How many electrons in an atom can have each of the following quantum number or s

Question

How many electrons in an atom can have each of the following quantum number or sublevel designations?

(a) n = 2, l = 1, ml = 0

(b) 5s

(c) n = 4, l = 2

Write the full ground-state electron configuration for the following. (Type your answer using the format 1s2 2s2 2p6 for 1s22s22p6.)

(a) Co

(b) Mg

(c) K

Starting with the previous noble gas, write the condensed ground-state electron configuration of an atom of the following elements. (Type your answer using the format [Ar] 3d10 4s2 for [Ar]3d104s2.)

(a) scandium

(b) antimony

(c) silver

(d) lead

Please answer all. I'm so confused.

Explanation / Answer

n is the PRINCIPAL QUANTUM NUMBER .it have values 1,2,3,4,5,6,7 which indicate shell number

l is the AZIMUTHAL QUANTUM NUMBER .It indicates the subshell present in a shell.

if l=0 then subshell is s .

if l=1 then subshell is p.

if l=2 then subshell is d.

if l=4 then subshell is f.

number of orbitals in each subshell is equal to 2l+1 and given by magnetic quantum number m.

m has values -l to +l

that means for s subshell since l=0 ,in this 2(0) + 1= 1 orbital . the value of m is 0

for p subshell ,since l=1, in this 2(1)+1 =2+1= 3 orbitals. the values of m are -1,0,+1.

for d subshell , since l=2, in this 2(2)+1= 4+1 = 5 orbitals . the values of m are -2,-1,0,+1,+2

for f subshell, since l=3, in this 2(3) +1 = 6+1 =7orbitals. the values of m are -3,-2,-1,0,+1,+2,+3

each orbital contains 2 electrons

so one s orbital contains 2 electrons

three p orbitals contain 6 electrons  

five d orbitals contain 10 electrons,

seven f orbitals contain 14 electrons .

now coming to questions

a) given n=2, l= 1,ml = 0

so the orbital is 2py and   it contains 2 electrons.

b) 5s orbital it is a single orbital and contains 2 electrons.

c) n=4 , l=2 orbital

orbital is 4d and contains 10 electrons.

q 2) ground state electronic configuration of given elements

a) Cobalt - Co -atomic number 27

electronic configuration = 1s22s22p63s23p64s23d7

b) magnesium -Mg atomic number 12

electronic configuration = 1s22s22p63s2

c) poatassium -K atomic number =19

electronic configuration =1s22s22p63s23p64s1

q3 ) condenced electronic configuration

a) scandium- Sc atomic number 21

electronic configuration = 1s22s22p63s23p64s23d1

by using nearest inert gas electronic configuration we can simply write electronic configuration of scandium as follows

the previous noble gas for scandium is argon with atomic number 18. so we need not write electronic configuration for 18 simply write [Ar] and continue the from19th electron . so

condenced electronic configuration = 1s22s22p63s23p64s23d1

= [Ar] 4s23d1

b) silver -Ag -atomic number 47

electronic configuration =1s22s22p63s23p64s23d104p65s24d9  ( expected)  

1s22s22p63s23p64s23d104p65s14d10 ( actual)

previous inert gas is krypton Kr with atomic number 36.

so we need not write  electronic configuration for 36simply writre [Kr] and continue from 37th electron as follows

Ag = 1s22s22p63s23p64s23d104p65s14d10

Ag = [Kr]5s14d10

c)lead Pb atomic number 82

electronic configuration = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2

nearest or previous inert gas is xenon with atomic number 54.

condenced electronic configuration= 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2

= [Xe]6s24f145d106p2

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