Problem 30
Question
Which of the following ions does not exist as octahedral aqua ions? (a) \(\mathrm{Al}\) (b) Ga (c) \(\mathrm{Tl}\) (d) \(\mathrm{B}\)
Step-by-Step Solution
Verified Answer
Boron (d) does not exist as an octahedral aqua ion.
1Step 1: Understanding Octahedral Aqua Ions
Octahedral aqua ions are complexes where a central metal ion is surrounded by six water (aqua) ligands in an octahedral geometry. Generally, they form with metal ions capable of forming coordination complexes by accepting pairs of electrons from ligand donor atoms.
2Step 2: Analyze Elements Given
Consider the elements Aluminum (Al), Gallium (Ga), Thallium (Tl), and Boron (B). We need to identify which elements can typically form complexes with six water molecules and which cannot. Metal ions that do not have the ability to expand their coordination number to six are unlikely to form such complexes.
3Step 3: Evaluate Boron's Chemistry
Boron generally forms covalent rather than ionic bonds due to its small size and high ionization energy. It does not easily form complexes like typical transition or post-transition metals. Furthermore, boron typically forms compounds with a maximum coordination number of 4, not 6.
4Step 4: Conclusion Based on Chemistry
Given that Boron (B) does not expand its coordination number to 6 and forms few complexes compared to the other elements, it is unlikely to exist as an octahedral aqua ion with water ligands.
Key Concepts
Octahedral GeometryCoordination ComplexesLigand Chemistry
Octahedral Geometry
In coordination chemistry, octahedral geometry is one of the most common geometric arrangements. It describes the shape of coordination complexes where the central atom is surrounded by six atoms, typically ligands, that form the vertices of an octahedron. This geometry is critical for understanding how metal ions interact with ligands.
The arrangement results in a symmetrical, three-dimensional structure with 90-degree bond angles between ligands. An essential feature of octahedral geometry is its versatility and ability to accommodate various ligands through different bonding interactions.
Coordination number, which in this case is 6, indicates the number of ligand bonding interactions with the central metal ion. This geometry is fundamental for many compounds in chemistry, especially those involving transition metals, as they often have the right size and electron configuration to allow for six-fold coordination.
Typically, octahedral geometry results in a stable configuration due to its symmetrical shape, improving the stability of the resulting coordination complex.
Coordination Complexes
Coordination complexes are structured chemistries wherein metal ions are centrally located and surrounded by molecules or ions called ligands. These complexes are an integral part of coordination chemistry, which studies the formation, properties, and reactions of these compounds.
- Central Metal Ion: This is the atom around which the complex is formed. It can be a transition metal or a post-transition metal capable of accepting electron pairs.
- Ligands: These are the molecules or ions that donate electron pairs to the central metal ion, forming a bond. They can vary from simple ions like chloride to larger molecules such as ethylenediamine.
Ligand Chemistry
Ligands play a crucial role in coordination chemistry as they are the atoms, ions, or molecules that bind to the central metal ion within a coordination complex. The nature and strength of these bindings influence the properties and functions of the resulting coordination complex.
- Types of Ligands: Ligands can be monodentate, bidentate, or polydentate depending on the number of donor atoms available for coordination to the metal.
- Electron Pair Donation: Ligands donate electron pairs to the central metal ion, forming coordination bonds crucial for the stability of the complex.
- Influence on Complexes: Different ligands can greatly affect the color, reactivity, and overall stability of metal complexes. They can be neutral molecules like water or ammonia, or charged anions like cyanide or nitrite.
Other exercises in this chapter
Problem 28
The electronic configuration of group III elements is (a) \(\mathrm{ns}^{1} \mathrm{np}^{2}\) (b) \(\mathrm{ns}^{1} \mathrm{np}^{3}\) (c) \(\mathrm{ns}^{2} \mat
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Which one of the following is not a Lewis acid? (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{FeCl}_{3}\) (c) \(\mathrm{BF}_{\underline{3}}\) (d) \(\mathr
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Ga is below Al in the periodic table, but atomic radius of Ga is less than Al. It is because of (a) lanthanoid contraction (b) greater screening effect (c) iner
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