Question

Which of the following has magnetic moment value of 5.9?
(a) Fe2+
(b) Fe3+
(c) Ni2+
(d) Cu2+

Answer 1

Answer:

THE ANSWER IS B.

Explanation:

Here the answer is B I am sure

Answer 2

Answer:

The correct answer is (b) $\bold{ Fe^{3+}}$

The magnetic moment of $Fe^{3+}$ is $5.9\ \text{BM}$

Explanation:

• Given, magnetic moment = $5.9\ \text{BM}$

The formula of magnetic moment = $\sqrt{n (n+2)}$

Here, n is the number of electrons on the valence shell.

Therefore;     $5.9$ $= \sqrt{n (n+2)}$

                      $5.9^{2} = n (n+2)$

                      $34.8 = n^{2} + 2n$

                      $0 = n^{2} + 2n -34.8$

• Now, Calculating the value of n by placing values of a, b and c in the quadratic formula :  
                          $n =\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}$

                                $n = \frac{-2\pm\sqrt{2^{2}-4(1) (-34.8) } }{2(1)}$

                                $n = \frac{-2\pm \sqrt{4+139} }{2}$

                                $n = \frac{-2\pm\sqrt{143} }{2}$

                                $n = \frac{-2\pm12}{2}$

                                $n = \frac{10}{2} \ \text{or} \ n = \frac{-14}{2}$

                                $n = 5 \ \text{or} \ n = -7$

Since the number of electrons cannot be negative, we can ignore the negative value of n. The value of n is 5.

• From the given options;
  $Fe^{3+}$ has an outer shell configuration of  $(3d^{5} 4s^{0} )$. Thus, $Fe^{3+} \text{ion}$ has five unpaired electrons in its valence shell.

Hence, $\bold{Fe^{3+} \text{ion}}$ has a magnetic moment value of $5.9\ \text{BM}$