Question

No of microstates for the term symbol 3F2
(a)2
(b)3
(c)4
(d)5
The ground state term for d2 configuration
(a)3F
(b)2F
(c)2D
(d)4F
Term symbol for L=2and S=1 is
(a)3D3
(b)3D2
(c)3D1
(d)all of above
Exact value of J for ground state of P2 configuration
(a)0
(b)1
(c)2
(d)3
No of microstates for term 1D
(a)1
(b)2
(c)4
(d)5
How many microstates are possible for d2 configuration including both weak and strong field limits?
(a)15
(b)45
(c)10
(d)90

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Answer 1

Explanation:

the ground state term D2 configuration

Answer 2

Given: $3F_2,d^2, P_2,1D$.

To find:

Number of microstates

• ground state term
• term symbols

Solution:

• $3F_2$ has spin s=1 as 2s+1 value is 3.

Now For term F, the value of L is 3.

The number of microstates will be equal to-

(2s+1)(2L+1)

• $(2 \times 1 + 1)(2 \times 3 + 1) \\ = 3 \times 7 \\ = 21$
• The total spin of $d^2$ is +1.

L value will be-

$+ 2 + 1 = + 3$

Thus the term will be F.

Now J value will be-

L+s to L-s.

4,3,2

The ground state term is $3F_2$.

• For s=1, 2s+1=3

L is equal to 2. The term is D.

The term symbol is $2^D$.

• For $P^2$ the spin is 1.

2s+1 is 3.

Now L is 1 for term P.

The value of J will be-

L+s to L-s

1+1,1-1

2,0

The exact value of J for the ground state of P2 configuration is 2.

• $1^D$ has spin s=0 as 2s+1 value is 1.
• Now For term D, the value of L is 2.
• The number of microstates will be equal to-
• (2s+1)(2L+1)
• $(2 \times 0 + 1)(2 \times 2 + 1) \\ = 1 \times 5 \\ = 5$.

The correct option is d

For $d^2$ the value of n is 10.

The value of r is 2.

The number of microstates will be-

n!/r!(n-r)!

10!/2!8!

$\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ \frac{10 \times 9}{2} \\ 5 \times 9 = 45$

The number of microstates is 45.