`[Cu(CN)_(4)^(2-)]` is a colourless complex. Explain. 

[Ti(H_(2)O)_(6)]^(3+) is coloured while [Sc(H_(2)O)_(6)]^(3+) is colourless. Explain.

K_(4)[Fe(CN)_(6)] is a

A solution of [Ni(H_(2)O)_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-) is colourless. Why?

[Fe(H_(2)O)_(6)]^(3+) is strongly paramagnetic whereas [Fe(CN)_(6)]^(3-) is weakly paramagnetic Explain.

CaCl is colourless, while CuSO_(4).5H_(2)O is coloured. Explain?

[Cu(NH_(3))_(4)]^(2+) is paramagnetic and square planar complex. Justify.

Calculate the overall complex dissociation equilibrium constant for the Cu(NH_(3))_(4)^(2+) ion, given that β4 for this complex is 2.1 xx 1013 .

Assertion: [Cu(en)_2]^(2+) is more stable than [Cu(NH_3)_4]^(2+) Reason: Both these complexes have a square planar shape.

Assertion: [Cu(CN)_4]^(2-) is more stable than [Cu(H_2O)_4]^(2+) Reason: On heating [Cu(H_2O)_4]^(2+) with NH_3, [Cu(NH_3)_4]^(2+) is formed.