Question

Covalent radius of $F$ is $07.2 nm$ Calculate its Allred-Rochow's electronegativity to the nearest integer.

Answer 1

We know Slater's pointers are selections that offer the qualities to the compelling atomic charge idea, or ${{Z}_{eff}}$ These principles depend upon trial info for electron advancement and ${{Z}_{eff}}$ ionization energies, and are resolved from this condition: ${{Z}_{eff}}=Z-S$
Where, $Z$ is atomic charge, ${{Z}_{eff}}$ is viable atomic charge, and $S$ is protected consistently.

Complete Step By Step Answer:
Given data contains, Radius of silicon $r=0.72$ we know that, Allred-Rochow equation,
$X=0.359\times \dfrac{{{Z}_{eff}}}{{{r}^{2}}}+0.744$
The effective nuclear charge is determined using the Slater’s rule, the electronic configuration of fluorine is: $1{{s}^{2}},2{{s}^{2}},2{{p}^{5}}$
${{Z}_{eff}}=(3\times 0.35)+(0.85\times 5)$ ………… [ here for s-orbital we have $0.35$ electrons whereas total there are $1+2$ that is $3$ s-orbital. Similarly for p-orbital we have $0.85$ electrons whereas total there are $5$ in s-orbital ]
${{Z}_{eff}}=5.30$
The electronegativity of fluorine can be calculate by substituting known value we get,
$\chi =0.359\times \dfrac{5.30}{{{0.72}^{2}}}+0.744$
On further simplification we get;
$\Rightarrow \chi =4.34$
Therefore, the electronegativity of fluorine is $4.34$

Note:
We have to recollect that Pauling came upon a tendency because of the "power" of an iota in an atom to attract electrons to itself. It’s a proportion of the molecule's capability to pull in an electron to itself whereas the electron is up to now connected to another particle. The higher the qualities, the just about certain that iota will pull electrons from another particle and into itself.
The occasional pattern for tendency by and large increments from left to right and diminishes because it goes down the gathering. The exemption is chemical element and also the respectable gases in hydrogen of the actual fact that the honourable gases hydrogen content with their filled peripheral shells, cannot stand to lose a valence electron dissimilar to the rest of the gathering $1$ metals.