Question

question 1 (0.0862 points)
how many valence electrons are in the nitrite ion?
22
18
26
17
question 2 (0.0862 points)
choose the pair of names and formulas that do not match.
copper(ii) nitrite: cu(no₂)₂
copper(ii) nitride: cu₃n₂
copper(ii) nitrate: cuno₃
copper(ii) nitrate: cu(no₃)₂

Explanation:

Question 1

Step1: Determine the formula of nitrite ion

The nitrite ion is $\ce{NO2^-}$.

Step2: Calculate valence electrons for each atom

• Nitrogen (N) is in group 15, so it has 5 valence electrons.
• Oxygen (O) is in group 16, so each O has 6 valence electrons. There are 2 O atoms, so total for O: $2\times6 = 12$.
• The ion has a -1 charge, which means it has gained 1 electron, so we add 1.

Step3: Sum the valence electrons

Total valence electrons = valence electrons of N + valence electrons of O + charge electrons.
So, $5 + 12 + 1 = 18$.

To determine the mismatched pair, we analyze the formulas based on ionic charges:

• Copper(II) has a charge of +2 ($\ce{Cu^{2+}}$).
• Nitrite ion is $\ce{NO2^-}$, so for copper(II) nitrite: $\ce{Cu(NO2)2}$ (since $2\times(+2) + 2\times(-1) = 0$) – matches.
• Nitride ion is $\ce{N^{3-}}$. For copper(II) nitride, the formula should balance charges: $3\ce{Cu^{2+}}$ and $2\ce{N^{3-}}$ give $\ce{Cu3N2}$ (since $3\times(+2) + 2\times(-3) = 0$) – matches.
• Nitrate ion is $\ce{NO3^-}$. For copper(II) nitrate, the formula should be $\ce{Cu(NO3)2}$ (since $+2 + 2\times(-1) = 0$ for $\ce{Cu^{2+}}$ and $\ce{NO3^-}$). The option "Copper(II) nitrate: $\ce{CuNO3}$" has an incorrect formula (charge imbalance: $+2 + (-1) = +1

eq 0$), while $\ce{Cu(NO3)2}$ is correct. So the mismatched pair is "Copper(II) nitrate: $\ce{CuNO3}$".

Answer:

18

Question 2