Amongst ${\mathrm{H}}_2\mathrm{O},<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{S},<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{Se}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{Te}$ the one with the highest boiling point is:

1. H₂O because of H-bonding.

2. H₂Te because of higher molecular weight.

3. H₂S because of H-bonding.

4. H₂Se because of lower molecular weight.

A molecule among the following with non-zero dipole moment is:

1. ${\mathrm{C}}_6{\mathrm{Cl}}_6$

2. ${\mathrm{C}}_2{\mathrm{Cl}}_6$

3. $\mathrm{C}{\left({\mathrm{CCl}}_3\right)}_4$

4. $:{\mathrm{CCl}}_2$

Which molecule is least likely to form hydrogen bonds? 

${\mathrm{Why\; does\; H₂O\; have\; a\; higher\; boiling\; point\; than\; H₂S?}}_{}$

1.  ${\mathrm{H₂O\; molecules\; are\; smaller\; and\; hence\; more\; closely\; packed}}_{}$

2. H₂O has a larger bond angle than H₂S, leading to tighter packing

3.  Due to intermolecular hydrogen bonding in liquid H₂O

4.  H₂O has a higher latent heat of vaporisation than H₂S

Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option. 

A molecule that has a complete octet is:

 1. BeCl₂(dimer)       

 2. BeH₂(dimer)

 3. BeH₂(s)               

 4. BeCl₂(s)

The maximum number of 90° angles between bond pair-bond pair of electrons is observed in:

1.  sp³d²-hybridization.            

2.  sp³d-hybridization.

3.  dsp²-hybridization.             

4.  dsp³-hybridization.

The molecule/s that will have non zero dipole moments is /are : 

1. 1,2-dichloro benzene.
2. Cis-3-hexene.
3. Trans-2-pentene.
4. All of the above.

Arrange the given oxides in decreasing order of their basic character:

1. Na₂O > MgO > Al₂O₃ > CuO

2. MgO > Al₂O₃ > CuO > Na₂O

3. Al₂O₃ > MgO > CuO > Na₂O

4. CuO > Na₂O > MgO > Al₂O₃

A pair in which both species are not likely to exist is: