The correct order of increasing ionic character in the molecules: LiF, K2O, and ClF3 is:
1.
2.
3.
4.
CH4 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 dsp2 hybridization. |
| 3. | Due to steric hindrance CH4 does not exhibit square planar geometry. |
| 4. | Carbon does not have d-orbitals to undergo d2sp3 hybridization. |
BeH2 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. |
Among and , _____ has higher dipole moment because -
1. NF3 ; the resultant bond moment of the N–H bonds add up with the lone pair while N – F bonds partly cancels the moment of the lone pair.
2. NH3 ; the resultant bond moment of the N–F bonds add up with lone pair while N - H bonds partly cancels the moment of the lone pair.
3. NF3 ; the resultant bond moment of the N–F bonds add up with the lone pair while N – H bonds partly cancels the moment of the lone pair.
4. NH3 ; the resultant bond moment of the N–H bonds add up with lone pair while N – F bonds partly cancels the moment of the lone pair.
The shapes of sp, , 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
The correct statement about the above reaction is:
| 1. | Hybridization of ‘B’ changes to sp2 from sp3 while there is no change in the hybridization of 'N'. |
| 2. | Hybridization of ‘N’ changes to sp3 from sp2 while there is no change in the hybridization of 'B'. |
| 3. | Hybridization of ‘N’ changes to sp2 from sp3 while there is no change in the hybridization of 'B'. |
| 4. | Hybridization of ‘B’ changes to sp3 from sp2 while there is no change in the hybridization of 'N'. |
The total number of sigma and pi bonds in C2H4 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.
The lone pairs of electrons can be defined as:
| 1. | Electron pairs that participate in bonding.
|
| 2. | Electron pairs that do not participate in bonding.
|
| 3. | Electron pairs that are present in inner most shell.
|
| 4. | Electron pairs that are present in valence shell of ions. |
The important condition/s required for the linear combination of atomic orbitals to form molecular orbitals is:
| 1. | The combining atomic orbitals must have the exact or nearly the same energy. |
| 2. | The combining atomic orbitals must have proper symmetry about the molecular axis. |
| 3. | The combining atomic orbitals must overlap to the maximum extent. |
| 4. | All of these. |
Be2 molecule does not exist. The best explanation to explain this on the basis of the molecular orbital theory is:
1. The bond order of Be2 is one.
2. The bond order of Be2 is negative.
3. The bond order of Be2 is zero.
4. The bond order of Be2 is two.
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