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ruslelena [56]
3 years ago
13

Why do some atoms form positive ions and others form negative ions?

Chemistry
2 answers:
arsen [322]3 years ago
8 0
Atoms form positive ions by losing some of the electrons making them more positive while atoms form negative ions when they gain electrons making them more negative. This happens due to their electron configuration for them to achieve a more stable state.
kenny6666 [7]3 years ago
5 0
It depends on the location on the periodic table and if it is easier to loose are gain electrons to fill the valence electron shell.

The electron affinity of the neutral atom of the element can show indicate whether or not the element is going the make positive or negative ions.  The higher the electron affinity is, the more the neutral atom wants to gain electrons.  The electron affinity for alkali metals and alkaline earth metals are low due to the fact that they are located on the left of the periodic table and only needs to loose 1 or 2 electrons to have a completed valence shell.  That means that they want to form cations (loose electrons).  Halogens have a high electron affinity since they are located on the right of the periodic table and only have to gain 1 electron to get a completed valence electron configuration.  That means that it forms anions.

I hope this helps.  Let me know if anything is unclear.   
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3 years ago
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Hope this helps:)

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In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic
Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

\bar \nu = \dfrac{1}{\lambda}

Thus, if λ = 3 µm,

\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

\bar \nu = \mathbf{\dfrac{f}{c}}

Thus, if f = 90 THz

\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

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Paul [167]
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