$\mathop {{H_3}C\mathop C\limits^ \ominus {H_2}}\limits_{(i)} \,,$$\mathop {{H_2}C = \mathop C\limits^ \ominus H}\limits_{(ii)} $ and $\mathop {H - C \equiv \mathop C\limits^ \ominus }\limits_{(iii)} $
is in the order of
Among alkane, alkene and alkyne are most acidic and alkanes are least acidic, so the order of base strengths
alkane $>$ alkene $>$ alkyne
$C H_{3} \stackrel{\ominus}{C} H_{2}>H_{2} C=\stackrel{\ominus}{C} H>H-C \equiv {C}^{\ominus}$
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$STATEMENT-2$: The higher oxidation states for the group $14$ elements are more stable for the heavier members of the group due to 'inert pair effect'.