Write the limiting forms (or Canonical forms) of the following ions:
1 answer:
Answer:
Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.
Explanation:
Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.
There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.
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Answer:
Four covalent bonds.
Explanation:
Hello,
In this case, given the attached picture in which you can find the Lewis dot structure for metanal (formaldehyde) we can see two C-H bonds and two C-O bonds via a double bond, thus, we can compute the type of each bond given the electronegativities of hydrogen, carbon and oxygen which are 2.1, 2.5 and 3.5 respectively:
Thus, since both electronegativity difference are less 1.7 we infer that all of them are covalent, therefore, it has four covalent bonds, two C-H bonds and a double C-O bond.
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The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>
The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
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