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The number of anions having perfect Oh geometry among
[SbCI6]3-, [SeCI6]2-,[TeBr6]2-,[BrF6]-,[IF6]-,[CIF6]-,[SeF6]2- is/are__________
[SbCI6]3-, [SeCI6]2-,[TeBr6]2-,[BrF6]-,[IF6]-,[CIF6]-,[SeF6]2- is/are__________
Correct answer is '5'. Can you explain this answer?
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The number of anions having perfect Oh geometry among[SbCI6]3-, [SeCI6...
Lesser the difference between size of central and surrounding atom lesser will be stereochemical activity of lone pair. Hence, perfect geometry.
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The number of anions having perfect Oh geometry among[SbCI6]3-, [SeCI6...
Number of Anions with Perfect Oh Geometry
In order to determine the number of anions with perfect Oh geometry, we need to analyze the molecular geometries of the given compounds. The Oh geometry refers to a shape with octahedral symmetry, where the central atom is surrounded by six ligands located at the vertices of an octahedron.
Let's examine each compound individually:
1. [SbCl6]3-
- This compound consists of a central antimony (Sb) atom bonded to six chloride (Cl) ligands.
- The presence of six ligands arranged symmetrically around the central atom corresponds to an octahedral geometry.
- Therefore, [SbCl6]3- has perfect Oh geometry.
2. [SeCl6]2-
- This compound contains a central selenium (Se) atom bonded to six chloride (Cl) ligands.
- Similar to [SbCl6]3-, [SeCl6]2- also exhibits octahedral geometry due to the arrangement of six ligands around the central atom.
- Hence, [SeCl6]2- has perfect Oh geometry.
3. [TeBr6]2-
- [TeBr6]2- consists of a central tellurium (Te) atom bonded to six bromine (Br) ligands.
- Like the previous compounds, [TeBr6]2- possesses octahedral geometry because of the symmetry of ligands around the central atom.
- Therefore, [TeBr6]2- has perfect Oh geometry.
4. [BrF6]-
- This anion comprises a central bromine (Br) atom bonded to six fluorine (F) ligands.
- The arrangement of the ligands does not meet the criteria for octahedral symmetry.
- Hence, [BrF6]- does not have perfect Oh geometry.
5. [IF6]-
- [IF6]- consists of a central iodine (I) atom bonded to six fluorine (F) ligands.
- Similar to [BrF6]-, [IF6]- does not exhibit octahedral geometry as the ligands are not arranged symmetrically.
- Thus, [IF6]- does not have perfect Oh geometry.
6. [ClF6]-
- [ClF6]- contains a central chlorine (Cl) atom bonded to six fluorine (F) ligands.
- The ligand arrangement in [ClF6]- does not fulfill the requirements for octahedral symmetry.
- Therefore, [ClF6]- does not possess perfect Oh geometry.
7. [SeF6]2-
- This compound consists of a central selenium (Se) atom bonded to six fluorine (F) ligands.
- The ligands in [SeF6]2- are arranged symmetrically around the central atom, resulting in octahedral geometry.
- Thus, [SeF6]2- has perfect Oh geometry.
Summary:
Out of the given compounds, the following anions have perfect Oh geometry:
- [SbCl6]3-
- [SeCl6]2-
- [TeBr6]2-
- [SeF6]2-
Therefore, the correct answer is '5'.
In order to determine the number of anions with perfect Oh geometry, we need to analyze the molecular geometries of the given compounds. The Oh geometry refers to a shape with octahedral symmetry, where the central atom is surrounded by six ligands located at the vertices of an octahedron.
Let's examine each compound individually:
1. [SbCl6]3-
- This compound consists of a central antimony (Sb) atom bonded to six chloride (Cl) ligands.
- The presence of six ligands arranged symmetrically around the central atom corresponds to an octahedral geometry.
- Therefore, [SbCl6]3- has perfect Oh geometry.
2. [SeCl6]2-
- This compound contains a central selenium (Se) atom bonded to six chloride (Cl) ligands.
- Similar to [SbCl6]3-, [SeCl6]2- also exhibits octahedral geometry due to the arrangement of six ligands around the central atom.
- Hence, [SeCl6]2- has perfect Oh geometry.
3. [TeBr6]2-
- [TeBr6]2- consists of a central tellurium (Te) atom bonded to six bromine (Br) ligands.
- Like the previous compounds, [TeBr6]2- possesses octahedral geometry because of the symmetry of ligands around the central atom.
- Therefore, [TeBr6]2- has perfect Oh geometry.
4. [BrF6]-
- This anion comprises a central bromine (Br) atom bonded to six fluorine (F) ligands.
- The arrangement of the ligands does not meet the criteria for octahedral symmetry.
- Hence, [BrF6]- does not have perfect Oh geometry.
5. [IF6]-
- [IF6]- consists of a central iodine (I) atom bonded to six fluorine (F) ligands.
- Similar to [BrF6]-, [IF6]- does not exhibit octahedral geometry as the ligands are not arranged symmetrically.
- Thus, [IF6]- does not have perfect Oh geometry.
6. [ClF6]-
- [ClF6]- contains a central chlorine (Cl) atom bonded to six fluorine (F) ligands.
- The ligand arrangement in [ClF6]- does not fulfill the requirements for octahedral symmetry.
- Therefore, [ClF6]- does not possess perfect Oh geometry.
7. [SeF6]2-
- This compound consists of a central selenium (Se) atom bonded to six fluorine (F) ligands.
- The ligands in [SeF6]2- are arranged symmetrically around the central atom, resulting in octahedral geometry.
- Thus, [SeF6]2- has perfect Oh geometry.
Summary:
Out of the given compounds, the following anions have perfect Oh geometry:
- [SbCl6]3-
- [SeCl6]2-
- [TeBr6]2-
- [SeF6]2-
Therefore, the correct answer is '5'.
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The number of anions having perfect Oh geometry among[SbCI6]3-, [SeCI6]2-,[TeBr6]2-,[BrF6]-,[IF6]-,[CIF6]-,[SeF6]2- is/are__________Correct answer is '5'. Can you explain this answer?
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