Question

draw the lewis structure for n₂cl₂ molecule. be sure to include all lone pair electrons and nonzero formal charges. click and drag to start drawing a structure.

Explanation:

Step1: Calculate valence - electrons

N has 5 valence electrons and Cl has 7 valence electrons. For \(N_2Cl_2\), the total number of valence electrons is \(2\times5 + 2\times7=24\).

Step2: Arrange atoms

The most likely structure has the two N atoms bonded to each other, with a Cl atom bonded to each N atom (\(Cl - N=N - Cl\)).

Step3: Distribute electrons

First, place single - bonds between atoms (\(Cl - N\) and \(N=N\)). Each single bond uses 2 electrons. Then, distribute the remaining electrons as lone pairs to satisfy the octet rule. The \(N=N\) bond is a double - bond (4 electrons). After placing the bonds (\(2\times Cl - N\) single bonds and 1 \(N=N\) double - bond, using \(2\times2 + 4=8\) electrons), we have \(24 - 8 = 16\) electrons left. Place 3 lone pairs on each Cl atom (\(2\times3\times2 = 12\) electrons used for Cl lone pairs) and 1 lone pair on each N atom (\(2\times2=4\) electrons used for N lone pairs).

Step4: Calculate formal charges

The formal charge formula is \(FC=V - N_b - \frac{N_{lp}}{2}\), where \(V\) is the number of valence electrons of the free atom, \(N_b\) is the number of bonding electrons, and \(N_{lp}\) is the number of non - bonding (lone - pair) electrons. For Cl: \(V = 7\), \(N_b=2\), \(N_{lp}=6\), \(FC = 7-2 - \frac{6}{2}=0\). For N: \(V = 5\), \(N_b = 4\) (2 in single - bond and 2 in double - bond), \(N_{lp}=2\), \(FC=5 - 4-\frac{2}{2}=0\).

Answer:

The Lewis structure has a double - bond between the two N atoms (\(N=N\)), with a Cl atom single - bonded to each N atom (\(Cl - N=N - Cl\)). Each Cl atom has 3 lone pairs of electrons and each N atom has 1 lone pair of electrons, and all atoms have zero formal charges.