Question

question 45 (0.0862 points)
the central atom in the nitrate ion has
3 single bonds and one lone pair of electrons
3 double bonds
one double bond and 2 lone pair of electrons
one double bond, 2 single bonds and no lone pair of electrons
question 46 (0.0862 points)
what are the coefficients for the following reaction when it is properly balanced?
potassium iodide + lead (ii) acetate → lead (ii) iodide + potassium acetate

Explanation:

Question 45

To determine the bonding of the central atom (N) in the nitrate ion ($\ce{NO_3^-}$), we analyze its Lewis structure and electron - domain geometry. The nitrate ion has a trigonal planar structure. The nitrogen atom forms one double bond with one oxygen atom and two single bonds with the other two oxygen atoms. Also, the nitrogen atom has no lone pairs of electrons. Let's analyze each option:

• Option 1: The central N in $\ce{NO_3^-}$ has no lone pairs, so this option is incorrect.
• Option 2: The nitrate ion does not have 3 double bonds. It has one double bond and two single bonds (with resonance), so this option is incorrect.
• Option 3: The central N has no lone pairs, so this option is incorrect.
• Option 4: This option correctly describes the bonding of the central N atom in the nitrate ion.

Step 1: Write the chemical formulas

The chemical formulas are: potassium iodide - $\ce{KI}$, lead (II) acetate - $\ce{Pb(CH_3COO)_2}$, lead (II) iodide - $\ce{PbI_2}$, potassium acetate - $\ce{KCH_3COO}$. So the unbalanced equation is $\ce{KI + Pb(CH_3COO)_2 -> PbI_2 + KCH_3COO}$.

Step 2: Balance the iodine atoms

On the left side, we have 1 I atom in $\ce{KI}$, and on the right side, we have 2 I atoms in $\ce{PbI_2}$. So we put a coefficient of 2 in front of $\ce{KI}$: $\ce{2KI + Pb(CH_3COO)_2 -> PbI_2 + KCH_3COO}$.

Step 3: Balance the potassium atoms

Now, on the left side, we have 2 K atoms (from 2$\ce{KI}$), and on the right side, we have 1 K atom in $\ce{KCH_3COO}$. So we put a coefficient of 2 in front of $\ce{KCH_3COO}$: $\ce{2KI + Pb(CH_3COO)_2 -> PbI_2 + 2KCH_3COO}$.

Now, let's check the other atoms. For lead (Pb), we have 1 Pb on the left and 1 Pb on the right. For acetate ($\ce{CH_3COO^-}$) groups, we have 2 on the left (from $\ce{Pb(CH_3COO)_2}$) and 2 on the right (from 2$\ce{KCH_3COO}$). The equation is now balanced.

Answer:

one double bond, 2 single bonds and no lone pair of electrons

Question 46