Question

Choose the correct ground state term symbol of high spin octahedral Ir3+

Answer 1

In quantum mechanics, the term symbol is an abbreviated description of the (total) angular momentum quantum numbers in a multi-electron atom (however, even a single electron can be described by a term symbol). The ground state term symbol is predicted by Hund's rules.

Answer 2

Final answer:

The ground state term symbol of high spin octahedral $Ir^{3+}$ $=$ $^{5}D$

Given that: We are given high spin octahedral $Ir^{3+}$.

To find: We have to find ground state term symbol of high spin octahedral $Ir^{3+}$.

Explanation:

• Term symbols give information about total spin and orbital angular momentum quantum numbers in a multi-electron atom.

• Term symbol is represented by: $^{2S+1}L$

Where,

$L$ = Term letter

$S$ = Total spin value

$2S+1$ = Spin multiplicity

• Subshell electronic configuration of $Ir^{3+} = [Xe]4f^{14}5d^{6}$

• Given that $Ir^{3+}$ is high spin octahedral.

• Electron distribution in $d^{6}$ orbitals of high spin octahedral:

           $d_{xy}$  $d_{yz}$  $d_{xz}$  $d_{x^{2}-y^{2}}$  $d_{z^{2}}$

            $2$      $1$      $1$       $1$      $1$

• The spin value ($m_{L}$) of d orbitals are:

          $d_{xy}$  $d_{yz}$  $d_{xz}$  $d_{x^{2}-y^{2}}$ $d_{z^{2}}$

           $2$     $1$     $0$      $-1$     $-2$

• Total spin, Σ$m_{L}$$=(2 \times 2)+(1 \times 1)+(1 \times0)+(1 \times(-1))+(1 \times(-2))$

                                $=4+1+0-1-2$

                                $=5-3\\=2$

• Spin 2 corresponds to the term letter $D$.

            $L=D$

• Total spin value of electron filled $d^{6}$:

           $S=2(-\frac{1}{2}+\frac{1}{2} )+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

              $=2$

• Spin multiplicity, $2S+1=(2\times 2)+1=5$

• Substitute the term letter$L$ and spin multiplicity $2S+1$ in the term symbol formula.

• Term symbol = $^{5}D$

Hence, the ground state term symbol of high spin octahedral $Ir^{3+}$ $=$ $^{5}D$

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