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${\mathrm{H}}_{3}{\mathrm{PO}}_{3}$ can be represented by structures 1 and 2 shown below.

These two structures cannot be taken as the canonical forms of the resonance hybrid, because :

1. The positions of the atoms have changed.

2. The positions of the atoms are constant.

3. H_{3}PO_{3} does not show resonance.

4. Two hydrogen atoms are missing.

Subtopic: Resonance & Nature of Compounds |

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The shape of ${\mathrm{H}}_{2}\mathrm{O}$ molecule is bent while that of ${\mathrm{CO}}_{2}$ is linear. The correct explanation is :

1. | The dipole moment of carbon dioxide is 3.84 D and water is zero |

2. | The dipole moment of water is zero and carbon dioxide is 1.84 D |

3. | The dipole moment of carbon dioxide is zero and water is 1.84 D |

4. | The dipole moment of carbon dioxide is 3.84 D and water is 1.84 D |

Subtopic: Polarity |

82%

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The application of dipole moment is/are :

1. | It is used to differentiate polar and non-polar bonds. |

2. | It is helpful in calculating the percentage ionic character of a molecule. |

3. | Both '1' and '2' |

4. | None of the above. |

Subtopic: Polarity |

87%

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The difference between electronegativity and electron gain enthalpy is :

1. | An element has a constant value of the electron gain enthalpy but not of electronegativity. |

2. | There is no difference between these two terms. |

3. | An element has a constant value of electronegativity but not of electron gain enthalpy. |

4. | None of the above. |

Subtopic: Lattice & Hydration Energy |

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The correct order of increasing ionic character in the molecules: LiF, K_{2}O, ${\mathrm{N}}_{2},{\mathrm{SO}}_{2}$ and ClF_{3} is -

1. ${\mathrm{N}}_{2}{\mathrm{ClF}}_{3}{\mathrm{SO}}_{2}{\mathrm{K}}_{2}\mathrm{O}\mathrm{LiF}$

2. ${\mathrm{N}}_{2}{\mathrm{ClF}}_{3}\mathrm{LiF}{\mathrm{SO}}_{2}{\mathrm{K}}_{2}\mathrm{O}$

3. ${\mathrm{SO}}_{2}<{\mathrm{K}}_{2}\mathrm{O}{\mathrm{ClF}}_{3}\mathrm{LiF}{\mathrm{N}}_{2}$

4. ${\mathrm{N}}_{2}{\mathrm{SO}}_{2}{\mathrm{ClF}}_{3}{\mathrm{K}}_{2}\mathrm{O}\mathrm{LiF}$

Subtopic: Polarity |

52%

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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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BeH_{2} molecule has a zero dipole moment because -

1. |
Dipole moments of each H–Be bond are unequal and are in opposite directions |

2. | Dipole moments of each H–Be bond are equal and are in opposite directions |

3. | Dipole moments of each H–Be bond are equal and are in same directions |

4. | Dipole moments of each H–Be bond are unequal and are in same directions |

Subtopic: Polarity |

89%

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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%

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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 |

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