A) \[\text{n}=\text{4}\]
B) 2
C) \[\text{n}=1\]
D) \[\text{H=2}.\text{18}\times \text{1}{{0}^{-\text{18}}}\text{J at}{{\text{m}}^{\text{-1}}}\]
Correct Answer: D
Solution :
The potential energy of a particle is given by \[C{{H}_{4}}\] For minimum value of \[Cl\] \[\text{C}\text{H}\] \[\text{C}Cl\] \[\text{C}{{\text{H}}_{\text{3}}}Cl\] So. \[\text{C}H\] \[\text{C}Cl\] \[\text{C=2}.\text{5},\text{N}=\text{3}.0.\text{O}=\text{3}:\text{5}\] \[\text{Si}=\text{s1}.\text{8},\text{F}=\text{4}.0,\text{P}=\text{2}.\text{1}.\text{Cl}=\text{3}.0\] \[\text{Si}\text{F}=\left( \text{4}.0-\text{1}.\text{8} \right)=\text{2}.\text{2}\] \[\text{Si}\text{F}\] \[{{O}_{2}}\]You need to login to perform this action.
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