Give the formula of the stable fluoride formed by \(\mathrm{Li}\) \(\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{N},\) and \(\mathrm{O} .\) For these fluorides, describe the variation in the bonding that occurs as we move from left to right across the period.

The formulas for the stable fluorides are as follows: \n\n- Lithium Fluoride (\(LiF\))\n- Beryllium Fluoride (\(BeF_2\))\n- Boron Trifluoride (\(BF_3\))\n- Carbon Tetrafluoride (\(CF_4\))\n- Nitrogen Trifluoride (\(NF_3\))\n- Oxygen Difluoride (\(OF_2\))\n\nThese fluorides are formed through different ionic and covalent bonds.

As we move from left to right across the period, the bonding in the element's stable fluoride changes. Lithium (Li) and Beryllium (Be) are metals and they form ionic bonds with Fluorine, thus Lithium Fluoride (\(LiF\)) and Beryllium Fluoride (\(BeF_2\)) are ionic compounds.\n\nFrom Boron (B) onwards, the elements are non-metals and they form covalent bonds. Thus, the compounds Boron Trifluoride (\(BF_3\)), Carbon Tetrafluoride (\(CF_4\)), Nitrogen Trifluoride (\(NF_3\)), and Oxygen Difluoride (\(OF_2\)) are covalently bonded. Furthermore, as you move from left to right, the number of fluorine atoms in the compounds tends to increase at first, then decrease, due to the variation in valence electrons of these elements. This is consistent with trends in the periodic table, specifically regarding atomic size, electronegativity, and ionization energy.