Which one of the following conversions involv | vedclass

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(C) Let us analyze each conversion:$(a)$ $CH_{4}$ ($sp^{3}$,tetrahedral) $\longrightarrow C_{2}H_{6}$ ($sp^{3}$,tetrahedral). No change in hybridisati

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$BF_{3} \longrightarrow BF_{4}^{-}$

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(C) Let us analyze each conversion:$(a)$ $CH_{4}$ ($sp^{3}$,tetrahedral) $\longrightarrow C_{2}H_{6}$ ($sp^{3}$,tetrahedral). No change in hybridisation or shape.$(b)$ $NH_{3}$ ($sp^{3}$,trigonal pyramidal) $\longrightarrow NH_{4}^{+}$ ($sp^{3}$,tetrahedral). Change in shape,but no change in hybridisation.$(c)$ $BF_{3}$ ($sp^{2}$,trigonal planar) $\longrightarrow BF_{4}^{-}$ ($sp^{3}$,tetrahedral). Both hybridisation and shape change.$(d)$ $H_{2}O$ ($sp^{3}$,bent) $\longrightarrow H_{3}O^{+}$ ($sp^{3}$,trigonal pyramidal). Change in shape,but no change in hybridisation.Thus,the conversion of $BF_{3}$ into $BF_{4}^{-}$ involves a change in both hybridisation and shape.

Label	Order	Property
$i.$	$NH_3 > H_2O > SO_2$	Bond angle
$ii.$	$H_2O > NH_3 > H_2S$	Dipole moment
$iii.$	$N_2 > O_2 > H_2$	Bond enthalpy
$iv.$	$NO^{+} > O_2 > O_2^{2-}$	Bond order

Which one of the following conversions involves a change in both hybridisation and shape?

Similar Questions

Which bond is shorter: $C=O$ or $N=O$? Why?

Observe the following sets:
Label Order Property
$i.$ $NH_3 > H_2O > SO_2$ Bond angle
$ii.$ $H_2O > NH_3 > H_2S$ Dipole moment
$iii.$ $N_2 > O_2 > H_2$ Bond enthalpy
$iv.$ $NO^{+} > O_2 > O_2^{2-}$ Bond order

Which of the above sets are correctly matched?

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