Question

part 3: use lewis dot structures to show the covalent bonding in the following pairs of elements. once you have determined the structure for the molecule, write its structural formula in the space provided; use a dash to represent a shared pair of electrons, and dots to show unshared electrons.

1. nitrogen triiodide (ni₃)

show work here. hint: nitrogen is in the middle!

1. carbon tetrabromide (cbr₄)

show work here. hint: carbon is in the middle!

1. dihydrogen monoxide (h₂o)

show work here. hint: oxygen is in the middle!

Explanation:

Step1: Determine valence electrons for $NI_3$

N has 5 valence - electrons and each I has 7 valence - electrons. Total valence electrons = 5+(3×7)=26. N is in the middle. N forms 3 single bonds with 3 I atoms, using 6 electrons. Remaining electrons = 26 - 6=20. Each I has 3 lone - pairs (6 electrons each) and N has 1 lone - pair. The Lewis structure is: N is in the center with 3 single bonds to I, and N has 1 lone - pair and each I has 3 lone - pairs. Structural formula: $N - I_3$ with lone - pairs shown as dots.

Step2: Determine valence electrons for $CBr_4$

C has 4 valence - electrons and each Br has 7 valence - electrons. Total valence electrons = 4+(4×7)=32. C is in the middle. C forms 4 single bonds with 4 Br atoms, using 8 electrons. Remaining electrons = 32 - 8 = 24. Each Br has 3 lone - pairs. The Lewis structure is: C in the center with 4 single bonds to Br, and each Br has 3 lone - pairs. Structural formula: $C - Br_4$ with lone - pairs shown as dots.

Step3: Determine valence electrons for $H_2O$

O has 6 valence - electrons and each H has 1 valence - electron. Total valence electrons = 6+(2×1)=8. O is in the middle. O forms 2 single bonds with 2 H atoms, using 4 electrons. Remaining electrons = 8 - 4 = 4. O has 2 lone - pairs. The Lewis structure is: O in the center with 2 single bonds to H and 2 lone - pairs. Structural formula: $H - O - H$ with lone - pairs shown as dots.

Answer:

1. Lewis structure of $NI_3$: N in the middle with 3 single bonds to I, N has 1 lone - pair and each I has 3 lone - pairs; Structural formula: $N - I_3$ with lone - pairs as dots.
2. Lewis structure of $CBr_4$: C in the middle with 4 single bonds to Br, each Br has 3 lone - pairs; Structural formula: $C - Br_4$ with lone - pairs as dots.
3. Lewis structure of $H_2O$: O in the middle with 2 single bonds to H and 2 lone - pairs; Structural formula: $H - O - H$ with lone - pairs as dots.