NEET Chemistry Chemical Bonding and Molecular Structure Questions Solved

(c) 

$$\begin{gathered} \because {\mathrm{SF}}_4 \mathrm{have} \mathrm{\mu } >0 \\ \therefore \mathrm{It} \mathrm{has} \mathrm{permanent} \mathrm{dipole} \mathrm{moment} \end{gathered}$$

(d) According to Fajan's rule,

Covalent character $$\begin{gathered} \propto \frac{1}{\mathrm{size} \mathrm{of} \mathrm{cation}} \\ \propto \mathrm{size} \mathrm{of} \mathrm{anion} \end{gathered}$$

In the given options, cation is same but anions are different. Among halogens the order of size is

                      F < Cl < Br < I

$\therefore$Order of covalent character is

                   MI > MBr > MCl > MF

(a) Isostructural Compounds having same structure and same hybridisation are known as isostructural species. e.g. XeF₂ and IF^_₂   are sp³d hybridised and both have linear shape.

F-I-F    F-Xe-F

(b) Molecules with zero bond order, do not exist.

According to molecular orbital theory,

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{Be}}_2^{+} (4+4-1=7) \\ = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^1 \\ \mathrm{Bond} \mathrm{order} \left(\mathrm{BO}\right) = \frac{4-3}{2} = 0.5 \\ \left(\mathrm{b}\right) {\mathrm{Be}}_2(4+4=8)= \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-4}{2}=0 \\ \left(\mathrm{c}\right) {\mathrm{B}}_2(5+5=10) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1 \approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1 \\ \left(\mathrm{d}\right) {\mathrm{Li}}_2 (3+3=6) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-2}{2}=1 \\ \mathrm{Thus}, {\mathrm{Be}}_2 \mathrm{does} \mathrm{not} \mathrm{exist} \mathrm{under} \mathrm{normal} \mathrm{conditions}. \end{gathered}$$

(c) Paramagnetic character is shown by those atoms or molecules which have unpaired electrons.

In the given compounds CO is not paramagnetic since, it does not have unpaired electrons. The configuration of CO molecule is 

CO(14) = $\sigma 1s^2, \stackrel{*}{\sigma }1s^2, \sigma 2s^2, \stackrel{*}{\sigma }2s^2, \sigma 2p_x^2, \pi 2p_y^2 \approx \pi 2p_z^2$

(a) In octahedral structure MX₆, the six hybrid orbitals (sp³d²) are directed towards the corners of a regular octahedral with an angle of 90$°$. According to following structure of MX₆, the number of X-M-X bonds at 180$°$ must be three. 

(d) In case of single bond, there is only one s-bond in case of double bond, there is one s and one p-bonds. Thus, angular shape of ozone (O₃) contains 2s and 1 p-bonds as

(a) SCl₄ (Seesaw structure)  is not isostructural with  Sicl₄ because shows tetrahedral structure due to involvement of repulsion between lone pair and bond pair of electrons.

SO₄^2- shows tetrahedral structure due to sp³ hybridisation.

PO₄^3- shows tetrahedral structure due to sp³ hybridisation.

NH₄⁺ shows tetrahedral structure due to sp³ hybridisation

(a) ClF₃ all bonds are not equal due to trigonal-bipyramidal (sp³d hybridisation) geometry of ClF₃ molecule.

BF₃ and AlF₃ show trigonal symmetric structure due to sp² hybridisation.

NF₃ shows pyramidal geometry due to sp³ hybridisation.

(c) According to VESPR theory repulsion order

  lp - lp > lp - bp > bp - bp

As the number of lone pairs of electrons increases, bond angle decreases due to repulsion between lp-lp, Moreover, as the electronegativity of central atom decreases, bond angle decreases.

Hence, the order of bond angle is

(Cl is less electronegative as compared to O).

clo2- =111,cl2O ,110, Clo2 =117

explanation is given, its can be explained on the multiple concept like valence electron,hybridisation,electro negativity, but fact is ,its a experiemental science , so we need to develop theory to explain the observation, this is basis of evolution of VBT to CFT.