MCQ
Highest stable $Mn$ fluoride is $MnF_4$ where as highest $Mn$ oxide is $Mn_2O_7$ due to
- A$O$ atom is smaller the $F$.
- ✓Oxygen has ability to form multiple bonds
- C$Mn^{7+}$ does not exist
- D$F$ can not-stabilise higher oxidation states
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column - $I$ | Column - $II$ |
| $(a)$ Copper | $(i)$ Non-metal |
| $(b)$ Fluorine | $(ii)$ Transition Metal |
| $(c)$ Silicon | $(iii)$ Lanthanoid |
| $(d)$ Cerium | $(iv)$ Metalloid |
Identify the correct match:

$C{{H}_{3}}COOEt+{{H}_{2}}O\xrightarrow{{{H}^{+}}}C{{H}_{3}}COOH+EtOH$
Reason : Low spin complexes have lesser number of unpaired electrons.