Find the sum of maximum number of electrons having +1 and -1 value of ' m ' in Ti. (Atomic number = 22)

Question

TestHub · Accepted Answer

To find the sum of the maximum number of electrons with $m_{l} = + 1$ and $m_{l} = - 1$ in Ti (Z=22), first write its electron configuration:

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}$ .

Electrons with $m_{l} = + 1$ or $m_{l} = - 1$ can exist in $p$ and $d$ orbitals.

For $p$ orbitals ( $l = 1$ ), $m_{l}$ can be $- 1, 0, + 1$ . Each $p$ subshell has one orbital for $m_{l} = + 1$ and one for $m_{l} = - 1$ , each holding 2 electrons.

For $d$ orbitals ( $l = 2$ ), $m_{l}$ can be $- 2, - 1, 0, + 1, + 2$ . Each $d$ subshell has one orbital for $m_{l} = + 1$ and one for $m_{l} = - 1$ , each holding 2 electrons.

In Ti:

$2 p^{6}$ : 2 electrons with $m_{l} = + 1$ , 2 electrons with $m_{l} = - 1$ .

$3 p^{6}$ : 2 electrons with $m_{l} = + 1$ , 2 electrons with $m_{l} = - 1$ .

$3 d^{2}$ : Due to Hund's rule, the two electrons will occupy different $d$ orbitals to maximize spin multiplicity. The most stable configuration would have them in orbitals with different $m_{l}$ values, for example, $m_{l} = - 2$ and $m_{l} = - 1$ , or $m_{l} = - 1$ and $m_{l} = 0$ , or $m_{l} = 0$ and $m_{l} = + 1$ , etc. To maximize the count for $m_{l} = + 1$ and $m_{l} = - 1$ , we can assign one electron to $m_{l} = + 1$ and one electron to $m_{l} = - 1$ . This is possible as per the question "maximum number". So, 1 electron with $m_{l} = + 1$ and 1 electron with $m_{l} = - 1$ .

Total electrons with $m_{l} = + 1$ : $2 (2 p) + 2 (3 p) + 1 (3 d) = 5$ .

Total electrons with $m_{l} = - 1$ : $2 (2 p) + 2 (3 p) + 1 (3 d) = 5$ .

Sum = $5 + 5 = 10$ .

The final answer is $10$ .

Chemistry - Atomic Structure Question with Solution | TestHub

Options:

Doubts & Discussion