Question

Number of electron exchanges present in d5 and d10 configuration are
1) 10,0
2) 20,10
3) 10,20 4) 15, 20​

Answer 1

Answer:

Ans. (3) 10 and 20

Explanation:

Let number of electrons with spin +1/2 be nA

and number of electrons with spin - 1/2 be nB

 

Answer 2

Answer:

The number of electron exchanges present in d⁵ configuration is $10$.

The number of electron exchanges present in d¹⁰ configuration is $20$.

Explanation:

As we know,

• The number of electron exchanges can be calculated by the equation given below:
• The number of electron exchanges = $\frac{n_{1} (n_{1} -1)}{2}$ + $\frac{n_{2} (n_{2} -1)}{2}$  ---equation (a)

Here,

• n₁ = The number of unpaired electrons of spin $+\frac{1}{2}$.
• n₂ = The number of unpaired electrons of spin $-\frac{1}{2}$.

1)  d⁵ configuration = ↑ ↑ ↑ ↑ ↑

Therefore,

• Number of electrons with spin $+\frac{1}{2}$ = 5
• Number of electrons with spin $-\frac{1}{2}$ = 0

After putting the value of n in equation (a), we get:

• Number of electron exchanges in d⁵ configuration =  $\frac{5 (5 -1)}{2} + \frac{0 (0 -1)}{2}$ = $10$

2) d¹⁰ configuration = ⇅ ⇅ ⇅ ⇅ ⇅

Therefore,

• Number of electrons with spin $+\frac{1}{2}$ = $5$
• Number of electrons with spin $-\frac{1}{2}$ = $5$

After putting the value of n in equation (a), we get:

• Number of electron exchanges in d¹⁰ configuration = $\frac{5 (5 -1)}{2} + \frac{5 (5 -1)}{2}$ = $20$.