JEE Main 202523 Jan, Evening shiftChemistryEasyd- and f-Block Elements158× since 2002Consider the following reactions K2Cr2O7→−H2OKOH[A]→−H2OH2SO4[B]+K2SO4\mathrm{K_2Cr_2O_7 \xrightarrow[-H_2O]{KOH} [A] \xrightarrow[-H_2O]{H_2SO_4} [B] + K_2SO_4}K2Cr2O7KOH−H2O[A]H2SO4−H2O[B]+K2SO4. The products [A] and [B], respectively are:AOption A. K2CrO4\mathrm{K_2CrO_4}K2CrO4 and Cr2O3\mathrm{Cr_2O_3}Cr2O3BOption B. K2CrO4\mathrm{K_2CrO_4}K2CrO4 and K2Cr2O7\mathrm{K_2Cr_2O_7}K2Cr2O7COption C. K2CrO4\mathrm{K_2CrO_4}K2CrO4 and CrO\mathrm{CrO}CrODOption D. K2Cr(OH)6\mathrm{K_2Cr(OH)_6}K2Cr(OH)6 and Cr2O3\mathrm{Cr_2O_3}Cr2O3CheckSkip · show solution
