When describing metallic bonding, scientists note that the metal ions seem to be floating in a sea of electrons. This arrangemen
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The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.
Answer:
Explanation:
Suppose an electron makes transition from n initial (ni) to n final (nf) then formula for wavelength is given by,
Where,
- λ is wavelength of photon
- Rʜ is rydberg constant, the value of Rʜ is
109690 Cm-¹ in <em>Puri Sharma Pathania</em> standard book of physical chemistry &
109737 Cm-¹ according to <em>Wikipedia</em>
- Z is the atomic number of atom, for hydrogen Z =1,
& according to given data, ni = 5, nf = 3
Solution:
Let's solve for wavelength,
Substituting all the given data in above formula,
Now we know that, 1 cm = 10000000 nm,
Converting the wavelength from Cm → Nm