MCQ
Identify the electronic configuration of an element whose atomic radii is determined by taking half the internuclear distance between like atoms.
- A[He]2s22p5
- B[Ar]4s2
- C[Ne]3s2
- D[Kr]4d55s2
Explanation:
The atomic radii of non-metallic elements are determined by taking half the internuclear distance between like atoms.
The only non-metal in the given options is option A, which is Flourine.
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$2CHI_3 + 6Ag \to 6AgI(s) + C_2H_2(g)$
$\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,} \\
{C{H_3} - C - CH - CH - CH - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,CHO\,\,}
\end{array}$
is :