Question

the ground term symbol for d1 configuration is​

Answer 1

Explanation:

4 configurations (d1, d4, d6, d9) give rise to D ground terms,

Answer 2

The ground term will be ${2}_{D_{\dfrac{3}{2}}}$.

Given: The given orbital is $d^1$.

To find: We have to find the ground term.

Solution:

$d^1$ has one electron so s=1/2. Here s is the spin.

l=+2. l is called azimuthal quantum number.

For l=2 the term is D.

Now the J value is given as l+s to l-s.

Thus the value of J will be-

$2 + \frac{1}{2} {\text to }2 - \frac{1}{2}$

The values of J are 5/2 and 3/2.

As $d^1$ has less than half-filled so, the ground state term will be the lower one.

Thus the ground term will be ${2}_{D_{\dfrac{3}{2}}}$.