Question

Question #6706a

Answer 1

`n=4, l=0, m=0, s=1/2`

Explanation:

For Chromium,

`Z=24`

Electronic configuration

`1s^2,2s^2,2p^6,3s^2,3p^6,3d^5,4s^1`

So, for the last electron,

`n=4`

`l=0`

`m=0`

`s=1/2`

Answer 2

The quantum numbers are `n = 3, l = 2, m_l = "+2", m_s = "+½"`.

Explanation:

Chromium has atomic number 24.

You can write its electron configuration as

`"[Cr] 1s"^2 color(white)(l)"2s"^2 "2p"^6 color(white)(l)"3s"^2 "3p"^6color(white)(l) "4s" color(white)(l)"3d"^5` or

`"[Cr] 1s"^2color(white)(l) "2s"^2 "2p"^6color(white)(l) "3s"^2 "3p"^6 "3d"^5color(white)(l)"4s"`

However you write it, you must remember that the `"3d"` orbitals have the highest energy and are filled last in the Aufbau process.

The quantum numbers of the first five `"3d"` electrons are

`bbul(ncolor(white)(m)lcolor(white)(m)m_lcolor(white)(m)m_s)`
`3color(white)(m)2color(white)(ml)"-2"color(white)(m)"+½"`
`3color(white)(m)2color(white)(ml)"-1"color(white)(m)"+½"`
`3color(white)(m)2color(white)(mll)0color(white)(m)"+½"`
`3color(white)(m)2color(white)(m)"+1"color(white)(m)"+½"`
`color(red)(3color(white)(m)2color(white)(m)"+2"color(white)(m)"+½")`

Thus, the quantum numbers of the last electron added to chromium are

`3, 2, "+2, +½"`.