Question

Ammonia forms a complex ion ${[Cu{(N{H_3})_4}]^{2 + }}$with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it?
A. Copper hydroxide is an amphoteric substance.
B. In acidic solutions, hydration protects copper ions.
C. In acidic solutions, protons coordinate with ammonia molecules making $N{H_4}^ +$ions and $N{H_3}$ molecules are no longer available for coordination to metal ions.
D. In alkaline solutions, insoluble$Cu{(OH)_2}$is precipitated out which is soluble in excess of any alkali.

Answer 1

Consider the reaction between the ligand ammonia and the protons from acidic medium. For the formation of complex, ligands should be in free (un-combined) state.

Complete step-by-step answer: In acidic solution proton coordinates with ammonia molecules to make $N{H_4}^ +$ions and $N{H_3}$molecules are not available for complex formation. Whereas ammonia molecules are available in alkaline solutions.
Ammonia$N{H_3}$ is a ligand which donates its pair of electrons to $C{u^{2 + }}$ions to make ${[Cu{(N{H_3})_4}]^{2 + }}$complex.
This can be done in basic medium. In acidic medium, the pair of electrons in ammonia is given to proton to make ammonium ion. Thus, ammonia ligand is not free and cannot make a complex with $C{u^{2 + }}$ions.
$C{u^{2 + }} + 4N{H_3}\xrightarrow{{{\text{alkaline medium}}}}{[Cu{(N{H_3})_4}]^{2 + }}$
$C{u^{2 + }} + 4N{H_3} + 4{H^ + }\xrightarrow{{{\text{acidic medium}}}}C{u^{2 + }} + 4N{H_4}^ +$

Hence, correct option is C.

Additional information:
The complex ion ${[Cu{(N{H_3})_4}]^{2 + }}$is a square planar and also its nature is paramagnetic.
Atomic number of Cu is 29. Electronic configuration is$4{s^1}{d^{10}}$. Therefore, having hybridization of $ds{p^2}$

Note: 1. $N{H_4}^ +$ion behaves as acidic in aqueous solution according to Bronsted Lowry Acid-Base Theory.
2. In acidic solution,$N{H_3}$makes a bond with ${H^ + }$ to give $N{H_4}^ +$ion which does not have a lone pair on N atoms. Hence ammonia in the given solution cannot act as a ligand.