Question

1. draw the lewis dot structure for each of the elements and identify how many lone pairs and bonding electrons it has

atom	lewis dot structures	# of lone pairs	# of bonding electrons
nitrogen

your answer:

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Explanation:

Step 1: Recall Lewis Dot Structure Rules

Lewis dot structures represent the valence electrons of an atom. Valence electrons are the outermost electrons. For main - group elements, the number of valence electrons is equal to the group number (for groups 1 - 2 and 13 - 18). Bromine (Br) is in group 17, so it has 7 valence electrons. Nitrogen (N) is in group 15, so it has 5 valence electrons.

Step 2: Lewis Dot Structure for Bromine

To draw the Lewis dot structure for Br:

• We place the symbol Br in the center. Then we distribute the 7 valence electrons as dots around it. We can represent the electrons as single dots first, then pair them. The Lewis dot structure for Br is \(\cdot\overset{\cdot\cdot}{\underset{\cdot\cdot}{Br}}\cdot\) (or more commonly, we can show the pairs and unpaired electrons. The correct way is to have 3 lone pairs (each pair is 2 electrons, so 3 pairs = 6 electrons) and 1 unpaired electron (which is the bonding electron, but for the atom alone, the unpaired electron is the one available for bonding). Wait, actually, for the bromine atom, the number of lone pairs: since there are 7 valence electrons, the number of lone pairs is \(\frac{7 - 1}{2}=3\) (because the unpaired electron is for bonding, and each lone pair has 2 electrons). The number of bonding electrons is 1 (the unpaired electron that can form a bond).

Step 3: Lewis Dot Structure for Nitrogen

For nitrogen (N), which has 5 valence electrons. The Lewis dot structure is \(\cdot\overset{\cdot\cdot}{N}\cdot\cdot\) (or in terms of lone pairs and bonding electrons: the number of lone pairs is \(\frac{5 - 3}{2}=1\)? Wait, no. Let's do it properly. The valence electrons of N are 5. We arrange the dots: we can have one lone pair (2 electrons) and three unpaired electrons? Wait, no. The correct Lewis dot structure for N is \(:\!N\!\cdot\cdot\cdot\) (three unpaired electrons and one lone pair? Wait, no. Wait, the number of valence electrons for N is 5. So we place 5 dots around N. The maximum number of lone pairs (each pair is 2 electrons) we can have: 2 electrons in a lone pair, then the remaining 3 electrons are unpaired. Wait, no, the formula for lone pairs (LP) and bonding electrons (BE) for an atom: \(LP=\frac{V - BE}{2}\), where \(V\) is the number of valence electrons, and \(BE\) is the number of bonding electrons (which is equal to the number of unpaired electrons, since each bond involves one electron from each atom). For a neutral atom, when considering its Lewis dot structure, the number of bonding electrons is the number of unpaired electrons, and lone pairs are the paired electrons.

For bromine (group 17, \(V = 7\)):

• We want to have 8 electrons (octet) in a bonded state, but as an atom, it has 7 valence electrons. So the Lewis dot structure has 3 lone pairs (6 electrons) and 1 unpaired electron (bonding electron). So \(LP = 3\), \(BE=1\).

For nitrogen (group 15, \(V = 5\)):

• The Lewis dot structure has 1 lone pair (2 electrons) and 3 unpaired electrons (bonding electrons). So \(LP = 1\), \(BE = 3\).

Answer:

ATOM	Lewis Dot Structures	# of lone pairs	# of bonding electrons
nitrogen	\(:\!N\!\cdot\cdot\cdot\)	1	3

(Note: The Lewis dot structure for bromine can also be represented as \(\text{Br}\) with 3 pairs of dots and 1 single dot, and for nitrogen as \(\text{N}\) with 1 pair of dots and 3 single dots. The key is the number of lone pairs (pairs of non - bonding electrons) and bonding electrons (unpaired electrons available for bonding).)