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Question: Find the number of reaction(s) in which coordinate bond is/are formed in product side: $BF_3 + F^- ...

Find the number of reaction(s) in which coordinate bond is/are formed in product side:

BF3+FBF4BF_3 + F^- \longrightarrow BF_4^- X

HF+SbF5HSbF6HF + SbF_5 \longrightarrow HSbF_6 X

PCl3+ClPCl4PCl_3 + Cl^- \longrightarrow PCl_4^-

NH3+H+NH4+NH_3 + H^+ \longrightarrow NH_4^+

H3O++H+H4O2+H_3O^+ + H^+ \longrightarrow H_4O_2^+ X

Answer

4

Explanation

Solution

A coordinate bond (or dative bond) is a type of covalent bond where both shared electrons are contributed by one atom (the donor) to another atom (the acceptor). This typically occurs in Lewis acid-base reactions.

Let's analyze each reaction:

  1. BF3+FBF4BF_3 + F^- \longrightarrow BF_4^-

    • BF3BF_3 (Boron trifluoride) is a Lewis acid because boron has an incomplete octet (6 valence electrons) and an empty p-orbital.
    • FF^- (Fluoride ion) is a Lewis base because it has lone pairs of electrons to donate.
    • In BF4BF_4^-, the fluoride ion donates a lone pair of electrons to the empty p-orbital of boron, forming a B-F bond. This is a coordinate bond.
    • Yes, a coordinate bond is formed.
  2. HF+SbF5HSbF6HF + SbF_5 \longrightarrow HSbF_6

    • This reaction involves the formation of a superacid, fluoroantimonic acid. The reaction can be viewed as HFHF acting as a source of FF^- and SbF5SbF_5 acting as a strong Lewis acid.
    • SbF5SbF_5 accepts a fluoride ion (FF^-) to form SbF6SbF_6^-. In this process, a lone pair from the fluoride ion is donated to an empty orbital of antimony.
    • The product HSbF6HSbF_6 is an ionic compound, (H+)(SbF6)(H^+)(SbF_6^-). The bond between Sb and the incoming F in SbF6SbF_6^- is a coordinate bond.
    • Yes, a coordinate bond is formed.
  3. PCl3+ClPCl4PCl_3 + Cl^- \longrightarrow PCl_4^-

    • PCl3PCl_3 (Phosphorus trichloride) has a lone pair on phosphorus and also empty d-orbitals, allowing it to act as a Lewis acid by expanding its octet.
    • ClCl^- (Chloride ion) is a Lewis base, donating a lone pair.
    • In PCl4PCl_4^-, the chloride ion donates a lone pair to an empty d-orbital of phosphorus, forming a P-Cl bond. This is a coordinate bond.
    • Yes, a coordinate bond is formed.
  4. NH3+H+NH4+NH_3 + H^+ \longrightarrow NH_4^+

    • NH3NH_3 (Ammonia) is a Lewis base because nitrogen has a lone pair of electrons.
    • H+H^+ (Proton) is a Lewis acid because it has an empty 1s orbital.
    • In NH4+NH_4^+ (Ammonium ion), the lone pair on the nitrogen atom is donated to the empty 1s orbital of the proton, forming an N-H bond. This is a coordinate bond.
    • Yes, a coordinate bond is formed.
  5. H3O++H+H4O2+H_3O^+ + H^+ \longrightarrow H_4O_2^+

    • H3O+H_3O^+ (Hydronium ion) has one lone pair on the oxygen atom. If it were to accept another H+H^+, it would form H4O2+H_4O^{2+}.
    • The species H4O2+H_4O^{2+} would have oxygen bonded to four hydrogen atoms with a formal charge of +2 on oxygen. This species is highly unstable and is not observed under normal chemical conditions.
    • The product given, H4O2+H_4O_2^+, is also chemically unusual and does not correspond to a stable, commonly recognized species formed from H3O+H_3O^+ and H+H^+.
    • Therefore, this reaction does not lead to the formation of a stable product via a coordinate bond.
    • No, a coordinate bond is not formed in a stable product.

Based on the analysis, a coordinate bond is formed in the product side for the first four reactions.

Explanation of the solution:

A coordinate bond forms when one atom donates both electrons for a shared pair.

  1. BF3+FBF4BF_3 + F^- \longrightarrow BF_4^-: FF^- donates a lone pair to electron-deficient B in BF3BF_3. Coordinate B-F bond forms.
  2. HF+SbF5HSbF6HF + SbF_5 \longrightarrow HSbF_6: FF^- from HFHF donates a lone pair to Lewis acidic SbF5SbF_5 to form SbF6SbF_6^-. Coordinate Sb-F bond forms.
  3. PCl3+ClPCl4PCl_3 + Cl^- \longrightarrow PCl_4^-: ClCl^- donates a lone pair to P in PCl3PCl_3 (using P's empty d-orbitals). Coordinate P-Cl bond forms.
  4. NH3+H+NH4+NH_3 + H^+ \longrightarrow NH_4^+: Lone pair on N in NH3NH_3 is donated to H+H^+. Coordinate N-H bond forms.
  5. H3O++H+H4O2+H_3O^+ + H^+ \longrightarrow H_4O_2^+: This reaction is chemically unsound. H3O+H_3O^+ accepting another H+H^+ would lead to highly unstable H4O2+H_4O^{2+}, not H4O2+H_4O_2^+. No stable coordinate bond is formed.

Thus, 4 reactions involve coordinate bond formation.