Question

decide whether these proposed lewis structures are reasonable.
proposed lewis structure is the proposed lewis structure reasonable?
yes.
no, it has the wrong number of valence electrons.
the correct number is:
no, it has the right number of valence electrons but doesnt satisfy the octet rule.
the symbols of the problem atoms are:*
yes.
no, it has the wrong number of valence electrons.
the correct number is:
no, it has the right number of valence electrons but doesnt satisfy the octet rule.
the symbols of the problem atoms are:*
yes.
no, it has the wrong number of valence electrons.
the correct number is:
no, it has the right number of valence electrons but doesnt satisfy the octet rule.
the symbols of the problem atoms are:*

• if two or more atoms of the same element dont satisfy the octet rule, just enter the chemical symbol as many times as necessary. for example, if two oxygen atoms dont satisfy the octet rule, enter \o,o\.

Explanation:

Step1: Calculate valence electrons for \( [\colon\ddot{N}-\ddot{N}=\ddot{N}\colon]^{-} \)

N has 5 valence electrons. There are 3 N atoms, so \( 3\times5 = 15 \) valence electrons. Plus 1 for the negative charge, total \( 15 + 1=16 \) valence electrons. In the structure: single bond (\( 2e^{-} \)), double bond (\( 4e^{-} \)), and lone pairs. First N: 3 lone pairs (\( 6e^{-} \)) + 2 in single bond (\( 2e^{-} \)) = \( 8e^{-} \). Second N: 2 lone pairs (\( 4e^{-} \)) + 2 in single bond (\( 2e^{-} \)) + 2 in double bond (\( 2e^{-} \)) = \( 8e^{-} \). Third N: 1 lone pair (\( 2e^{-} \)) + 4 in double bond (\( 4e^{-} \)) + 2 (formal charge consideration, but total electrons count: \( 6 + 4+6 = 16e^{-} \)). But octet rule: all N should have 8 electrons around. First N: \( 6 + 2=8 \), second N: \( 4+4 = 8 \), third N: \( 2+6=8 \). Wait no - check again. Wait formula for valence electrons: \( V = N_{valence\ of\ atoms}+charge \). For \( N_{3}^{-} \), \( V=3\times5 + 1=16 \). The structure has correct electron count. But octet? Each N has 8 (lone pairs + bond pairs). So first structure is okay? Wait no - wait another approach. Wait for \( N_{3}^{-} \), the correct Lewis structure is \( [N\equiv N - \ddot{N}\colon]^{-} \). But the given structure \( [\colon\ddot{N}-\ddot{N}=\ddot{N}\colon]^{-} \): check octet. First N: 3 lone pairs (\( 6e^{-} \)) + 1 bond (\( 2e^{-} \)) = \( 8e^{-} \). Second N: 2 lone pairs (\( 4e^{-} \)) + 1 single (\( 2e^{-} \)) + 1 double (\( 4e^{-} \)) = \( 10e^{-} \) (violates octet). So wrong due to octet.

Step2: Calculate valence electrons for \( H - C\equiv N\colon \)

H has 1, C has 4, N has 5. Total \( 1 + 4+5=10 \). In structure: single bond (\( 2e^{-} \)), triple bond (\( 6e^{-} \)), N has 2 lone pairs (\( 4e^{-} \)). Total \( 2 + 6+4 = 12 \) (wrong electron count. Correct count is 10. Let's recount: \( H(1)+C(4)+N(5)=10 \). In correct \( H - C\equiv N\colon \), single bond (\( 2 \)), triple bond (\( 6 \)), N has 1 lone pair (\( 2 \)). \( 2 + 6+2=10 \). The given structure has N with 2 lone pairs (\( 4e^{-} \)) → wrong electron count.

Step3: Calculate valence electrons for \( H - \ddot{C}-\ddot{C}-H \)

H: 2 atoms (\( 2\times1 = 2 \)), C: 2 atoms (\( 2\times4 = 8 \)). Total \( 2 + 8=10 \). In structure: each C has 3 lone pairs (\( 6e^{-} \)) + 1 single bond (\( 2e^{-} \)) (for each C). Total \( 2\times(6 + 2)+2\times1= 14 \) (wrong). Correct for \( C_{2}H_{2} \) (if it's \( H - C\equiv C - H \)): \( 2\times1+2\times4=10 \). In \( H - C\equiv C - H \): triple bond (\( 6e^{-} \)) per C - C, single bonds (\( 2e^{-} \) each C - H). Each C has \( 2 + 6=8e^{-} \) (octet). The given structure has each C with \( 6 + 2=8 \) (octet - but electron count wrong.

Answer:

• For \( [\colon\ddot{N}-\ddot{N}=\ddot{N}\colon]^{-} \): No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are: \( N \) (the middle N has 10 electrons).
• For \( H - C\equiv N\colon \): No, it has the wrong number of valence electrons. The correct number is: \( 10 \).
• For \( H - \ddot{C}-\ddot{C}-H \): No, it has the wrong number of valence electrons. The correct number is: \( 10 \).