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3 years ago

Which pair of elements would form an ionic bond?

Chemistry

2 answers:

serg [7]3 years ago

5 0

Option (2) Strontium and chlorine.

Ionic bond is formed by complete transfer of electron/s by an atom of electropositive element to an electronegative element. Thus, atoms of two elements having difference in electronegativity will form ionic compound. More is the difference, stronger will be the ionic compound.

In the Pauling scale, if the electronegativity difference between atoms of two elements is more than 2.0, then the bond formed between them is ionic.

1. Carbon and oxygen: According to Pauling scale, the electronegtaivity of carbon and oxygen is 2.5 and 3.5 respectively. Thus, difference will be:

$\Delta E=3.5-2.5=1.0$

Since, the difference is less than 2.0 thus, the bond formed is not ionic.

2. Strontium and chlorine:According to Pauling scale, the electronegtaivity of strontium and chlorine is 0.95 and 3.0 respectively. Thus, difference will be:

$\Delta E=3.0-0.95=2.05$

Since, the difference is more than 2.0 thus, the bond formed is ionic.

3. Cesium and germanium: According to Pauling scale, the electronegtaivity of cesium and germanium is 0.79 and 2.01 respectively. Thus, difference will be:

$\Delta E=2.01-0.79=1.22$

Since, the difference is less than 2.0 thus, the bond formed is not ionic.

4. Magnesium and aluminium: According to Pauling scale, the electronegtaivity of magnesium and aluminium is 1.31 and 1.61 respectively. Thus, difference will be:

$\Delta E=1.61-1.31=0.3$

Since, the difference is less than 2.0 thus, the bond formed is not ionic.

Therefore, the pair of strontium and chlorine will form ionic bond.

VMariaS [17]3 years ago

3 0

The pair of elements that will form an ionic bond are Strontium and Chlorine.

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Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

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Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

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Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

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