MCQ
For $n = 3$ energy level, the number of possible orbitals (all kinds) are
- A$1$
- B$3$
- C$4$
- ✓$9$
No. of electrons $ = 2{n^2}$
Hence no. of orbital $ = \frac{{2{n^2}}}{2} = {n^2}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Comment on optical activity of the products
$(A)$ The $pH$ of $1 \times 10^{-8}\,M\,HCl$ solution is $8$.
$(B)$ The conjugate base $H _2 PO _4^{-}$is $HPO _4{ }^{2-}$.
$(C)$ $K _{ w }$ increases with increase in temperature.
$(D)$ When a solution of weak monoprotic acid is titrated against a strong base at half neutralisation point, $pH =\frac{1}{2} pK _{ a }$
Choose the correct answer from the option given below.
