Chemistry MCQs for NEET — Practice Questions with Answers

(c) Key concept: The species that have same number of electrons have same bond order.

use H =( V+M -C+A ) / 2

WHERE V, = valance electron of central atom,

              M is no of mono valent atom , ie halogen or alkali metal

            C/A = charge on cation or anion ..

a = c and N acts as central atom , o is divalent ,so no need to count. 

(6+2 ) /2= 4 , ( 7+ 1)/2 = 4,same for option d

(a) and (d)

Hence, both options (a) and (d) are correct.

(d) According to the postulate of VSEPR theory, a lone pair occupies more space than a bonding pair, since it lies closer to the central atom. This means that the repulsion between the different electron pairs follow the order.

               lp-lp>lp-bp>bp-bp

$$\begin{gathered} \left(\mathrm{c}\right) \mathrm{Hybridisation}=\frac{\mathrm{Number} \mathrm{of} \mathrm{\sigma }- \mathrm{electrons}}{2} \\ = \frac{2+2 (\mathrm{negative} \mathrm{ion})}{2} = 2=sp \end{gathered}$$

Hence, in the carbanion CH3C$\equiv$C$°$, pair of electron as (-)ve charge is present in sp-hybridised-orbital.

(a) As the number of lone pair of electrons on central element increases, repulsion between those lone pair of electrons increases and therefore, bond angle decreases.
          Molecules                            Bond angle
CH₄ (no lone pair of electrons)           109.5°
NH₃ (one lone pair of electrons)         107.5°
H₂O (two lone pair of electrons)         104.45