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Tpy6a [65]
3 years ago
8

What kind of bond will form between the atoms?

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

ionic bond:

Bond formed when an atom donates its electron and other atom receives those electrons.

polar covalent:

Bond formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.

Nonpolar covalent:

Bond formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.

Metallic:

Formed between two metals.

So the bond between :

Phosphorus and chlorine-polar covalent bond as it is formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.

Potassium and oxygen -ionic bond as here potassium donates its electron and oxygen receives those electrons

Fluorine and fluorine -Non polar covalent bond as formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.

Copper and aluminum-metallic bond as Formed between two metals.

Carbon and fluorine -polar covalent bond as it is formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.

Carbon and hydrogen --Non polar covalent bond as formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.

Aluminum and oxygen--ionic bond as here aluminum donates its electron and oxygen receives those electrons

Silver and copper --metallic bond as Formed between two metals.

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2 years ago
In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,
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The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}

\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}

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