what is the formal charge of phosphorus in the lewis structure shown here? options: -2, -1, +1, 0 (with corresponding colored boxes)

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# Explanation:
## Step1: Recall formal charge formula
The formula for formal charge ($FC$) is $FC = V - N - \frac{B}{2}$, where $V$ is the number of valence electrons in the free atom, $N$ is the number of non - bonding electrons, and $B$ is the number of bonding electrons.

For phosphorus ($P$):
- The number of valence electrons in a free $P$ atom, $V = 5$ (since $P$ is in group 15 of the periodic table).
- From the Lewis structure, we can see the number of non - bonding electrons ($N$): Looking at the Lewis structure of the compound (which has $P$ bonded to $O$ and $F$ atoms), the $P$ atom has no non - bonding electrons, so $N = 0$.
- The number of bonding electrons ($B$): Let's count the bonds. $P$ is double - bonded to $O$ (2 bonds, so 4 electrons) and single - bonded to 3 $F$ atoms (3 bonds, so 6 electrons). So the total number of bonding electrons $B=4 + 6=10$? Wait, no, wait. Wait, in the Lewis structure, for the $P$ atom: the double bond with $O$ contributes 2 bonds (4 electrons) and the three single bonds with $F$ contribute 3 bonds (6 electrons). But when calculating formal charge, we consider the number of bonding electron pairs? Wait, no, the formula is $FC=V - (N + \frac{B}{2})$, where $B$ is the number of bonding electrons. Wait, actually, the correct formula is $FC = \text{Valence electrons in free atom}-\text{Non - bonding electrons}-\text{Bonding electrons}/2$.

Wait, let's re - examine the Lewis structure. The Lewis structure shows $P$ bonded to one $O$ (double bond) and three $F$ atoms (single bonds). So:
- Valence electrons of $P$ ($V$): 5.
- Non - bonding electrons on $P$ ($N$): 0 (since all electrons around $P$ are in bonds).
- Bonding electrons ($B$): In a double bond, there are 4 bonding electrons, and in each single bond, there are 2 bonding electrons. So for $P$, the number of bonding electrons is $4+3\times2 = 10$? Wait, no, that can't be. Wait, no, the formula is $FC=V-(N + \frac{B}{2})$, where $B$ is the number of bonding electrons. Wait, actually, the correct formula is $FC=\text{Number of valence electrons in neutral atom}-\text{Number of non - bonding electrons}-\frac{\text{Number of bonding electrons}}{2}$.

Wait, let's take the atoms:
- For $O$: Valence electrons $V = 6$. Non - bonding electrons: 4 (the two lone pairs). Bonding electrons: 4 (double bond). So $FC_O=6 - 4-\frac{4}{2}=6 - 4 - 2 = 0$.
- For $F$: Valence electrons $V = 7$. Non - bonding electrons: 6 (three lone pairs). Bonding electrons: 2 (single bond). So $FC_F=7 - 6-\frac{2}{2}=7 - 6 - 1 = 0$.

Now for $P$:
Valence electrons $V = 5$. Non - bonding electrons $N = 0$ (no lone pairs on $P$). Bonding electrons $B$: The double bond with $O$ has 4 electrons, and each single bond with $F$ has 2 electrons. So total bonding electrons $B = 4+3\times2=10$? Wait, no, that's wrong. Wait, the number of bonds: $P$ has a double bond (2 bonds) with $O$ and three single bonds (3 bonds) with $F$. So total number of bonds is $2 + 3=5$, and the number of bonding electrons is $5\times2 = 10$? Wait, no, each bond has 2 electrons. So for $P$, the number of bonding electrons is $10$? Then $FC_P=5-0-\frac{10}{2}=5 - 5=0$? Wait, that can't be. Wait, maybe I made a mistake. Wait, the formula is $FC = V - N - \frac{B}{2}$, where $B$ is the number of bonding electrons. Wait, let's check the overall charge of the ion. The ion is $POF_3$? Wait, no, looking at the Lewis structure, the overall charge of the ion: let's calculate the sum of formal charges.

Wait, maybe the ion is $POF_3$ with a charge? Wait, no, the Lewis structure shows $P$ double - bonded to $O$ and single - bonded to three $F$s. Let's recalculate the formal charge of $P$ correctly.

Valence electrons of $P$: 5.

Non - bonding electrons on $P$: 0 (all electrons around $P$ are in bonds).

Bonding electrons: The double bond with $O$ contributes 4 electrons (2 bonds), and each single bond with $F$ contributes 2 electrons (1 bond per $F$). So total bonding electrons $B = 4+3\times2 = 10$. Then $FC=5 - 0-\frac{10}{2}=5 - 5 = 0$? Wait, but that seems off. Wait, maybe the compound is $POF_3$ and the overall charge? Wait, no, maybe I misread the Lewis structure. Wait, the Lewis structure has $P$ in the center, double - bonded to $O$ (with two lone pairs on $O$) and single - bonded to three $F$s (each with three lone pairs). So let's calculate the formal charge of $P$ again.

The formula for formal charge is:

$FC=\text{Valence electrons of the atom in its free state}-\text{Number of non - bonding electrons}-\text{Number of bonding electron pairs}$

Wait, another way to write the formula is $FC = V - (N + \frac{B}{2})$, where $B$ is the number of bonding electrons.

Valence electrons of $P$ ($V$): 5.

Non - bonding electrons ($N$): 0 (no lone pairs on $P$).

Bonding electrons ($B$): In the double bond with $O$, there are 4 bonding electrons, and in each single bond with $F$, there are 2 bonding electrons. So $B = 4+3\times2=10$.

Then $FC = 5-(0+\frac{10}{2})=5 - 5 = 0$. Wait, but that would mean the formal charge of $P$ is 0. But let's check the overall charge. The $O$ has formal charge 0 (as calculated before: $FC_O = 6-4 - \frac{4}{2}=0$), each $F$ has formal charge 0 ($FC_F=7 - 6-\frac{2}{2}=0$). So the overall charge of the molecule (or ion) would be $0 + 0+3\times0+0 = 0$? But maybe the ion is $POF_3^-$ or something else? Wait, no, the options are - 2, - 1, + 1, 0.

Wait, maybe I made a mistake in the number of bonding electrons. Wait, the double bond: a double bond has 2 bond pairs (4 electrons), and each single bond has 1 bond pair (2 electrons). So the number of bond pairs around $P$ is $2 + 3=5$. The formula can also be written as $FC=V - N - \text{number of bond pairs}$.

So $FC = 5-0 - 5=0$. So the formal charge of $P$ is 0.

## Step2: Confirm with the formula
Using the formal charge formula $FC = V - N-\frac{B}{2}$, where:
- $V = 5$ (valence electrons of $P$)
- $N = 0$ (non - bonding electrons on $P$)
- $B = 10$ (bonding electrons: 4 from double bond, 6 from three single bonds)

$FC=5 - 0-\frac{10}{2}=5 - 5 = 0$

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Step1: Recall formal charge formula

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