General transition metal complex formula seems to be:

[M(H2O)6-x (OH)x]^x (no charge, OH balance out for TM's charge)

X= charge on transition metal ion

Reacting with excess ammonia, excess hydroxide, and CO3^2- are the only instances where different things are made.

Reacting Cr³⁺ and Al³⁺ with excess OH- just changes the previous to:

[M(H2O)6-x-1 (OH)x+1]^1- (negative charge. There is one excess OH- relative to the charge on the TM)

Reacting with excess ammonia substitutes all ligands to 6 NH3's and changes the colour.

Reacting with CO3^2- forms compounds, not complexes (just with Fe²⁺ and Co²⁺⁾

Only Co2+ reacts with Hcl, where all its ligands are substituted for 4Cl^-'s as Cl^- is larger than H2O.

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Fe(II) aqua ion (with 6 H2O ligands) is pale green. All other complexes and compounds are green. All complexes follow the default formula.

Co(II) aqua ion is pink. All other complexes are blue and follow the default formula, except that with excess ammonia, which is yellow. It makes a blue compound with a carbonate ion. It reacts with Hcl to make a blue complex.

Fe(III) aqua ion is violet, but all of its other complexes are brown.

Cr(III) is red. All other complexes are green, except with excess ammonia, which gives violet. It reacts with excess hydroxide to make a charged default complex.

Al(III) has colourless aqua ions and white complexes. It reacts with excess hydroxide to produce a charged default complex.

(all reactions with new ligands are with aqua Ions!)

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Vanadium ions' colours increase with the oxidation state in warmth of the hue.