Answer
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Step 1:: Determine the total number of valence electrons in the NH^4 + ion.
5 \ \text{valence electrons (N)} + 4 \ \text{valence electrons (H)} - 1 \ \text{positive charge} = 8 \ \text{valence electrons}
The nitrogen atom (N) is in the 5th period of the periodic table and has 5 valence electrons. Each hydrogen atom (H) in the 1st period has 1 valence electron. Since there are 4 hydrogen atoms and one nitrogen atom, and the charge is + 1, we have:
Step 2:: Place the atoms in their positions and distribute the valence electrons.
Place nitrogen in the center with hydrogen atoms surrounding it. Draw a single bond between the nitrogen and each hydrogen atom, using 2 electrons from the nitrogen for each bond.
Step 3:: Add the remaining valence electrons to the nitrogen atom as lone pairs.
Since there are 8 valence electrons in total, and we've used 4 electrons for the single bonds, we have 4 electrons left. Place these 4 electrons as 2 lone pairs on the nitrogen atom.
Step 4:: Check if the central atom has an octet.
The nitrogen atom has 5 valence electrons from the periodic table, and we've added 4 more electrons (2 lone pairs) to it. This means nitrogen has a total of 9 electrons around it. However, in this case, we consider the + 1 charge on the NH^4 + ion, which means the nitrogen atom has donated 1 electron, effectively completing the octet.
Step 5:: Draw the Lewis structure for NH^4 +.
\chemfig{N:-[2]H-[6]H-[0]H-[0]H^{+}}
The final Lewis structure for NH^4 + is:
Final Answer
The Lewis structure for NH^4 + is shown above, with the nitrogen atom in the center and single bonds to each hydrogen atom. The nitrogen atom has 2 lone pairs of electrons and a positive charge, representing the + 1 charge on the ion.
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