MCQ
Product $(C)$ is


- ✓$\begin{matrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
C{{H}_{3}}-C{{H}_{2}}-C-H \\
\end{matrix}$ - B

- C

- D





$(B)$ $\begin{matrix}
C{{H}_{3}}-C{{H}_{2}}-CH-Cl \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,| \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\
\end{matrix}$
$(C)$ $C{{H}_{3}}-C{{H}_{2}}-CHO$
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$A.$ For 1 s orbital, the probability density is maximum at the nucleus.
$B.$ For $2 s$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.
$D.$ $p$ and d-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of p-orbital is zero at the nucleus.