Match the ionization processes listed in column I with the changes observed as listed in column II. For this use the codes given below:

Note: Here change in magnetic property refers to the change from diamagnetic to paramagnetic
or paramagnetic to diamagnetic. 
 

A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10 ^-10 m , then the fraction of an electronic charge on each atom will be:

1. 42%
2. 52%
3. 37%
4. 25%

Arrange the following molecules from most to least polar.

$C{H}_4 C{F}_{2}C{l}_2 C{F}_2{H}_2 CC{l}_4 CC{l}_2{H}_2$
$I II III IV V$

(1) III>V>II > IV=I                                 

(2) II>IV>III>IV>I

(3) III>II>V>IV>I                                 

(4) V>III>II>IV=I    

Which of the following statements is/are correct?

The increasing order of the strength of hydrogen bond in the following mentioned linkages
is

(i) O — H — S     (ii) O — H  — O    (iii) F – H – F   (iv) F – H – O

(1) (i) < (ii) < (iv) < (iii)                                                     

(2) (ii) < (i) < (iv) < (iii)

(3) (i) < (ii) < (iii) < (iv)                                                     

(4) (ii) < (i) < (iii) < (iv)

The boiling point of $CC{l}_4$ is higher than that of $CHC{l}_3$ because :

(1) The dipole moment of $CC{l}_4$ is greater than that of $CHC{l}_3$

(2) $CC{l}_4$ forms hydrogen bonds.

(3) $CC{l}_4$ has higher molecular mass than that of $CHC{l}_3$

(4) $CC{l}_4$ is more ionic than $CHC{l}_3$

Assertion : Compounds of ${\mathrm{Hg}}^{2+}$ ions having an ionic radius of 116 pm are move covalent in character

                   than those of ${\mathrm{Ca}}^{2+}$ ions with almost identical size (114 pm) and the same charge.

Reason : For two ions of the same size and charge, one with an (n-1)${\mathrm{d}}^{10}{\mathrm{ns}}^0.$ electronic configuration

                 will be more polarizing than a cation with an (n-1)${\mathrm{s}}^2$ (n-1)${\mathrm{p}}^6{\mathrm{ns}}^0,$ electronic configuration.

Assertion : In tetrahedral hybridisation i.e., in ${\mathrm{sp}}^3$ hybridisation all p-orbitals are involved and no

                   p-orbital is left for coming with $\mathrm{\pi }$-bonds.

Reason : Central atom can not form double bonds in the molecules or species having ${\mathrm{sp}}^3$hybridisation.

Assertion : F bond angle P = bond angle Q but not precisely $90°.$

Reason : The molecule is slight bent T-shaped and there is repulsion between lone pairs of electrons.