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aivan3 [116]
3 years ago
7

Rutherford investigated atoms by firing small, positively charged particles at a thin sheet of gold foil. What results would he

expect to see if he believed Thomson’s model of the atom to be correct?
Chemistry
1 answer:
Bezzdna [24]3 years ago
6 0
If Thomson’s model of the atom were correct, Rutherford should have seen a complete reflection of those particles when fired at the foil. Those particles would have hit a solid object (Thomson model) and would have been reflected back at the emitter. However, Rutherford discovered that most of those particles passed through the sheet, either continuing straight through or having the angle change a small amount (bouncing off the nucleus). Only a small number of the particles were reflected back, having hit the nucleus straight on. This led him to believe that there was a small, dense portion of the atom that resided within a larger, emptier space.
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"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"
arlik [135]

Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

5 0
3 years ago
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svlad2 [7]

Answer:

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lara [203]

Explanation:

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Answer:It is false

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I took a quiz with this question in it and I chose true but I got it wrong

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