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 |

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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 |

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

$1.1\mathrm{s}\mathrm{and}2\mathrm{s}\phantom{\rule{0ex}{0ex}}2.1\mathrm{s}\mathrm{and}2{\mathrm{p}}_{\mathrm{x}}\phantom{\rule{0ex}{0ex}}3.2{\mathrm{p}}_{\mathrm{y}}\mathrm{and}2{\mathrm{p}}_{\mathrm{y}}\phantom{\rule{0ex}{0ex}}4.1\mathrm{s}\mathrm{and}1\mathrm{s}$

Subtopic: Hybridisation |

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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 |

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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 |

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

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'.

From NCERT

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 |

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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 |

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The incorrect statement from the following is

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

2. s and p orbitals are combined to form a sigma bond as well as a pie bond.

3. Hybrid orbitals only form sigma bonds.

4. Sigma bonds are stronger than pie bonds.

Subtopic: Hybridisation |

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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 |

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