How many lone pairs are associted with in `IF_(7)` ?

Explanation

To determine how many lone pairs are associated with the iodine atom in the molecule IF₇, we can follow these steps:

### Step 1: Identify the Valence Electrons
Iodine (I) has an atomic number of 53, and its electronic configuration is [Kr] 4d¹⁰ 5s² 5p⁵. This means iodine has 7 valence electrons (2 from the 5s subshell and 5 from the 5p subshell).

### Step 2: Determine the Number of Fluorine Atoms
In the molecule IF₇, there are 7 fluorine (F) atoms. Each fluorine atom has 7 valence electrons (1s² 2s² 2p⁵) and needs 1 additional electron to complete its octet.

### Step 3: Calculate Total Electrons Used in Bonding
Each fluorine atom will form a single bond with iodine, using one of iodine's valence electrons. Since there are 7 fluorine atoms, they will use 7 of iodine's valence electrons to form 7 bonds.

### Step 4: Determine Remaining Electrons
Since iodine starts with 7 valence electrons and uses all 7 to bond with the 7 fluorine atoms, there are no electrons left unbonded.

### Step 5: Identify Lone Pairs
Lone pairs are pairs of valence electrons that are not involved in bonding. Since all 7 of iodine's valence electrons are used in bonding with fluorine, there are no lone pairs associated with the iodine atom in IF₇.

### Conclusion
Thus, the number of lone pairs associated with the iodine atom in IF₇ is **0**.

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