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i62 💰:9.4

🏆 i71 💰:50.0

i70 💰:25.0

i70 💰:125.0

i68 💰:125.0

🏆 i67 💰:25.0

i67 💰:62.5

i66 💰:12.5

i66 💰:75.0

i65 💰:125.0

i63 💰:62.5

i61 💰:75.0

i60 💰:9.4

i60 💰:37.5

i58 💰:62.5

i57 💰:22.5

i57 💰:37.5

i56 💰:45.0

i55 💰:37.5

i54 💰:62.5

i52 💰:35.0

i51 💰:37.5

i50 💰:1.3

i49 💰:15.0

i49 💰:62.5

i48 💰:11.3

i47 💰:7.5

i44 💰:25.0

i40 💰:5.0

i39 💰:100.0

i38 💰:37.5

i37 💰:11.0

i34 💰:12.5

i33 💰:37.5

i30 💰:1.0

i29 💰:1.5

i28 💰:150.0

i18 💰:1.0

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Responses

The statement is **false** because the Lewis Electron Dot Structure (LEDS) for the ammonium ion \([N...

Answers to: explain why it is false

The statement is **false** because the Lewis Electron Dot Structure (LEDS) for the ammonium ion \([NH_4^+]\) does not accurately represent the valence electrons involved.

In a correct LEDS for \([NH_4^+]\):

1. **Total Valence Electrons**: The nitrogen atom contributes 5 valence electrons, and each hydrogen atom contributes 1 (totaling 4 for 4 hydrogens), which initially gives 9. However, since the ammonium ion has a positive charge, it has lost one electron, resulting in a total of 8 valence electrons to distribute.

2. **Bonding**: The nitrogen is connected to four hydrogen atoms through single bonds, utilizing all 8 valence electrons efficiently.

3. **Formal Charges**: In the correct structure, nitrogen has no formal charge while each hydrogen also has no formal charge.

In the picture you provided, if it simply represents a structure that does not accommodate the above points, then it fails to depict the correct Lewis structure.

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