- A$XeF_2$
- B$XeF_4$
- ✓$XeOF_4$
- D$XeO_3$
$Xe{F_6} + 2{H_2}O\, \to Xe{O_2}{F_2}+ \,4HF$
$Si{O_2}\, + \,2Xe{F_6}\, \to \mathop {\,2XeO{F_4}\, + \,Si{F_4}}\limits_{(Xenoneoxy\,\,\,tetra\,\,fluoride)} $
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(Given that : $SrCO_{3(s)} \rightleftharpoons SrO_{(s)}+ CO_{2(g)} \,, K_p=1.6\,atm$)
$C _2 H _4( g )+3 O _2( g ) \rightarrow 2 CO _2( g )+2 H _2 O ( l )$ the amount of heat produced as measured in bomb calorimeter is $1406\,kJ\,mol { }^{-1}$ at $300\,K$. The minimum value of $T \Delta S$ needed to reach equilibrium is $(-).......\,kJ$. (Nearest integer) Given : $R =8.3\,JK ^{-1} mol ^{-1}$
$(A)$ Homolytic bond cleavage
$(B)$ Heterolytic bond cleavage
$(C)$ Free radical formation
$(D)$ Primary free radical
$(E)$ Secondary free radical
Choose the correct answer from the options given below: