Electronegativity and Bond polatity
@superhelp101

Question

Electronegativity and Bond polatity
@superhelp101

abhisar · Answer

$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \rm F& \rm O& \rm Cl& \rm N& \rm Br& \rm C& \rm S& \rm H& \rm Na& \rm K\
\hline
4.0&3.5&3.0&3.0&2.8&2.5&2.5&2.1&0.9&0.8\
\hline\end{array}$

abhisar · Answer

$\bullet$ If the electronegativity difference (usually called ΔEN) is less than 0.5, then the bond is nonpolar covalent.
$\bullet$ If the ΔEN is between 0.5 and 1.6, the bond is considered polar covalent
If the ΔEN is greater than 2.0, then the bond is ionic.
$\bullet$ If the ΔEN is between 1.6 and 2.0 and if a metal is involved, then the bond is considered ionic. If only nonmetals are involved, the bond is considered polar covalent.

abhisar · Answer

Let's see a few examples
$\Large\text{Methane (CH$_4$)}$
In C-H bond ΔEN (Electronegativity difference) = 2.5-2.1=0.4. Now since it is < 0.5 it will be non-poar (Rule1). $\color{blue}{\text{Thus valence electrons are shared equally between C & H}}$

abhisar · Answer

$\Large\text{Water(H$_2$O)}$
In O-H bond ΔEN = 3.5-2.1=1.4. Now since 0.5<1.4<1.6 it will be a polar bond (Rule2).  $\color{blue}{\text{Thus valence electrons are not equally shared and spend more time near} \ \text{oxygen atom because it has higher electrronegativity value.}}$

abhisar · Answer

$\Large\text{Hydrogen Fluoride(HF) & Sodium Bromide(NaBr)}$
In H-F bond ΔEN = 3-2.1=0.9. In Na_Br bond ΔEN=0.9-2.8=1.9. Now since in both ΔEN = 1.9>1.6<2.0. Rule 4 will be applied. Since both H & F are non-metals HF will be Polar covalent. Na is metal so NaBr will be ionic.

superhelp101 · Answer

oh sorry. Thank you

abhisar · Answer

$\color{red}{\huge\bigstar}\huge\text{You're Most Welcome!  }\color{red}\bigstar$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}$

eric_d · Answer

This explanation is helpful to me....
@Abhisar

abhisar · Answer

That's great to hear
$\color{green}{\huge\ddot\smile}$

anonymous · Answer

Great work @abhisar :)

abb0t · Answer

Is this a question?