Chemistry · Chemical Bonding · NEET
After it forms, YES - it is exactly like any other covalent bond. Two electrons are shared between two atoms. The ONLY difference is where the electrons came from at the start. In a normal covalent bond each atom gives one electron. In a coordinate bond ONE atom gives both electrons. Once formed, you cannot tell them apart. This matters for NEET because questions test the FORMATION, not the final strength.
Think about who pays. Normal covalent bond: atom A gives 1 electron, atom B gives 1 electron, total 2 shared. Coordinate (dative) bond: atom A gives BOTH electrons (a full lone pair), atom B gives none but has an empty orbital to receive them. So a coordinate bond needs a donor with a lone pair AND an acceptor with an empty orbital. That is the rule to remember.
In ammonia (NH3) nitrogen has 3 normal N-H bonds and ONE lone pair. When an H+ ion (a bare proton with an empty orbital and NO electrons) comes near, nitrogen donates its lone pair to form the 4th N-H bond. That 4th bond is the coordinate bond. But here is the trap: all four N-H bonds become IDENTICAL after forming. You cannot point to one and say 'this is the coordinate one.' NEET loves this fact.
The atom that already HAS a lone pair and is more electron-rich is the donor. It donates INTO an atom that has an empty orbital (electron-deficient). Examples: In NH3 to BF3, nitrogen (lone pair) donates to boron (empty orbital, only 6 electrons). In O3, the central O donates a lone pair to a terminal O. The donor becomes slightly positive; the acceptor becomes slightly negative.
No, it is basically the same strength once formed, because it IS a covalent bond. Do not fall for the idea that 'dative = weaker' or 'dative = stronger.' The word 'dative' only describes how the bond was MADE (both electrons from one atom). This is why the arrow notation is used only to show origin, not to show extra or less strength.
Look for these signs: (1) A central atom that started with fewer electrons than its octet and got 'topped up' (like B in BF3 accepting from NH3). (2) A species where an atom clearly has more bonds than its normal valency (like N making 4 bonds in NH4+, or O making 3 bonds in H3O+ and O3). (3) Metal complexes where water or NH3 ligands donate lone pairs to a metal ion, such as [Al(H2O)6]3+. In each case one atom supplied the whole shared pair.
For the ozone (O3) Lewis structure with a central O atom (1) doubly bonded to one terminal O atom (2) and singly bonded to the other terminal O atom (3), the correct formal charges on the oxygen atoms numbered 2, 1 and 3 respectively are:
Aluminium chloride in acidified aqueous solution forms a complex 'A', in which the hybridisation state of Al is 'B'. What are 'A' and 'B', respectively?
Which of the following statements is NOT correct about diborane (B2H6)?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The arrow points FROM the donor atom TO the acceptor atom. For example, in NH3 to BF3 you draw H3N to BF3, with the arrow starting at nitrogen (the lone-pair donor) and pointing to boron (the empty-orbital acceptor). The arrow only shows the direction the electron pair came from, not extra strength.
Yes. A coordinate bond always needs a donor with a lone pair AND an acceptor with an empty orbital to receive that pair. If the acceptor has no empty orbital, the lone pair has nowhere to go and no bond forms. This is why electron-deficient species like BF3, H+, and metal ions are common acceptors.
Yes. Water (H2O) has two lone pairs on oxygen. When it accepts an H+ (which has an empty orbital and no electrons), oxygen donates one lone pair to form a third O-H bond. That third bond is a coordinate bond, and all three O-H bonds then become identical.
Learn these standard examples: NH4+ (N donates to H+), H3O+ (O donates to H+), O3 / ozone (central O donates to terminal O), NH3-BF3 adduct (N donates to B), CO (has a dative bond), and metal complexes like [Al(H2O)6]3+ or [Cu(NH3)4]2+ where ligands donate lone pairs to the metal.
It links three high-weightage areas: octet-rule exceptions (electron-deficient acceptors), formal charge (donor gets +, acceptor gets -), and coordination compounds (metal-ligand bonds are all coordinate bonds). Recognising a coordinate bond quickly helps you solve formal-charge and hybridisation questions faster.
