Question

Find the value of z in H₄P₂O₈.

z = $\frac{x+d}{a+c+b}$

Total number of σ bond = x Total number of lone pair of e⁻s are = a Total number of sp³ hybridised atoms are b Total number of pπ-pπ bonds are c Total number of pπ-dπ bonds are d

Answer 1

z = $\frac{5}{8}$

Answer 2

To find the value of z, we first need to determine the structure of H₄P₂O₈ and then calculate the values of x, a, b, c, and d.

1. Determine the structure of H₄P₂O₈: First, calculate the oxidation state of Phosphorus (P) in H₄P₂O₈. Let the oxidation state of P be S. 4(+1) + 2(S) + 8(-2) = 0 4 + 2S - 16 = 0 2S = 12 S = +6 Since the maximum oxidation state of Phosphorus is +5, an oxidation state of +6 indicates the presence of a peroxide (O-O) linkage in the molecule. Thus, H₄P₂O₈ is peroxodiphosphoric acid. Its structure is:

Simplified structural representation:

      O     O
      ||    ||
HO -- P -- O -- O -- P -- OH
      |          |
      OH         OH

2. Calculate x (Total number of σ bonds):

• P=O bonds: 2 (each has 1 σ bond) = 2 σ bonds
• P-OH bonds: 4 (each has 1 σ bond) = 4 σ bonds
• P-O (in P-O-O linkage) bonds: 2 (each has 1 σ bond) = 2 σ bonds
• O-O bond: 1 (1 σ bond) = 1 σ bond
• O-H bonds: 4 (each has 1 σ bond) = 4 σ bonds

Total σ bonds (x) = 2 + 4 + 2 + 1 + 4 = 13. So, x = 13.

3. Calculate a (Total number of lone pairs of electrons):

• Phosphorus (P) atoms: Each P atom is in a +5 oxidation state and forms 5 bonds (1 double, 3 single). All 5 valence electrons are involved in bonding, so no lone pairs on P atoms.
• Oxygen (O) atoms: There are 8 oxygen atoms in total.
  • Two P=O oxygen atoms: Each forms one double bond (2 bonds). With 6 valence electrons, 2 are used in bonding, leaving 4 non-bonding electrons (2 lone pairs) on each. (2 O atoms * 2 lone pairs/O = 4 lone pairs)
  • Four P-OH oxygen atoms: Each forms two single bonds (one to P, one to H). With 6 valence electrons, 2 are used in bonding, leaving 4 non-bonding electrons (2 lone pairs) on each. (4 O atoms * 2 lone pairs/O = 8 lone pairs)
  • Two P-O-O oxygen atoms: Each forms two single bonds (one to P, one to O). With 6 valence electrons, 2 are used in bonding, leaving 4 non-bonding electrons (2 lone pairs) on each. (2 O atoms * 2 lone pairs/O = 4 lone pairs)

Total lone pairs (a) = 4 + 8 + 4 = 16. So, a = 16.

4. Calculate b (Total number of sp³ hybridised atoms): Hybridization is determined by the steric number (number of sigma bonds + number of lone pairs).

• Phosphorus (P) atoms: Each P atom forms 4 sigma bonds (1 from P=O, 3 from P-O single bonds) and has 0 lone pairs. Steric number = 4 + 0 = 4. Hence, each P atom is sp³ hybridized. (2 P atoms * 1 sp³ /P = 2 sp³ atoms)
• Oxygen (O) atoms:
  • Two P=O oxygen atoms: Each forms 1 sigma bond and has 2 lone pairs. Steric number = 1 + 2 = 3. Hence, sp² hybridized. (Not sp³)
  • Four P-OH oxygen atoms: Each forms 2 sigma bonds and has 2 lone pairs. Steric number = 2 + 2 = 4. Hence, sp³ hybridized. (4 O atoms * 1 sp³ /O = 4 sp³ atoms)
  • Two P-O-O oxygen atoms: Each forms 2 sigma bonds and has 2 lone pairs. Steric number = 2 + 2 = 4. Hence, sp³ hybridized. (2 O atoms * 1 sp³ /O = 2 sp³ atoms)

Total sp³ hybridised atoms (b) = 2 (from P) + 4 (from P-OH O) + 2 (from P-O-O O) = 8. So, b = 8.

5. Calculate c (Total number of pπ-pπ bonds): In oxoacids of phosphorus, the π-bond in P=O is formed by the overlap between the 2p orbital of oxygen and the vacant 3d orbital of phosphorus. This is a pπ-dπ bond. There are no other atoms forming pπ-pπ bonds. So, c = 0.

6. Calculate d (Total number of pπ-dπ bonds): There are two P=O double bonds in the molecule. Each P=O double bond consists of one σ bond and one pπ-dπ bond. So, d = 2.

7. Calculate z: z = $\frac{x+d}{a+c+b}$ Substitute the calculated values: z = $\frac{13+2}{16+0+8}$ z = $\frac{15}{24}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3: z = $\frac{15 \div 3}{24 \div 3}$ z = $\frac{5}{8}$

The final answer is $\boxed{\frac{5}{8}}$.

Explanation of the solution: The molecule H₄P₂O₈ is peroxodiphosphoric acid, confirmed by the +6 oxidation state of P, indicating an O-O peroxide linkage.

1. σ bonds (x): Count all single bonds and the sigma component of double bonds. (2 P=O + 4 P-OH + 2 P-O + 1 O-O + 4 O-H = 13).
2. Lone pairs (a): P has no lone pairs. Each O atom (P=O, P-OH, P-O-O) has 2 lone pairs. (22 + 42 + 2*2 = 16).
3. sp³ atoms (b): Each P is sp³ (4 sigma bonds). P-OH and P-O-O oxygens are sp³ (2 sigma bonds, 2 lone pairs). P=O oxygens are sp². (2 P + 4 P-OH O + 2 P-O-O O = 8).
4. pπ-pπ bonds (c): None, as P=O involves pπ-dπ bonding.
5. pπ-dπ bonds (d): Two P=O bonds, each contributing one pπ-dπ bond.
6. Calculate z: Substitute values into the given formula: z = (13+2)/(16+0+8) = 15/24 = 5/8.