Chemistry, 24.01.2020 23:31 Leftie500

Predict whether each of the following bonds is ionic, polar, covalent, or nonpolar covalent:

a) si--o;
b) k--cl;
c) s--f;
d) p--br;
e) li--o;
f) n--p.

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Answer from: sierravick123owr441

Explanation:

It is known that in order to determine a compound to be ionic, polar, covalent or non-polar covalent it is necessary to determine the electronegativity difference of the combining atoms.

So, when electronegativity difference is from 0.0 to 0.4 then bond formed between the two atoms is non-polar covalent in nature.

When electronegativity difference is greater than 0.4 and less than 1.8 then bond between the two atoms is a polar covalent bond.

When electronegativity difference is 1.8 or greater than the bond formed is ionic in nature.

Electronegativity difference of the given molecules is as follows.

Si-O = (1.90 - 3.44) = 1.54

K-Cl = (0.82 - 3.16) = 2.34

S-F = (2.58 - 3.98) = 1.4

P-Br = (2.19 - 2.96) = 0.77

Li-O = (0.98 - 3.44) = 2.46

N-P = (3.04 - 2.19) = 0.85

Therefore, given compounds are classified as follows.

Si-O, S-F, P-Br, and N-P all have a polar covalent bond. Whereas K-Cl and Li-O are ionic in nature.

Answer from: Quest

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Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.