A common form of elemental phosphorus is the white phosphorus, where four \(\mathrm{P}\) atoms are arranged in a tetrahedron. All four phosphorus atoms are equivalent. White phosphorus reacts spontaneously with the oxygen in air to form \(\mathrm{P}_{4} \mathrm{O}_{6} .\) (a) How many valance electron pairs are in the \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule? (b) When $\mathrm{P}_{4} \mathrm{O}_{6}\( is dissolved in water, it produces a \)\mathrm{H}_{3} \mathrm{PO}_{3}\(, molecule. \)\mathrm{H}_{3} \mathrm{PO}_{3}$ has two forms, \(\mathrm{P}\) forms 3 covalent bonds in the first form and \(\mathrm{P}\) forms 5 covalent bonds in the second form. Draw two possible Lewis structures of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). (c) Which structure obeys the octet rule?

Step by step solution

01

(a) Counting valence electron pairs in P4O6

To determine the number of valence electron pairs in P4O6, we first need to determine the number of valence electrons. There are 4 phosphorus (P) atoms and 6 oxygen (O) atoms in the molecule. Phosphorus has 5 valence electrons, while oxygen has 6 valence electrons. Total valence electrons in P4O6 = (4 * 5) + (6 * 6) = 20 + 36 = 56 Now we will divide the total valence electrons by 2 to find the number of valence electron pairs: Number of valence electron pairs = 56 / 2 = 28 There are 28 valence electron pairs in the P4O6 molecule.

02

(b) Drawing two possible Lewis structures of H3PO3

(i) First form: P forming 3 covalent bonds In this form, the phosphorus atom is surrounded by three oxygen atoms. Two oxygen atoms are bonded to hydrogen atoms, and the third oxygen has a double bond with phosphorus and two lone pairs. The Lewis structure is: O || H - O - P - O - H | H (ii) Second form: P forming 5 covalent bonds In this form, the phosphorus atom is surrounded by three oxygen atoms, two of these oxygen atoms are bonded to hydrogen atoms, and one hydrogen atom is bonded directly to the phosphorus atom. There are two double bonds between phosphorus and two of the oxygen atoms. The Lewis structure is: O || H - O - P - O - H || H

03

(c) Determining which structure obeys the octet rule

Octet rule states that atoms tend to achieve eight electrons in their outer shell, similar to noble gases' electron configuration. In the first form, the central phosphorus atom has 5 electrons in its outer shell (3 bonds and 1 lone pair), making a total of 8 electrons, which obeys the octet rule. In second form, the central phosphorus atom has 10 electrons in its outer shell (5 bonds and no lone pairs), making a total of 10 electrons, which does not obeys the octet rule. Therefore, the first form (P forming 3 covalent bonds) of H3PO3 obeys the octet rule.

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