MCQ
The hybridisation of $S$ in $SF_4$ is
- A$dsp^2$
- B$sp^3$
- ✓$sp^3d$
- D$sp^3d^2$
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(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}{CaCO}_3: 100, \mathrm{MgCO}_3: 84)$
$\begin{array}{|l|l|} \hline \,\,List\,\,\,\,I\,\,(Compound) & \,\,List\,\,II\,\,(Oxidation\,\,state\,\,of\,\,N) \\ \hline (A)\,\,N{{O}_{2}} & (1)\,\,\,\,+\,\,5 \\ \hline (B)\,\,HNO & (2)\,\,-\,\,3 \\ \hline (C)\,\,N{{H}_{3}} & (3)\,\,+\,\,4 \\ \hline (D)\,\,{{N}_{2}}{{O}_{5}} & (4)\,\,+\,\,1 \\ \hline \end{array}$
code : $A \,\, B \,\,C \,\,D$
[At. nos. $Zn = 30, Sc = 21, $$Ti = 22, Cr = 24$]
Reason : Graphite is a good conductor.