Magnetic momentum of an ion `M^(+3)` is `sqrt( 35)` BM. Then, number of electron in d orbital of M element will be `"___________"`

Magnetic moment of M^(x+) is sqrt(24)BM . The number of unpaired electrons in M^(x+) is.

The magnetic moment of an ion is sqrt(24) B.M. Then that ion may be

List the quantum numbers m_l and l of electron in 3d orbital.

Magnetic moment of Mn^(+2) is 5.8 B.M. Hence the number of unpaired electrons presenting the metal ion is

Spin magnetic moment of X^(n+) (Z=26) is sqrt(24) B.M. Hence number of unpaired electrons and value of n respectively are:

Spin magnetic moment of X^(n+) (Z=26) is sqrt(24) B.M. Hence number of unpaired electrons and value of n respectively are:

The magnetic moment of M^(x+) (atomic number = 25 is sqrt(15) BM .The number of unpaired electron and the value of x respectively are

The magnetic moment of M^(x+) (atomic number of M=25) is sqrt(15)BM . The number of unpaired electrons and the value of x respectively are-