A) nitrite ion
B) nitrate ion
C) nitroniumion
D) nitrogen peroxide
Correct Answer: C
Solution :
\[\underset{1-butene-3-yne}{\mathop{\underset{s{{p}^{2}}}{\mathop{\overset{1}{\mathop{C}}\,{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{\overset{2}{\mathop{C}}\,H}}\,-\underset{sp}{\mathop{\overset{3}{\mathop{C}}\,}}\,\equiv \underset{sp}{\mathop{\overset{4}{\mathop{C}}\,H}}\,}}\,\][ \[\because \] \[C-C\] \[\Rightarrow \]\[s{{p}^{3}},\]\[C=C\]\[\Rightarrow \]\[s{{p}^{2}},\] \[C\equiv C\]\[\Rightarrow \]\[sp\]] Thus, 1-butene-3yne contains only \[sp\]and \[s{{p}^{2}}\] hybridised carbons.
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