Question

In which of the following option/options, the order of arrangement does not agree with the variation of property indicated against it?
(a) BF\(_3\)>NF\(_3\)>NH\(_3\) (Dipole moment)
(b) HgCl\(_2\)>NH\(_4\)\(^+\)>SF\(_4\) (Bond angle)
(c) NH\(_3\)<H\(_2\)O<HF (Strength of intermolecular hydrogen bonding)
(d) H - I>H - Br>H – Cl (Bond length)

• A. a, b and c
• B. a only
• C. c and d only
• D. d only

Hint:

For dipole moment questions, always consider both bond polarity and molecular geometry. Symmetrical molecules (like BF\(_3\), CCl\(_4\), CO\(_2\)) have a net dipole moment of zero even if their bonds are polar.

Answer 1

The Correct Option is B

Step 1: Analyze each statement.
(a) Dipole moment: BF\(_3\)>NF\(_3\)>NH\(_3\)

• BF\(_3\): Symmetrical trigonal planar geometry. The individual B-F bond dipoles cancel each other out. The net dipole moment (\(\mu\)) is zero. \(\mu\) = 0 D.
  
• NF\(_3\): Trigonal pyramidal geometry. The dipole moment of the N lone pair opposes the resultant dipole moment of the three N-F bonds. This results in a small net dipole moment. \(\mu\) \(\approx\) 0.2 D.
  
• NH\(_3\): Trigonal pyramidal geometry. The dipole moment of the N lone pair is in the same direction as the resultant dipole moment of the three N-H bonds, leading to a large net dipole moment. \(\mu\) \(\approx\) 1.47 D.
  

The correct order of dipole moments is NH\(_3\)>NF\(_3\)>BF\(_3\). The given order is incorrect.
(b) Bond angle: HgCl\(_2\)>NH\(_4\)\(^+\)>SF\(_4\)

• HgCl\(_2\): Linear molecule (sp hybridization). Bond angle = 180\(^\circ\).
  
• NH\(_4\)\(^+\): Tetrahedral ion (sp\(^3\) hybridization) with no lone pairs. Bond angle = 109.5\(^\circ\).
  
• SF\(_4\): See-saw shape (sp\(^3\)d hybridization) with one lone pair. The equatorial bond angle is \(\approx\) 102\(^\circ\) and the axial angle is \(\approx\) 173\(^\circ\). The smaller angle is typically considered for comparison.
  

The order 180\(^\circ\)>109.5\(^\circ\)>102\(^\circ\) is correct.
(c) Strength of intermolecular hydrogen bonding: NH\(_3\)<H\(_2\)O<HF
The strength of an individual hydrogen bond is determined by the electronegativity difference between the H atom and the atom it's bonded to (F, O, or N). Since electronegativity follows the order F>O>N, the strength of an individual H-bond is HF>H\(_2\)O>NH\(_3\). The given order is the reverse of this. However, if the question means the extent of H-bonding, water is highest due to its ability to form 4 H-bonds per molecule. Let's assume it refers to individual H-bond strength. In that case, the order should be reversed. But if it's overall intermolecular force strength, order is H2O>HF>NH3. Let's assume the question refers to the strength of the H-X bond polarity which dictates H-bond strength. In that case the order is correct. Let's assume this is considered correct.
Wait, the order given is NH\(_3\)<H\(_2\)O<HF. This order reflects the increasing electronegativity of N, O, F and thus the increasing strength of a single hydrogen bond. This statement is correct under that interpretation.
(d) Bond length: H-I>H-Br>H-Cl
Bond length increases with the size of the atoms. The atomic radius of halogens increases down the group: I>Br>Cl. Therefore, the H-X bond length also increases in the same order. The given order is correct.
Step 2: Conclusion.
Only the order in statement (a) is definitively incorrect. The orders in (b), (c), and (d) are correct based on standard chemical principles.
Step 3: Final Answer:
The option with the incorrect order is (a).