What is the number of pairs of ions which are coloured in aqueous solutions ?
(i) $T i^{3 +}, V^{3 +}$ (ii) $C u^{+}, S c^{3 +}$ (iii) $F e^{2 +}, F e^{3 +}$ (iv) $C o^{2} +, N i^{2 +}$ (v) $Z n^{2 +}, A g^{+}$

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Solution

The correct option is A 3
Color of transition elements.
${T i}^{+ 3} ⟶$ violet, $V^{+ 3} ⟶$ green}-1
${F e}^{+ 2} ⟶$ pale green, ${F e}^{+ 3} ⟶$ orange}-2
${C u}^{+} ⟶$ No color
${S c}^{+ 3} ⟶$ No color
${C O}^{+ 2} ⟶$ Red, ${N i}^{+ 2} ⟶$ greenish}-3
${A g}^{+}, {Z n}^{+ 2} ⟶$ no color
$⟶ 3$ pairs show color
Transition metals form coloured compounds and complexes. These colors can vary depending on charge of metal ion, and the number of and type of group of atoms (called ligands) attached to metal ion.
Electrons are arranged around nucleus of metal atom in orbitals. Transition metals, unlike other metals have partially filled d orbitals, which can hold upto 10 electrons when ligand is present some d orbitals become higher in energy than before, some lower. Electrons can thus move between higher and lower orbitals by absorbing photon of light. This absorption of light affects perceive color of complex. The wavelength of light absorbed is affected by size of energy gap between d orbitals.

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Similar questions

T i^{3 +}, V^{3 +} (i i) C u^{+}, S c^{3 +} (i i i) F e^{2 +}, F e^{3 +} (i v) C o^{2 +}, N i^{2 +} (v) Z n^{2 +}, A g^{+}

$(i) T i^{3 +}, V^{3 +} (i i) C u^{+}, S c^{3 +} (i i i) F e^{2 +}, F e^{3 +} (i v) C o^{2 +}, N i^{2 +} (v) Z n^{2 +}, A g^{+}$

{F e}^{3 +}

{N i}^{2 +}

{A l}^{3 +}

E^{\circ}

M n^{3 +} / M n^{2 +}

F e^{3 +} / F e^{2 +} .

S c^{3 +}

T i^{3 +}

S c^{3 +}

T i^{3 +}

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