Home
Class 12
CHEMISTRY
Complex [ML(5)] can exhibit trigonal bip...

Complex `[ML_(5)]` can exhibit trigonal bipyramidal and square pyramidal geometry. Determine total number of `180^@, 90^@ & 120^@` L-M-L bond angles.

Text Solution

AI Generated Solution

The correct Answer is:
To determine the total number of `180^@`, `90^@`, and `120^@` L-M-L bond angles in the complex `[ML_(5)]`, which can exhibit both trigonal bipyramidal and square pyramidal geometries, we will analyze each geometry step by step. ### Step 1: Analyze Trigonal Bipyramidal Geometry 1. **Identify the Geometry**: In trigonal bipyramidal geometry, there are 5 ligands around the central metal atom. The arrangement consists of 3 ligands in a plane (equatorial) and 2 ligands above and below this plane (axial). 2. **Bond Angles**: - **120° Angles**: The equatorial ligands form bond angles of `120°` with each other. There are 3 pairs of equatorial ligands, contributing to 3 bond angles of `120°`. - **90° Angles**: Each axial ligand forms a `90°` angle with each of the equatorial ligands. Since there are 3 equatorial ligands, each axial ligand contributes `3` angles of `90°`. Since there are 2 axial ligands, the total number of `90°` angles is `3 (from one axial) + 3 (from the other axial) = 6`. - **180° Angles**: There is only one `180°` angle formed between the two axial ligands. Thus, there is `1` angle of `180°`. 3. **Total for Trigonal Bipyramidal**: - `120°`: 3 angles - `90°`: 6 angles - `180°`: 1 angle ### Step 2: Analyze Square Pyramidal Geometry 1. **Identify the Geometry**: In square pyramidal geometry, there is a square base formed by 4 ligands and one ligand at the apex. 2. **Bond Angles**: - **90° Angles**: All four ligands in the square base form `90°` angles with each other. This contributes `4` angles. Additionally, the apex ligand forms `90°` angles with each of the 4 base ligands, contributing another `4` angles. Thus, the total number of `90°` angles is `4 (base) + 4 (apex) = 8`. - **120° Angles**: There are no `120°` angles in square pyramidal geometry. Thus, the count is `0`. - **180° Angles**: There are two `180°` angles formed between the ligands across the square base. Thus, there are `2` angles of `180°`. 3. **Total for Square Pyramidal**: - `90°`: 8 angles - `120°`: 0 angles - `180°`: 2 angles ### Step 3: Combine Results Now, we sum the angles from both geometries: - **Trigonal Bipyramidal**: - `120°`: 3 - `90°`: 6 - `180°`: 1 - **Total**: 3 + 6 + 1 = 10 angles - **Square Pyramidal**: - `90°`: 8 - `120°`: 0 - `180°`: 2 - **Total**: 8 + 0 + 2 = 10 angles ### Final Calculation Adding the totals from both geometries gives us: - Total angles = 10 (from trigonal bipyramidal) + 10 (from square pyramidal) = **20 angles**. ### Conclusion The total number of `180°`, `90°`, and `120°` bond angles in the complex `[ML_(5)]` is **20**. ---
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise CHEMISTRY|146 Videos
  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise QUESTION|1 Videos

Similar Questions

Explore conceptually related problems

In a regular trigonal bipyramidal MX_(5) , the number of X - M - X bonds at 180^(@) is

Among the following statements:- I. PCl_(5) is trigonal bipyramidal wheras IF_(5) is square pyramidal. II. Bond enthalpy of O-H bond in water and ethanol is different. III. All carbon atoms have same hybridisation in carbon suboxide (C_(3)O_(2)) Find out the correct statements.

A coordination complex of type MX_(2)Y_(2) (M-metal ion: X, Y-monodentate lingads), can have either a tetrahedral or a square planar geometry. The maximum number of posible isomers in these two cases are respectively:

A coordination complex of type MX_(2)Y_(2) (M-metal ion: X, Y-monodentate lingads), can have either a tetrahedral or a square planar geometry. The maximum number of posible isomers in these two cases are respectively:

Consider the following covalent compounds in their solid state and find the value of expression (X+Y+Z) . N_(2)O_(5),Cl_(2)O_(6),PC l_(),I_(2)Cl_(6),XeF_(6),PB r_(5) Where X=total number of compounds in which central atom of cationic or anionic part is sp^(3) hybridized. Y=Total number of compounds having 90^(@) bond angle either in cationic or anionic part. Z=Total number of compounds having 109^(@)28' bond angle either in cationic or anionic part.

