The incorrect statement about NH_{3} and H_{2}O is :

1. The bond angle in NH_{3} is less than in H_{2}O.

2. Both have distorted tetrahedral geometries.

3. The bond angle in H_{2}O is less than in NH_{3}.

4. Both are sp^{3} hybridized.

Subtopic: Hybridisation |

69%

From NCERT

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${\mathrm{AlCl}}_{3}$ $+$ ${\mathrm{Cl}}^{-}$ $\to $ ${\mathrm{AlCl}}_{4}^{-}$

The change in the hybridization of the Al atom in the above reaction is:

1. | \(s p^{2} \text { to } s p^{3}\) | 2. | \(\mathrm{sp}^{3} \text { to } s p^{2}\) |

3. | \(\mathrm{sp}^{3} \text { to } \mathrm{dsp}{ }^{2}\) | 4. | \(\operatorname{sp}^{2} \text { to } d s p^{2}\) |

Subtopic: Hybridisation |

83%

From NCERT

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Considering the X-axis as the internuclear axis, a sigma bond will not be formed in:

$1.$ $1\mathrm{s}$ $\mathrm{and}$ $2\mathrm{s}$

$2.$ $1\mathrm{s}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{x}}$

$3.$ $2{\mathrm{p}}_{\mathrm{y}}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{y}}$

$4.$ $1\mathrm{s}$ $\mathrm{and}$ $1\mathrm{s}$

Subtopic: Hybridisation |

77%

From NCERT

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CH_{4} does not exhibit square planar geometry because:

1. | For square planar geometry, 5 bonds are required. |

2. | Carbon does not have d-orbitals to undergo dsp^{2 }hybridization. |

3. | Due to steric hindrance CH_{4} does not exhibit square planar geometry. |

4. | Carbon does not have d-orbitals to undergo d^{2}sp^{3 }hybridization. |

Subtopic: Hybridisation |

66%

From NCERT

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The shapes of sp, ${\mathrm{sp}}^{2}$, ${\mathrm{sp}}^{3}$ orbitals formed due to hybridization of atomic orbitals would be respectively-

1. Trigonal planar, linear, tetrahedral

2. Linear, trigonal planar, tetrahedral

3. Linear, tetrahedral, trigonal planar

4. Tetrahedral, trigonal planar, linear

Subtopic: Hybridisation |

94%

From NCERT

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${\mathrm{BF}}_{3}+{\mathrm{NH}}_{3}\to {\mathrm{F}}_{3}\mathrm{B}.{\mathrm{NH}}_{3}$

The correct statement about the above reaction is -

1. | Hybridization of ‘B’ changes to sp^{2} from sp^{3} while there is no change in the hybridization of 'N' |

2. | Hybridization of ‘N’ changes to sp^{3} from sp^{2} while there is no change in the hybridization of 'B' |

3. | Hybridization of ‘N’ changes to sp^{2} from sp^{3} while there is no change in the hybridization of 'B' |

4. | Hybridization of ‘B’ changes to sp^{3} from sp^{2} while there is no change in the hybridization of 'N'. |

Subtopic: Hybridisation |

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The total number of sigma and pi bonds in C_{2}H_{4 }is

1. 6 sigma bonds and 1 pi-bond.

2. 3 sigma bonds and 3 pi-bonds.

3. 5 sigma bonds and 1 pi-bond.

4. 2 sigma bonds and 2 pi-bonds.

Subtopic: Hybridisation |

87%

From NCERT

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The axial bonds are longer as compared to equatorial bonds in PCl_{5} because -

1. | Axial bond pairs suffer more repulsion from the equatorial bond pairs |

2. | Equatorial bond pairs suffer more repulsion from the axial bond pairs |

3. | Both 1, and 2 |

4. | None of the above |

Subtopic: Hybridisation |

74%

From NCERT

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The incorrect statement among the following is:

1. | The sigma bond forms via head-on overlap and the pie bond forms via sidewise overlapping of orbitals. |

2. | s and p orbitals combine to form a sigma bond as well as a pie bond. |

3. | Hybrid orbitals form sigma bonds only. |

4. | Sigma bonds are stronger than pie bonds. |

Subtopic: Hybridisation |

64%

From NCERT

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Hybridization of C_{1}, C_{2} , and C_{3} in the structure given below are

1. ${\mathrm{C}}_{1}={\mathrm{sp}}^{2},{\mathrm{C}}_{2}={\mathrm{sp}}^{2},{\mathrm{C}}_{3}={\mathrm{sp}}^{2}$

2. ${\mathrm{C}}_{1}={\mathrm{sp}}^{3},{\mathrm{C}}_{2}={\mathrm{sp}}^{3},{\mathrm{C}}_{3}={\mathrm{sp}}^{2}$

3. ${\mathrm{C}}_{1}={\mathrm{sp}}^{3},{\mathrm{C}}_{2}={\mathrm{sp}}^{2},{\mathrm{C}}_{3}={\mathrm{sp}}^{2}$

4. ${\mathrm{C}}_{1}={\mathrm{sp}}^{3},{\mathrm{C}}_{2}={\mathrm{sp}}^{2},{\mathrm{C}}_{3}={\mathrm{sp}}^{3}$

Subtopic: Hybridisation |

89%

From NCERT

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