Question

bonding occurs when a metal transfers one or more electrons to a nonmetal in an effort to attain a stable octet of electrons. for example, the transfer of an electron from sodium to chlorine can be depicted by a lewis dot diagram.
na· + :cl: → na⁺cl⁻
calcium would need two chlorine atoms to get rid of its two valence electrons.
:cl· + ·ca· + :cl: → ca²⁺cl₂⁻
show the transfer of electrons in the following combinations.

1. k + f

k· :f· k⁺ f⁻

1. mg + i

mg· :i: → mg²⁺ i⁻²

1. be + s

:be· :s: → be²⁺ s²⁻

1. na + o

na· :o: na· na²⁺ o²⁻

1. al + br

·al· :br·

Explanation:

Step1: Analyze K and F Valence Electrons

Potassium (K) has 1 valence electron, fluorine (F) has 7 valence electrons. K will transfer its 1 valence electron to F.
Lewis dot: $\text{K} \cdot + :\overset{..}{ \underset{..}{ \text{F} }} :
ightarrow \text{K}^+ + :\overset{...}{ \underset{..}{ \text{F} }}^-$
Formula: $\text{KF}$

Step2: Analyze Mg and I (Iodine)

Magnesium (Mg) has 2 valence electrons, iodine (I) has 7 valence electrons. Mg needs to transfer 2 electrons, so it will react with 2 I atoms.
Lewis dot: $\cdot \text{Mg} \cdot + 2 \times :\overset{..}{ \underset{..}{ \text{I} }} :
ightarrow \text{Mg}^{2+} + 2 \times :\overset{...}{ \underset{..}{ \text{I} }}^-$
Formula: $\text{MgI}_2$

Step3: Analyze Be and S (Sulfur)

Beryllium (Be) has 2 valence electrons, sulfur (S) has 6 valence electrons. Be transfers 2 electrons to S.
Lewis dot: $\cdot \text{Be} \cdot + :\overset{..}{ \underset{..}{ \text{S} }} :
ightarrow \text{Be}^{2+} + :\overset{...}{ \underset{..}{ \text{S} }}^{2-}$
Formula: $\text{BeS}$

Step4: Analyze Na and O (Oxygen)

Sodium (Na) has 1 valence electron, oxygen (O) has 6 valence electrons. O needs 2 electrons, so 2 Na atoms will react with 1 O atom.
Lewis dot: $2 \times \text{Na} \cdot + :\overset{..}{ \underset{..}{ \text{O} }} :
ightarrow 2 \text{Na}^+ + :\overset{...}{ \underset{..}{ \text{O} }}^{2-}$
Formula: $\text{Na}_2\text{O}$

Step5: Analyze Al and Br (Bromine)

Aluminum (Al) has 3 valence electrons, bromine (Br) has 7 valence electrons. Al transfers 3 electrons, so it will react with 3 Br atoms.
Lewis dot: $\cdot \text{Al} \cdot \cdot + 3 \times :\overset{..}{ \underset{..}{ \text{Br} }} :
ightarrow \text{Al}^{3+} + 3 \times :\overset{...}{ \underset{..}{ \text{Br} }}^-$
Formula: $\text{AlBr}_3$

Answer:

1. $\text{KF}$ (Lewis: $\text{K}^+ :\text{F}^-$)
2. $\text{MgI}_2$ (Lewis: $\text{Mg}^{2+} 2\text{I}^-$)
3. $\text{BeS}$ (Lewis: $\text{Be}^{2+} \text{S}^{2-}$)
4. $\text{Na}_2\text{O}$ (Lewis: $2\text{Na}^+ \text{O}^{2-}$)
5. $\text{AlBr}_3$ (Lewis: $\text{Al}^{3+} 3\text{Br}^-$)