To determine which pairs of ions are colorless, we need to analyze the electronic configurations of the ions in each pair. The key points to remember are:
1. Transition metal ions with incompletely filled d orbitals are typically colored.
2. Transition metal ions with completely filled d orbitals or vacant d orbitals are colorless.
Let's analyze each pair step by step:
### Step 1: Analyze the first pair (Ti³⁺ and Cu²⁺)
- **Ti³⁺ (Titanium ion)**:
- Atomic number of Titanium (Ti) = 22
- Electronic configuration: [Ar] 3d² 4s²
- For Ti³⁺, we remove 3 electrons: [Ar] 3d¹ (1 electron in d orbital, incompletely filled)
- **Cu²⁺ (Copper ion)**:
- Atomic number of Copper (Cu) = 29
- Electronic configuration: [Ar] 3d¹⁰ 4s¹
- For Cu²⁺, we remove 2 electrons: [Ar] 3d⁹ (9 electrons in d orbital, incompletely filled)
**Conclusion**: Both Ti³⁺ and Cu²⁺ are colored due to incompletely filled d orbitals.
### Step 2: Analyze the second pair (Sc³⁺ and Zn²⁺)
- **Sc³⁺ (Scandium ion)**:
- Atomic number of Scandium (Sc) = 21
- Electronic configuration: [Ar] 3d¹ 4s²
- For Sc³⁺, we remove 3 electrons: [Ar] (no electrons in d orbital, vacant d orbital)
- **Zn²⁺ (Zinc ion)**:
- Atomic number of Zinc (Zn) = 30
- Electronic configuration: [Ar] 3d¹⁰ 4s²
- For Zn²⁺, we remove 2 electrons: [Ar] 3d¹⁰ (10 electrons in d orbital, completely filled)
**Conclusion**: Both Sc³⁺ and Zn²⁺ are colorless due to a vacant d orbital and a completely filled d orbital.
### Step 3: Analyze the third pair (Co²⁺ and Fe³⁺)
- **Co²⁺ (Cobalt ion)**:
- Atomic number of Cobalt (Co) = 27
- Electronic configuration: [Ar] 3d⁷ 4s²
- For Co²⁺, we remove 2 electrons: [Ar] 3d⁷ (7 electrons in d orbital, incompletely filled)
- **Fe³⁺ (Iron ion)**:
- Atomic number of Iron (Fe) = 26
- Electronic configuration: [Ar] 3d⁶ 4s²
- For Fe³⁺, we remove 3 electrons: [Ar] 3d⁵ (5 electrons in d orbital, incompletely filled)
**Conclusion**: Both Co²⁺ and Fe³⁺ are colored due to incompletely filled d orbitals.
### Step 4: Analyze the fourth pair (Ni²⁺ and V)
- **Ni²⁺ (Nickel ion)**:
- Atomic number of Nickel (Ni) = 28
- Electronic configuration: [Ar] 3d⁸ 4s²
- For Ni²⁺, we remove 2 electrons: [Ar] 3d⁸ (8 electrons in d orbital, incompletely filled)
- **V (Vanadium ion)**:
- Atomic number of Vanadium (V) = 23
- Electronic configuration: [Ar] 3d³ 4s²
- For V, we consider its common oxidation state, which is +5: [Ar] 3d⁰ (0 electrons in d orbital, vacant d orbital)
**Conclusion**: Ni²⁺ is colored due to an incompletely filled d orbital, while V (in +5 oxidation state) is colorless due to a vacant d orbital.
### Final Answer:
The colorless pair of ions is **Sc³⁺ and Zn²⁺**.
