The 'S' atom among the following that does not have \(\mathrm{sp}^3\) hybridization is

1.	\(\mathrm{SO}^{2-}_4\)	2.	\(\mathrm{SF}_4\)
3.	\(\mathrm{SF}_2\)	4.	\(\mathrm{S}_8\)

molecular shapes of ${\mathrm{SF}}_4,<mspace\; width="1em"></mspace>{\mathrm{CF}}_4<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{XeF}}_4$ are 

1.  Same with 2, 0 and 1 lone pairs of electrons respectively

2.  The same with 1, 1 and 1 lone pairs of electrons respectively 

3.  Different with 0, 1 and 2 lone pairs of electrons respectively 

4.  Different with 1, 0 and 2 lone pairs of electrons respectively

Which of the following statements are correct about bond order?

The correct statement is:

1. When O₂ is converted into $O_2^{+}$ then bond distance increases

2. When N₂ is converted into $N_2^{+}$ then bond distance increases

3. When CO is converted into CO⁺ then bond distance increases

4. All of the above

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

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

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