p6 3s1 and chlorine being 1s2 2s2 2p6 3s2 3p5 but I’m not sure how that means chlorine has one more occupied electron shell?
Here is the question: https://ibb.co/k9wHhP4
Thank you!
Please could I have some help on these questions?
5) I thought there would be 4 lone pairs of electrons, carbon would have 2 and oxygen would have 2 for a full outer shell but that is incorrect?
Here is the question: https://ibb.co/g6kPVZb
12) i thought the answer would be d but the answer is A. I worked out the electron configuration of sodium being 1s2 2s2 2p6 3s1 and chlorine being 1s2 2s2 2p6 3s2 3p5 but I’m not sure how that means chlorine has one more occupied electron shell?
Here is the question: https://ibb.co/k9wHhP4
Thank you!
Reply 1
Hi,
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
Reply 2
13
Original post
by moodengHi,
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
Hi, thank you so much this makes a lot more sense. For the second one, please could you explain why d would be wrong, I understand option a better now.
Reply 3
Original post
by moodengHi,
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
For the first one, I think the answer is 3. There are 14 valence electrons (4 in carbon, 4 in carbon, and 6 in oxygen). The question tells us there are four double bonds, C=C=O, therefore 8 out of the 14 electrons are involved in bonding. This leaves 6 electrons, which will form 3 lone pairs.
For the second one, I think you need to consider the fact the question is asking about ions. Sodium would lose it's outer electron in the 3s1 and become a positively charged ion. The chlorine would gain an electron to complete the 3p shell. This means the sodium ion would have an electronic configuration of 1s2 2s2 2p6, where as the chloride ion would have an electronic configuration of 1s2 2s2 2p6 3s2 3p6. Therefore, the chloride ion would have one more electron shell occupied 🙂
Hope this helps and let me know if you have any questions!
You are correct on both counts.
You could calculate the number of lone pairs that way, but I personally looked at each atom and concluded that since each double bond formed by an atom uses two of its outer electrons up, the middle carbon has no lone pairs (since all 4 of its outer electrons must be used in bonding), the other carbon must have one lone pair (since it has 2 e^- left that aren’t used to make any bonds) and the oxygen must have two lone pairs (since it only uses 2 of its 6 outer e^- to form the one bond it has).
Your method is perfectly valid still, though. It depends on whether you are better at doing a single calculation or prefer to just do it by inspection.
My suggestion to the OP would be to highlight key words and information in the question. It would have helped them a lot in the second question imo, as they appear to have missed that ionic radii were of interest and so the loss and gain of electrons needed to be factored in. Of course, the question being accessed digitally probably didn’t help, but this would be good to do in actual exams.
(edited 1 year ago)
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