A) \[C{{H}_{4}}\]
B) \[N{{H}_{3}}\]
C) HCl
D) \[Ca{{H}_{2}}\]
Correct Answer: D
Solution :
\[\overset{+2}{\mathop{Ca}}\,\overset{x}{\mathop{{{H}_{2}}}}\,\] i.e., \[2+2x=0\], \[x=-1\] \[2x=-2\] or \[x=\frac{-2}{2}=-1\]
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