Question

consider the species n₂⁻, n₂, and n₂⁺.which of these species will be paramagnetic?
n₂ and n₂⁻
n₂⁺ and n₂
n₂⁺ and n₂⁻
n₂⁻, n₂, and n₂⁺
none of the answers is correct.
question 20
1 pts
for a polyatomic ion having the general formula ab₃ⁿ⁺,where a is an atom from group 6a, b is an atom from group 7a, and the ionic charge is n±, what charge must the ion have in order to yield a molecular geometry that is trigonal pyramidal?
3-
1-
1+
3+
5+

Explanation:

Step1: Analyze paramagnetism of nitrogen species

Paramagnetism is due to un - paired electrons. The molecular orbital configuration of $N_2$ is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^2$, all electrons are paired, so $N_2$ is diamagnetic. $N_2^-$ has one extra electron in an antibonding $\pi^*$ orbital, so it has an un - paired electron and is paramagnetic. $N_2^+$ has one less electron from the bonding $\sigma_{2p_z}$ orbital, so it has an un - paired electron and is paramagnetic.

Step2: Determine molecular geometry of $AB_3^{n\pm}$

For a molecule with the general formula $AB_3^{n\pm}$, to have a trigonal - pyramidal geometry, the central atom A must have one lone pair of electrons. Group 6A atoms have 6 valence electrons and Group 7A atoms have 7 valence electrons. Let's consider the valence - shell electron - pair repulsion (VSEPR) theory. For $AB_3^{n\pm}$, the number of valence electrons of A is 6, and each B contributes 7 electrons. The total number of valence electrons from A and B is $6 + 3\times7=27$. For a trigonal - pyramidal geometry, the central atom A should have 4 electron - pairs (3 bonding pairs and 1 lone pair). Using the formula for the number of valence electrons in a polyatomic ion $V = 8\times$ (number of electron - pairs around the central atom), we want 4 electron - pairs around A. The number of valence electrons in the ion should be such that after accounting for the bonding and lone pairs, we get the correct geometry. If the ion is $AB_3^-$, the total number of valence electrons is $6+3\times7 + 1=28$. The central atom A has 4 electron - pairs (3 bonding and 1 lone), which gives a trigonal - pyramidal geometry.

Answer:

For the first question: C. $N_2^+$ and $N_2^-$
For the second question: B. 1 -