Hint: There are four types of quantum numbers.
Correct Answer: Option $$C$$
Explanation for correct option :
(C): The principal energy level N can have a maximum of 32 electrons:
The principal energy level N can have a maximum of $$32$$ electrons.
The maximum number of electrons present in a particular shell is calculated by
$$\text Formula = 2\times n^{2}$$
Here " $$\mathrm{n}$$ " represents the shell number.
For instance, $$\mathrm{K}$$ shell is the first shell and it can hold up to $$2(1)^{2}=2$$ electrons.
Similarly, the $$L$$ shell is the second shell and it can hold up to $$2(2)^{2}=8$$ electrons.
This formula helps to calculate the maximum number of electrons that an orbit can accommodate.
Principal energy level can have maximum $$2 \mathrm{n}^{2}$$ electron.
For $$\mathrm{n}=4$$, it have maximum electron of $$2 * 4^{2}=32$$.
Hence, the statement is true
Explanation for incorrect options:
(A): There are seven principal electron energy levels
There are infinite principal electron energy levels.
Hence, the statement is False.
(B): The second principal energy level has four sub- energy levels and contains a maximum of eight electrons
The second principal energy level has only two sub- energy levels and contains a maximum of eight electrons.
For $$\mathrm{n}=2$$, it have maximum electron of $$2 *2 ^{2}=8$$.
Hence, the statement is False.
(D): The $$4s$$ sub-energy level has high energy than the $$3d$$ sub energy level.
Energy of the orbits is decided by $$(n+l)$$ rule.
$$n$$ is called principal quantum number. $$n=(1,2,3,4 \ldots . .)$$
$$l$$ is known as azimuthal quantum number. $$l=(0,1,2 \ldots \ldots n-1)$$
Solve the question step by step-
$$4 s ; n=4$$ and for $$s, l=0$$
Energy level of $$4 s$$ is decided by $$(n+l)$$ rule.
So, energy level is $4+0 \Rightarrow 4$
For $$3 d ; n=3$$ and $$l=2$$
Energy level is $$(n+l) \Rightarrow(3+2) \Rightarrow 5$$
$$3 d$$ is at a higher energy level than $$4 s$$.
Hence, the statement is false.