Coordination Compounds

To determine the IUPAC name for the complex [Pt(NH₃)₂Cl(NO₂)], we need to follow the rules for naming coordination compounds:

• Identify the ligands and their order: The ligands are listed alphabetically in the compound's name, ignoring any prefixes. Here, we have two ligands: Ammine (NH₃) and Chlorido (Cl) along with Nitrito, which is a bidentate ligand and specifically bound via the nitrogen atom in this context, hence Nitrito-N.
• Determine the oxidation state of the metal: Platinum (Pt) is the central metal in this complex. The ammonia (NH₃) ligands are neutral, chlorido (Cl) is -1, and nitrito-N (NO₂) is -1. The overall charge for the molecule is neutral, thus: Ptn+(2×0)+(1×-1)+(1×-1)=0, giving us n = +2 for platinum. Hence, it is platinum(II).
• Name the complex: Combine the ligand names with the metal in the following manner: "Diamminechloridonitrito-N-platinum(II)". Note the use of "Diammine" for two amine groups, "chlorido" for Cl, and "nitrito-N" indicating the nitrogen atom as the donor atom.

Thus, the correct IUPAC name is: Diamminechloridonitrito-N-platinum(II).

1. Lewis Acid: A Lewis acid is defined as a chemical species that can accept an electron pair. In coordination chemistry, the central atom or ion acts as a Lewis acid because it can accept electron pairs from ligands, which are Lewis bases that donate electron pairs.
2. Coordination Compounds: These compounds consist of a central atom/ion surrounded by molecules or ions called ligands. The central metal ion attracts these ligands to form a stable structure due to its ability to accept electron pairs.
3. Example: In a coordination compound like [Cu(NH₃)₄]²⁺, copper (Cu) is the central ion, and it acts as a Lewis acid by accepting electron pairs from ammonia (NH₃) molecules.

To determine which compound will be repelled in an external magnetic field, we need to assess the magnetic properties of each compound. Generally, compounds are diamagnetic or paramagnetic:

• Diamagnetic compounds have no unpaired electrons and are repelled by a magnetic field.
• Paramagnetic compounds have unpaired electrons and are attracted by a magnetic field.

Let's analyze the given options:

1. : The central atom is , which usually has unpaired electrons, making it paramagnetic.
2. : The central atom is , known for having unpaired electrons, making it paramagnetic.
3. : The central atom is . Cadmium typically does not have unpaired electrons in its common oxidation state, making it diamagnetic.
4. : The central atom is , which typically has unpaired electrons, making it paramagnetic.

Based on the above analysis, is the compound that will be repelled by an external magnetic field due to its diamagnetic nature.

The coordination number of a metal ion in a complex refers to the number of ligand donor atoms bonded to the metal. In the complex [Fe(C₂O₄)₃]^3-, the ligand is oxalate (C₂O₄^2-), which is a bidentate ligand. This means each oxalate ligand can form two bonds with the metal ion.

Let's calculate:

1. The complex contains three oxalate ligands: [Fe(C₂O₄)₃]^3-.
2. Each oxalate ligand coordinates with two donor sites.
3. Therefore, the total number of coordination sites is 3 ligands × 2 donor sites per ligand = 6.

Thus, the coordination number of the metal (Fe) in the complex is six.

To determine the correct order of increasing boiling points among n-Butane, Ethoxyethane (Diethyl ether), Pentanal, and Pentan-1-ol, we have to consider intermolecular forces which include Van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

• n-Butane: It is an alkane with weak Van der Waals forces and no significant dipole moments or hydrogen bonding, resulting in a low boiling point.
• Ethoxyethane (Diethyl ether): It has stronger Van der Waals forces than n-butane and can participate in dipole-dipole interactions because of its ether group, leading to a higher boiling point compared to n-butane.
• Pentanal: As an aldehyde, it possesses a polar carbonyl group, which means it exhibits stronger dipole-dipole interactions compared to ethoxyethane, thus having a higher boiling point than ethoxyethane.
• Pentan-1-ol: This compound has the ability to form hydrogen bonds due to the presence of a hydroxyl group (-OH), which drastically increases its boiling point compared to the others.

Thus, the order of increasing boiling points is: n-Butane, Ethoxyethane, Pentanal, Pentan-1-ol.