Solve this:

Solve this: 16. (i) (iii) Arrange the following overlapping in increasing order of bond strength : 2s—2s. 2p—2p and 2p—2s. Arrange the following in increasing order of repulsion : bpibp (iii) bp. Arrange the following in increasing order of bond angle •

Dear Student,

(i) The order of increasing bond strength is s-s<s-p<p-pThe p-p bond is the strongest as p-orbital can overlap more effectively than s orbitals of the same shell. Thus, p-p bonds are stronger than s-s bonds. Hence, the correct order is 2s-2s < 2p-2s < 2p-2p(ii) The increasing order of repulsion is :b.p-b.p < l.p-b.p < l.p-l.pThis is because line pairs areshared by one atom only and so are closer to the nucleus as compared to bond pair which are shared by two atoms, thus create stronger electrostatic repulsion. (iii) The order of increasing bond angle is :H2O <NH3 <NH4+<H3O+

 

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Bond angle of NH3 = 107degree. NH4+ = 109degree. H2O = 104.5degree. H3O+ = 109.5degree. Therefore, inc. order of bond angle is H2O, NH3, NH4+, H3O+
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