Question

When an electron in an excited state of Mo atoms falls from L to K-shell, an X-ray is emitted. These X-rays are diffracted at angle of ${7.75^ \circ }$ by planes with a separation of $2.94{A^ \circ }$ .What is the difference in energy between k-shell and L-shell in Mo, assuming of first order diffraction? $(sin{7.75^ \circ } = 0.1349)$

Answer 1

We can use Bragg's equation to know the wavelength of X-Ray produce and use the energy equation to know the energy difference.
Formula Used:
Braggs eqn. ${\text{n}}\lambda {\text{ = 2dSin}}\theta$ and energy =$\dfrac{{{\text{hc}}}}{\lambda }$.

Complete step by step answer:
Let’s start by writing and analysing the chemical formula of reactant and the product given to us in the question. We are given two reactants which are diazonium salt and fluoroboric acid. Diazonium salt are group of compounds having a common $R - {N_2}^ + {X^ - }$ group attached to them, R is an organic compound such as alkyl or aryl and X is a halogen attached with nitrogen cation. The second reactant is fluoroboric acid which is also known as tetrafluoroboric acid is an inorganic acid having chemical formula $HB{F_4}$. It dissociates as ${H^ + }$ and $B{F_4}^ -$. Now coming to the product which is aryl fluoride are aromatic compounds in which one or more hydrogen on an aryl ring is replaced by fluorine atom.
There are many ways to produce aryl fluorine, but according to the reactants given the answer to this question will be the Balz-Schiemann Reaction. In the Balz-Schiemann Reaction the fluoroboric acid acts as a fluorine ion donor which replaces the whole ${N_2}X$ group present in diazonium salt and forms aryl fluoride. The reaction is as follows:
$Ar - {N_2}X{\text{ }} + {\text{ }}HB{F_4}{\text{ }}\xrightarrow{{}}{\text{ }}Ar - {N_{2}}B{F_4}{\text{ }}\xrightarrow{\Delta }{\text{ }}Ar - F{\text{ }} + {\text{ }}B{F_{3}}{\text{ }} + {\text{ }}{N_2}$
Hence, the answer to this question is option B. Balz-Schiemann Reaction

Note:
The other reactions which also produce aryl halides are direct substitution, Gattermann, Sandmeyer, etc. All these follow a different procedure such as in case of direct substitution we directly remove the hydrogen from the benzene ring and attach a halide to it forming an aryl halide. Similarly in case of Sandmeyer reaction we form a diazonium ion which is then reacted with a compound containing halide to give aryl halide.