- ACyanamide ion $(CN_2\,^{2-})$ is isoelectronic with $CO_2$ and has the same linear structure
- B$Mg_2C_3$ reacts with water to form propyne
- C$CaC_2$ has $NaCl$ type lattice
- ✓All of the above
In $(CN_2^{2-})$, we have $22(6+7+7+2=22)$ electrons. Both $CO_2$ and $(CN_2^{2-})$ have linear structures. Thus, statement $(a)$ is correct.
$M{{g}_{2}}{{C}_{3}}+4{{H}_{2}}O\to 2Mg{{(OH)}_{2}}+\underset{\Pr opyne}{\mathop{C{{H}_{3}}C\equiv CH}}\,$
i.e., statement $(b)$ is also correct.
The structure of $CaC_2$ is of $NaCl$ type
i.e., statement $(c)$ is also correct.
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Statement $I$: IUPAC name of $\mathrm{HO}-\mathrm{CH}_2-\left(\mathrm{CH}_2\right)_3-$ $\mathrm{CH}_2-\mathrm{COCH}_3$ is $7$-hydroxyheptan-2-one.
Statement $II$: $2$-oxoheptan-$7$-ol is the correct IUPAC name for above compound.In the light of the above statements. choose the most appropriate answer from the options given below: