# Finding number of orbitals in nickel whose |m| = 1 [closed]

I am new to quantum mechanics, and I need to find out total number of orbitals in nickel , which have $\vert m \vert = 1$ and at least one electron is present.

m = magnetic quantum number

I need to understand whole thing, so I need full explanation.

• Write all possibilites and count them. Jul 2, 2017 at 1:11
• I'm voting to close this question as off-topic because itcan be considered homework with no input from OP Jul 2, 2017 at 13:04
• @Mithoron How can be considered a homework? What's the meaning of no input from OP?
– Fghj
Jul 2, 2017 at 15:11

Conventionally, the location of an electron is written in the following way,

$n,~ l,~m_l,~m_s$

$n$ represents the principal quantum number,

$l$ represents the azimuthal quantum number,

$m_l$ represents the magnetic quantum number

$m_s$ represents the spin quantum number.

The absolute value of the magnetic quantum number must be less than or equal to that of the azimuthal quantum number. This means that the azimuthal quantum number must be greater than or equal to 1 (according to your question). Because the azimuthal quantum number must be less than that of the principal quantum number, the principal quantum number must be greater than or equal to 2.

Remember that the azimuthal quantum number corresponds to the subshell.

0 - $s$ subshell, 1 - $p$ subshell, 2 - $d$ subshell, 3 - $f$ subshell,

The ground state electron configuration of nickel is $1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^8$

Now, time to find electrons in orbitals that fit our requirements. First, $n ≥ 2$, this gives us:

$2s~ 2p ~3s ~3p~ 4s ~3d$

Second, $l ≥1$, this gives us:

$2p ~3p~ 3d$

Because we want the magnetic quantum number to be only 1 or -1, there are two possibilities for orbitals each that fulfill the requirements.

This gives us 2 + 2 + 2 or

6 possibilities.