The calculated spin only magnetic moment of $C r^{2 +}$ is :

3.87 B M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.92 B M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.84 B M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.90 B M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $4.90 B M$
Electronic configuration
$C r^{2 +} : [A r] 3 d^{4}$
No. of unpaired electron = 4
$μ = \sqrt{n (n + 2)} B M$
$μ = \sqrt{4 (4 + 2)} B M$
$μ \approx 4.90 B M$

Suggest Corrections

Similar questions

C r^{2 +}

F e^{2 +}

M^{2 +} (a q)

M^{2 +} (a q)

(Z = 27)

Join BYJU'S Learning Program