which of the given statement(s) is /are false ?
(P) orbital angular momentum of the azimuthal quantum number as lowest for any principle quantum number is `(h)/(pi)`,
(Q) if n=3 ,l=0 ,m=0 for the last valence shell electron ,them the possible atomic number may be 12 or 13 .
(R ) total spin of electrons for the atom `_(25)` Mn is `+-(7)/(2)` .
(S ) spin magnetic moment of inert gas is zero .

Options

A. I, II and III
B. II and III only
C. I and IV only
D. none of these

Answer

To determine which of the given statements are false, we will analyze each statement one by one.

### Step 1: Analyze Statement P
**Statement P:** "Orbital angular momentum of the azimuthal quantum number as lowest for any principle quantum number is \( \frac{h}{\pi} \)."

- The formula for orbital angular momentum \( L \) is given by:
\[ L = \sqrt{l(l+1)} \cdot \frac{h}{2\pi} \]
where \( l \) is the azimuthal quantum number. The lowest value of \( l \) is 0, which gives:
\[ L = \sqrt{0(0+1)} \cdot \frac{h}{2\pi} = 0 \]
- Therefore, the statement that the lowest value is \( \frac{h}{\pi} \) is incorrect. The correct lowest value is 0.

**Conclusion for P:** False

### Step 2: Analyze Statement Q
**Statement Q:** "If \( n=3 \), \( l=0 \), \( m=0 \) for the last valence shell electron, then the possible atomic number may be 12 or 13."

- For \( n=3 \) and \( l=0 \), we are referring to the 3s subshell. The maximum electrons in the 3s subshell is 2.
- The electron configuration for atomic number 12 (Magnesium) is \( 1s^2 2s^2 2p^6 3s^2 \) and for atomic number 13 (Aluminum) is \( 1s^2 2s^2 2p^6 3s^2 3p^1 \).
- The last valence electron in the case of atomic number 12 is in the 3s subshell, and for atomic number 13, the last valence electron is in the 3p subshell.

**Conclusion for Q:** False (the possible atomic numbers are 11 or 12, not 12 or 13)

### Step 3: Analyze Statement R
**Statement R:** "Total spin of electrons for the atom \( _{25}Mn \) is \( \pm \frac{7}{2} \)."

- The electron configuration of Manganese (Mn) is \( [Ar] 4s^2 3d^5 \).
- The total spin \( S \) for unpaired electrons is calculated as:
\[ S = \frac{n}{2} \]
where \( n \) is the number of unpaired electrons. In Mn, there are 5 unpaired electrons in the 3d subshell.
- Therefore, \( S = \frac{5}{2} \), and the total spin can be \( +\frac{5}{2} \) or \( -\frac{5}{2} \).

**Conclusion for R:** False (the correct total spin is \( \pm \frac{5}{2} \))

### Step 4: Analyze Statement S
**Statement S:** "Spin magnetic moment of inert gas is zero."

- Inert gases have completely filled electron shells, which means there are no unpaired electrons.
- The spin magnetic moment \( \mu \) is given by:
\[ \mu = \sqrt{s(s+1)} \cdot \frac{h}{2\pi} \]
Since \( s = 0 \) for inert gases, the spin magnetic moment is indeed zero.

**Conclusion for S:** True

### Final Conclusion
- Statements P, Q, and R are false.
- Statement S is true.

### Summary of False Statements:
- P: False
- Q: False
- R: False
- S: True

### Answer:
The false statements are P, Q, and R.