A) 0
B) \[\text{a}\text{. b }\!\!\times\!\!\text{ c}\]
C) \[\text{a}\text{. c }\!\!\times\!\!\text{ b}\]
D) \[\text{3a}\text{. b }\!\!\times\!\!\text{ c}\]
Correct Answer: B
Solution :
\[{{\left[ \text{Fe}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]}^{\text{3+}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{2}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\] \[\because \] \[{{\text{E}}_{\text{1}}}\text{=}{{\text{E}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}}\text{.}\frac{\text{q}}{{{\text{r}}^{\text{2}}}}\] \[{{E}_{R}}=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos {{60}^{\circ }}}\] \[=\sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}^{2}\times \frac{1}{2}=\sqrt{3}{{E}_{1}}}\] \[\therefore \] \[{{E}_{R}}=\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\tau =qEL\sin \theta \]
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