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Question

Chemistry Question on Atomic Structure

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

A

n=4,l=2,m=1,s=+12n = 4, \, l = 2, \, m = -1, \, s = +\frac{1}{2}

B

n=4,l=0,m=0,s=+12n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}

C

n=3,l=0,m=1,s=+12n = 3, \, l = 0, \, m = 1, \, s = +\frac{1}{2}

D

n=2,l=0,m=0,s=+12n = 2, \, l = 0, \, m = 0, \, s = +\frac{1}{2}

Answer

n=4,l=0,m=0,s=+12n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}

Explanation

Solution

The electron configuration for potassium (K\text{K}, atomic number 19) is:

K:1s2,2s2,2p6,3s2,3p6,4s1.\text{K} : 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1.

The outermost orbital is the 4s4s orbital. The quantum numbers for the outermost electron are:

n=4,l=0,m=0,s=+12.n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}.