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9 May, 21:56

1. A neutral atom of fluorine (F) has 9 electrons and an electron configuration of 1s2 2s2 2p5. How will it ionize to achieve an octet in its valence shell?

A. It will gain one electron in its 2nd energy level to have a valance configuration of 2s2 2p6.

B. It will lose 5 electrons from its 2nd energy level to have a valance configuration of 2s2

C. It will lose 7 electrons form it 2nd energy level to have a valance configuration of 1s2

D. It will no ionize, it already has an octet of electrons in its valance shell.

2. A neutral atom of neon has 10 electrons and an electron configuration of 1s2 2s2 2p6. How will it ionize to satisfy the octet rule?

A. It will lose 6 electrons from the 2nd energy level to have a valence configuration of 1s2 2s2.

B. It will lose 8 electrons from the 2nd energy level to have a valence configuration of 1s2.

C. It will gain 8 electrons in a 3rd energy level to have a valence configuration of 1s2 2s2

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Answers (1)
  1. 9 May, 22:01
    0
    These answers dont make sense

    1. the 2s2 orbital will give one of its electrons to the 2p5 orbital so the configuration would be 1s22s12p6 (2s1 is half filled and 2p6 is completely filled which is a much more stable configuration)

    2. Neon does not need to ionize it is a noble gas
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