Why generally metals are good conductor and non metals are bad conductor of electricity

1 answer:

3 0

Metals - The close packing of the metal ions in the lattice and the delocalised electrons ensure it can carry kinetic energy through the lattice.

Non - metals have fewer free electrons and electrons are held tightly, meaning they're not allowed to move freely.

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Based on the replacement reaction, what would the products of the reaction be?

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Answer:

$\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ . The missing ion would be $\rm OH^{-}$ .

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

$\rm BeSO_4$ , and
$\rm NH_4 OH$ .

In particular,

$\rm BeSO_4$ is made up of $\rm Be^{2+}$ ions and $\rm {SO_4}^{2-}$ ions, while
$\rm NH_4 OH$ is made up of $\rm {NH_4}^{+}$ ions and $\rm OH^{-}$ ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

$\rm Be^{2+}$ is a cation. In $\rm BeSO_4$ , $\rm Be^{2+}$ was bounded $\rm {SO_4}^{2-}$ anions. During the reaction, it bonds with $\rm OH^{-}$ anions to produce $\rm Be(OH)_2$ .

$\rm {NH_4}^{+}$ is also a cation. In $\rm NH_4 OH$ , $\rm {NH_4}^{+}$ was bounded to $\rm OH^{-}$ ions. During the reaction, it bonds with $\rm {SO_4}^{2-}$ anions to produce $\rm (NH_4)_2 SO_4$ .

Hence, the two products will be $\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ .

Note that charges on the ions must balance. For example, a $\rm Be^{2+}$ ion carries twice as much charge as an $\rm {NH_4}^{+}$ ion. As a result, each $\rm Be^{2+}$ ion would bond with twice as many $\rm OH^{-}$ ions as $\rm {NH_4}^{+}$ would in $\rm NH_4 OH$ .

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