MCQ
The $IUPAC$ name of $Xe[Pt F_6 ]$ is
- AHexafluoroplatinate $(VI)$ xenon
- BXenonhexafluoroplatinate $(V)$
- CXenonhexafluoroplatinate $(VI)$
- ✓Xenoniumhexafluoroplatinum $(V)$
$(+1) 4+x+(-1) 5+(-1) 1=0$
$x=+2$
$\left( B \right)\,N{a_4}\left[ {\mathop {Fe{O_4}}\limits^y } \right]$
$(+1) 4+y+(-2) 4=0$
$y=+4$
$\left( C \right)\,\left[ {F\mathop {{e_2}}\limits^z {{\left( {CO} \right)}_9}} \right]$
$2 z+0 \times 9=0$
$z=0$
so $( x + y + z )=+2+4+0$
$=6$
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Reason : Phosphorus has lower electronegativity than nitrogen.
