${\mathrm{H}}_{3}{\mathrm{PO}}_{3}$ 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 |

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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 |

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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 above

From NCERT

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 |

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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 |

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

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'.

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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