Total number of lone pair of electrons in I$$_3^-$$ ion is:

First, we recall the basic idea of a Lewis structure: we must distribute all the valence electrons of the species in the form of bonding pairs and lone (non-bonding) pairs such that every atom attains a stable electronic configuration.

Iodine belongs to Group 17, so one isolated I atom possesses $$7$$ valence electrons. The species under consideration is $$\mathrm{I_3^-},$$ which contains three iodine atoms and an extra negative charge (one extra electron).

Therefore, the total number of valence electrons is obtained by the formula

$$\text{Total valence e}^- = ( \text{number of I atoms} ) \times ( \text{valence e}^- \text{ per I} ) + ( \text{extra electrons from charge} ).$$

Substituting the required numbers, we have

$$\text{Total valence e}^- = 3 \times 7 + 1 = 21 + 1 = 22 \text{ electrons}.$$

Since each pair of electrons consists of two electrons, the total number of electron pairs available is

$$\frac{22}{2} = 11 \text{ pairs}.$$

Now we decide on the skeletal structure. Experimentally and from VSEPR considerations, two iodine atoms are placed terminally and one iodine atom is chosen as the central atom, so the skeleton is $$\mathrm{I - I - I}.$$ To connect the three iodine atoms, we require two single bonds:

$$\mathrm{I\,{:}\!\!-\!\!:I\,{:}\!\!-\!\!:I}.$$

Each single bond utilizes one pair of electrons. Hence the bonding requirement consumes

$$2 \text{ bonds} \times 1 \text{ pair per bond} = 2 \text{ pairs}.$$

The number of pairs remaining for allocation as lone pairs is now

$$11 - 2 = 9 \text{ pairs}.$$

Next, we satisfy the octet of the two terminal iodine atoms. Each terminal iodine already shares one pair in its I-I bond, so it still needs three additional lone pairs to complete an octet (total of four pairs around it).

Thus for the two terminal atoms we allot

$$2 \text{ terminals} \times 3 \text{ pairs per terminal} = 6 \text{ pairs}.$$

Subtracting these from the pool of remaining pairs, we obtain

$$9 - 6 = 3 \text{ pairs}.$$

These 3 pairs are finally placed on the central iodine atom as lone pairs because it already has two bonding pairs and can expand its octet (iodine is a period-5 element and has available d-orbitals).

Counting all the lone-pair allocations we have just made, we see

$$\text{Total lone pairs} = (3 \text{ lone pairs on each of 2 terminal I atoms}) + (3 \text{ lone pairs on the central I atom})$$

$$= 2 \times 3 + 3 = 6 + 3 = 9 \text{ lone pairs}.$$

Hence, the correct answer is Option D.