Solveeit Logo
Login

Question

Question: Which of the bond length order data is INCORRECT -...

Which of the bond length order data is INCORRECT -

A

P-CI > P-F in PCl3F2PCl_3F_2

B

S-F(axial) > S-F(eq) in SF6SF_6

C

S-F (axial) > S-F(eq) in SF4SF_4

D

NO2NO_2 < NO3NO_3 (N-O)

Answer

B

Explanation

Solution

The question asks to identify the incorrect bond length order data among the given options. Let's analyze each option:

(A) P-Cl > P-F in PCl₃F₂:

  • PCl3F2PCl_3F_2 has a trigonal bipyramidal (TBP) geometry.
  • According to Bent's Rule, more electronegative atoms prefer to occupy axial positions (which have less s-character), and less electronegative atoms prefer equatorial positions (which have more s-character).
  • Fluorine is more electronegative than chlorine, so fluorine atoms occupy the axial positions, and chlorine atoms occupy the equatorial positions.
  • While axial bonds in TBP are generally longer than equatorial bonds (when comparing the same type of bond, e.g., P-Cl axial vs P-Cl equatorial), here we are comparing P-Cl (equatorial) with P-F (axial).
  • The inherent bond length of P-Cl is significantly longer than P-F due to the larger atomic size of chlorine and weaker orbital overlap.
    • Typical P-Cl bond length ≈ 204 pm.
    • Typical P-F bond length ≈ 157 pm.
  • In PCl3F2PCl_3F_2, the experimental bond lengths are: P-Cl(equatorial) ≈ 204.3 pm and P-F(axial) ≈ 159.3 pm.
  • Therefore, P-Cl (equatorial) > P-F (axial) is correct.

(B) S-F(axial) > S-F(eq) in SF₆:

  • SF6SF_6 has an ideal octahedral geometry.
  • In an ideal octahedron, all six S-F bonds are equivalent in length due to the high symmetry of the molecule. There are no distinct axial and equatorial positions in terms of bond length.
  • Therefore, the statement S-F(axial) > S-F(eq) is incorrect. All S-F bond lengths in SF6SF_6 are identical (approximately 156.4 pm).

(C) S-F (axial) > S-F(eq) in SF₄:

  • SF4SF_4 has a seesaw geometry, which is derived from a trigonal bipyramidal arrangement with one lone pair occupying an equatorial position.
  • The lone pair occupies an equatorial position to minimize lone pair-bond pair repulsions (LP-BP > BP-BP).
  • The presence of the lone pair in an equatorial position significantly influences the bond lengths. The axial S-F bonds experience greater repulsion from the lone pair and are pushed away, making them longer than the equatorial S-F bonds.
  • Experimental values: S-F(axial) ≈ 164.6 pm and S-F(equatorial) ≈ 154.2 pm.
  • Therefore, S-F(axial) > S-F(eq) is correct.

(D) NO₂ < NO₃ (N-O):

  • This compares the average N-O bond length in NO2NO_2 (nitrogen dioxide) and NO3⁻NO_3^⁻ (nitrate ion).
  • NO2NO_2: The N-O bonds in NO2NO_2 are equivalent due to resonance. The bond order is approximately 1.5 (between a single and a double bond). The experimental N-O bond length is approximately 119.7 pm.
  • NO3⁻NO_3^⁻: The N-O bonds in the nitrate ion are equivalent due to resonance. Each N-O bond is a hybrid of one double bond and two single bonds distributed over three positions, resulting in a bond order of (2+1+1)/3 = 4/3 ≈ 1.33. The experimental N-O bond length is approximately 121.8 pm.
  • A higher bond order corresponds to a shorter bond length. Since the bond order of N-O in NO2NO_2 (1.5) is greater than in NO3⁻NO_3^⁻ (1.33), the N-O bond length in NO2NO_2 should be shorter than in NO3⁻NO_3^⁻.
  • Therefore, NO2NO_2 < NO3NO_3 (N-O) is correct.

Based on the analysis, option (B) presents an incorrect bond length order.