question_answer

The state of hybridization of \[\text{mo}{{\text{l}}^{\text{-1}}}\] and \[1.77\times {{10}^{-5}}\]is in the following sequence                                                                                                            [AIPMT (S) 2009]

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

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

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

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

If number of\[{{[Sc{{({{H}_{2}}O)}_{3}}{{(N{{H}_{3}})}_{3}}]}^{3+}}\] bonds = 4; hybridisation is \[s{{p}^{3}}\]

Double and triple bonds are not considered while finding hybridisation.