Question

0 p₄ + 0 f₂ → 0 pf₃

Explanation:

Step1: Balance phosphorus atoms

On left: 4 P in $\text{P}_4$; on right: 1 P in $\text{PF}_3$. Multiply $\text{PF}_3$ by 4:
$\text{P}_4 + \text{F}_2
ightarrow 4\text{PF}_3$

Step2: Balance fluorine atoms

On right: $4 \times 3 = 12$ F; on left: 2 F in $\text{F}_2$. Multiply $\text{F}_2$ by 6:
$\text{P}_4 + 6\text{F}_2
ightarrow 4\text{PF}_3$

Step3: Verify atom counts

Left: 4 P, $6 \times 2 = 12$ F; Right: $4 \times 1 = 4$ P, $4 \times 3 = 12$ F. All atoms balanced.

Answer:

Left box: $1$ $\text{P}_4$ + $6$ $\text{F}_2$
Right box: $4$ $\text{PF}_3$
Balanced equation: $\text{P}_4 + 6\text{F}_2
ightarrow 4\text{PF}_3$