Question

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

(a) n = 2, l = 1, m_l = 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.

Answer 1

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,m_l = 0

so the orbital is 2p_y 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