(i) Ni²⁺ (3d⁸) - has two unpaired electrons, promotion of d orbital by absorption of energy possible - shows colour.
(ii) Zn²⁺ (3d¹⁰) - has no unpaired electron, no promotion of d orbitals possible - shows white colour.

Transition elements have strong tendency to form:
complexes because of two reasons. They are
(i) Small size and high positive charge density.
(ii) Presence of cant (n-1) d orbitals which are of appropriate energy to accept lone pair and unshared pair of electrons from the ligands for bonding with them.

CrO < Cr₂O₃ < CrO₃
∴ Higher the oxidation state, more will be the acidic character.

Fe³⁺ has the electronic configuration
[Ar] 4s⁰ 3d⁵
It is 5 unpaired electron
μ = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5+2)}\) = \(\sqrt{35}\) = 5.9 BM

Transition metal ions are smaller in case of size and have higher ionic charge, therefore they have high enthalpy of hydration.

Ce⁴⁺ gains electrons to form Ce³⁺ ion therefore it is considered a good oxidising agent

Cr(24): [Ar] 3d⁵4s¹

+3

This is because the last electron enters into the (n-2) f-subshell.

5f^1-14 6d^0-1 7s²

Cu, Ag and Au which have been used in making of coins in ancient times are called coinage metals

Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ ⟶ 2Cr³⁺ + 6Fe³⁺ + 7H₂O

[Rn]⁸⁸ 5⁺¹⁴ 6d⁰7S²

(i) The compounds of Cu²⁺ are coloured as it has one free electron its valence shell which absorb I radiation of visible region and get excited to emit its complementary colour.
(ii) Zn has no free electron it has fully filled shells. Due to extra stable orbitals electron can't be excited by radiations of visible light, hence its compounds are colourless.

In Gd⁺³ there are 64 electrons. Hence electronic configuration will be [Xe]4f⁷ 5d¹ 6s². Hence no electrons are there in outer d - orbital. Due to this it is colourless.