ICI_3 is an orange colored solid that dimerizes in solid state as I_2Cl_6 Based on VSEPR theory , number of 90 degree Cl-l-Cl bond angles is …….. In the dimeric species.

The concept of hybridisation has been introduced to explain the shapes of molecules. It involves the intermixing of two or more atomic orbitals belonging to same atom but in or more atomic orbitals belonging to same atom but in different sub-shells so as to intermix and redistribute energies to from equivalent orbitals called hybrid orbitals. Depending upon total number and nature of the orbitals involved, the hybridisation may be divided into sp (linear), sp^(2) (trigonal), sp^(3) (tetrahedral), sp^(3)d (trigonal bipyramidal), sp^(3)d^(3) (octahedral) and sp^(3)d^(3) (pentagonal bipyramidal) types. it may be noted that the orbitals of isolated atoms never hybridize and they do so at the time of bond formation. A hybrid orbital from s-and p-orbitals can contribute to

Consider the following compounds (1) H_(3)CF (2) H_(2)CF_(2) (3) CH_(4) (4) H_(3)C CF_(3) (5) CH_(3)CH_(2)+ (6) C_(2)H_(4) and calculate value of Y+X, (where X is the total number of compounds which have H-C-H bond angles equal to 109^(@)28' and Y is the total number of compounds which have H-C-H bond angles greater than 109^(@)28' and less than 120^(@)

Consider the following compounds (1) H_(3)CF (2) H_(2)CF_(2) (3) CH_(4) (4) H_(3)C CF_(3) (5) CH_(3)CH_(3) (6) C_(2)H_(4) and calculate value of Y+X, (where X is the total number of compounds which have H-C-H bond angles equal to 109^(@)28' and Y is the total number of compounds which have H-C-H bond angles greater than 109^(@)28' and less than 120^(@)

Total number of F-l-F bond angles which are 90^(@) present in lF_(7) is

JEE MAINS PREVIOUS YEAR ENGLISH-JEE MAIN-CHEMISTRY
  1. The increasing order of the atomic radii of the following element...

    Text Solution

    |

  2. How many atoms lie in the same plane in the major product (C) ? (Wher...

    Text Solution

    |

  3. Complex [ML(5)] can exhibit trigonal bipyramidal and square pyramidal ...

    Text Solution

    |

  4. At constant volume 4 mol of an ideal gas when heated form 300k to 50...

    Text Solution

    |

  5. NaClO(3) is used even is spacecrafts to produce O(2) . The daily con...

    Text Solution

    |

  6. Given: E(Sn^(2+)//Sn) ^0 = -0.14V , E(Pb^(2+)//Pb) ^0 = -0.13V. Determ...

    Text Solution

    |

  7. The electronic configuration of bivalent Europium and trivalent cerium...

    Text Solution

    |

  8. If the magnetic moment of a dioxygen species is 1.73 B.M, it may be:

    Text Solution

    |

  9. The major product (Y) in the following reactions is: CH(3)-overset(C...

    Text Solution

    |

  10. The major product Z obtained in the following reaction scheme is:

    Text Solution

    |

  11. The increasing order of basicity for the following intermediates is (f...

    Text Solution

    |

  12. Complex Cr(H2 O)6 Cln shows geometrical isomerism and also reacts with...

    Text Solution

    |

  13. According to the following diagram, A reduces BO(2) when the temperatu...

    Text Solution

    |

  14. A, B, C and D are four artificial sweetners. (i) A and D give positi...

    Text Solution

    |

  15. [PdFClBrI]^(2-) Number of Geometrical Isomers = n. For [Fe(CN)6]^(n-6)...

    Text Solution

    |

  16. Rate of reaction in absence of catalyst at 700 K is same as in presenc...

    Text Solution

    |

  17. Which of the following oxides are acidic, basic and amphoteric Respect...

    Text Solution

    |

  18. A substance 'X' having low melting point, does not conduct electricity...

    Text Solution

    |

  19. Identify (A) in the following reaction sequence :

    Text Solution

    |

  20. Ksp of PbCl2 = 1.6 × 10^(-5) On mixing 300 mL, 0.134M Pb(NO3)2(aq.) + ...

    Text Solution

    |