Home
Class 11
CHEMISTRY
The geometries of the molecules BF3 and ...

The geometries of the molecules `BF_3` and `NF_3` are trigonal planar and trigonal pyramidal, respectively. Which statement best accounts for the difference? : N is more electronegative than B, `BF_3` is ionic, `NF_3` is covalent, B utilise `sp^2` hybridization, while N does not, N is `sp^3` hybridised and has non-bonding pair of valence electrons, while B does not

A

N is more electronegative than B

B

`BF_3` is ionic, `NF_3` is covalent

C

B utilise `sp^2` hybridization, while N does not

D

N is `sp^3` hybridised and has non-bonding pair of valence electrons, while B does not

Text Solution

Verified by Experts

The correct Answer is:
D
N is `sp^3` hybridised and has non-bonding pair of valence electrons, while B does not
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL BONDING

    BRILLIANT PUBLICATION|Exercise LEVEL-II (ASSERTION-REASON TYPE)|20 Videos

  • CHEMICAL BONDING

    BRILLIANT PUBLICATION|Exercise QUESTIONS|6 Videos

  • CHEMICAL BONDING

    BRILLIANT PUBLICATION|Exercise LEVEL-I|120 Videos

  • CHEMICAL AND IONIC EQUILIBRIUM

    BRILLIANT PUBLICATION|Exercise Level - III (Linked Comprehension Type Questions)|1 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    BRILLIANT PUBLICATION|Exercise QUESTION (LEVEL -ll)|42 Videos

Similar Questions

Explore conceptually related problems

Account for the following: i) 'NF_3' is pyramidal while 'BF_3' is trigonal planar. ii) Bond angle in 'H_2 O' is larger than bond angle in 'H_2 S'.

Choose the correct molecules from the given clues: H_2O , SF_6 , BF_3 , a) Clue-1 The central atom is in sp^2 hybridised state and the molecule has trigonal planar in shape. Clue-2 The bond angle is 120°

Hybridisation of the nitrogen atom and electron geometry around nitrogen atom in pyridine is sp^3 , pyramidal, sp^2 , trigonal planar, sp^2 , linear, sp^3 , tetrahedral

Which one of the following statements is incorrect in relation to ionization enthalpy? (1)Ionization enthalpy increases for each successive electron.(2)The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration (3)End of valence electrons is marked by a big jump in ionization enthalpy (4)Removal of electron from orbitals bearing lower n value is higher than from orbital having higher 'n' value.

Presence of unpaired electrons makes a species paramagnetic. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is considerably important. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the spin-only formula, i.e., mu = sqrt(n(n+2)) where n=number of unpaired electrons and mu =magnetic momentin (in Bohr magneton, B.M.). If n=1, then mu = 1.73 B.M . The magnetic moment of a transition metal ion is found to be 3.87 B.M. The number of unpaired electrons present in it is:

BRILLIANT PUBLICATION-CHEMICAL BONDING-LEVEL-II
  1. All of the following lists include at least one ionic compound except

    Text Solution

    |

  2. When forming covalent bonds which atom can have more than eight valenc...

    Text Solution

    |

  3. The geometries of the molecules BF3 and NF3 are trigonal planar and tr...

    Text Solution

    |

  4. All of the following molecules have their atoms lying in the same plan...

    Text Solution

    |

  5. CO2 is a gas at room temperature while SiO2 is a high melting solid. T...

    Text Solution

    |

  6. The nitrite ion NO2^(-) may be respresented by two major resonance for...

    Text Solution

    |

  7. Among the following species, identify the isostructural pairs : NF3, N...

    Text Solution

    |

  8. Arrange the following compounds in the order of increasing dipole mome...

    Text Solution

    |

  9. The cyanide ions CN^(ѳ) and N2 are isoelectronic, but in contrast to C...

    Text Solution

    |

  10. The correct order of hybridisation of the central atom in the followin...

    Text Solution

    |

  11. The common features among the species CN^(ѳ), CO and NO^(o+) are : ...

    Text Solution

    |

  12. The percentage of 'p character in the orbitals forming P-P bonds in 'P...

    Text Solution

    |

  13. For which of the following molecules, mu ne 0?

    Text Solution

    |

  14. The ionic radii of isoelectronic species 'N^(3-), O^(2-)' and 'F^-' in...

    Text Solution

    |

  15. Element X is strongly electropositive and element Y is strongly electr...

    Text Solution

    |

  16. Polarising action of Cd^(2+) on anions is stronger than that of Ca^(2+...

    Text Solution

    |

  17. PCl5 has a shape of trigonal bipyramid whereas IF5 has the shape of a ...

    Text Solution

    |

  18. Amolecule of the type AX5 has square pyramidal geometry. Hence, number...

    Text Solution

    |

  19. Hybridization of the nitrogen atom and electron geometry around nitrog...

    Text Solution

    |

  20. H2O molecule is dipolar, whereas BeF2 is non polar. It is because

    Text Solution

    |