A) tetraaquadichlorochromate
B) tetraaquadichlorochromium (III) chloride
C) tetra aquadichlorochromium (II) chloride
D) dichlorotetraaquachromium (III) chloride
Correct Answer: B
Solution :
Alphabetically ligand is written first with ending at \[O\] followed by metal ending at ium and then negative end of coordination compound. \[\operatorname{tetra}\,\,aqua\,\,\xleftarrow{{}}\,\,for\,\,4{{H}_{2}}O\] \[dichloro\,\,\xleftarrow{{}}\,\,for\,\,2\,\,Cl\] \[chromium\,\,(III)\,\,\xleftarrow{{}}\,\,for\,\,C{{r}^{3\,+}}\] \[chloride\,\,\xleftarrow{{}}\,\,for\,\,C{{l}^{\Theta }}\] IUPAC name = tetraaquadichlrochromium (III) chlorideYou need to login to perform this action.
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