Question

which orbital and electron can be designated for n=3, l=1, m=-1, s=-1/2
a 3pz, spinning anti- clockwise
b 3px, spinning clockwise
c 3d, spinning clockwise
d 3py, , spinning anti-clockwise
question 9 of 9

Explanation:

1. Analyze the quantum numbers: \( n = 3 \) (principal quantum number, energy level), \( l = 1 \) (azimuthal quantum number, so orbital is \( p \) orbital, since \( l = 0 \) is \( s \), \( l = 1 \) is \( p \), \( l = 2 \) is \( d \), etc.). So option C (3d) is invalid as \( l = 2 \) for \( d \) orbitals, but here \( l = 1 \).
2. For \( p \) orbitals (\( l = 1 \)), the magnetic quantum number \( m \) can be \( - 1,0,+ 1 \). The spin quantum number \( s=-\frac{1}{2} \) corresponds to anti - clockwise spin ( \( s =+\frac{1}{2} \) is clockwise).
3. Now, check the options:

• Option A: 3pz, spinning anti - clockwise. But the \( m=-1 \) for \( p \) orbitals: the \( p_x \) orbital can be associated with \( m = - 1 \) (depending on the convention, but more importantly, let's check spin. \( s=-\frac{1}{2} \) is anti - clockwise. Wait, no, let's re - evaluate. Wait, the spin quantum number \( s =-\frac{1}{2} \) is anti - clockwise (by convention, \( s =+\frac{1}{2} \) is clockwise). Now, for the \( p \) orbitals, when \( l = 1 \), the \( m \) values: for \( p_x \), \( p_y \), \( p_z \), the \( m \) values can be assigned as \( m=-1 \) for \( p_x \) (or \( p_y \) depending on the axis convention, but the key is spin. Wait, no, let's correct. The spin \( s =-\frac{1}{2} \) is anti - clockwise. Now, option D: 3py, spinning anti - clockwise. Wait, but let's check the orbital type first. Since \( l = 1 \), it's a \( p \) orbital, so \( n = 3 \), so 3p orbital. Now, the spin \( s=-\frac{1}{2} \) is anti - clockwise. Now, let's check the options again. Option C is out (3d). Option B has clockwise spin (\( s =+\frac{1}{2} \) would be clockwise), but we have \( s=-\frac{1}{2} \), so B is out. Option A: 3pz, anti - clockwise. But wait, the \( m=-1 \): for \( p \) orbitals, the \( m \) values: typically, \( p_x \) can be \( m=-1 \), \( p_y \) \( m = 0 \)? No, maybe my axis convention was wrong. Wait, actually, the magnetic quantum number \( m \) for \( l = 1 \) can be \( - 1,0,+1 \). The spin \( s=-\frac{1}{2} \) is anti - clockwise. Now, looking at the options, option D: 3py, spinning anti - clockwise. Wait, but maybe the correct answer is D? Wait, no, let's re - check. Wait, the spin quantum number \( s =-\frac{1}{2} \) is anti - clockwise (by definition, \( s =+\frac{1}{2} \) is clockwise). The orbital is 3p (since \( n = 3 \), \( l = 1 \)). Now, among the options, option D is 3py, spinning anti - clockwise. Wait, but let's check the options again. Option A: 3pz, anti - clockwise. Option D: 3py, anti - clockwise. Wait, maybe the key is the spin. Since \( s=-\frac{1}{2} \), it's anti - clockwise. Now, the orbital: \( l = 1 \) is \( p \), so 3p. Now, the options with 3p and anti - clockwise spin are A and D. But maybe the \( m=-1 \) corresponds to \( p_y \) or \( p_x \). Wait, maybe I made a mistake earlier. Let's recall: the principal quantum number \( n = 3 \), azimuthal \( l = 1 \) (p - orbital), magnetic \( m=-1 \), spin \( s =-\frac{1}{2} \) (anti - clockwise). Now, option D is 3py, spinning anti - clockwise. So that's the correct option.

Answer:

D. 3py, spinning anti - clockwise