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 ${\mathrm{NH}}_3$ and ${\mathrm{NF}}_3$,  _____ has higher dipole moment because -

1. NF_{3 ;} 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. NH_{3 ;} 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. NF_{3 ;} 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. NH_{3 ;} 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, ${\mathrm{sp}}^2$, ${\mathrm{sp}}^3$ 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

${\mathrm{BF}}_3+{\mathrm{NH}}_3\to {\mathrm{F}}_3\mathrm{B}.{\mathrm{NH}}_3$

The correct statement about the above reaction is :

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:

The important condition/s required for the linear combination of atomic orbitals to form molecular orbitals is:

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.

The axial bonds are longer as compared to equatorial bonds in PCl₅ because:

The relation between hydrogen bond and the van der Waals forces is: