A) x + 2y + 4z = 12
B) 4x + 2y + z = 12
C) x + 2y + 4z = 3
D) 4x + 2y + z = 3
Correct Answer: B
Solution :
Let equation of the plane is \[A{{g}_{2}}C{{O}_{3}}\] Then, \[=\frac{2.76\times 216}{276}=2.16g\] \[\begin{align} & C{{H}_{3}}-CH-C{{H}_{2}}C{{H}_{2}}OH\xleftarrow{{{H}_{2}}O} \\ & | \\ & C{{H}_{3}} \\ & \text{3-meethyl butanol} \\ \end{align}\] and \[\begin{align} & C{{H}_{3}}-CH-C{{H}_{2}}C{{H}_{2}}OMgBr \\ & | \\ & C{{H}_{3}} \\ \end{align}\] are the points on the coordinate axes. Since, the centroid of the triangle is (1, 2, 4). \[Ph-C\equiv C-C{{H}_{3}}+{{H}_{2}}O\xrightarrow{H{{g}^{2+}}/{{H}^{+}}}\] \[\begin{align} & OH \\ & | \\ & Ph-C=CH-C{{H}_{3}} \\ \end{align}\] \[\begin{align} & OOH \\ & ||| \\ & Ph-C-C{{H}_{2}}C{{H}_{3}}\xleftarrow[-{{H}_{2}}O]{}Ph-C-C{{H}_{2}}C{{H}_{3}} \\ & | \\ & OH \\ \end{align}\] and \[\xleftarrow[{{H}_{2}}O]{H{{g}^{2+}}/{{H}^{+}}}\] \[\alpha -hydroxy\] The equation of the plane is \[\alpha -hydroxy\] \[F{{e}^{2+}}=\left[ Ar \right]3{{d}^{6}}4{{s}^{0}}\Rightarrow \text{4 unpaired electrons}\] 4x + 2y + z = 12
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