Question

How many orbitals and electrons are associated with n=4?

• A. 32, 64
• B. 16, 32
• C. 4, 16
• D. 8, 16

Answer 1

Answer: (B)

16, 32

For $n^{th}$ orbital possible values of azimuthal quantum number (subshell), l are from 0 to (n-1). Total of 'n' values.

In $n=4$, $l=0,1,2,3$ thus there are 4 subshells i.e.$s,p,d,f$ respectively.

Magnetic quantum number $m_l$ have values from -l to +l and total of $2l+1$ values.

For $n=4$, possible values of $l$ and $m_l$ are:

$m_l=0$ for l=0; total $m_l$ values =1

$m_l=-1,0,1$ for l=1; total $m_l$ values =3

$m_l=-2,-1,0,1,2$ for l=2; total $m_l$ values =5

$m_l=-3,-2,-1,0,1,2,3$ for l=3; total $m_l$ values =7

Total number of orbitals = total values of $m_l$

for n=4,

$\therefore 1+3+5+7=16$ orbitals

Each orbital can occupy maximum of two electron

Number of electrons $=2\times 16=32$

Therefore in n=4, number of subshells=4, orbitals=16 and number of electrons =32.