question_answer

The state of hybridization of \[\text{mo}{{\text{l}}^{\text{-1}}}\]and \[1.77\times {{10}^{-5}}\]is in the following sequence

A) \[sp,\,s{{p}^{3}},\,s{{p}^{2}}\] and \[s{{p}^{3}}\]

B) \[s{{p}^{3}},\,s{{p}^{2}},\,s{{p}^{2}}\] and \[sp\]

C) \[sp,\,s{{p}^{2}},\,s{{p}^{2}}\] and \[s{{p}^{3}}\]

D) \[sp,\,s{{p}^{2}},\,s{{p}^{3}}\] and \[s{{p}^{2}}\]

Correct Answer: A

Solution :

Key Idea Count number of a bond and then find hybridisation as follows. If number of \[1.3\times {{10}^{4}}g\] bonds =2; hybridisation is sp, If number of \[CC{{l}_{3}}CHO\]bonds = 3; hybridisation is \[(At.no.Zn=30,Sc=21,Ti=22,Cr=24)\] If number of\[{{[Sc{{({{H}_{2}}O)}_{3}}{{(N{{H}_{3}})}_{3}}]}^{3+}}\] bonds = 4; hybridisation is \[{{[Ti{{(en)}_{2}}{{(N{{H}_{3}})}_{2}}]}^{4+}}\] Double and triple bonds are not considered while finding hybridisation.