The magnetic moment of K3[Fe(CN)6] is found to be 1.7 BM . How many unpaired electron (s) is/are present per molecule ?

Question

The magnetic moment of K3[Fe(CN)6] is found to be 1.7 BM . How many unpaired electron (s) is/are present per molecule ? (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Given, magnetic moment of K3[Fe(CN)6]=1.7 BM Magnetic moment =√n(n+2) n= number of unpaired electrons present in molecule 1.7=√n(n+2) -n2+2n-2.89=0 Then n=0.97 or 1

Q. The magnetic moment of $K_{3} [F e (CN)_{6}]$ is found to be $1.7 BM$ . How many unpaired electron $(s)$ is/are present per molecule ?

1

2

3

4

Given, magnetic moment of
$K_{3} [F e (CN)_{6}] = 1.7 BM$
Magnetic moment $= n (n + 2)$
$n =$ number of unpaired electrons present in molecule $1.7 = n (n + 2)$
$- n^{2} + 2 n - 2.89 = 0$
Then $n = 0.97$ or $1$

Q. The magnetic moment of K3​[Fe(CN)6​] is found to be 1.7BM . How many unpaired electron (s) is/are present per molecule ?

1

2

3

4

Given, magnetic moment of K3​[Fe(CN)6​]=1.7BM Magnetic moment =n(n+2)​ n= number of unpaired electrons present in molecule 1.7=n(n+2)​ −n2+2n−2.89=0 Then n=0.97 or 1

Q. The magnetic moment of $K_{3} [F e (CN)_{6}]$ is found to be $1.7 BM$ . How many unpaired electron $(s)$ is/are present per molecule ?

Given, magnetic moment of
$K_{3} [F e (CN)_{6}] = 1.7 BM$
Magnetic moment $= n (n + 2)$
$n =$ number of unpaired electrons present in molecule $1.7 = n (n + 2)$
$- n^{2} + 2 n - 2.89 = 0$
Then $n = 0.97$ or $1$