- A${10^9}$
- B${10^{ - 12}}$
- C${10^{ - 15}}$
- ✓${10^{ - 21}}$
$1$ zepto $ = {10^{ - 21}}$
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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.

$\mathrm{HF}, \mathrm{H}_2, \mathrm{H}_2 \mathrm{~S}, \mathrm{CO}_2, \mathrm{NH}_3, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{CHCl}_3, \mathrm{SiF}_4$, $\mathrm{H}_2 \mathrm{O}, \mathrm{BeF}_2$
