Question

Draw the electron dot diagram of the ionic bonding in the
following compounds.
Hints: (Atomic number Na = 11, F = 9, Mg =12)
Sodium fluoride (NaF)
Magnesium fluoride (MgF2)

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Answer 1

Given:

Given ionic compounds are NaF and Mg$F_{2}$

To Find:

The electron dot diagram of the ionic bonding in the

following compounds.

Sodium fluoride (NaF)

Magnesium fluoride (MgF2)

Solution:

Electronic configuration of Na = $1s^{2} 2s^{2} 2p^{6} 3s^{1}$

Electronic configuration of Mg = $1s^{2} 2s^{2} 2p^{6} 3s^{2}$

Electronic configuration of F = $1s^{2} 2s^{2} 2p^{5}$

So, the Na atom will lose an electron and become $Na^{+}$ and Mg loses two electrons and becomes $Mg^{2+}$.

Now, the F atom will gain an electron and forms $F^{-}$.

Therefore ionic bonds will be formed between Na, F, and Mg, F.

Reactions are given below

1. Na - e -> $Na^{+}$

2. F+e -> $F^-$

3. Mg+2e -> $Mg^{2+}$

4. $Na^+ +F^- = NaF$

5. $Mg^{2+} +2 F^- =MgF_{2}$

Please check the attached image for Lewis dot diagram

Answer 2

Answer:

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