Q.vi. The conjugate base of $\mathrm{[Zn(H_2O)_4]^{2+}}$ is a. $\mathrm{[Zn(H_2O)_4]^{2-}NH_3}$
b. $\mathrm{[Zn(H_2O)_3]^{2-}}$
c. $\mathrm{[Zn(H_2O)_3OH]^{+}}$
d. $\mathrm{[Zn(H_2O)H]^{3+}}$

[!NOTE]
Options a and b are transcribed exactly as the textbook prints them -- including the $2-$ charge on their brackets (the book's circled minus, confirmed on a zoomed read of the printed page). The correct answer, option c, is unaffected.

Step 1. By the Bronsted-Lowry theory (section 3.3.2), the conjugate base of an acid is what remains after that acid donates (loses) exactly one proton, H+.

Step 2. The given species, $[Zn(H_2O)_4]^{2+}$, can act as a Bronsted acid because one of its coordinated water molecules can donate a proton -- this is a standard behaviour of aquated metal cations, which are weakly acidic exactly because the metal's positive charge polarizes the O-H bonds of its coordinated water.

Step 3. Removing one H+ from one of the four coordinated H2O ligands converts that ligand into a coordinated OH- (hydroxide) ligand, while the other three water ligands are unaffected: $[Zn(H_2O)_4]^{2+}\rightarrow[Zn(H_2O)_3(OH)]^++H^+$.

Step 4. The overall charge also drops by one (losing a proton, i.e. a +1 charge, from a 2+ species leaves a 1+ species), consistent with $[Zn(H_2O)_3OH]^+$, which is option (c). …

Apply the conjugate-base definition (lose exactly one H+) to the given complex ion, tracking both the ligan …

• Removing an entire H2O ligand instead of just one H+ from it -- these are chemically different processes (loss of a full ligand vs. deprotonation of one ligand). …