Question

1. choose the best answer using the periodic table.

what is the correct lewis structure for $nf_3$?

Explanation:

Step1: Determine valence electrons

Nitrogen (\(N\)) has \(5\) valence electrons (\(2s^{2}2p^{3}\)), and each fluorine (\(F\)) has \(7\) valence electrons (\(2s^{2}2p^{5}\)). For \(NF_{3}\), the total number of valence electrons is \(5+(3\times7)=26\).

Step2: Analyze bonding and lone - pairs

Nitrogen forms single bonds with three fluorine atoms. Each single bond (\(N - F\)) uses \(2\) electrons. So, \(3\) single bonds use \(3\times2 = 6\) electrons. The remaining electrons are \(26 - 6=20\). These remaining electrons are distributed as lone - pairs. Each \(F\) atom gets \(3\) lone - pairs (\(3\times6 = 18\) electrons for \(F\) atoms), and nitrogen gets \(1\) lone - pair (\(2\) electrons).

Step3: Check for double bonds and incorrect structures

• In the second option (\(F - N - F - F\)), the formula is \(NF_{4}\) which is not \(NF_{3}\), so it's wrong.
• In the third option (\(F = N - F\) with another \(F\)), a double bond (\(N = F\)) would require more electron redistribution than needed for \(NF_{3}\) (and \(N\) in \(NF_{3}\) follows the octet rule with single bonds and a lone - pair).
• The first and fourth options: The first option has the correct \(NF_{3}\) formula (\(N\) bonded to three \(F\) atoms with appropriate lone - pairs on \(N\) and \(F\) atoms.

Answer:

The first option (the structure with \(N\) bonded to three \(F\) atoms, each \(F\) having three lone - pairs and \(N\) having one lone - pair) is the correct Lewis structure for \(NF_{3}\).